2010 Pearson Education, Inc. Upper Saddle River, NJ. All Rights Reserved.

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1 pplication of forces to the handles of these wrenches will produce a tendenc to rotate each wrench about its end. It is important to know how to calculate this effect and, in some cases, to be able to simplif this sstem to its resultants Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved.

2 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. orce Sstem Resultants HPTER JETIVES To discuss the concept of the moment of a force and show how to calculate it in two and three dimensions. To provide a method for finding the moment of a force about a specified ais. d To define the moment of a couple. To present methods for determining the resultants of nonconcurrent force sstems. To indicate how to reduce a simple distributed loading to a resultant force having a specified location. (a) d.1 Moment of a orce Scalar ormulation d d sin u u When a force is applied to a bod it will produce a tendenc for the bod to rotate about a point that is not on the line of action of the force. This tendenc to rotate is sometimes called a torque, but most often it is called the moment of a force or simpl the moment. or eample, consider a wrench used to unscrew the bolt in ig. 1a. If a force is applied to the handle of the wrench it will tend to turn the bolt about point (or the ais). The magnitude of the moment is directl proportional to the magnitude of and the perpendicular distance or moment arm d. The larger the force or the longer the moment arm, the greater the moment or turning effect. Note that if the force is applied at an angle u Z 90, ig. 1b, then it will be more difficult to turn the bolt since the moment arm d =d sin u will be smaller than d. If is applied along the wrench, ig. 1c, its moment arm will be ero since the line of action of will intersect point (the ais).s a result, the moment of about is also ero and no turning can occur. (b) (c) ig

3 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 118 HPTER RE S YSTEM R E S U LT N T S Moment ais M We can generalie the above discussion and consider the force and point which lie in the shaded plane as shown in ig. 2a. The moment M about point, or about an ais passing through and perpendicular to the plane, is a vector quantit since it has a specified magnitude and direction. d Magnitude. The magnitude of is M (a) Sense of rotation M = d ( 1) d M where d is the moment arm or perpendicular distance from the ais at point to the line of action of the force. Units of moment magnitude consist of force times distance, e.g., N # m or lb # ft. (b) ig. 2 Direction. The direction of M is defined b its moment ais, which is perpendicular to the plane that contains the force and its moment arm d. The right-hand rule is used to establish the sense of direction of M. ccording to this rule, the natural curl of the fingers of the right hand, as the are drawn towards the palm, represent the tendenc for rotation caused b the moment. s this action is performed, the thumb of the right hand will give the directional sense of M, ig. 2a. Notice that the moment vector is represented three-dimensionall b a curl around an arrow. In two dimensions this vector is represented onl b the curl as in ig. 2b. Since in this case the moment will tend to cause a counterclockwise rotation, the moment vector is actuall directed out of the page. 2 d 2 M 2 M 1 d 1 d 3 M 3 3 ig. 3 1 Resultant Moment. or two-dimensional problems, where all the forces lie within the plane, ig. 3, the resultant moment (M R ) about point (the ais) can be determined b finding the algebraic sum of the moments caused b all the forces in the sstem. s a convention, we will generall consider positive moments as counterclockwise since the are directed along the positive ais (out of the page). lockwise moments will be negative. Doing this, the directional sense of each moment can be represented b a plus or minus sign. Using this sign convention, the resultant moment in ig. 3 is therefore a+ (M R ) = d; (M R ) = 1 d 1-2 d d 3 If the numerical result of this sum is a positive scalar, (M R ) will be a counterclockwise moment (out of the page); and if the result is negative, (M R ) will be a clockwise moment (into the page).

4 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..1 MMENT RE SLR RMULTIN 119 EXMPLE.1 or each case illustrated in ig., determine the moment of the force about point. SLUTIN (SLR NLYSIS) The line of action of each force is etended as a dashed line in order to establish the moment arm d. lso illustrated is the tendenc of rotation of the member as caused b the force. urthermore, the orbit of the force about is shown as a colored curl. Thus, ig. a M = 1100 N212 m2 = 200 N # m b ns. 100 N ig. b M = 10 N210.7 m2 = 37. N # m b ns. ig. c M = 10 lb21 ft + 2 cos 30 ft2 = 229 lb # ft b ns. ig. d M = 160 lb211 sin ft2 = 2. lb # ft d ns. ig. e M = 17 kn21 m - 1 m2 = 21.0 kn # m d ns. 2 m (a) 2 ft 2 m 0 lb 0.7 m 0 N ft 2 cos ft (b) (c) 2 m 1 m 7 kn 3 ft 1 ft 1 sin ft m 60 lb (d) (e) ig.

5 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 120 HPTER RE S YSTEM R E S U LT N T S EXMPLE.2 Determine the resultant moment of the four forces acting on the rod shown in ig. about point. 0 N SLUTIN ssuming that positive moments act in the counterclockwise, we have +k direction, i.e., 2 m 2 m 60 N 3 m 20 N a+m R = d; M R = -0 N12 m N N13 sin 30 m2-0 N1 m + 3 cos 30 m2 0 N M R = -33 N # m = 33 N # m b ns. ig. or this calculation, note how the moment-arm distances for the 20-N and 0-N forces are established from the etended (dashed) lines of action of each of these forces. H M d d N s illustrated b the eample problems, the moment of a force does not alwas cause a rotation. or eample, the force tends to rotate the beam clockwise about its support at with a moment M = d. The actual rotation would occur if the support at were removed. The abilit to remove the nail will require the moment of H about point to be larger than the moment of the force about that is needed to pull the nail out. N

6 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..2 RSS PRDUT ross Product The moment of a force will be formulated using artesian vectors in the net section. efore doing this, however, it is first necessar to epand our knowledge of vector algebra and introduce the cross-product method of vector multiplication. The cross product of two vectors and ields the vector, which is written = * ( 2) and is read equals cross. Magnitude. The magnitude of is defined as the product of the magnitudes of and and the sine of the angle u between their tails 10 u Thus, = sin u. Direction. Vector has a direction that is perpendicular to the plane containing and such that is specified b the right-hand rule; i.e., curling the fingers of the right hand from vector (cross) to vector, the thumb points in the direction of, as shown in ig. 6. Knowing both the magnitude and direction of, we can write = * = 1 sin u2u ( 3) where the scalar sin u defines the magnitude of and the unit vector u defines the direction of. The terms of Eq. 3 are illustrated graphicall in ig. 6. u u ig. 6

7 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 122 HPTER RE S YSTEM R E S U LT N T S Laws of peration. The commutative law is not valid; i.e., * Z *. Rather, * = - * This is shown in ig. 7 b using the right-hand rule. The cross product * ields a vector that has the same magnitude but acts in the opposite direction to ; i.e., * = -. If the cross product is multiplied b a scalar a, it obes the associative law; ig. 7 a1 * 2 = 1a2 * = * 1a2 = 1 * 2a This propert is easil shown since the magnitude of the resultant vector 1 ƒ a ƒ sin u2 and its direction are the same in each case. The vector cross product also obes the distributive law of addition, * 1 + D2 = 1 * * D2 k i j The proof of this identit is left as an eercise (see Prob. 1). It is important to note that proper order of the cross products must be maintained, since the are not commutative. i ig. 8 j artesian Vector ormulation. Equation 3 ma be used to find the cross product of an pair of artesian unit vectors. or eample, to find i * j, the magnitude of the resultant vector is 1i21j21sin 90 2 = = 1, and its direction is determined using the right-hand rule. s shown in ig. 8, the resultant vector points in the +k direction. Thus, i * j = 112k. In a similar manner, i * j = k i * k = -j i * i = 0 j * k = i j * i = -k j * j = 0 k * i = j k * j = -i k * k = 0 j i ig. 9 k These results should not be memoried; rather, it should be clearl understood how each is obtained b using the right-hand rule and the definition of the cross product. simple scheme shown in ig. 9 is helpful for obtaining the same results when the need arises. If the circle is constructed as shown, then crossing two unit vectors in a counterclockwise fashion around the circle ields the positive third unit vector; e.g., k * i = j. rossing clockwise, a negative unit vector is obtained; e.g., i * k = -j.

8 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..2 RSS PRDUT 123 Let us now consider the cross product of two general vectors and which are epressed in artesian vector form. We have * = 1 i + j + k2 * 1 i + j + k2 = 1i * i2 + 1i * j2 + 1i * k2 + 1j * i2 + 1j * j2 + 1j * k2 + 1k * i2 + 1k * j2 + 1k * k2 arring out the cross-product operations and combining terms ields * = 1-2i - 1-2j + 1-2k ( ) This equation ma also be written in a more compact determinant form as i j k * = 3 3 ( ) Thus, to find the cross product of an two artesian vectors and, it is necessar to epand a determinant whose first row of elements consists of the unit vectors i, j, and k and whose second and third rows represent the,, components of the two vectors and, respectivel.* * determinant having three rows and three columns can be epanded using three minors, each of which is multiplied b one of the three terms in the first row. There are four elements in each minor, for eample, definition, this determinant notation represents the terms , which is simpl the product of the two elements intersected b the arrow slanting downward to the right minus the product of the two elements intersected b the arrow slanting downward to the left or a 3 * 3 determinant, such as Eq., the three minors can be generated in accordance with the following scheme: or element i: or element j: i j k i j k Remember the negative sign or element k: i j k dding the results and noting that the j element must include the minus sign ields the epanded form of * given b Eq..

9 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 12 HPTER RE S YSTEM R E S U LT N T S Moment ais.3 Moment of a orce Vector ormulation r M The moment of a force about point, or actuall about the moment ais passing through and perpendicular to the plane containing and, ig. 10a, can be epressed using the vector cross product, namel, M = r * ( 6) r u (a) u Moment ais d r M Here r represents a position vector directed from to an point on the line of action of. We will now show that indeed the moment M, when determined b this cross product, has the proper magnitude and direction. Magnitude. The magnitude of the cross product is defined from Eq. 3 as M = r sin u, where the angle u is measured between the tails of r and. To establish this angle, r must be treated as a sliding vector so that u can be constructed properl, ig. 10b. Since the moment arm d = r sin u, then M = r sin u = 1r sin u2 = d (b) which agrees with Eq. 1. r3 Line of action r2 ig. 10 M r 1 r 2 r 3 r 1 Direction. The direction and sense of M in Eq. 6 are determined b the right-hand rule as it applies to the cross product. Thus, sliding r to the dashed position and curling the right-hand fingers from r toward, r cross, the thumb is directed upward or perpendicular to the plane containing r and and this is in the same direction as M, the moment of the force about point, ig. 10b. Note that the curl of the fingers, like the curl around the moment vector, indicates the sense of rotation caused b the force. Since the cross product does not obe the commutative law, the order of r * must be maintained to produce the correct sense of direction for. M Principle of Transmissibilit. The cross product operation is often used in three dimensions since the perpendicular distance or moment arm from point to the line of action of the force is not needed. In other words, we can use an position vector r measured from point to an point on the line of action of the force, ig. 11. Thus, M = r 1 * = r 2 * = r 3 * ig. 11 Since can be applied at an point along its line of action and still create this same moment about point, then can be considered a sliding vector. This propert is called the principle of transmissibilit of a force.

10 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..3 MMENT RE VETR RMULTIN 12 artesian Vector ormulation. If we establish,, coordinate aes, then the position vector r and force can be epressed as artesian vectors, ig. 12a. ppling Eq. we have Moment ais M i j k M = r * = 3 r r r 3 ( 7) r where r, r, r represent the,, components of the position vector drawn from point to an point on the line of action of the force (a),, represent the,, components of the force vector If the determinant is epanded, then like Eq. we have M = 1r - r 2i - 1r - r 2j + 1r - r 2k ( 8) r r r r The phsical meaning of these three moment components becomes evident b studing ig. 12b. or eample, the i component of M can be determined from the moments of,, and about the ais. The component does not create a moment or tendenc to cause turning about the ais since this force is parallel to the ais. The line of action of passes through point, and so the magnitude of the moment of about point on the ais is r. the right-hand rule this component acts in the negative i direction. Likewise, passes through point and so it contributes a moment component of r i about the ais. Thus, 1M 2 = 1r - r 2 as shown in Eq. 8. s an eercise, establish the j and k components of M in this manner and show that indeed the epanded form of the determinant, Eq. 8, represents the moment of about point. nce M is determined, realie that it will alwas be perpendicular to the shaded plane containing vectors r and, ig. 12a. (b) ig r 3 2 M R r 1 Resultant Moment of a Sstem of orces. If a bod is acted upon b a sstem of forces, ig. 13, the resultant moment of the forces about point can be determined b vector addition of the moment of each force. This resultant can be written smbolicall as r 2 M R = 1r * 2 ( 9) ig. 13

11 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 126 HPTER RE S YSTEM R E S U LT N T S EXMPLE.3 Determine the moment produced b the force in ig. 1a about point. Epress the result as a artesian vector. 2 kn 12 m SLUTIN r s shown in ig. 1a, either or can be used to determine the moment about point. These position vectors are r = 12k6 m and r = i + 12j6 m r 12 m u m orce epressed as a artesian vector is i + 12j - 12k6 m = u = 2 knc 21 m m m2 d 2 = 0.88i j k6 kn (a) Thus i j k M = r * = = [0(-1.376) - 12(1.376)]i - [0(-1.376) - 12(0.88)] j + [0(1.376) - 0(0.88)]k = -16.i +.1j6 kn m ns. or r M i j k M = r * = = [12(-1.376) - 0(1.376)]i - [(-1.376) - 0(0.88)] j + [(1.376) - 12(0.88)]k r = -16.i +.1j6 kn m ns. (b) ig. 1 NTE: s shown in ig. 1b, M acts perpendicular to the plane that contains, r, and r. Had this problem been worked using M = d, notice the difficult that would arise in obtaining the moment arm d.

12 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..3 MMENT RE VETR RMULTIN 127 EXMPLE. Two forces act on the rod shown in ig. 1a. Determine the resultant moment the create about the flange at. Epress the result as a artesian vector. 1 { 60i 0j 20k} lb 1 ft 2 ft ft 2 {80i 0j 30k} lb r r 2 (b) (a) SLUTIN Position vectors are directed from point to each force as shown in ig. 1b. These vectors are r = j6 ft r = i + j - 2k6 ft The resultant moment about is therefore M R = 1r * 2 = r * 1 + r * 3 i j k i j k = = [ ]i - [0]j + [ ()1-602]k M R {30i 0j 60k} lb ft g 39.8 b 121 a 67. (c) ig. 1 + [ ]i - [ (-2)1802]j + [ ]k = 30i - 0j + 60k6 lb # ft ns. NTE: This result is shown in ig. 1c. The coordinate direction angles were determined from the unit vector for M R. Realie that the two forces tend to cause the rod to rotate about the moment ais in the manner shown b the curl indicated on the moment vector.

13 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 128 HPTER RE S YSTEM R E S U LT N T S 1 2 r ig 16. Principle of Moments concept often used in mechanics is the principle of moments, which is sometimes referred to as Varignon s theorem since it was originall developed b the rench mathematician Varignon ( ). It states that the moment of a force about a point is equal to the sum of the moments of the components of the force about the point.this theorem can be proven easil using the vector cross product since the cross product obes the distributive law. or eample, consider the moments of the force and two of its components about point. ig. 16. Since = we have M = r * = r * = r * 1 + r * 2 or two-dimensional problems, ig. 17, we can use the principle of moments b resolving the force into its rectangular components and then determine the moment using a scalar analsis. Thus, M = - d M This method is generall easier than finding the same moment using M = d. ig. 17 Important Points d The moment of the applied force about point is eas to determine if we use the principle of moments. It is simpl M = d. The moment of a force creates the tendenc of a bod to turn about an ais passing through a specific point. Using the right-hand rule, the sense of rotation is indicated b the curl of the fingers, and the thumb is directed along the moment ais, or line of action of the moment. The magnitude of the moment is determined from M = d, where d is called the moment arm, which represents the perpendicular or shortest distance from point to the line of action of the force. In three dimensions the vector cross product is used to determine the moment, i.e., M = r *. Remember that r is directed from point to an point on the line of action of. The principle of moments states that the moment of a force about a point is equal to the sum of the moments of the force s components about the point. This is a ver convenient method to use in two dimensions.

14 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved.. PRINIPLE MMENTS 129 EXMPLE. Determine the moment of the force in ig. 18a about point. d 7 d 3 cos m ( kn) cos 3 m kn d 3 sin m (a) ( kn) sin (b) SLUTIN I The moment arm d in ig. 18a can be found from trigonometr. d = (3 m) sin 7 = m Thus, M = d = (kn)(2.898 m2 = 1. kn # m b ns. Since the force tends to rotate or orbit clockwise about point, the moment is directed into the page. SLUTIN II The and components of the force are indicated in ig. 18b. onsidering counterclockwise moments as positive, and appling the principle of moments, we have a+ M = - d - d = -1 cos kn213 sin 30 m2-1 sin kn213 cos 30 m2 = -1. kn # m = 1. kn # mb ns. SLUTIN III The and aes can be set parallel and perpendicular to the rod s ais as shown in ig. -18c. Here produces no moment about point since its line of action passes through this point. Therefore, a+ M = - d = -( sin 7 kn)(3 m) = -1. kn # m = 1. kn # m b ns. ( kn) sin 7 3 m ( kn) sin 7 (c) ig. 18

15 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 130 HPTER RE S YSTEM R E S U LT N T S EXMPLE.6 orce acts at the end of the angle bracket shown in ig. 19a. Determine the moment of the force about point. SLUTIN I (SLR NLYSIS) 0.2 m The force is resolved into its and components as shown in ig. 19b, then a+m = 00 sin 30 N10.2 m2-00 cos 30 N10. m2 0. m (a) = 00 N or = N # m = 98.6 N # m b M = -98.6k6 N # m ns. 0.2 m SLUTIN II (VETR NLYSIS) Using a artesian vector approach, the force and position vectors shown in ig. 19c are 0. m 00 sin N r = 0.i - 0.2j6 m (b) 00 cos N = 00 sin 30 i - 00 cos 30 j6 N = 200.0i - 36.j6 N The moment is therefore r 0.2 m i j k M = r * = = 0i - 0j + [ ]k 0. m (c) ig. 19 = -98.6k6 N # m ns. NTE: It is seen that the scalar analsis (Solution I) provides a more convenient method for analsis than Solution II since the direction of the moment and the moment arm for each component force are eas to establish. Hence, this method is generall recommended for solving problems displaed in two dimensions, whereas a artesian vector analsis is generall recommended onl for solving three-dimensional problems.

16 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved.. PRINIPLE MMENTS 131 UNDMENTL PRLEMS 1. Determine the moment of the force about point. 600 lb. Determine the moment of the force about point. ft ft 3 ft ft 600 lb 1 ft 1 2. Determine the moment of the force about point. 100 N 3. Determine the moment of the force about point. Neglect the thickness of the member. 100 mm 0 N 60 2 m m 200 mm mm 3. Determine the moment of the force about point. 300 N 6. Determine the moment of the force about point. 00 N 0.3 m 3 m 0. m 3 6

17 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 132 HPTER RE S YSTEM R E S U LT N T S 7. Determine the resultant moment produced b the forces about point. 10. Determine the moment of force about point. Epress the result as a artesian vector. 00 N 300 N 1 m 2 m 2. m 00 N 3 m m 0.2 m 600 N 0.12 m 7 8. Determine the resultant moment produced b the forces about point. 0.3 m N N 9. Determine the resultant moment produced b the forces about point lb 6 ft Determine the moment of force about point. Epress the result as a artesian vector. ft 1 ft ft If 1 = 100i - 120j + 7k6 lb and 2 = -200i + 20j + 100k6 lb, determine the resultant moment produced b these forces about point. Epress the result as a artesian vector lb 2 ft lb 6 ft ft ft ft 9 12

18 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved.. PRINIPLE MMENTS 133 PRLEMS 1. If,, and D are given vectors, prove the distributive law for the vector cross product, i.e., : ( + D) = ( : ) + ( : D). 2. Prove the triple scalar product identit # : = : #. 3. Given the three nonero vectors,, and, show that if # ( : ) = 0, the three vectors must lie in the same plane. *. Two men eert forces of = 80 lb and P = 0 lb on the ropes. Determine the moment of each force about. Which wa will the pole rotate, clockwise or counterclockwise?. If the man at eerts a force of P = 30 lb on his rope, determine the magnitude of the force the man at must eert to prevent the pole from rotating, i.e., so the resultant moment about of both forces is ero. * 8. The handle of the hammer is subjected to the force of = 20 lb. Determine the moment of this force about the point. 9. In order to pull out the nail at, the force eerted on the handle of the hammer must produce a clockwise moment of 00 lb # in. about point. Determine the required magnitude of force. in. 18 in. P 6 ft 12 ft 3 Probs. 8/9 10. The hub of the wheel can be attached to the ale either with negative offset (left) or with positive offset (right). If the tire is subjected to both a normal and radial load as shown, determine the resultant moment of these loads about point on the ale for both cases. Probs. / 6. If u =, determine the moment produced b the -kn force about point. 7. If the moment produced b the -kn force about point is 10 kn # m clockwise, determine the angle u, where 0 u m 0. m u kn 0.0 m 0. m kn 0.0 m 0. m 800 N 800 N kn Probs. 6/7 ase 1 ase 2 Prob. 10

19 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 13 HPTER RE S YSTEM R E S U LT N T S 11. The member is subjected to a force of = 6 kn. If u =, determine the moment produced b about point. * 12. Determine the angle u (0 u 180 ) of the force so that it produces a maimum moment and a minimum moment about point. lso, what are the magnitudes of these maimum and minimum moments? 13. Determine the moment produced b the force about point in terms of the angle u. Plot the graph of M versus u, where 0 u The chilles tendon force of t = 60 N is mobilied when the man tries to stand on his toes. s this is done, each of his feet is subjected to a reactive force of N f = 00 N. Determine the resultant moment of t and N f about the ankle joint. * 16. The chilles tendon force t is mobilied when the man tries to stand on his toes.s this is done, each of his feet is subjected to a reactive force of N t = 00 N. If the resultant moment produced b forces t and N t about the ankle joint is required to be ero, determine the magnitude of. t t 1. m u 6 kn 6 m 200 mm Probs. 11/12/13 1. Serious neck injuries can occur when a football plaer is struck in the face guard of his helmet in the manner shown, giving rise to a guillotine mechanism. Determine the moment of the knee force P = 0 lb about point. What would be the magnitude of the neck force so that it gives the counterbalancing moment about? 6 mm 100 mm Probs. 1/16 N f 00 N 17. The two bos push on the gate with forces of = 30 lb and as shown. Determine the moment of each force about. Which wa will the gate rotate, clockwise or counterclockwise? Neglect the thickness of the gate. 18. Two bos push on the gate as shown. If the bo at eerts a force of = 30 lb, determine the magnitude of the force the bo at must eert in order to prevent the gate from turning. Neglect the thickness of the gate. 2 in. P 0 lb 60 in. 6 ft 3 ft in. Probs. 1 Probs. 17/18

20 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved.. PRINIPLE MMENTS The tongs are used to grip the ends of the drilling pipe P. Determine the torque (moment) M P that the applied force = 10 lb eerts on the pipe about point P as a function of u. Plot this moment M P versus u for 0 u 90. * 20. The tongs are used to grip the ends of the drilling pipe P. If a torque (moment) of M P = 800 lb # ft is needed at P to turn the pipe, determine the cable force that must be applied to the tongs. Set u = 30. * 2. In order to raise the lamp post from the position shown, force is applied to the cable. If = 200 lb, determine the moment produced b about point. 2. In order to raise the lamp post from the position shown, the force on the cable must create a counterclockwise moment of 100 lb # ft about point. Determine the magnitude of that must be applied to the cable. u P 6 in ft 3 in. M P 10 ft Probs. 19/20 Probs. 2/2 21. Determine the direction u for 0 u 180 of the force so that it produces the maimum moment about point. alculate this moment. 22. Determine the moment of the force about point as a function of u. Plot the results of M (ordinate) versus u (abscissa) for 0 u Determine the minimum moment produced b the force about point. Specif the angle u (0 u 180 ). 00 N u 26. The foot segment is subjected to the pull of the two plantarfleor muscles. Determine the moment of each force about the point of contact on the ground lb lb 60 in. 2 m 3 m 1 in. 3. in. Probs. 21/22/23 Prob. 26

21 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 136 HPTER RE S YSTEM R E S U LT N T S 27. The 70-N force acts on the end of the pipe at. Determine (a) the moment of this force about point, and (b) the magnitude and direction of a horiontal force, applied at, which produces the same moment.take u = The rod on the power control mechanism for a business jet is subjected to a force of 80 N. Determine the moment of this force about the bearing at. * 28. The 70-N force acts on the end of the pipe at. Determine the angles u 10 u of the force that will produce maimum and minimum moments about point. What are the magnitudes of these moments? N 10 mm m 70 N u 0.3 m 0.7 m Probs. 27/28 Prob Determine the moment of each force about the bolt located at. Take = 0 lb, = 0 lb. 30. If = 30 lb and = lb, determine the resultant moment about the bolt located at. * 32. The towline eerts a force of P = kn at the end of the 20-m-long crane boom. If u = 30, determine the placement of the hook at so that this force creates a maimum moment about point. What is this moment? 33. The towline eerts a force of P = kn at the end of the 20-m-long crane boom. If = 2 m, determine the position u of the boom so that this force creates a maimum moment about point. What is this moment? P kn u 20 m 1. m Probs. 29/30 Probs. 32/33

22 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved.. PRINIPLE MMENTS In order to hold the wheelbarrow in the position shown, force must produce a counterclockwise moment of 200 N # m about the ale at. Determine the required magnitude of force. 3. The wheelbarrow and its contents have a mass of 0 kg and a center of mass at G. If the resultant moment produced b force and the weight about point is to be ero, determine the required magnitude of force. * 0. Determine the moment produced b force about point. Epress the result as a artesian vector. 1. Determine the moment produced b force about point. Epress the result as a artesian vector. 2. Determine the resultant moment produced b forces and about point. Epress the result as a artesian vector. * 36. The wheelbarrow and its contents have a center of mass at G. If = 100 N and the resultant moment produced b force and the weight about the ale at is ero, determine the mass of the wheelbarrow and its contents. 0.6 m 6 m 20 N 780 N 0. m G 2 m 2. m 0.3 m 1.2 m 3 m Prob. 3/3/36 Probs. 0/1/2 37. Determine the moment produced b 1 about point. Epress the result as a artesian vector. 38. Determine the moment produced b 2 about point. Epress the result as a artesian vector. 39. Determine the resultant moment produced b the two forces about point. Epress the result as a artesian vector. 3 ft 2 { 10i 30j 0k} lb 2 ft 2 ft Probs. 37/38/39 1 ft 1 { 20i 10j 30k} lb 3. Determine the moment produced b each force about point located on the drill bit. Epress the results as artesian vectors. 10 mm 300 mm 600 mm { 0i 100j 60k} N Prob mm { 0i 120j 60k} N *. force of = 6i - 2j + 1k6 kn produces a moment of M = i + j - 1k6 kn # m about the origin of coordinates, point. If the force acts at a point having an coordinate of = 1 m, determine the and coordinates.

23 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 138 HPTER RE S YSTEM R E S U LT N T S. The pipe assembl is subjected to the 80-N force. Determine the moment of this force about point. 6. The pipe assembl is subjected to the 80-N force. Determine the moment of this force about point. * 8. orce acts perpendicular to the inclined plane. Determine the moment produced b about point. Epress the result as a artesian vector. 9. orce acts perpendicular to the inclined plane. Determine the moment produced b about point. Epress the result as a artesian vector. 00 mm 300 mm 3 m 3 m 00 N 20 mm 200 mm m 80 N 0 Probs. 8/9 Probs. /6 7. The force = 6i + 8j + 10k6 N creates a moment about point of M = -1i + 8j + 2k6 N # m. If the force passes through a point having an coordinate of 1 m, determine the and coordinates of the point. lso, realiing that M = d, determine the perpendicular distance d from point to the line of action of N horiontal force is applied perpendicular to the handle of the socket wrench. Determine the magnitude and the coordinate direction angles of the moment created b this force about point. P 200 mm d M 1 m 7 mm 20 N 1 Prob. 7 Prob. 0

24 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved.. MMENT RE UT SPEIIED XIS 139. Moment of a orce about a Specified is Sometimes, the moment produced b a force about a specified ais must be determined. or eample, suppose the lug nut at on the car tire in ig. 20a needs to be loosened. The force applied to the wrench will create a tendenc for the wrench and the nut to rotate about the moment ais passing through ; however, the nut can onl rotate about the ais. Therefore, to determine the turning effect, onl the component of the moment is needed, and the total moment produced is not important. To determine this component, we can use either a scalar or vector analsis. u d M M Moment is d Scalar nalsis. To use a scalar analsis in the case of the lug nut in ig. 20a, the moment arm perpendicular distance from the ais to the line of action of the force is d = d cos u. Thus, the moment of about the ais is M = d = (d cos u). ccording to the right-hand rule, M is directed along the positive ais as shown in the figure. In general, for an ais a, the moment is (a) ig. 20 M a = d a ( 10) If large enough,the cable force on the boom of this crane can cause the crane to topple over. To investigate this, the moment of the force must be calculated about an ais passing through the base of the legs at and.

25 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 10 HPTER RE S YSTEM R E S U LT N T S u r j M 0 r (b) ig. 20 u M Vector nalsis. To find the moment of force in ig. 20b about the ais using a vector analsis, we must first determine the moment of the force about an point on the ais b appling Eq. 7, M = r *. The component M along the ais is the projection of M onto the ais. It can be found using the dot product discussed in hapter 2, so that M = j # M = j # (r * ), where j is the unit vector for the ais. We can generalie this approach b letting u a be the unit vector that specifies the direction of the a ais shown in ig. 21. Then the moment of about the ais is M a = u a # (r * ). This combination is referred to as the scalar triple product. If the vectors are written in artesian form, we have M a = [u a i + u a j + u a k] # 3 i j k r r r 3 = u a (r - r ) - u a (r - r ) + u a (r - r ) This result can also be written in the form of a determinant, making it easier to memorie.* u a u a u a M a = u a # 1r * 2 = 3 r r r 3 ( 11) M a u a a r is of projection ig. 21 M r where M a u a, u a, u a r, r, r,, represent the,, components of the unit vector defining the direction of the a ais represent the,, components of the position vector etended from an point on the a ais to an point on the line of action of the force represent the,, components of the force vector. When is evaluated from Eq. 11, it will ield a positive or negative scalar. The sign of this scalar indicates the sense of direction of M a along the a ais. If it is positive, then M a will have the same sense as u a, whereas if it is negative, then M a will act opposite to u a. nce M a is determined, we can then epress M a as a artesian vector, namel, M a = M a u a ( 12) The eamples which follow illustrate numerical applications of the above concepts. *Take a moment to epand this determinant, to show that it will ield the above result.

26 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved.. MMENT RE UT SPEIIED XIS 11 Important Points The moment of a force about a specified ais can be determined provided the perpendicular distance d a from the force line of action to the ais can be determined. M a = d a. If vector analsis is used, M a = u a # 1r * 2, where u a defines the direction of the ais and r is etended from an point on the ais to an point on the line of action of the force. M a M a If is calculated as a negative scalar, then the sense of direction of is opposite to u a. The moment M a epressed as a artesian vector is determined from M a = M a u a. EXMPLE.7 Determine the resultant moment of the three forces in ig. 22 about the ais, the ais, and the ais. SLUTIN force that is parallel to a coordinate ais or has a line of action that passes through the ais does not produce an moment or tendenc for turning about that ais. Therefore, defining the positive direction of the moment of a force according to the right-hand rule, as shown in the figure, we have 3 0 lb 2 0 lb 1 60 lb M = (60 lb)(2 ft) + (0 lb)(2 ft) + 0 = 220 lb # ft M = 0 - (0 lb)(3 ft) - (0 lb)(2 ft) = -230 lb # ft ns. ns. 2 ft 2 ft ig ft 3 ft M = (0 lb)(2 ft) = -80 lb # ft ns. The negative signs indicate that M and M act in the - and directions, respectivel. -

27 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 12 HPTER RE S YSTEM R E S U LT N T S EXMPLE.8 M Determine the moment produced b the force in ig. 23a, which tends to rotate the rod about the ais. = 300 N 0.6 m 0.2 m 0.3 m 0. m (a) SLUTIN vector analsis using M = u # 1r * 2 will be considered for the solution rather than tring to find the moment arm or perpendicular distance from the line of action of to the ais. Each of the terms in the equation will now be identified. Unit vector u defines the direction of the ais of the rod, ig. 23b, where u = r {0.i + 0.2j} m = = 0.89i j r 210. m m2 Vector r is directed from an point on the ais to an point on the line of action of the force. or eample, position vectors r and r D are suitable, ig. 23b. (lthough not shown, r or r D can also be used.) or simplicit, we choose r D, where r D = 0.6i6 m The force is r M = -300k6 N Substituting these vectors into the determinant form and epanding, we have D r D u (b) ig M = u # 1rD * 2 = = 0.89[ ] [ ] = 80.0 N # m + 0[ ] This positive result indicates that the sense of direction as u. M is in the same Epressing M as a artesian vector ields M = M u = N # m210.89i j2 = 72.0i j6 N # m The result is shown in ig. 23b. ns. NTE: If ais is defined using a unit vector directed from toward, then in the above formulation -u would have to be used.this would lead to M = N # m. onsequentl, M = M 1-u 2, and the same result would be obtained.

28 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved.. MMENT RE UT SPEIIED XIS 13 EXMPLE.9 Determine the magnitude of the moment of force about segment of the pipe assembl in ig. 2a. SLUTIN The moment of about the ais is determined from M = u # (r * ), where r is a position vector etending from an point on the ais to an point on the line of action of. s indicated in ig. 2b, either r D, r, r D, or r can be used; however, r D will be considered since it will simplif the calculation. The unit vector, which specifies the direction of the ais, is u D 0. m 0. m 300 N 0.3 m u = r r = 0.3i + 0.j6 m = 0.6i + 0.8j m m2 0. m (a) 0.2 m 0.1 m and the position vector r D is r D = 0.i + 0.k6 m The force epressed as a artesian vector is D = a r D r D b = (300 N) 0.i - 0.j + 0.2k6 m 2(0. m) 2 + (-0. m) 2 + (0.2 m) 2 R = {200i - 200j + 100k} N r D r D u r r Therefore, (b) M = u # 1rD * 2 ig = = 0.6[ ( 0. )1-2002] - 0.8[ ( 0. )12002] + 0 = 100 N# m ns.

29 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 1 HPTER RE S YSTEM R E S U LT N T S UNDMENTL PRLEMS 13. Determine the magnitude of the moment of the force = 300i - 200j + 10k6 N about the ais. Epress the result as a artesian vector. 16. Determine the magnitude of the moment of the force about the ais. {30i 20j 0k} N 1. Determine the magnitude of the moment of the force = 300i - 200j + 10k6 N about the ais. Epress the result as a artesian vector. 2 m 0.3 m m 3 m 0. m 0.2 m Determine the moment of the force = 0i - 0j + 20k6 lb about the ais. Epress the result as a artesian vector. 13/1 1. Determine the magnitude of the moment of the 200-N force about the ais. 2 ft 3 ft ft Determine the moment of force about the, the, and the aes. Use a scalar analsis. 0.3 m N 60 3 m N 0.2 m 2 m 2 m 1 18

30 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved.. MMENT RE UT SPEIIED XIS 1 PRLEMS 1. Determine the moment produced b force about the diagonal of the rectangular block. Epress the result as a artesian vector. * 2. Determine the moment produced b force about the diagonal D of the rectangular block. Epress the result as a artesian vector.. Determine the magnitude of the moments of the force about the,, and aes. Solve the problem (a) using a artesian vector approach and (b) using a scalar approach.. Determine the moment of the force about an ais etending between and. Epress the result as a artesian vector. { 6i 3j 10k} N D 1. m G 3 m ft 3 m Probs. 1/2 3 ft 2 ft {i 12j 3k} lb 3. The tool is used to shut off gas valves that are difficult to access. If the force is applied to the handle, determine the component of the moment created about the ais of the valve. Probs. / * 6. Determine the moment produced b force about segment of the pipe assembl. Epress the result as a artesian vector. 0.2 m { 60i 20j 1k} N { 20i 10j 1k} N 0. m m 3 m m Prob. 3 Prob. 6

31 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 16 HPTER RE S YSTEM R E S U LT N T S 7. Determine the magnitude of the moment that the force eerts about the ais of the shaft. Solve the problem using a artesian vector approach and using a scalar approach. * 60. Determine the magnitude of the moment produced b the force of = 200 N about the hinged ais (the ais) of the door. 0. m 200 mm 20 mm 0 mm 1 m 2 m 200 N 2. m 1 16 N Prob. 60 Prob If = 0 N, determine the magnitude of the moment produced b this force about the ais. 9. The friction at sleeve can provide a maimum resisting moment of 12 N # m about the ais. Determine the largest magnitude of force that can be applied to the bracket so that the bracket will not turn. 61. If the tension in the cable is = 10 lb, determine the magnitude of the moment produced b this force about the hinged ais, D, of the panel. 62. Determine the magnitude of force in cable in order to produce a moment of 00 lb # ft about the hinged ais D, which is needed to hold the panel in the position shown ft ft 300 mm 100 mm 10 mm 6 ft D 6 ft 6 ft Probs. 8/9 Probs. 61/62

32 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved.. MMENT RE UT SPEIIED XIS The -frame is being hoisted into an upright position b the vertical force of = 80 lb. Determine the moment of this force about the ais passing through points and when the frame is in the position shown. * 6. The -frame is being hoisted into an upright position b the vertical force of = 80 lb. Determine the moment of this force about the ais when the frame is in the position shown. 6. The -frame is being hoisted into an upright position b the vertical force of = 80 lb. Determine the moment of this force about the ais when the frame is in the position shown. 3 ft 3 ft The fle-headed ratchet wrench is subjected to a force of P = 16 lb, applied perpendicular to the handle as shown. Determine the moment or torque this imparts along the vertical ais of the bolt at. 67. If a torque or moment of 80 lb # in. is required to loosen the bolt at, determine the force P that must be applied perpendicular to the handle of the fle-headed ratchet wrench. 6 ft Probs. 63/6/6 * 68. The pipe assembl is secured on the wall b the two brackets. If the flower pot has a weight of 0 lb, determine the magnitude of the moment produced b the weight about the ais. 69. The pipe assembl is secured on the wall b the two brackets. If the frictional force of both brackets can resist a maimum moment of 10 lb # ft, determine the largest weight of the flower pot that can be supported b the assembl without causing it to rotate about the ais. 60 ft ft 3 ft Probs. 68/ vertical force of = 60 N is applied to the handle of the pipe wrench. Determine the moment that this force eerts along the ais ( ais) of the pipe assembl. oth the wrench and pipe assembl lie in the - plane. Suggestion: Use a scalar analsis. 71. Determine the magnitude of the vertical force acting on the handle of the wrench so that this force produces a component of moment along the ais ( ais) of the pipe assembl of (M ) = -i6 N # m. oth the pipe assembl and the wrench lie in the - plane. Suggestion: Use a scalar analsis. 3 ft in. P 00 mm 10 mm 0.7 in. 200 mm Probs. 66/67 Probs. 70/71

33 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 18 HPTER RE S YSTEM R E S U LT N T S.6 Moment of a ouple d ig. 2 r couple is defined as two parallel forces that have the same magnitude, but opposite directions, and are separated b a perpendicular distance d, ig. 2. Since the resultant force is ero, the onl effect of a couple is to produce a rotation or tendenc of rotation in a specified direction. or eample, imagine that ou are driving a car with both hands on the steering wheel and ou are making a turn. ne hand will push up on the wheel while the other hand pulls down, which causes the steering wheel to rotate. The moment produced b a couple is called a couple moment. We can determine its value b finding the sum of the moments of both couple forces about an arbitrar point. or eample, in ig. 26, position vectors r and r are directed from point to points and ling on the line of action of - and. The couple moment determined about is therefore M = r * + r * - = (r - r ) * r r However r = r + r or r = r - r, so that M = r * ( 13) ig. 26 This result indicates that a couple moment is a free vector, i.e., it can act at an point since M depends onl upon the position vector r directed between the forces and not the position vectors r and r, directed from the arbitrar point to the forces. This concept is unlike the moment of a force, which requires a definite point (or ais) about which moments are determined. M Scalar ormulation. The moment of a couple, M, ig. 27, is defined as having a magnitude of M = d ( 1) d where is the magnitude of one of the forces and d is the perpendicular distance or moment arm between the forces. The direction and sense of the couple moment are determined b the right-hand rule, where the thumb indicates this direction when the fingers are curled with the sense of rotation caused b the couple forces. In all cases, M will act perpendicular to the plane containing these forces. ig. 27 Vector ormulation. The moment of a couple can also be epressed b the vector cross product using Eq. 13, i.e., M = r * ( 1) pplication of this equation is easil remembered if one thinks of taking the moments of both forces about a point ling on the line of action of one of the forces. or eample, if moments are taken about point in ig. 26, the moment of - is ero about this point, and the moment of is defined from Eq. 1.Therefore, in the formulation r is crossed with the force to which it is directed.

34 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..6 MMENT UPLE N 0 N 0. m 0.3 m 0 N 30 N ig. 28 Equivalent ouples. If two couples produce a moment with the same magnitude and direction, then these two couples are equivalent. or eample, the two couples shown in ig. 28 are equivalent because each couple moment has a magnitude of M = 30 N(0. m) = 0 N(0.3 m) = 12 N # m, and each is directed into the plane of the page. Notice that larger forces are required in the second case to create the same turning effect because the hands are placed closer together. lso, if the wheel was connected to the shaft at a point other than at its center, then the wheel would still turn when each couple is applied since the 12 N # m couple is a free vector. Resultant ouple Moment. Since couple moments are vectors, their resultant can be determined b vector addition. or eample, consider the couple moments M 1 and M 2 acting on the pipe in ig. 29a. Since each couple moment is a free vector, we can join their tails at an arbitrar point and find the resultant couple moment, M R = M 1 + M 2 as shown in ig. 29b. If more than two couple moments act on the bod, we ma generalie this concept and write the vector resultant as M 2 M 1 (a) M R = 1r * 2 ( 16) M 2 M 1 These concepts are illustrated numericall in the eamples that follow. In general, problems projected in two dimensions should be solved using a scalar analsis since the moment arms and force components are eas to determine. (b) M R ig. 29

35 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 10 HPTER RE S YSTEM R E S U LT N T S Important Points couple moment is produced b two noncollinear forces that are equal in magnitude but opposite in direction. Its effect is to produce pure rotation, or tendenc for rotation in a specified direction. Steering wheels on vehicles have been made smaller than on older vehicles because power steering does not require the driver to appl a large couple moment to the rim of the wheel. couple moment is a free vector, and as a result it causes the same rotational effect on a bod regardless of where the couple moment is applied to the bod. The moment of the two couple forces can be determined about an point. or convenience, this point is often chosen on the line of action of one of the forces in order to eliminate the moment of this force about the point. In three dimensions the couple moment is often determined using the vector formulation, M = r *, where r is directed from an point on the line of action of one of the forces to an point on the line of action of the other force. resultant couple moment is simpl the vector sum of all the couple moments of the sstem. EXMPLE lb d 2 3 ft lb d 1 ft lb d 3 ft Determine the resultant couple moment of the three couples acting on the plate in ig. 30. SLUTIN s shown the perpendicular distances between each pair of couple forces are d 1 = ft, d 2 = 3 ft, and d 3 = ft. onsidering counterclockwise couple moments as positive, we have 2 0 lb lb lb a+m R = M; M R = - 1 d d 2-3 d 3 = (-200 lb)( ft) + (0 lb)(3 ft) - (300 lb)( ft) ig. 30 = -90 lb # ft = 90 lb # ft b ns. The negative sign indicates that M R has a clockwise rotational sense.

36 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..6 MMENT UPLE 11 EXMPLE.11 Determine the magnitude and direction of the couple moment acting on the gear in ig. 31a. 600 N 600 sin N 600 N 600 cos N 0.2 m 0.2 m 600 cos N 600 N (a) 600 N 600 sin N (b) SLUTIN The easiest solution requires resolving each force into its components as shown in ig. 31b. The couple moment can be determined b summing the moments of these force components about an point, for eample, the center of the gear or point. If we consider counterclockwise moments as positive, we have a+m = M ; M = (600 cos 30 N)(0.2 m) - (600 sin 30 N)(0.2 m) = 3.9 N# m d ns. or a+m = M ; M = (600 cos 30 N)(0.2 m) - (600 sin 30 N)(0.2 m) = 3.9 N# m d ns. 600 N This positive result indicates that M has a counterclockwise rotational sense, so it is directed outward, perpendicular to the page. NTE: The same result can also be obtained using M = d, where d is the perpendicular distance between the lines of action of the couple forces, ig. 31c. However, the computation for d is more involved. Realie that the couple moment is a free vector and can act at an point on the gear and produce the same turning effect about point. d 600 N (c) ig. 31

37 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 12 HPTER RE S YSTEM R E S U LT N T S EXMPLE.12 Determine the couple moment acting on the pipe shown in ig. 32a. Segment is directed 30 below the plane. r 2 lb r 2 lb 2 lb 8 in. 6 in. 2 lb (b) (a) 2 lb (c) r 2 lb SLUTIN I (VETR NLYSIS) The moment of the two couple forces can be found about an point. If point is considered, ig. 32b, we have M = r * 1-2k2 + r * 12k2 = 18j2 * 1-2k cos 30 i + 8j - 6 sin 30 k2 * 12k2 = -200i j + 200i = -130j6 lb # in. ns. It is easier to take moments of the couple forces about a point ling on the line of action of one of the forces, e.g., point, ig. 32c. In this case the moment of the force at is ero, so that M = r * 12k2 = 16 cos 30 i - 6 sin 30 k2 * 12k2 = -130j6 lb # in. ns. 2 lb 6 in. d (d) ig lb SLUTIN II (SLR NLYSIS) lthough this problem is shown in three dimensions, the geometr is simple enough to use the scalar equation M = d. The perpendicular distance between the lines of action of the couple forces is d = 6 cos 30 =.196 in., ig. 32d. Hence, taking moments of the forces about either point or point ields M = d = 2 lb in.2 = lb # in. ppling the right-hand rule, M acts in the -j direction. Thus, M = -130j6 lb # in. ns.

38 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..6 MMENT UPLE 13 EXMPLE.13 Replace the two couples acting on the pipe column in ig. 33a b a resultant couple moment. 12 N 3 D M N m 3 M R M m 0. m 10 N 12 N 10 N 3 M 1 60 N m M 1 (a) (b) ig. 33 (c) SLUTIN (VETR NLYSIS) The couple moment M 1, developed b the forces at and, can easil be determined from a scalar formulation. M 1 = d = 10 N10. m2 = 60 N # m the right-hand rule, acts in the +i direction, ig. 33b. Hence, M 1 M 1 = 60i6 N # m Vector analsis will be used to determine M 2, caused b forces at and D. If moments are computed about point D, ig. 33a, M 2 = r D *, then M 2 = r D * = 10.3i2 * 12 j kd = 10.3i2 * [100j - 7k] = 301i * j2-22.1i * k2 = 22.j + 30k6 N # m Since M 1 and M 2 are free vectors, the ma be moved to some arbitrar point and added vectoriall, ig. 33c. The resultant couple moment becomes M R = M 1 + M 2 = 60i + 22.j + 30k6 N # m ns.

39 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 1 HPTER RE S YSTEM R E S U LT N T S UNDMENTL PRLEMS 19. Determine the resultant couple moment acting on the beam. 00 N 00 N 22. Determine the couple moment acting on the beam. 10 kn N 0.2 m 200 N 3 m 2 m 300 N 300 N Determine the resultant couple moment acting on the triangular plate. 200 lb 10 lb m 1 m 1 m 3 10 kn Determine the resultant couple moment acting on the pipe assembl. (M c ) 1 0 lb ft ft ft (M c ) lb ft 1. ft 2 ft 3. ft 200 lb 10 lb 2 ft 2 ft ft 300 lb 300 lb Determine the magnitude of so that the resultant couple moment acting on the beam is 1. kn # m clockwise. 0.9 m 2 kn 0.3 m 2 kn (M c ) 2 20 lb ft Determine the couple moment acting on the pipe assembl and epress the result as a artesian vector. 0 N 3 0. m 0.3 m 3 0 N 21 2

40 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..6 MMENT UPLE 1 PRLEMS * 72. The frictional effects of the air on the blades of the standing fan creates a couple moment of M = 6 N # m on the blades. Determine the magnitude of the couple forces at the base of the fan so that the resultant couple moment on the fan is ero. 7. The caster wheel is subjected to the two couples. Determine the forces that the bearings eert on the shaft so that the resultant couple moment on the caster is ero. 00 N M 0 mm 100 mm mm 0.1 m 0.1 m 0 mm 00 N Prob. 72 Prob Determine the required magnitude of the couple 7. If = 200 lb, determine the resultant couple moments M 2 and M 3 so that the resultant couple moment moment. is ero. * 76. Determine the required magnitude of force if the resultant couple moment on the frame is 200 lb # ft, clockwise. M 2 2 ft 2 ft 3 2 ft 10 lb M 3 2 ft 10 lb 3 Prob. 73 M N m 2 ft Probs. 7/76

41 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 16 HPTER RE S YSTEM R E S U LT N T S 77. The floor causes a couple moment of M = 0 N # m and M = 30 N # m on the brushes of the polishing machine. Determine the magnitude of the couple forces that must be developed b the operator on the handles so that the resultant couple moment on the polisher is ero. What is the magnitude of these forces if the brush at suddenl stops so that M = 0? * 80. Two couples act on the beam. Determine the magnitude of so that the resultant couple moment is 0 lb # ft, counterclockwise. Where on the beam does the resultant couple moment act? 0.3 m M M 200 lb 1. ft 1.2 ft 200 lb 2 ft Prob. 80 Prob If u = 30, determine the magnitude of force so that the resultant couple moment is 100 N # m, clockwise. 79. If = 200 N, determine the required angle u so that the resultant couple moment is ero. 81. The cord passing over the two small pegs and of the square board is subjected to a tension of 100 N. Determine the required tension P acting on the cord that passes over pegs and D so that the resultant couple produced b the two couples is 1 N # m acting clockwise. Take u = The cord passing over the two small pegs and of the board is subjected to a tension of 100 N. Determine the minimum tension P and the orientation u of the cord passing over pegs and D, so that the resultant couple moment produced b the two cords is 20 N # m, clockwise. 300 mm u 300 N P u 300 mm 100 N 1 u 300 N N D u 300 mm P Probs. 78/79 Probs. 81/82

42 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..6 MMENT UPLE device called a rolamite is used in various was to replace slipping motion with rolling motion. If the belt, which wraps between the rollers, is subjected to a tension of 1 N, determine the reactive forces N of the top and bottom plates on the rollers so that the resultant couple acting on the rollers is equal to ero. 8. Determine the resultant couple moment acting on the beam. Solve the problem two was: (a) sum moments about point ; and (b) sum moments about point. N 2 mm T 1 N T 1 N 2 mm 8 kn 0.3 m 1. m 1.8 m 2 kn 8 kn 2 kn N Prob. 8 Prob. 83 * 8. Two couples act on the beam as shown. Determine the magnitude of so that the resultant couple moment is 300 lb # ft counterclockwise. Where on the beam does the resultant couple act? 86. Two couples act on the cantilever beam. If = 6 kn, determine the resultant couple moment. 87. Determine the required magnitude of force, if the resultant couple moment on the beam is to be ero. 3 m 3 m 1. ft 200 lb kn 3 0. m 200 lb 0. m 3 kn Prob. 8 Probs. 86/87

43 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 18 HPTER RE S YSTEM R E S U LT N T S * 88. Two couples act on the frame. If the resultant couple moment is to be ero, determine the distance d between the 0-lb couple forces. 89. Two couples act on the frame. If d = ft, determine the resultant couple moment. ompute the result b resolving each force into and components and (a) finding the moment of each couple (Eq. 13) and (b) summing the moments of all the force components about point. 90. Two couples act on the frame. If d = ft, determine the resultant couple moment. ompute the result b resolving each force into and components and (a) finding the moment of each couple (Eq. 13) and (b) summing the moments of all the force components about point. 3 ft 60 lb ft 0 lb 1 ft 3 d 3 60 lb 93. If = 80 N, determine the magnitude and coordinate direction angles of the couple moment. The pipe assembl lies in the plane. 9. If the magnitude of the couple moment acting on the pipe assembl is 0 N # m, determine the magnitude of the couple forces applied to each wrench. The pipe assembl lies in the plane. 200 mm 200 mm 300 mm 300 mm 300 mm Probs. 93/9 2 ft 0 lb Probs. 88/89/ If M1 = 00 N# m, M2 = 600 N# m, and M 3 = 0 N# m, determine the magnitude and coordinate direction angles of the resultant couple moment. * 92. Determine the required magnitude of couple moments M 1, M 2, and M 3 so that the resultant couple moment is M R = -300i + 0j - 600k6 N # m. 9. rom load calculations it is determined that the wing is subjected to couple moments M = 17 kip # ft and M = 2 kip # ft. Determine the resultant couple moments created about the and aes.the aes all lie in the same horiontal plane. M 3 M M 2 2 M M 1 Probs. 91/92 Prob. 9

44 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..6 MMENT UPLE 19 * 96. Epress the moment of the couple acting on the frame in artesian vector form. The forces are applied perpendicular to the frame. What is the magnitude of the couple moment? Take = 0 N. 97. In order to turn over the frame, a couple moment is applied as shown. If the component of this couple moment along the ais is M = -20i6 N # m, determine the magnitude of the couple forces. * 100. If M1 = 180 lb# ft, M2 = 90 lb# ft, and M 3 = 120 lb# ft, determine the magnitude and coordinate direction angles of the resultant couple moment Determine the magnitudes of couple moments M 1, M 2, and M 3 so that the resultant couple moment is ero. 10 lb ft 3 m 1 ft M 3 2 ft 2 ft 2 ft 1. m Probs. 96/97 M 2 Probs. 100/101 M 1 3 ft 98. Determine the resultant couple moment of the two couples that act on the pipe assembl.the distance from to is d = 00 mm. Epress the result as a artesian vector. 99. Determine the distance d between and so that the resultant couple moment has a magnitude of. M R = 20 N # m 102. If 1 = 100 lb and 2 = 200 lb, determine the magnitude and coordinate direction angles of the resultant couple moment Determine the magnitude of couple forces 1 and 2 so that the resultant couple moment acting on the block is ero. {3k} N { 0i} N 20 mm d { 3k} N {0i} N 30 mm 20 lb 2 ft 3 ft ft lb 2 Probs. 98/99 Probs. 102/103

45 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 160 HPTER RE S YSTEM R E S U LT N T S.7 Simplification of a orce and ouple Sstem Sometimes it is convenient to reduce a sstem of forces and couple moments acting on a bod to a simpler form b replacing it with an equivalent sstem, consisting of a single resultant force acting at a specific point and a resultant couple moment. sstem is equivalent if the eternal effects it produces on a bod are the same as those caused b the original force and couple moment sstem. In this contet, the eternal effects of a sstem refer to the translating and rotating motion of the bod if the bod is free to move, or it refers to the reactive forces at the supports if the bod is held fied. or eample, consider holding the stick in ig. 3a, which is subjected to the force at point. If we attach a pair of equal but opposite forces and at point, which is on the line of action of, ig. 3b, we observe that at and at will cancel each other, leaving onl at, ig. 3c. orce has now been moved from to without modifing its eternal effects on the stick; i.e., the reaction at the grip remains the same. This demonstrates the principle of transmissibilit, which states that a force acting on a bod (stick) is a sliding vector since it can be applied at an point along its line of action. We can also use the above procedure to move a force to a point that is not on the line of action of the force. If is applied perpendicular to the stick, as in ig. 3a, then we can attach a pair of equal but opposite forces and to, ig. 3b. orce is now applied at, and the other two forces, at and at, form a couple that produces the couple moment M = d, ig. 3c. Therefore, the force can be moved from to provided a couple moment M is added to maintain an equivalent sstem. This couple moment is determined b taking the moment of about. Since M is actuall a free vector, it can act at an point on the stick. In both cases the sstems are equivalent which causes a downward force and clockwise couple moment M = d to be felt at the grip. (a) (b) ig. 3 (c) d m d (a) (b) ig. 3 (c)

46 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..7 SIMPLIITIN RE ND UPLE SYSTEM 161 Sstem of orces and ouple Moments. Using the above method, a sstem of several forces and couple moments acting on a bod can be reduced to an equivalent single resultant force acting at a point and a resultant couple moment. or eample, in ig. 36a, is not on the line of action of 1, and so this force can be moved to point provided a couple moment M 1 = r 1 * is added to the bod. Similarl, the couple moment M 2 = r 2 * 2 should be added to the bod when we move 2 to point. inall, since the couple moment M is a free vector, it can just be moved to point. doing this, we obtain the equivalent sstem shown in ig. 36b, which produces the same eternal effects (support reactions) on the bod as that of the force and couple sstem shown in ig. 36a. If we sum the forces and couple moments, we obtain the resultant force R = and the resultant couple moment (M R ) = M + M 1 + M 2, ig. 36c. Notice that R is independent of the location of point ; however, (M R ) depends upon this location since the moments M 1 and M 2 are determined using the position vectors r 1 and r 2. lso note that (M R ) is a free vector and can act at an point on the bod, although point is generall chosen as its point of application. We can generalie the above method of reducing a force and couple sstem to an equivalent resultant force R acting at point and a resultant couple moment (M R ) b using the following two equations. (a) (b) M 2 M r 1 r 2 2 M 2 r M 1 r 1 1 R R = (M R ) = M + M ( 17) u M R The first equation states that the resultant force of the sstem is equivalent to the sum of all the forces; and the second equation states that the resultant couple moment of the sstem is equivalent to the sum of all the couple moments M plus the moments of all the forces M about point. If the force sstem lies in the plane and an couple moments are perpendicular to this plane, then the above equations reduce to the following three scalar equations. (c) ig. 36 ( R ) = ( R ) = (M R ) = M + M ( 18) Here the resultant force is determined from the vector sum of its two components and ( R ). ( R )

47 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 162 HPTER RE S YSTEM R E S U LT N T S d 1 d2 (M R ) W 1 W 2 W R The weights of these traffic lights can be replaced b their equivalent resultant force W R = W 1 + W 2 and a couple moment (M R ) = W 1 d 1 + W 2 d 2 at the support,. In both cases the support must provide the same resistance to translation and rotation in order to keep the member in the horiontal position. Procedure for nalsis The following points should be kept in mind when simplifing a force and couple moment sstem to an equivalent resultant force and couple sstem. Establish the coordinate aes with the origin located at point and the aes having a selected orientation. orce Summation. If the force sstem is coplanar, resolve each force into its and components. If a component is directed along the positive or ais, it represents a positive scalar; whereas if it is directed along the negative or ais, it is a negative scalar. In three dimensions, represent each force as a artesian vector before summing the forces. Moment Summation. When determining the moments of a coplanar force sstem about point, it is generall advantageous to use the principle of moments, i.e., determine the moments of the components of each force, rather than the moment of the force itself. In three dimensions use the vector cross product to determine the moment of each force about point. Here the position vectors etend from to an point on the line of action of each force.

48 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..7 SIMPLIITIN RE ND UPLE SYSTEM 163 EXMPLE.1 Replace the force and couple sstem shown in ig. 37a b an equivalent resultant force and couple moment acting at point. 3 kn (3 kn)sin 0.1 m 0.1 m 0.1 m 0.1 m (3 kn)cos 3 ( kn) 0.2 m 0.3 m kn 3 kn 0.2 m 0.3 m kn (a) ig. 37 SLUTIN orce Summation. The 3 kn and kn forces are resolved into their and components as shown in ig. 37b. We have (b) ( kn) : + ( R ) = ; ( R ) = (3 kn)cos ( kn) =.98 kn : + c( R ) = ; Using the Pthagorean theorem, ig. 37c, the magnitude of R = 21 R R 2 2 Its direction u is u = tan -1 a ( R) b = tan kn a ( R ).98 kn b = 9.3 R is ns. Moment Summation. The moments of 3 kn and kn about point will be determined using their and components. Referring to ig. 37b, we have a+ (M R ) = M ; ( R ) = (3 kn)sin 30 - ( kn) - kn (M R ) = (3 kn)sin 30 (0.2 m) - (3 kn)cos 30 (0.1 m) + 3 ( kn) (0.1 m) = -2.6 kn # m = 2.6 kn # m This clockwise moment is shown in ig. 37c. = kn kn2 2 = 8.8 kn - ( kn) (0. m) - ( kn)(0.2 m) NTE: Realie that the resultant force and couple moment in ig. 37c will produce the same eternal effects or reactions at the supports as those produced b the force sstem, ig 37a. b = -6.0 kn = 6.0 knt ns. ns. (M R ) 2.6 kn m ( R ).98 kn u R ( R ) 6.0 kn (c)

49 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 16 HPTER RE S YSTEM R E S U LT N T S EXMPLE.1 Replace the force and couple sstem acting on the member in ig. 38a b an equivalent resultant force and couple moment acting at point. 70 N 1 m 1.2 m 1.2 m (a) 3 00 N 200 N 1 m 200 N (M R ) 37. N m u ( R ) 300 N ( R ) 30 N R (b) SLUTIN orce Summation. Since the couple forces of 200 N are equal but opposite, the produce a ero resultant force, and so it is not necessar to consider them in the force summation. The 00-N force is resolved into its and components, thus, : + ( R ) = ; ig. 38 ( R ) = 3 (00 N) = 300 N : + c( R ) = ; ( R ) = (00 N) - 70 N = -30 N = 30 NT rom ig. 1b, the magnitude of R is R = 2( R ( R 2 2 nd the angle u is = 2(300 N) 2 + (30 N) 2 = 61 N u = tan -1 a ( R) ( R ) b = tan -1 a 30 N 300 N b = 9. ns. ns. Moment Summation. Since the couple moment is a free vector, it can act at an point on the member. Referring to ig. 38a, we have a + (M R ) = M + M c ; (M R ) = (00 N) (2. m) - (00 N) 3 (1 m) - (70 N)(1.2 m) N# m = -37. N# m = 37. N # m This clockwise moment is shown in ig. 38b. b ns.

50 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..7 SIMPLIITIN RE ND UPLE SYSTEM 16 EXMPLE.16 The structural member is subjected to a couple moment M and forces 1 and 2 in ig. 39a. Replace this sstem b an equivalent resultant force and couple moment acting at its base, point. SLUTIN (VETR NLYSIS) The three-dimensional aspects of the problem can be simplified b using a artesian vector analsis. Epressing the forces and couple moment as artesian vectors, we have 1 = -800k6 N M 00 N m 3 r N 0.1 m r N 0.1 m 2 = 1300 N2u 1 m = 1300 N2a r r b {-0.1i + 0.1j} m = 300 Nc d = -29.6i j6 N m m2 (a) M = -00 j k = -00j + 300k6 N # m orce Summation. R = ; R = = -800k i j = -20i + 166j - 800k6 N ns. M R Moment Summation. M R = M + M M R = M + r * 1 + r * 2 (b) R M R = 1-00j + 300k2 + 11k2 * 1-800k2+ 3 i j k ig. 39 = 1-00j + 300k i j2 = -166i - 60j + 300k6 N # m ns. The results are shown in ig. 39b.

51 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 166 HPTER RE S YSTEM R E S U LT N T S UNDMENTL PRLEMS 2. Replace the loading sstem b an equivalent resultant force and couple moment acting at point. 100 lb 28. Replace the loading sstem b an equivalent resultant force and couple moment acting at point. 100 lb 3 1 ft 0 lb ft 3 ft 3 ft 3 10 lb 3 ft 3 ft 10 lb 200 lb Replace the loading sstem b an equivalent resultant force and couple moment acting at point. 26. Replace the loading sstem b an equivalent resultant force and couple moment acting at point. 0 N 30 N N m 1 m 2 m 2 { 0k} N 1 { 300i 10j 200k} N 1. m 3 m 3 m N 27. Replace the loading sstem b an equivalent resultant force and couple moment acting at point Replace the loading sstem b an equivalent resultant force and couple moment acting at point N M c 7 N m 900 N 300 N 0.3 m N 300 N m 0. m 0. m 0.7 m 0.7 m 0.7 m 0.7 m 27 30

52 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..7 SIMPLIITIN RE ND UPLE SYSTEM 167 PRLEMS * 10. Replace the force sstem acting on the truss b a resultant force and couple moment at point Replace the two forces b an equivalent resultant force and couple moment at point. Set = 20 lb. * 108. Replace the two forces b an equivalent resultant force and couple moment at point. Set. = 1 lb 200 lb 10 lb 100 lb 2 ft 2 ft 2 ft 2 ft 3 6 in. 20 lb 1. in. 3 6 ft 00 lb 0 Prob in. Probs. 107/ Replace the force sstem acting on the beam b an equivalent force and couple moment at point Replace the force sstem acting on the post b a resultant force and couple moment at point Replace the force sstem acting on the beam b an equivalent force and couple moment at point. 0. m 2. kn 1. kn 3 3 kn 1 m 1 m 00 N 0.2 m 3 20 N 300 N 2 m m 2 m 1 m Probs. 10/106 Prob. 109

53 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 168 HPTER RE S YSTEM R E S U LT N T S 110. Replace the force and couple moment sstem acting on the overhang beam b a resultant force and couple moment at point. * 112. Replace the two forces acting on the grinder b a resultant force and couple moment at point. Epress the results in artesian vector form. 0.3 m 30 kn kn m 26 kn m 20 mm 1 {10i 1j 0k} N 2 m 1 m 1 m Prob m 100 mm 10 mm 2 mm 2 { 1i 20j 30k} N 0 mm Prob Replace the force sstem b a resultant force and couple moment at point Replace the two forces acting on the post b a resultant force and couple moment at point. Epress the results in artesian vector form. 00 N 70 N N D 7 kn kn 1.2 m 1.2 m 1 m 200 N 6 m 2 m D 3 m 6 m 8 m Prob. 111 Prob. 113

54 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..7 SIMPLIITIN RE ND UPLE SYSTEM The three forces act on the pipe assembl. If 1 = 0 N and 2 = 80 N, replace this force sstem b an equivalent resultant force and couple moment acting at. Epress the results in artesian vector form. * 116. Replace the force sstem acting on the pipe assembl b a resultant force and couple moment at point. Epress the results in artesian vector form. 2 { 10i 2j 20k} lb 180 N 1 { 20i 10j 2k}lb 1.2 m 2 2 ft 1. ft 1 0. m 0.7 m 2 ft 2 ft Prob. 11 Prob Handle forces 1 and are applied to the electric drill. Replace this force sstem b an equivalent resultant force and couple moment acting at point. Epress the results in artesian vector form The slab is to be hoisted using the three slings shown. Replace the sstem of forces acting on slings b an equivalent force and couple moment at point. The force 1 is vertical. 2 {2j k} N 0.1 m 1 {6i 3j 10k} N 3 kn 0.2 m 2 kn kn 0.3 m 6 m 2 m 2 m Prob. 11 Prob. 117

55 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 170 HPTER RE S YSTEM R E S U LT N T S.8 urther Simplification of a orce and ouple Sstem In the preceding section, we developed a wa to reduce a force and couple moment sstem acting on a rigid bod into an equivalent resultant force R acting at a specific point and a resultant couple moment (M R ).The force sstem can be further reduced to an equivalent single resultant force provided the lines of action of R and (M R ) are perpendicular to each other. ecause of this condition, onl concurrent, coplanar, and parallel force sstems can be further simplified. oncurrent orce Sstem. Since a concurrent force sstem is one in which the lines of action of all the forces intersect at a common point, ig. 0a, then the force sstem produces no moment about this point. s a result, the equivalent sstem can be represented b a single resultant force R = acting at, ig. 0b. 3 R 2 (a) 2 ig. 0 (b) oplanar orce Sstem. In the case of a coplanar force sstem, the lines of action of all the forces lie in the same plane, ig. 1a, and so the resultant force R = of this sstem also lies in this plane. urthermore, the moment of each of the forces about an point is directed perpendicular to this plane. Thus, the resultant moment (M R ) and resultant force R will be mutuall perpendicular, ig. 1b. The resultant moment can be replaced b moving the resultant force R a perpendicular or moment arm distance d awa from point such that R produces the same moment (M R ) about point, ig. 1c. This distance d can be determined from the scalar equation (M R ) = R d = M or d = (M R ) > R.

56 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..8 URTHER SIMPLIITIN RE ND UPLE SYSTEM R d R (M R ) 1 (a) (b) (c) ig. 1 Parallel orce Sstem. The parallel force sstem shown in ig. 2a consists of forces that are all parallel to the ais. Thus, the resultant force R = at point must also be parallel to this ais, ig. 2b. The moment produced b each force lies in the plane of the plate, and so the resultant couple moment, (M R ), will also lie in this plane, along the moment ais a since R and (M R ) are mutuall perpendicular. s a result, the force sstem can be further reduced to an equivalent single resultant force R, acting through point P located on the perpendicular b ais, ig. 2c. The distance d along this ais from point requires (M R ) = R d = M or d = M > R. 1 2 R 3 a (M R ) a d R P (a) (b) b (c) b ig. 2

57 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 172 HPTER RE S YSTEM R E S U LT N T S R The four cable forces are all concurrent at point on this bridge tower. onsequentl the produce no resultant moment there, onl a resultant force R. Note that the designers have positioned the cables so that R is directed along the bridge tower directl to the support, so that it does not cause an bending of the tower. Procedure for nalsis The technique used to reduce a coplanar or parallel force sstem to a single resultant force follows a similar procedure outlined in the previous section. Establish the,,, aes and locate the resultant force an arbitrar distance awa from the origin of the coordinates. orce Summation. R The resultant force is equal to the sum of all the forces in the sstem. or a coplanar force sstem, resolve each force into its and components. Positive components are directed along the positive and aes, and negative components are directed along the negative and aes. Moment Summation. The moment of the resultant force about point is equal to the sum of all the couple moments in the sstem plus the moments of all the forces in the sstem about. This moment condition is used to find the location of the resultant force from point.

58 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..8 URTHER SIMPLIITIN RE ND UPLE SYSTEM 173 d 1 d2 d W 1 W 2 W R Here the weights of the traffic lights are replaced b their resultant force W R = W 1 + W 2 which acts at a distance d = (W 1 d 1 + W 2 d 2 )> W R from. oth sstems are equivalent. Reduction to a Wrench In general, a three-dimensional force and couple moment sstem will have an equivalent resultant force R acting at point and a resultant couple moment (M R ) that are not perpendicular to one another, as shown in ig. 3a. lthough a force sstem such as this cannot be further reduced to an equivalent single resultant force, the resultant couple moment (M R ) can be resolved into components parallel and perpendicular to the line of action of R, ig. 3a. The perpendicular component M can be replaced if we move R to point P, a distance d from point along the b ais, ig. 3b. s seen, this ais is perpendicular to both the a ais and the line of action of R. The location of P can be determined from d = M > R. inall, because M is a free vector, it can be moved to point P, ig. 3c. This combination of a resultant force R and collinear couple moment M will tend to translate and rotate the bod about its ais and is referred to as a wrench or screw. wrench is the simplest sstem that can represent an general force and couple moment sstem acting on a bod. R M M R (M R ) a M b a d P R b a P M b (a) ig. 3 (b) (c)

59 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 17 HPTER RE S YSTEM R E S U LT N T S EXMPLE.17 Replace the force and couple moment sstem acting on the beam in ig. a b an equivalent resultant force, and find where its line of action intersects the beam, measured from point. kn 1 kn m 8 kn 3 0. m d ( R ) 2.0 kn u R (R).80 kn 1. m 1. m 1. m 1. m (a) (b) ig. SLUTIN orce Summation. : + ( R ) = ; + c( R ) = ; Summing the force components, ( R ) = 8 kn 3 =.80 kn : ( R ) = - kn + 8 kn = 2.0 knc rom ig. b, the magnitude of R is R = kn kn2 2 =.37 kn ns. The angle u is u = tan kn a.80 kn b = 26.6 ns. Moment Summation. We must equate the moment of R about point in ig. b to the sum of the moments of the force and couple moment sstem about point in ig. a. Since the line of action of ( R ) acts through point, onl ( R ) produces a moment about this point. Thus, a +(MR ) = M ; 2.0 kn(d) = -( kn)(1. m) - 1 kn# m -[8 kn 3 ] (0. m) + [8 kn ](. m) d = 2.2 m ns.

60 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..8 URTHER SIMPLIITIN RE ND UPLE SYSTEM 17 EXMPLE.18 The jib crane shown in ig. a is subjected to three coplanar forces. Replace this loading b an equivalent resultant force and specif where the resultant s line of action intersects the column and boom. 3 ft ft 3 ft 6 ft 3 17 lb SLUTIN orce Summation. Resolving the 20-lb force into and components and summing the force components ields : + R = ; R = -20 lb 3-17 lb = -32 lb = 32 lb ; ft 60 lb (a) 20 lb + c R = ; R = -20 lb - 60 lb = -260 lb = 260 lbt s shown b the vector addition in ig. b, 32 lb R = 2(32 lb) 2 + (260 lb) 2 = 16 lb u = tan lb a 32 lb b = 38.7 u ns. ns. R 260 lb Moment Summation. Moments will be summed about point. ssuming the line of action of R intersects at a distance from, ig. b, we have 32 lb u R 260 lb a+m R = M ; 32 lb lb 102 = 17 lb 1 ft2-60 lb 13 ft lb ft2-20 lb 18 ft2 (b) ig. = 2.29 ft ns. the principle of transmissibilit, R can be placed at a distance where it intersects, ig. b. In this case we have a+m R = M ; 32 lb 111 ft2-260 lb 12 = 17 lb 1 ft2-60 lb 13 ft lb ft2-20 lb 18 ft2 = 10.9 ft ns.

61 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 176 HPTER RE S YSTEM R E S U LT N T S EXMPLE N 00 N 00 N 100 N m m 2 m The slab in ig. 6a is subjected to four parallel forces. Determine the magnitude and direction of a resultant force equivalent to the given force sstem and locate its point of application on the slab. 8 m P(, ) R (a) ig. 6 (b) SLUTIN (SLR NLYSIS) orce Summation. rom ig. 6a, the resultant force is + c R = ; - R = -600 N N - 00 N - 00 N = -100 N = 100 NT ns. Moment Summation. We require the moment about the ais of the resultant force, ig. 6b, to be equal to the sum of the moments about the ais of all the forces in the sstem, ig. 6a.The moment arms are determined from the coordinates since these coordinates represent the perpendicular distances from the ais to the lines of action of the forces. Using the right-hand rule, we have (M R ) = M ; N2 = 600 N N1 m2-00 N110 m N = -300 = 2.0 m ns. In a similar manner, a moment equation can be written about the ais using moment arms defined b the coordinates of each force. (M R ) = M ; 1100 N2 = 600 N18 m2-100 N16 m N N = 200 = 3 m ns. NTE: force of R = 100 N placed at point P(3.00 m, 2.0 m) on the slab, ig. 6b, is therefore equivalent to the parallel force sstem acting on the slab in ig. 6a.

62 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..8 URTHER SIMPLIITIN RE ND UPLE SYSTEM 177 EXMPLE.20 Replace the force sstem in ig. 7a b an equivalent resultant force and specif its point of application on the pedestal. SLUTIN orce Summation. Summing forces, R = ; Here we will demonstrate a vector analsis. R = + + = -300k6 lb + -00k6 lb + 100k6 lb = -700k6 lb ns. 300 lb 100 lb in. r r r in. 00 lb 2 in. in. Location. Moments will be summed about point. The resultant force is assumed to act through point P (,, 0), ig. 7b. Thus R (a) (M R ) = M ; r P * R = (r * ) + (r * ) + (r * ) 1i + j2 * 1-700k2 = [1i2 * 1-300k2] + [1-i + 2j2 * 1-00k2] + [(-j) * (100k)] -700(i * k) - 700( j * k) = -1200(i * k) (i * k) ( j * k) - 00( j * k) 700j - 700i = 1200j j i - 00i r P P R { 700k} lb Equating the i and j components, -700 = -100 = 2 in. 700 = -800 = -1.1 in. (1) ns. (2) ns. (b) ig. 7 The negative sign indicates that the coordinate of point P is negative. NTE: It is also possible to establish Eq. 1 and 2 directl b summing moments about the and aes. Using the right-hand rule, we have (M R ) = M ; (M R ) = M ; -700 = -100 lb( in.) - 00 lb(2 in.) 700 = 300 lb1 in.2-00 lb1 in.)

63 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 178 HPTER RE S YSTEM R E S U LT N T S UNDMENTL PRLEMS 31. Replace the loading sstem b an equivalent resultant force and specif where the resultant s line of action intersects the beam measured from. 3. Replace the loading sstem b an equivalent resultant force and specif where the resultant s line of action intersects the member measured from. 0. m 1. m 00 lb 00 lb 20 lb 0. m 0. m 8 kn 6 kn 3 kn 3 ft 3 ft 3 ft 3 ft 3 m Replace the loading sstem b an equivalent resultant force and specif where the resultant s line of action intersects the member measured from. 200 lb 3 ft 3 ft 3 ft lb 0 lb 3 3. Replace the loading shown b an equivalent single resultant force and specif the and coordinates of its line of action. 00 N 100 N m 3 m 00 N m 33. Replace the loading sstem b an equivalent resultant force and specif where the resultant s line of action intersects the member measured from. 20 kn 2 m 2 m 2 m 1 kn 3 2 m Replace the loading shown b an equivalent single resultant force and specif the and coordinates of its line of action. 200 N 2 m 1 m 3 m 3 m 200 N 100 N 3 m 2 m 100 N 1 m 33 36

64 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..8 URTHER SIMPLIITIN RE ND UPLE SYSTEM 179 PRLEMS 118. The weights of the various components of the truck are shown. Replace this sstem of forces b an equivalent resultant force and specif its location measured from The sstem of four forces acts on the roof truss. Determine the equivalent resultant force and specif its location along, measured from point The weights of the various components of the truck are shown. Replace this sstem of forces b an equivalent resultant force and specif its location measured from point. 27 lb ft 300 lb ft 200 lb 10 lb ft 3 ft 300 lb 00 lb 1 ft 6 ft 2 ft 170 lb Prob. 121 Probs. 118/ Replace the force and couple sstem acting on the frame b an equivalent resultant force and specif where the resultant s line of action intersects member, measured from. * 120. The sstem of parallel forces acts on the top of the Warren truss. Determine the equivalent resultant force of the sstem and specif its location measured from point Replace the force and couple sstem acting on the frame b an equivalent resultant force and specif where the resultant s line of action intersects member, measured from. 2 kn 2 ft 1 kn 00 N 00 N 00 N 1 m 1 m 1 m 1 m 00 lb ft 10 lb 3 ft 0 lb 3 ft Prob. 120 Probs. 122/123

65 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 180 HPTER RE S YSTEM R E S U LT N T S * 12. Replace the force and couple moment sstem acting on the overhang beam b a resultant force, and specif its location along measured from point Replace the force sstem acting on the post b a resultant force, and specif where its line of action intersects the post measured from point. * 128. Replace the force sstem acting on the post b a resultant force, and specif where its line of action intersects the post measured from point. 0.3 m 30 kn kn m 26 kn m 1 m 00 N 0. m 0.2 m 3 20 N 2 m 1 m 1 m 2 m 1 m 300 N Prob m Probs. 127/ Replace the force sstem acting on the frame b an equivalent resultant force and specif where the resultant s line of action intersects member, measured from point Replace the force sstem acting on the frame b an equivalent resultant force and specif where the resultant s line of action intersects member, measured from point The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specif its location (, ) on the slab. Take 1 = 30 kn, 2 = 0 kn The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specif its location (, ) on the slab. Take 1 = 20 kn, 2 = 0 kn. 3 lb ft 20 lb 2 ft 20 kn 0 kn 1 3 ft 2 lb 2 2 ft 3 m 8 m 6 m m 2 m Probs. 12/126 Probs. 129/130

66 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..8 URTHER SIMPLIITIN RE ND UPLE SYSTEM The tube supports the four parallel forces. Determine the magnitudes of forces and D acting at and D so that the equivalent resultant force of the force sstem acts through the midpoint of the tube. 13. If = 0 kn and = 3 kn, determine the magnitude of the resultant force and specif the location of its point of application (, ) on the slab. 13. If the resultant force is required to act at the center of the slab, determine the magnitude of the column loadings and and the magnitude of the resultant force. D 600 N 00 mm D 00 mm 00 N 200 mm 200 mm 2. m 0.7 m 0.7 m 3 m 30 kn 0.7 m 90 kn 2. m 3 m 0.7 m 20 kn Prob. 131 Probs. 13/13 * 132. Three parallel bolting forces act on the circular plate. Determine the resultant force, and specif its location (, ) on the plate. = 200 lb, = 100 lb, and = 00 lb. * 136. Replace the parallel force sstem acting on the plate b a resultant force and specif its location on the plane The three parallel bolting forces act on the circular plate. If the force at has a magnitude of = 200 lb, determine the magnitudes of and so that the resultant force R of the sstem has a line of action that coincides with the ais. Hint: This requires M = 0 and M = 0. 1 m 0. m 2 kn kn 1. ft 1 m 1 m 0. m 3 kn Probs. 132/133 Prob. 136

67 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 182 HPTER RE S YSTEM R E S U LT N T S 137. If = 7 kn and = kn, represent the force sstem acting on the corbels b a resultant force, and specif its location on the plane Determine the magnitudes of and so that the resultant force passes through point of the column. * 10. Replace the three forces acting on the plate b a wrench. Specif the magnitude of the force and couple moment for the wrench and the point P(, ) where its line of action intersects the plate. 6 kn 10 mm 100 mm 60 mm 600 mm 70 mm 8kN 700 mm 100 mm 10 mm 12 ft P { 60j} lb 12 ft Probs. 137/138 { 80k}lb { 0i} lb Prob Replace the force and couple moment sstem acting on the rectangular block b a wrench. Specif the magnitude of the force and couple moment of the wrench and where its line of action intersects the plane. 11. Replace the three forces acting on the plate b a wrench. Specif the magnitude of the force and couple moment for the wrench and the point P(, ) where its line of action intersects the plate. {00i} N {800k} N ft 600 lb ft 0 lb 600 lb P 2 ft 3 ft m 6 m {300j} N 300 lb Prob. 139 Prob. 11

68 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..9 REDUTIN SIMPLE DISTRIUTED LDING Reduction of a Simple Distributed Loading p Sometimes, a bod ma be subjected to a loading that is distributed over its surface. or eample, the pressure of the wind on the face of a sign, the pressure of water within a tank, or the weight of sand on the floor of a storage container, are all distributed loadings. The pressure eerted at each point on the surface indicates the intensit of the loading. It is measured using pascals Pa (or N/m 2 ) in SI units or lb/ft 2 in the U.S. ustomar sstem. b L R (a) p p() Uniform Loading long a Single is. The most common tpe of distributed loading encountered in engineering practice is generall uniform along a single ais.* or eample, consider the beam (or plate) in ig. 8a that has a constant width and is subjected to a pressure loading that varies onl along the ais. This loading can be described b the function p = p() N/m 2. It contains onl one variable, and for this reason, we can also represent it as a coplanar distributed load. To do so, we multipl the loading function b the width b m of the beam, so that w() = p()b N/m, ig. -8b. Using the methods of Sec..8, we can replace this coplanar parallel force sstem with a single equivalent resultant force R acting at a specific location on the beam, ig. 8c. w w d d d w w() L (b) Magnitude of Resultant orce. rom Eq R = 2, the magnitude of R is equivalent to the sum of all the forces in the sstem. In this case integration must be used since there is an infinite number of parallel forces d acting on the beam, ig. 8b. Since d is acting on an element of length d, and w() is a force per unit length, then d = w12 d = d. In other words, the magnitude of d is determined from the colored differential area d under the loading curve. or the entire length L, R L (c) ig. 8 +T R = ; R = Lw12 d = L d = ( 19) Therefore, the magnitude of the resultant force is equal to the total area under the loading diagram, ig. 8c. *The more general case of a nonuniform surface loading acting on a bod is considered in Sec. 9..

69 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 18 HPTER RE S YSTEM R E S U LT N T S p R p p() Location of Resultant orce. ppling Eq. 17 1M R = M 2, the location of the line of action of R can be determined b equating the moments of the force resultant and the parallel force distribution about point (the ais). Since d produces a moment of d = w12 d about, ig. 8b, then for the entire length, ig. 8c, b a+ (M R ) = M ; - R = - L w12 d L Solving for, using Eq. 19, we have w (a) d d w w() w12 d = L w12 d L = L d d L ( 20) w d L (b) R L (c) This coordinate, locates the geometric center or centroid of the area under the distributed loading. In other words, the resultant force has a line of action which passes through the centroid (geometric center) of the area under the loading diagram, ig. 8c. Detailed treatment of the integration techniques for finding the location of the centroid for areas is given in hapter 9. In man cases, however, the distributed-loading diagram is in the shape of a rectangle, triangle, or some other simple geometric form. The centroid location for such common shapes does not have to be determined from the above equation but can be obtained directl from the tabulation given on the inside back cover. nce is determined, R b smmetr passes through point 1, 02 on the surface of the beam, ig. 8a. Therefore, in this case the resultant force has a magnitude equal to the volume under the loading curve p = p12 and a line of action which passes through the centroid (geometric center) of this volume. a b 2 b R w 0 The beam supporting this stack of lumber is subjected to a uniform loading of w 0. The resultant force is therefore equal to the area under the loading diagram R = w 0 b. It acts trough the centroid or geometric center of this area, b>2 from the support. Important Points oplanar distributed loadings are defined b using a loading function w = w12 that indicates the intensit of the loading along the length of a member. This intensit is measured in N>m or lb>ft. The eternal effects caused b a coplanar distributed load acting on a bod can be represented b a single resultant force. This resultant force is equivalent to the area under the loading diagram, and has a line of action that passes through the centroid or geometric center of this area.

70 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..9 REDUTIN SIMPLE DISTRIUTED LDING 18 EXMPLE.21 Determine the magnitude and location of the equivalent resultant force acting on the shaft in ig. 9a. w w w (60 2 )N/m 20 N/m R 160 N d w d 2 m d 1. m (a) (b) ig. 9 SLUTIN Since w = w12 is given, this problem will be solved b integration. The differential element has an area d = w d = 60 2 d. ppling Eq. 19, +T R = ; 2 m R = d = 60 2 d = 60 3 L L 0 3 ` 0 = 160 N ns. The location of R measured from, ig. 9b, is determined from Eq. 20. = L d d L = L 0 2 m d 160 N = 2 m 60 ` 2 m N = = N = 1. m ns. NTE: These results can be checked b using the table on the inside back cover, where it is shown that for an eparabolic area of length a, height b, and shape shown in ig. 9a, we have = ab 3 2 m120 N>m2 = 3 = 160 N and = 3 a = 3 12 m2 = 1. m

71 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 186 HPTER RE S YSTEM R E S U LT N T S EXMPLE.22 distributed loading of p = (800) Pa acts over the top surface of the beam shown in ig. 0a. Determine the magnitude and location of the equivalent resultant force. p = 800 Pa 7200 Pa p 0.2 m 9 m (a) w w 160 N/m 9 m (b) 10 N/m SLUTIN Since the loading intensit is uniform along the width of the beam (the ais), the loading can be viewed in two dimensions as shown in ig. 0b. Here w = 1800 N>m m2 = N>m t = 9 m, note that w = 10 N>m. lthough we ma again appl Eqs. 19 and 20 as in the previous eample, it is simpler to use the table on the inside back cover. The magnitude of the resultant force is equivalent to the area of the triangle. R 6.8 kn R = m2110 N>m2 = 680 N = 6.8 kn ns. 6 m (c) ig. 0 3 m The line of action of Hence, R The results are shown in ig. 0c. passes through the centroid of this triangle. = 9 m m2 = 6 m ns. NTE: We ma also view the resultant R as acting through the centroid of the volume of the loading diagram p = p12 in ig. 0a. Hence R intersects the plane at the point (6 m, 0). urthermore, the magnitude of R is equal to the volume under the loading diagram; i.e., R = V = N>m2 219 m210.2 m2 = 6.8 kn ns.

72 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..9 REDUTIN SIMPLE DISTRIUTED LDING 187 EXMPLE.23 The granular material eerts the distributed loading on the beam as shown in ig. 1a. Determine the magnitude and location of the equivalent resultant of this load. 100 lb/ft SLUTIN The area of the loading diagram is a trapeoid, and therefore the solution can be obtained directl from the area and centroid formulas for a trapeoid listed on the inside back cover. Since these formulas are not easil remembered, instead we will solve this problem b using composite areas. Here we will divide the trapeoidal loading into a rectangular and triangular loading as shown in ig. 1b. The magnitude of the force represented b each of these loadings is equal to its associated area, 0 lb/ft 0 lb/ft 9 ft (a) lb/ft 1 = ft210 lb>ft2 = 22 lb 2 = 19 ft210 lb>ft2 = 0 lb The lines of action of these parallel forces act through the centroid of their associated areas and therefore intersect the beam at ft (b) 1 = ft2 = 3 ft R The two parallel forces 1 and 2 can be reduced to a single resultant R. The magnitude of is R 2 = ft2 =. ft +T R = ; R = = 67 lb ns. (c) We can find the location of R with reference to point, ig. 1b and 1c. We require c+ M R = M ; 1672 = = ft ns. 100 lb/ft lb/ft NTE: The trapeoidal area in ig. 1a can also be divided into two triangular areas as shown in ig. 1d. In this case and 3 = ft21100 lb>ft2 = 0 lb = ft210 lb>ft2 = 22 lb 3 = ft2 = 3 ft 9 ft (d) ig. 1 = 9 ft ft2 = 6 ft NTE: Using these results, show that again R = 67 lb and = ft.

73 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 188 HPTER RE S YSTEM R E S U LT N T S UNDMENTL PRLEMS 37. Determine the resultant force and specif where it acts on the beam measured from. 0. Determine the resultant force and specif where it acts on the beam measured from. 6 kn/m 9 kn/m 3 kn/m 10 lb/ft 200 lb/ft 00 lb 1. m 3 m 1. m 6 ft 3 ft 3 ft Determine the resultant force and specif where it acts on the beam measured from. 1. Determine the resultant force and specif where it acts on the beam measured from. 10 lb/ft 6 kn/m 3 kn/m 6 ft 8 ft. m 1. m Determine the resultant force and specif where it acts on the beam measured from. 2. Determine the resultant force and specif where it acts on the beam measured from. 6 kn/m w 160 N/m w m 6 m m 39 2

74 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..9 REDUTIN SIMPLE DISTRIUTED LDING 189 PRLEMS 12. Replace the distributed loading with an equivalent resultant force, and specif its location on the beam measured from point. 1. Replace the distributed loading with an equivalent resultant force, and specif its location on the beam measured from point. 1 kn/m w 0 w 0 10 kn/m 3 m 3 m Prob m 13. Replace the distributed loading with an equivalent resultant force, and specif its location on the beam measured from point. L 2 L 2 Prob The distribution of soil loading on the bottom of a building slab is shown. Replace this loading b an equivalent resultant force and specif its location, measured from point. 8 kn/m kn/m 0 lb/ft 100 lb/ft 800 N/m 3 m 3 m Prob. 13 * 1. Replace the distributed loading b an equivalent resultant force and specif its location, measured from point. 200 N/m 300 lb/ft 12 ft 9 ft Prob Determine the intensities w 1 and w 2 of the distributed loading acting on the bottom of the slab so that this loading has an equivalent resultant force that is equal but opposite to the resultant of the distributed loading acting on the top of the plate. 3 ft 6 ft 300 lb/ft 1. ft 2 m 3 m w 1 w 2 Prob. 1 Prob. 17

75 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 190 HPTER RE S YSTEM R E S U LT N T S * 18. The bricks on top of the beam and the supports at the bottom create the distributed loading shown in the second figure. Determine the required intensit w and dimension d of the right support so that the resultant force and couple moment about point of the sstem are both ero. 10. The beam is subjected to the distributed loading. Determine the length b of the uniform load and its position a on the beam such that the resultant force and couple moment acting on the beam are ero. a b 0 lb/ft 0. m 3 m d 200 N/m 60 lb/ft 10 ft 6 ft 7 N/m 0. m w Prob m d Prob The wind pressure acting on a triangular sign is uniform. Replace this loading b an equivalent resultant force and couple moment at point. 1.2 m 0.1 m 11. urrentl eight-five percent of all neck injuries are caused b rear-end car collisions. To alleviate this problem, an automobile seat restraint has been developed that provides additional pressure contact with the cranium. During dnamic tests the distribution of load on the cranium has been plotted and shown to be parabolic. Determine the equivalent resultant force and its location, measured from point. 10 Pa 1.2 m 0. ft 12 lb/ft w w 12(1 2 2 ) lb/ft 1 m 18 lb/ft Prob. 19 Prob. 11

76 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..9 REDUTIN SIMPLE DISTRIUTED LDING 191 * 12. Wind has blown sand over a platform such that the intensit of the load can be approimated b the function w = N>m. Simplif this distributed loading to an equivalent resultant force and specif its magnitude and location measured from. 1. Replace the distributed loading with an equivalent resultant force, and specif its location on the beam measured from point. w 00 N/m 8 kn/m w w 1 ( ) 2 2 w (0. 3 ) N/m m 10 m Prob. 1 Prob Wet concrete eerts a pressure distribution along the wall of the form. Determine the resultant force of this distribution and specif the height h where the bracing strut should be placed so that it lies through the line of action of the resultant force. The wall has a width of m. 1. Replace the loading b an equivalent resultant force and couple moment at point. * 16. Replace the loading b an equivalent resultant force and couple moment acting at point. p 0 lb/ft 0 lb/ft m p ( 1 /2 ) kpa ft h 8 kpa 6 ft 100 lb/ft 60 Prob. 13 Probs. 1/16

77 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 192 HPTER RE S YSTEM R E S U LT N T S 17. The lifting force along the wing of a jet aircraft consists of a uniform distribution along, and a semiparabolic distribution along with origin at. Replace this loading b a single resultant force and specif its location measured from point. * 160. The distributed load acts on the beam as shown. Determine the magnitude of the equivalent resultant force and specif its location, measured from point lb/ft w w ( ) lb/ft w w ( ) lb/ft 1 1 lb/ft 2 lb/ft 12 ft 2 ft Prob ft Prob The distributed load acts on the beam as shown. Determine the magnitude of the equivalent resultant force and specif where it acts, measured from point. 19. The distributed load acts on the beam as shown. Determine the maimum intensit w ma. What is the magnitude of the equivalent resultant force? Specif where it acts, measured from point If the distribution of the ground reaction on the pipe per foot of length can be approimated as shown, determine the magnitude of the resultant force due to this loading. 2 lb/ft w w ( ) lb/ft u 2. ft 0 lb/ft w 2 (1 cos u) lb/ft ft Probs. 18/19 Prob. 161

78 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. HPTER REVIEW 193 HPTER REVIEW Moment of orce Scalar Definition Moment ais force produces a turning effect or moment about a point that does not lie on its line of action. In scalar form, the moment magnitude is the product of the force and the moment arm or perpendicular distance from point to the line of action of the force. M = d d M The direction of the moment is defined using the right-hand rule. M alwas acts along an ais perpendicular to the plane containing and d, and passes through the point. Rather than finding d, it is normall easier to resolve the force into its and components, determine the moment of each component about the point, and then sum the results. This is called the principle of moments. M = d = - d Moment of a orce Vector Definition Since three-dimensional geometr is generall more difficult to visualie, the vector cross product should be used to determine the moment. Here M = r *, where r is a position vector that etends from point to an point,, or on the line of action of. If the position vector r and force are epressed as artesian vectors, then the cross product results from the epansion of a determinant. M = r * = r * = r * i j k M = r * = 3 r r r 3 M r r r

79 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 19 HPTER RE S YSTEM R E S U LT N T S Moment about an is If the moment of a force is to be determined about an arbitrar ais a, then the projection of the moment onto the ais must be obtained. Provided the distance d a that is perpendicular to both the line of action of the force and the ais can be found, then the moment of the force about the ais can be determined from a scalar equation. M a = d a a M a r a d a Note that when the line of action of intersects the ais then the moment of about the ais is ero.lso, when the line of action of is parallel to the ais, the moment of about the ais is ero. M a In three dimensions, the scalar triple product should be used. Here u a is the unit vector that specifies the direction of the ais, and r is a position vector that is directed from an point on the ais to an point on the line of action of the force. If M a is calculated as a negative scalar, then the sense of direction of M a is opposite to u a. u a u a u M a = u a # 1r * 2 = 3 r r r 3 u a is of projection a r ouple Moment couple consists of two equal but opposite forces that act a perpendicular distance d apart. ouples tend to produce a rotation without translation. M = d d The magnitude of the couple moment is M = d, and its direction is established using the right-hand rule. r If the vector cross product is used to determine the moment of a couple, then r etends from an point on the line of action of one of the forces to an point on the line of action of the other force that is used in the cross product. M = r *

80 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. HPTER REVIEW 19 Simplification of a orce and ouple Sstem n sstem of forces and couples can be reduced to a single resultant force and resultant couple moment acting at a point. The resultant force is the sum of all the forces in the sstem, R =, and the resultant couple moment is equal to the sum of all the moments of the forces about the point and couple moments. = M + M. M R 2 M 1 M R R r 1 r 2 u urther simplification to a single resultant force is possible provided the force sstem is concurrent, coplanar, or parallel. To find the location of the resultant force from a point, it is necessar to equate the moment of the resultant force about the point to the moment of the forces and couples in the sstem about the same point. b a R M R b a b d M R R a P R b a If the resultant force and couple moment at a point are not perpendicular to one another, then this sstem can be reduced to a wrench, which consists of the resultant force and collinear couple moment. M R u R b a d M R P a b oplanar Distributed Loading simple distributed loading can be represented b its resultant force, which is equivalent to the area under the loading curve. This resultant has a line of action that passes through the centroid or geometric center of the area or volume under the loading diagram. w w w() L L R

81 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 196 HPTER RE S YSTEM R E S U LT N T S REVIEW PRLEMS 162. The beam is subjected to the parabolic loading. Determine an equivalent force and couple sstem at point. * 16. Determine the coordinate direction angles a, b, g of, which is applied to the end of the pipe assembl, so that the moment of about is ero. 16. Determine the moment of the force about point. The force has coordinate direction angles of a = 60, b = 120, g =. Epress the result as a artesian vector. w 00 lb/ft 20 lb w (2 2 )lb/ft ft 6 in. 10 in. Prob in. 6 in. Probs. 16/ Two couples act on the frame. If the resultant couple moment is to be ero, determine the distance d between the 100-lb couple forces The snorkel boom lift is etended into the position shown. If the worker weighs 160 lb, determine the moment of this force about the connection at. 3 ft 100 lb 30 d 3 ft 10 lb 3 2 ft 2 ft lb ft 0 10 lb 3 Prob. 163 Prob. 166

82 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. REVIEW PRLEMS Determine the moment of the force about the door hinge at. Epress the result as a artesian vector. * 168. Determine the magnitude of the moment of the force about the hinged ais aa of the door Replace the force at b an equivalent resultant force and couple moment at point P. Epress the results in artesian vector form. 2. m 1. m 20 N a P 10 ft 6 ft ft 120 lb 8 ft 6 ft 8 ft a 1 m 0. m Prob. 171 Probs. 167/ Epress the moment of the couple acting on the pipe assembl in artesian vector form. Solve the problem (a) using Eq. 13 and (b) summing the moment of each force about point. Take = 2k6 N If the couple moment acting on the pipe has a magnitude of 00 N # m, determine the magnitude of the vertical force applied to each wrench. * 172. The horiontal 30-N force acts on the handle of the wrench. Determine the moment of this force about point. Specif the coordinate direction angles a, b, g of the moment ais The horiontal 30-N force acts on the handle of the wrench. What is the magnitude of the moment of this force about the ais? 300 mm 200 mm 10 mm 200 mm 30 N 00 mm 200 mm 0 mm 10 mm Probs. 169/170 Probs. 172/173

83 The crane is subjected to its weight and the load it supports. In order to calculate the support reactions on the crane, it is necessar to appl the principles of equilibrium Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved.

84 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. Equilibrium of a Rigid od HPTER JETIVES To develop the equations of equilibrium for a rigid bod. To introduce the concept of the free-bod diagram for a rigid bod. To show how to solve rigid-bod equilibrium problems using the equations of equilibrium..1 onditions for Rigid-od Equilibrium In this section, we will develop both the necessar and sufficient conditions for the equilibrium of the rigid bod in ig. 1a. s shown, this bod is subjected to an eternal force and couple moment sstem that is the result of the effects of gravitational, electrical, magnetic, or contact forces caused b adjacent bodies. The internal forces caused b interactions between particles within the bod are not shown in this figure because these forces occur in equal but opposite collinear pairs and hence will cancel out, a consequence of Newton s third law. 1 M M 1 (a) ig. 1

85 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 200 HPTER EQ U I L I R I U M R IGID DY 1 M 2 Using the methods of the previous chapter, the force and couple moment sstem acting on a bod can be reduced to an equivalent resultant force and resultant couple moment at an arbitrar point on or off the bod, ig. 1b. If this resultant force and couple moment are both equal to ero, then the bod is said to be in equilibrium. Mathematicall, the equilibrium of a bod is epressed as 3 M 1 (a) 2 R = = 0 (M R ) = M = 0 ( 1) (M R ) 0 (b) (M R ) 0 r (c) ig. 1 R 0 R 0 The first of these equations states that the sum of the forces acting on the bod is equal to ero. The second equation states that the sum of the moments of all the forces in the sstem about point, added to all the couple moments, is equal to ero. These two equations are not onl necessar for equilibrium, the are also sufficient. To show this, consider summing moments about some other point, such as point in ig. 1c. We require M = r * R + (M R ) = 0 Since r Z 0, this equation is satisfied onl if Eqs. 1 are satisfied, namel R = 0 and (M R ) = 0. When appling the equations of equilibrium, we will assume that the bod remains rigid. In realit, however, all bodies deform when subjected to loads. lthough this is the case, most engineering materials such as steel and concrete are ver rigid and so their deformation is usuall ver small. Therefore, when appling the equations of equilibrium, we can generall assume that the bod will remain rigid and not deform under the applied load without introducing an significant error. This wa the direction of the applied forces and their moment arms with respect to a fied reference remain unchanged before and after the bod is loaded. EQUILIRIUM IN TW DIMENSINS W G 2T ig. 2 R In the first part of the chapter, we will consider the case where the force sstem acting on a rigid bod lies in or ma be projected onto a single plane and, furthermore, an couple moments acting on the bod are directed perpendicular to this plane. This tpe of force and couple sstem is often referred to as a two-dimensional or coplanar force sstem. or eample, the airplane in ig. 2 has a plane of smmetr through its center ais, and so the loads acting on the airplane are smmetrical with respect to this plane. Thus, each of the two wing tires will support the same load T, which is represented on the side (two-dimensional) view of the plane as 2T.

86 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..2 REE-DY DIGRMS ree-od Diagrams Successful application of the equations of equilibrium requires a complete specification of all the known and unknown eternal forces that act on the bod. The best wa to account for these forces is to draw a free-bod diagram.this diagram is a sketch of the outlined shape of the bod, which represents it as being isolated or free from its surroundings, i.e., a free bod. n this sketch it is necessar to show all the forces and couple moments that the surroundings eert on the bod so that these effects can be accounted for when the equations of equilibrium are applied. thorough understanding of how to draw a free-bod diagram is of primar importance for solving problems in mechanics. Support Reactions. efore presenting a formal procedure as to how to draw a free-bod diagram, we will first consider the various tpes of reactions that occur at supports and points of contact between bodies subjected to coplanar force sstems. s a general rule, If a support prevents the translation of a bod in a given direction, then a force is developed on the bod in that direction. If rotation is prevented, a couple moment is eerted on the bod. or eample, let us consider three was in which a horiontal member, such as a beam, is supported at its end. ne method consists of a roller or clinder, ig. 3a. Since this support onl prevents the beam from translating in the vertical direction, the roller will onl eert a force on the beam in this direction, ig. 3b. The beam can be supported in a more restrictive manner b using a pin, ig. 3c. The pin passes through a hole in the beam and two leaves which are fied to the ground. Here the pin can prevent translation of the beam in an direction f, ig. 3d, and so the pin must eert a force on the beam in this direction. or purposes of analsis, it is generall easier to represent this resultant force b its two rectangular components and, ig. 3e. If and are known, then and f can be calculated. The most restrictive wa to support the beam would be to use a fied support as shown in ig. 3f. This support will prevent both translation and rotation of the beam. To do this a force and couple moment must be developed on the beam at its point of connection, ig. 3g. s in the case of the pin, the force is usuall represented b its rectangular components and. Table 1 lists other common tpes of supports for bodies subjected to coplanar force sstems. (In all cases the angle u is assumed to be known.) arefull stud each of the smbols used to represent these supports and the tpes of reactions the eert on their contacting members. roller (a) (b) pin (c) f or (d) fied support (f) (e) M (g) ig. 3

87 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 202 HPTER EQ U I L I R I U M R IGID DY (1) TLE 1 Supports for Rigid odies Subjected to Two-Dimensional orce Sstems Tpes of onnection Reaction Number of Unknowns u u ne unknown. The reaction is a tension force which acts awa from the member in the direction of the cable. cable (2) u weightless link u or u ne unknown. The reaction is a force which acts along the ais of the link. (3) ne unknown. The reaction is a force which acts perpendicular to the surface at the point of contact. u roller u () u u or u ne unknown. The reaction is a force which acts perpendicular to the slot. roller or pin in confined smooth slot () u rocker u ne unknown. The reaction is a force which acts perpendicular to the surface at the point of contact. (6) u smooth contacting surface u ne unknown. The reaction is a force which acts perpendicular to the surface at the point of contact. (7) u u u member pin connected to collar on smooth rod or ne unknown. The reaction is a force which acts perpendicular to the rod. continued

88 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..2 REE-DY DIGRMS 203 TLE 1 ontinued Tpes of onnection Reaction Number of Unknowns (8) u smooth pin or hinge or f Two unknowns. The reactions are two components of force, or the magnitude and direction f of the resultant force. Note that f and u are not necessaril equal [usuall not, unless the rod shown is a link as in (2)]. (9) member fied connected to collar on smooth rod M Two unknowns. The reactions are the couple moment and the force which acts perpendicular to the rod. (10) M or M f Three unknowns. The reactions are the couple moment and the two force components, or the couple moment and the magnitude and direction f of the resultant force. fied support Tpical eamples of actual supports are shown in the following sequence of photos. The numbers refer to the connection tpes in Table 1. This concrete girder rests on the ledge that is assumed to act as a smooth contacting surface. (6) The cable eerts a force on the bracket in the direction of the cable. (1) The rocker support for this bridge girder allows horiontal movement so the bridge is free to epand and contract due to a change in temperature. () This utilit building is pin supported at the top of the column. (8) The floor beams of this building are welded together and thus form fied connections. (10)

89 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 20 HPTER EQ U I L I R I U M R IGID DY Internal orces. s stated in Sec..1, the internal forces that act between adjacent particles in a bod alwas occur in collinear pairs such that the have the same magnitude and act in opposite directions (Newton s third law). Since these forces cancel each other, the will not create an eternal effect on the bod. It is for this reason that the internal forces should not be included on the free-bod diagram if the entire bod is to be considered. or eample, the engine shown in ig. a has a free-bod diagram shown in ig. b. The internal forces between all its connected parts such as the screws and bolts, will cancel out because the form equal and opposite collinear pairs. nl the eternal forces T 1 and T 2, eerted b the chains and the engine weight W, are shown on the free-bod diagram. T 2 T 1 G (a) W ig. Weight and the enter of Gravit. When a bod is within a gravitational field, then each of its particles has a specified weight. It was shown in Sec..8 that such a sstem of forces can be reduced to a single resultant force acting through a specified point. We refer to this force resultant as the weight W of the bod and to the location of its point of application as the center of gravit. The methods used for its determination will be developed in hapter 9. In the eamples and problems that follow, if the weight of the bod is important for the analsis, this force will be reported in the problem statement. lso, when the bod is uniform or made from the same material, the center of gravit will be located at the bod s geometric center or centroid; however, if the bod consists of a nonuniform distribution of material, or has an unusual shape, then the location of its center of gravit G will be given. Idealied Models. When an engineer performs a force analsis of an object, he or she considers a corresponding analtical or idealied model that gives results that approimate as closel as possible the actual situation. To do this, careful choices have to be made so that selection of the tpe of supports, the material behavior, and the object s dimensions can be justified. This wa one can feel confident that an design or analsis will ield results which can be trusted. In comple (b)

90 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..2 REE-DY DIGRMS 20 cases this process ma require developing several different models of the object that must be analed. In an case, this selection process requires both skill and eperience. The following two cases illustrate what is required to develop a proper model. In ig. a, the steel beam is to be used to support the three roof joists of a building. or a force analsis it is reasonable to assume the material (steel) is rigid since onl ver small deflections will occur when the beam is loaded. bolted connection at will allow for an slight rotation that occurs here when the load is applied, and so a pin can be considered for this support. t a roller can be considered since this support offers no resistance to horiontal movement. uilding code is used to specif the roof loading so that the joist loads can be calculated. These forces will be larger than an actual loading on the beam since the account for etreme loading cases and for dnamic or vibrational effects. inall, the weight of the beam is generall neglected when it is small compared to the load the beam supports. The idealied model of the beam is therefore shown with average dimensions a, b, c, and d in ig. b. s a second case, consider the lift boom in ig. 6a. inspection, it is supported b a pin at and b the hdraulic clinder, which can be approimated as a weightless link. The material can be assumed rigid, and with its densit known, the weight of the boom and the location of its center of gravit G are determined. When a design loading P is specified, the idealied model shown in ig. 6b can be used for a force analsis. verage dimensions (not shown) are used to specif the location of the loads and the supports. Idealied models of specific objects will be given in some of the eamples throughout the tet. It should be realied, however, that each case represents the reduction of a practical situation using simplifing assumptions like the ones illustrated here. (a) a b c d (b) ig. P G (a) ig. 6 (b)

91 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 206 HPTER EQ U I L I R I U M R IGID DY Procedure for nalsis To construct a free-bod diagram for a rigid bod or an group of bodies considered as a single sstem, the following steps should be performed: Draw utlined Shape. Imagine the bod to be isolated or cut free from its constraints and connections and draw (sketch) its outlined shape. Show ll orces and ouple Moments. Identif all the known and unknown eternal forces and couple moments that act on the bod. Those generall encountered are due to (1) applied loadings, (2) reactions occurring at the supports or at points of contact with other bodies (see Table 1), and (3) the weight of the bod. To account for all these effects, it ma help to trace over the boundar, carefull noting each force or couple moment acting on it. Identif Each Loading and Give Dimensions. The forces and couple moments that are known should be labeled with their proper magnitudes and directions. Letters are used to represent the magnitudes and direction angles of forces and couple moments that are unknown. Establish an, coordinate sstem so that these unknowns,,, etc., can be identified. inall, indicate the dimensions of the bod necessar for calculating the moments of forces. Important Points No equilibrium problem should be solved without first drawing the free-bod diagram, so as to account for all the forces and couple moments that act on the bod. If a support prevents translation of a bod in a particular direction, then the support eerts a force on the bod in that direction. If rotation is prevented, then the support eerts a couple moment on the bod. Stud Table 1. Internal forces are never shown on the free-bod diagram since the occur in equal but opposite collinear pairs and therefore cancel out. The weight of a bod is an eternal force, and its effect is represented b a single resultant force acting through the bod s center of gravit G. ouple moments can be placed anwhere on the free-bod diagram since the are free vectors. orces can act at an point along their lines of action since the are sliding vectors.

92 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..2 REE-DY DIGRMS 207 EXMPLE.1 Draw the free-bod diagram of the uniform beam shown in ig. 7a. The beam has a mass of 100 kg. 2 m 1200 N (a) 6 m SLUTIN The free-bod diagram of the beam is shown in ig. 7b. Since the support at is fied, the wall eerts three reactions on the beam, denoted as,, and M. The magnitudes of these reactions are unknown, and their sense has been assumed. The weight of the beam, W = N = 981 N, acts through the beam s center of gravit G, which is 3 m from since the beam is uniform. 2 m 1200 N Effect of applied force acting on beam Effect of fied support acting on beam M 3 m G 981 N Effect of gravit (weight) acting on beam (b) ig. 7

93 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 208 HPTER EQ U I L I R I U M R IGID DY EXMPLE.2 Draw the free-bod diagram of the foot lever shown in ig. 8a. The operator applies a vertical force to the pedal so that the spring is stretched 1. in. and the force in the short link at is 20 lb. 1. in. 1 in. k 20 lb/in. in. (b) (a) 20 lb 30 lb 1. in. ig. 8 1 in. in. (c) SLUTIN inspection of the photo the lever is loosel bolted to the frame at. The rod at is pinned at its ends and acts as a short link. fter making the proper measurements, the idealied model of the lever is shown in ig. 8b. rom this, the free-bod diagram is shown in ig. 8c. The pin support at eerts force components and on the lever. The link at eerts a force of 20 lb, acting in the direction of the link. In addition the spring also eerts a horiontal force on the lever. If the stiffness is measured and found to be k = 20 lb>in., then since the stretch s = 1. in., using Eq. 3 2, s = ks = 20 lb>in. 11. in.2 = 30 lb. inall, the operator s shoe applies a vertical force of on the pedal. The dimensions of the lever are also shown on the free-bod diagram, since this information will be useful when computing the moments of the forces. s usual, the senses of the unknown forces at have been assumed. The correct senses will become apparent after solving the equilibrium equations.

94 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..2 REE-DY DIGRMS 209 EXMPLE.3 Two smooth pipes, each having a mass of 300 kg, are supported b the forked tines of the tractor in ig. 9a. Draw the free-bod diagrams for each pipe and both pipes together. 0.3 m 0.3 m Effect of sloped blade acting on Effect of acting on R (b) T 293 N Effect of gravit (weight) acting on (c) Effect of sloped fork acting on (a) R 293 N (d) P SLUTIN The idealied model from which we must draw the free-bod diagrams is shown in ig. 9b. Here the pipes are identified, the dimensions have been added, and the phsical situation reduced to its simplest form. The free-bod diagram for pipe is shown in ig. 9c. Its weight is W = N = 293 N. ssuming all contacting surfaces are smooth, the reactive forces T,, R act in a direction normal to the tangent at their surfaces of contact. The free-bod diagram of pipe is shown in ig. 9d. an ou identif each of the three forces acting on this pipe? In particular, note that R, representing the force of on, ig. 9d, is equal and opposite to R representing the force of on, ig. 9c. This is a consequence of Newton s third law of motion. The free-bod diagram of both pipes combined ( sstem ) is shown in ig. 9e. Here the contact force R, which acts between and, is considered as an internal force and hence is not shown on the freebod diagram. That is, it represents a pair of equal but opposite collinear forces which cancel each other. T 293 N (e) ig N P

95 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 210 HPTER EQ U I L I R I U M R IGID DY EXMPLE. Draw the free-bod diagram of the unloaded platform that is suspended off the edge of the oil rig shown in ig. 10a.The platform has a mass of 200 kg. 70 G 1 m 1.0 m 0.8 m (b) (a) ig. 10 T SLUTIN 70 G 1.0 m 0.8 m 1962 N (c) 1 m The idealied model of the platform will be considered in two dimensions because b observation the loading and the dimensions are all smmetrical about a vertical plane passing through its center, ig. 10b.The connection at is considered to be a pin, and the cable supports the platform at. The direction of the cable and average dimensions of the platform are listed, and the center of gravit G has been determined. It is from this model that we have drawn the free-bod diagram shown in ig. 10c. The platform s weight is = 1962 N. The force components and along with the cable force T represent the reactions that both pins and both cables eert on the platform, ig. 10a. onsequentl, after the solution for these reactions, half their magnitude is developed at and half is developed at.

96 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..2 REE-DY DIGRMS 211 PRLEMS 1. Draw the free-bod diagram of the 0-kg paper roll which has a center of mass at G and rests on the smooth blade of the paper hauler. Eplain the significance of each force acting on the diagram. (See ig. 7b.) *. Draw the free-bod diagram of the beam which supports the 80-kg load and is supported b the pin at and a cable which wraps around the pulle at D. Eplain the significance of each force on the diagram. (See ig. 7b.) 3 mm D G 3 Prob Draw the free-bod diagram of member, which is supported b a roller at and a pin at. Eplain the significance of each force on the diagram. (See ig. 7b.) 2 m 2 m E 1. m 390 lb 800 lb ft Prob. 8 ft Prob. 2 ft 3 ft 3. Draw the free-bod diagram of the dumpster D of the truck, which has a weight of 000 lb and a center of gravit at G. It is supported b a pin at and a pin-connected hdraulic clinder (short link). Eplain the significance of each force on the diagram. (See ig. 7b.). Draw the free-bod diagram of the truss that is supported b the cable and pin. Eplain the significance of each force acting on the diagram. (See ig. 7b.) D 3 m G m 1 m 2 m 3 kn kn 2 m 2 m 2 m Prob. 3 Prob.

97 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 212 HPTER EQ U I L I R I U M R IGID DY 6. Draw the free-bod diagram of the crane boom which has a weight of 60 lb and center of gravit at G. The boom is supported b a pin at and cable. The load of 120 lb is suspended from a cable attached at. Eplain the significance of each force acting on the diagram. (See ig. 7b.) 18 ft ft G 9. Draw the free-bod diagram of the bar, which has a negligible thickness and smooth points of contact at,, and. Eplain the significance of each force on the diagram. (See ig. 7b.) 3 in. in. 8 in. 10 lb Prob Draw the free-bod diagram of the spanner wrench subjected to the 20-lb force. The support at can be considered a pin, and the surface of contact at is smooth. Eplain the significance of each force on the diagram. (See ig. 7b.) 1 in. 20 lb Prob Draw the free-bod diagram of the winch, which consists of a drum of radius in. It is pin-connected at its center, and at its outer rim is a ratchet gear having a mean radius of 6 in. The pawl serves as a two-force member (short link) and prevents the drum from rotating. Eplain the significance of each force on the diagram. (See ig. 7b.) 6 in. Prob. 7 * 8. Draw the free-bod diagram of member which is supported b a smooth collar at, roller at, and short link D. Eplain the significance of each force acting on the diagram. (See ig. 7b.) 2. kn D 3 in. 2 in. 6 in. 60 kn m 3 m in. 00 lb m 6 m Prob. 8 Prob. 10

98 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..2 REE-DY DIGRMS 213 NEPTUL PRLEMS P 1. Draw the free-bod diagram of the uniform trash bucket which has a significant weight. It is pinned at and rests against the smooth horiontal member at. Show our result in side view. Label an necessar dimensions. P 3. Draw the free-bod diagram of the wing on the passenger plane. The weights of the engine and wing are significant. The tires at are free to roll. P 3 P 1 P 2. Draw the free-bod diagram of the outrigger used to support a backhoe. The top pin is connected to the hdraulic clinder, which can be considered to be a short link (two-force member), the bearing shoe at is smooth, and the outrigger is pinned to the frame at. *P. Draw the free-bod diagram of the wheel and member used as part of the landing gear on a jet plane. The hdraulic clinder D acts as a two-force member, and there is a pin connection at. D P 2 P

99 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 21 HPTER EQ U I L I R I U M R IGID DY.3 Equations of Equilibrium 3 In Sec..1 we developed the two equations which are both necessar and sufficient for the equilibrium of a rigid bod, namel, = 0 and M = 0. When the bod is subjected to a sstem of forces, which all lie in the plane, then the forces can be resolved into their and components. onsequentl, the conditions for equilibrium in two dimensions are 2 = 0 = 0 M = 0 ( 2) 1 (a) M R R Here and represent, respectivel, the algebraic sums of the and components of all the forces acting on the bod, and M represents the algebraic sum of the couple moments and the moments of all the force components about the ais, which is perpendicular to the plane and passes through the arbitrar point. (b) lternative Sets of Equilibrium Equations. lthough Eqs. 2 are most often used for solving coplanar equilibrium problems, two alternative sets of three independent equilibrium equations ma also be used. ne such set is R = 0 M = 0 M = 0 ( 3) (c) ig. 11 When using these equations it is required that a line passing through points and is not parallel to the ais.to prove that Eqs. 3 provide the conditions for equilibrium, consider the free-bod diagram of the plate shown in ig. 11a. Using the methods of Sec..8, all the forces on the free-bod diagram ma be replaced b an equivalent resultant force R =, acting at point, and a resultant couple moment M R = M, ig. 11b. If M = 0 is satisfied, it is necessar that M R = 0. urthermore, in order that R satisf = 0, it must have no component along the ais, and therefore R must be parallel to the ais, ig. 11c. inall, if it is required that M = 0, where does not lie on the line of action of R, then R = 0. Since Eqs. 3 show that both of these resultants are ero, indeed the bod in ig. 11a must be in equilibrium.

100 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..3 EQUTINS EQUILIRIUM 21 second alternative set of equilibrium equations is M = 0 M = 0 M = 0 ( ) Here it is necessar that points,, and do not lie on the same line.to prove that these equations, when satisfied, ensure equilibrium, consider again the free-bod diagram in ig. 11b. If M = 0 is to be satisfied, then M R = 0. M = 0 is satisfied if the line of action of R passes through point as shown in ig. 11c. inall, if we require M = 0, it is necessar that R = 0, and so the plate in ig. 11a must then be in equilibrium. Procedure for nalsis oplanar force equilibrium problems for a rigid bod can be solved using the following procedure. ree-od Diagram. Establish the, coordinate aes in an suitable orientation. Draw an outlined shape of the bod. Show all the forces and couple moments acting on the bod. Label all the loadings and specif their directions relative to the or ais. The sense of a force or couple moment having an unknown magnitude but known line of action can be assumed. Indicate the dimensions of the bod necessar for computing the moments of forces. Equations of Equilibrium. ppl the moment equation of equilibrium, M = 0, about a point () that lies at the intersection of the lines of action of two unknown forces. In this wa, the moments of these unknowns are ero about, and a direct solution for the third unknown can be determined. When appling the force equilibrium equations, = 0 and = 0, orient the and aes along lines that will provide the simplest resolution of the forces into their and components. If the solution of the equilibrium equations ields a negative scalar for a force or couple moment magnitude, this indicates that the sense is opposite to that which was assumed on the free-bod diagram.

101 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 216 HPTER EQ U I L I R I U M R IGID DY EXMPLE. Determine the horiontal and vertical components of reaction on the beam caused b the pin at and the rocker at as shown in ig. 12a. Neglect the weight of the beam. 600 N 0.2 m D 200 N 600 sin N 600 cos N 0.2 m D 200 N 2 m 3 m 2 m 2 m 3 m 2 m (a) 100 N ig. 12 (b) 100 N SLUTIN ree-od Diagram. Identif each of the forces shown on the freebod diagram of the beam, ig. 12b. (See Eample.1.) or simplicit, the 600-N force is represented b its and components as shown in ig. 12b. Equations of Equilibrium. Summing forces in the direction ields : + = 0; 600 cos N - = 0 = 2 N ns. direct solution for can be obtained b appling the moment equation M = 0 about point. a+ M = 0; 100 N12 m sin N21 m cos N210.2 m2-17 m2 = 0 = 319 N ns. Summing forces in the direction, using this result, gives + c = 0; 319 N sin N N N + = 0 = 0 N ns. NTE: We can check this result b summing moments about point. a+ M = 0; sin N212 m cos N210.2 m N21 m N217 m m2 = 0 = 0 N ns.

102 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..3 EQUTINS EQUILIRIUM 217 EXMPLE.6 The cord shown in ig. 13a supports a force of 100 lb and wraps over the frictionless pulle. Determine the tension in the cord at and the horiontal and vertical components of reaction at pin. 0. ft u 100 lb (a) ig. 13 SLUTIN ree-od Diagrams. The free-bod diagrams of the cord and pulle are shown in ig. 13b. Note that the principle of action, equal but opposite reaction must be carefull observed when drawing each of these diagrams: the cord eerts an unknown load distribution p on the pulle at the contact surface, whereas the pulle eerts an equal but opposite effect on the cord. or the solution, however, it is simpler to combine the free-bod diagrams of the pulle and this portion of the cord, so that the distributed load becomes internal to this sstem and is therefore eliminated from the analsis, ig. 13c. 100 lb p T (b) p Equations of Equilibrium. Summing moments about point to eliminate and, ig. 13c, we have a+ M = 0; 100 lb 10. ft2 - T10. ft2 = 0 T = 100 lb ns. Using the result, : + = 0; sin 30 lb = 0 = 0.0 lb ns. 0. ft u + c = 0; lb cos 30 lb = 0 = 187 lb ns. 100 lb (c) T NTE: It is seen that the tension remains constant as the cord passes over the pulle. (This of course is true for an angle u at which the cord is directed and for an radius r of the pulle.)

103 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 218 HPTER EQ U I L I R I U M R IGID DY EXMPLE.7 The member shown in ig. 1a is pin-connected at and rests against a smooth support at. Determine the horiontal and vertical components of reaction at the pin. 0.7 m (a) 1 m 60 N 0. m 90 N m N 0.7 m 1 m 90 N m (b) 60 N ig. 1 SLUTIN ree-od Diagram. s shown in ig. 1b, the reaction N is perpendicular to the member at. lso, horiontal and vertical components of reaction are represented at. Equations of Equilibrium. Summing moments about, we obtain a direct solution for N, a+ M = 0; -90 N # m - 60 N11 m2 + N 10.7 m2 = 0 Using this result, : + = 0; + c = 0; N = 200 N sin 30 N = 0 = 100 N cos 30 N - 60 N = 0 = 233 N ns. ns.

104 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..3 EQUTINS EQUILIRIUM 219 EXMPLE.8 The bo wrench in ig. 1a is used to tighten the bolt at. If the wrench does not turn when the load is applied to the handle, determine the torque or moment applied to the bolt and the force of the wrench on the bolt. SLUTIN ree-od Diagram. The free-bod diagram for the wrench is shown in ig. 1b. Since the bolt acts as a fied support, it eerts force components and and a moment on the wrench at. Equations of Equilibrium. :+ = 0; 300 mm 00 mm M N + 30 cos 60 N = =.00 N ns. 2 N 30 N (a) + c = 0; N - 30 sin 60 N = 0 = 7.0 N ns. a + M = 0; M M ND 10.3 m2-130 sin 60 N210.7 m2 = 0 M = 32.6 N # m ns. Note that must be included in this moment summation. This couple moment is a free vector and represents the twisting resistance of the bolt on the wrench. Newton s third law, the wrench eerts an equal but opposite moment or torque on the bolt. urthermore, the resultant force on the wrench is M 0.3 m 0. m N 30 N (b) ig. 1 = = 7.1 N ns. NTE: lthough onl three independent equilibrium equations can be written for a rigid bod, it is a good practice to check the calculations using a fourth equilibrium equation. or eample, the above computations ma be verified in part b summing moments about point : a + M = 0; ND 10. m N # m N10.7 m2 = N # m N # m N # m = 0

105 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 220 HPTER EQ U I L I R I U M R IGID DY EXMPLE.9 Determine the horiontal and vertical components of reaction on the member at the pin, and the normal reaction at the roller in ig. 16a. SLUTIN ree-od Diagram. The free-bod diagram is shown in ig. 16b. The pin at eerts two components of reaction on the member, and. 70 lb 70 lb 3 ft 3 ft 3 ft 3 ft 2 ft 2 ft (a) (b) N ig. 16 Equations of Equilibrium. The reaction N can be obtained directl b summing moments about point since and produce no moment about. a+ M = 0; [N cos 30 ](6 ft) - [N sin 30 ](2 ft) - 70 lb(3 ft) = 0 N = 36.2 lb = 36 lb ns. Using this result, : + = 0; - (36.2 lb ) sin 30 = 0 = 268 lb ns. + c = 0; + (36.2 lb) cos lb = 0 = 286 lb ns.

106 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..3 EQUTINS EQUILIRIUM 221 EXMPLE.10 The uniform smooth rod shown in ig. 17a is subjected to a force and couple moment. If the rod is supported at b a smooth wall and at and either at the top or bottom b rollers, determine the reactions at these supports. Neglect the weight of the rod. 2 m 300 N m 000 N m 2 m 2 m m 000 N m 2 m 300 N 2 m (a) SLUTIN ree-od Diagram. s shown in ig. 17b, all the support (b) reactions act normal to the surfaces of contact since these surfaces are smooth. The reactions at and are shown acting in the positive ig. 17 direction. This assumes that onl the rollers located on the bottom of the rod are used for support. Equations of Equilibrium. Using the, coordinate sstem in ig. 17b, we have : + = 0; sin 30 + sin 30 - = 0 (1) + c = 0; -300 N + cos 30 + cos 30 = 0 (2) a+ M = 0; - 12 m N # m - 16 m cos 30 N218 m2 = 0 (3) When writing the moment equation, it should be noted that the line of action of the force component 300 sin 30 N passes through point, and therefore this force is not included in the moment equation. Solving Eqs. 2 and 3 simultaneousl, we obtain = N = -1 kn ns. = 136. N = 1.3 kn ns. Since is a negative scalar, the sense of is opposite to that shown on the free-bod diagram in ig. 17b. Therefore, the top roller at serves as the support rather than the bottom one. Retaining the negative sign for (Wh?) and substituting the results into Eq. 1, we obtain 136. sin 30 N + ( sin 30 N) - = 0 = 173 N ns.

107 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 222 HPTER EQ U I L I R I U M R IGID DY EXMPLE.11 The uniform truck ramp shown in ig. 18a has a weight of 00 lb and is pinned to the bod of the truck at each side and held in the position shown b the two side cables. Determine the tension in the cables. (a) SLUTIN The idealied model of the ramp, which indicates all necessar dimensions and supports, is shown in ig. 18b. Here the center of gravit is located at the midpoint since the ramp is considered to be uniform. ree-od Diagram. Working from the idealied model, the ramp s free-bod diagram is shown in ig. 18c. Equations of Equilibrium. Summing moments about point will ield a direct solution for the cable tension. Using the principle of moments, there are several was of determining the moment of T about. If we use and components, with T applied at, we have 2 ft G a+ M = 0; -T cos sin 30 ft2 + T sin cos 30 ft2 ft lb 1 cos 30 ft2 = 0 (b) T = 12 lb The simplest wa to determine the moment of T about is to resolve it into components along and perpendicular to the ramp at.then the moment of the component along the ramp will be ero about, so that a+ M = 0; -T sin ft lb 1 cos 30 ft2 = 0 20 T T = 12 lb 2 ft G lb ft Since there are two cables supporting the ramp, T = T = 712 lb 2 ns. ig. 18 (c) NTE: s an eercise, show that = 1339 lb and = 887. lb.

108 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..3 EQUTINS EQUILIRIUM 223 EXMPLE.12 Determine the support reactions on the member in ig. 19a. The collar at is fied to the member and can slide verticall along the vertical shaft. 900 N 900 N 1. m 1. m 1 m 1. m 1. m 1 m 00 N m M 00 N m SLUTIN (a) ig. 19 (b) N ree-od Diagram. The free-bod diagram of the member is shown in ig. 19b. The collar eerts a horiontal force and moment M on the member. The reaction N of the roller on the member is vertical. Equations of Equilibrium. The forces and N can be determined directl from the force equations of equilibrium. : + = 0; = 0 ns. + c = 0; N N = 0 N N ns. The moment M can be determined b summing moments either about point or point. a+ M = 0; M N(1. m) - 00 N # m N [3 m + (1 m) cos ] = 0 or a + M = 0; M = -186 N # m = 1.9 kn # m b ns. M N [1. m + (1 m) cos ] - 00 N # m = 0 M = -186 N # m = 1.9 kn # m b ns. The negative sign indicates that M has the opposite sense of rotation to that shown on the free-bod diagram.

109 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 22 HPTER EQ U I L I R I U M R IGID DY The bucket link on the back-hoe is a tpical eample of a two-force member since it is pin connected at its ends and, provided its weight is neglected, no other force acts on this member. The link used for this railroad car brake is a three-force member. Since the force in the tie rod at and from the link at are parallel, then for equilibrium the resultant force at the pin must also be parallel with these two forces. W. Two- and Three-orce Members The solutions to some equilibrium problems can be simplified b recogniing members that are subjected to onl two or three forces. Two-orce Members s the name implies, a two-force member has forces applied at onl two points on the member. n eample of a twoforce member is shown in ig. 20a. To satisf force equilibrium, and must be equal in magnitude, = =, but opposite in direction ( = 0), ig. 20b. urthermore, moment equilibrium requires that and share the same line of action, which can onl happen if the are directed along the line joining points and ( M = 0 or M = 0), ig. 20c. Therefore, for an two-force member to be in equilibrium, the two forces acting on the member must have the same magnitude, act in opposite directions, and have the same line of action, directed along the line joining the two points where these forces act. (a) (b) Two-force member ig. 20 Three-orce Members If a member is subjected to onl three forces, it is called a three-force member. Moment equilibrium can be satisfied onl if the three forces form a concurrent or parallel force sstem. To illustrate, consider the member subjected to the three forces 1, 2, and 3, shown in ig. 21a. If the lines of action of 1 and 2 intersect at point, then the line of action of 3 must also pass through point so that the forces satisf M = 0.s a special case, if the three forces are all parallel, ig. 21b, the location of the point of intersection,, will approach infinit. (c) 2 2 The boom on this lift is a three-force member, provided its weight is neglected. Here the lines of action of the weight of the worker, W, and the force of the two-force member (hdraulic clinder) at,, intersect at. or moment equilibrium, the resultant force at the pin,, must also be directed towards. 3 1 (a) 1 Three-force member ig. 21 (b) 3

110 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved.. TW- ND THREE-RE MEMERS 22 EXMPLE.13 The lever is pin supported at and connected to a short link D as shown in ig. 22a. If the weight of the members is negligible, determine the force of the pin on the lever at. 00 N SLUTIN 0. m ree-od Diagrams. s shown in ig. 22b, the short link D is a two-force member, so the resultant forces at pins D and must be equal, opposite, and collinear. lthough the magnitude of the force is unknown, the line of action is known since it passes through and D. Lever is a three-force member, and therefore, in order to satisf moment equilibrium, the three nonparallel forces acting on it must be concurrent at, ig. 22c. In particular, note that the force on the lever at is equal but opposite to the force acting at on the link. Wh? The distance must be 0. m since the lines of action of and the 00-N force are known. 0.2 m 0.2 m D 0.1 m (a) Equations of Equilibrium. requiring the force sstem to be concurrent at, since M = 0, the angle u which defines the line of action of can be determined from trigonometr, u = tan -1 a b = 60.3 D (b) 0. m Using the, aes and appling the force equilibrium equations, 00 N : + = 0; + c = 0; cos cos + 00 N = 0 sin sin = 0 0. m Solving, we get = 1.07 kn ns. 0.2 m u = 1.32 kn NTE: We can also solve this problem b representing the force at b its two components and and appling M = 0, = 0, = 0 to the lever. nce and are determined, we can get and u. 0.1 m (c) ig m

111 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 226 HPTER EQ U I L I R I U M R IGID DY UNDMENTL PRLEMS ll problem solutions must include an D. 1. Determine the horiontal and vertical components of reaction at the supports. Neglect the thickness of the beam.. Determine the components of reaction at the fied support. Neglect the thickness of the beam. 200 N 200 N 200 N 00 lb lb ft 3 m 1 m 1 m 1 m 00 N ft ft ft Determine the horiontal and vertical components of reaction at the pin and the reaction on the beam at. kn 1. m 1. m 1. m. The 2-kg bar has a center of mass at G. If it is supported b a smooth peg at, a roller at, and cord, determine the reactions at these supports. 0. m 0.2 m G 0.3 m D D The truss is supported b a pin at and a roller at. Determine the support reactions. 10 kn 6. Determine the reactions at the smooth contact points,, and on the bar. 2 m m kn 2 m 20 N m 0. m 0.1 m 0.2 m 3 6

112 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved.. TW- ND THREE-RE MEMERS 227 PRLEMS ll problem solutions must include an D. 11. Determine the normal reactions at and in Prob. 1. * 12. Determine the tension in the cord and the horiontal and vertical components of reaction at support of the beam in Prob Determine the horiontal and vertical components of reaction at and the tension in the cable for the truss in Prob.. 1. Determine the horiontal and vertical components of reaction at and the tension in cable on the boom in Prob Determine the horiontal and vertical components of reaction at and the normal reaction at on the spanner wrench in Prob. 7. * 16. Determine the normal reactions at and and the force in link D acting on the member in Prob Determine the normal reactions at the points of contact at,, and of the bar in Prob Determine the horiontal and vertical components of reaction at pin and the force in the pawl of the winch in Prob ompare the force eerted on the toe and heel of a 120-lb woman when she is wearing regular shoes and stiletto heels. ssume all her weight is placed on one foot and the reactions occur at points and as shown. * 20. The train car has a weight of lb and a center of gravit at G.It is suspended from its front and rear on the track b si tires located at,, and. Determine the normal reactions on these tires if the track is assumed to be a smooth surface and an equal portion of the load is supported at both the front and rear tires. ft ft 21. Determine the horiontal and vertical components of reaction at the pin and the tension developed in cable used to support the steel frame. G Prob ft 120 lb 120 lb 60 kn 1 m 1 m 1 m 30 kn m 3 m 3.7 in. 1.2 in. 0.7 in. 3.7 in. Prob. 19 Prob. 21

113 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 228 HPTER EQ U I L I R I U M R IGID DY 22. The articulated crane boom has a weight of 12 lb and center of gravit at G. If it supports a load of 600 lb, determine the force acting at the pin and the force in the hdraulic clinder when the boom is in the position shown. 2. The 300-lb electrical transformer with center of gravit at G is supported b a pin at and a smooth pad at. Determine the horiontal and vertical components of reaction at the pin and the reaction of the pad on the transformer. 1. ft ft 1 ft G 8 ft 3 ft G 0 1 ft Prob The airstroke actuator at D is used to appl a force of = 200 N on the member at. Determine the horiontal and vertical components of reaction at the pin and the force of the smooth shaft at on the member. * 2. The airstroke actuator at D is used to appl a force of on the member at. The normal reaction of the smooth shaft at on the member is 300 N. Determine the magnitude of and the horiontal and vertical components of reaction at pin. Prob skeletal diagram of a hand holding a load is shown in the upper figure. If the load and the forearm have masses of 2 kg and 1.2 kg, respectivel, and their centers of mass are located at G 1 and G 2, determine the force developed in the biceps D and the horiontal and vertical components of reaction at the elbow joint. The forearm supporting sstem can be modeled as the structural sstem shown in the lower figure. D mm G 1 G 2 D 200 mm 600 mm Probs. 23/2 60 D G mm G 2 13 mm Prob mm

114 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved.. TW- ND THREE-RE MEMERS s an airplane s brakes are applied, the nose wheel eerts two forces on the end of the landing gear as shown. Determine the horiontal and vertical components of reaction at the pin and the force in strut. 29. The mass of 700 kg is suspended from a trolle which moves along the crane rail from d = 1.7 m to d = 3. m. Determine the force along the pin-connected knee strut (short link) and the magnitude of force at pin as a function of position d. Plot these results of and (vertical ais) versus d (horiontal ais) mm 2 m d 600 mm 1. m 2 kn Prob kn Prob. 27 * 28. The 1.-Mg drainpipe is held in the tines of the fork lift. Determine the normal forces at and as functions of the blade angle u and plot the results of force (vertical ais) versus u (horiontal ais) for 0 u If the force of = 100 lb is applied to the handle of the bar bender, determine the horiontal and vertical components of reaction at pin and the reaction of the roller on the smooth bar. 31. If the force of the smooth roller at on the bar bender is required to be 1. kip, determine the horiontal and vertical components of reaction at pin and the required magnitude of force applied to the handle. 0 in. 0. m 60 u in. Prob. 28 Probs. 30/31

115 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 230 HPTER EQ U I L I R I U M R IGID DY * 32. The jib crane is supported b a pin at and rod. If the load has a mass of 2 Mg with its center of mass located at G, determine the horiontal and vertical components of reaction at the pin and the force developed in rod on the crane when = m. 33. The jib crane is supported b a pin at and rod. The rod can withstand a maimum tension of 0 kn. If the load has a mass of 2 Mg, with its center of mass located at G, determine its maimum allowable distance and the corresponding horiontal and vertical components of reaction at. 3. The framework is supported b the member which rests on the smooth floor. When loaded, the pressure distribution on is linear as shown. Determine the length d of member and the intensit w for this case. ft m 7 ft 800 lb 3.2 m 0.2 m d w D G Prob. 3 Probs. 32/33 3. Determine the horiontal and vertical components of reaction at the pin and the normal force at the smooth peg on the member. * 36. utriggers and are used to stabilie the crane from overturning when lifting large loads. If the load to be lifted is 3 Mg, determine the maimum boom angle u so that the crane does not overturn. The crane has a mass of Mg and center of mass at G, whereas the boom has a mass of 0.6 Mg and center of mass at. G. m 0. m G m u G 0. m 600 N 0.7 m 2.3 m 2.8 m Prob. 3 Prob. 36

116 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved.. TW- ND THREE-RE MEMERS The wooden plank resting between the buildings deflects slightl when it supports the 0-kg bo. This deflection causes a triangular distribution of load at its ends, having maimum intensities of w and w. Determine w and w, each measured in N>m, when the bo is standing 3 m from one end as shown. Neglect the mass of the plank. * 0. The platform assembl has a weight of 20 lb and center of gravit at G 1. If it is intended to support a maimum load of 00 lb placed at point G 2, determine the smallest counterweight W that should be placed at in order to prevent the platform from tipping over. G 2 w w 3 m 6 m 0. m 0.3 m 2 ft 6 ft G 1 Prob ft 38. Spring D remains in the horiontal position at all times due to the roller at D. If the spring is unstretched when u = 0 and the bracket achieves its equilibrium position when u = 30, determine the stiffness k of the spring and the horiontal and vertical components of reaction at pin. 1 ft 6 ft Prob. 0 1 ft D 39. Spring D remains in the horiontal position at all times due to the roller at D. If the spring is unstretched when u = 0 and the stiffness is k = 1. kn>m, determine the smallest angle u for equilibrium and the horiontal and vertical components of reaction at pin. 1. Determine the horiontal and vertical components of reaction at the pin and the reaction of the smooth collar on the rod. D k 0.6 m u 0. m 300 lb 0 lb D 300 N 2 ft 1 ft 1 ft ft Probs. 38/39 Prob. 1

117 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 232 HPTER EQ U I L I R I U M R IGID DY 2. Determine the support reactions of roller and the smooth collar on the rod. The collar is fied to the rod, but is allowed to slide along rod D. *. Determine the horiontal and vertical components of force at the pin and the reaction at the rocker of the curved beam. 900 N 00 N 200 N 1 m 1 m 2 m N m D 2 m Prob. 2 Prob. 3. The uniform rod has a weight of 1 lb. Determine the force in the cable when the rod is in the position shown.. The floor crane and the driver have a total weight of 200 lb with a center of gravit at G. If the crane is required to lift the 00-lb drum, determine the normal reaction on both the wheels at and both the wheels at when the boom is in the position shown. 6. The floor crane and the driver have a total weight of 200 lb with a center of gravit at G. Determine the largest weight of the drum that can be lifted without causing the crane to overturn when its boom is in the position shown. ft 12 ft 3 ft 10 D 6 ft T G 2.2 ft 1. ft E 8. ft Prob. 3 Probs. /6

118 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved.. TW- ND THREE-RE MEMERS The motor has a weight of 80 lb. Determine the force that each of the chains eerts on the supporting hooks at,, and. Neglect the sie of the hooks and the thickness of the beam. 1 ft ft 80 lb 1. ft The winch cable on a tow truck is subjected to a force of T = 6 kn when the cable is directed at u = 60. Determine the magnitudes of the total brake frictional force for the rear set of wheels and the total normal forces at both front wheels and both rear wheels for equilibrium. The truck has a total mass of Mg and mass center at G. 1. Determine the minimum cable force T and critical angle u which will cause the tow truck to start tipping, i.e., for the normal reaction at to be ero. ssume that the truck is braked and will not slip at. The truck has a total mass of Mg and mass center at G. 1.2 m G u T 3 m 2 m 2. m 1. m Prob. 7 Probs. 0/1 * 8. Determine the force P needed to pull the 0-kg roller over the smooth step.take u = Determine the magnitude and direction u of the minimum force P needed to pull the 0-kg roller over the smooth step. * 2. Three uniform books, each having a weight W and length a, are stacked as shown. Determine the maimum distance d that the top book can etend out from the bottom one so the stack does not topple over. P u 0.1 m 0.6 m a d 20 Probs. 8/9 Prob. 2

119 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 23 HPTER EQ U I L I R I U M R IGID DY 3. Determine the angle u at which the link is held in equilibrium if member D moves 2 in. to the right. The springs are originall unstretched when u = 0. Each spring has the stiffness shown. The springs remain horiontal since the are attached to roller guides.. The horiontal beam is supported b springs at its ends. Each spring has a stiffness of k = kn>m and is originall unstretched so that the beam is in the horiontal position. Determine the angle of tilt of the beam if a load of 800 N is applied at point as shown. E k 100 lb/ft u 6 in. k E 00 lb/ft 6 in. D * 6. The horiontal beam is supported b springs at its ends. If the stiffness of the spring at is k = kn>m, determine the required stiffness of the spring at so that if the beam is loaded with the 800 N it remains in the horiontal position. The springs are originall constructed so that the beam is in the horiontal position when it is unloaded. 800 N 1 m 3 m Prob. 3 Probs. /6. The uniform rod has a weight of 1 lb and the spring is unstretched when u = 0. If u = 30, determine the stiffness k of the spring. 7. The smooth disks D and E have a weight of 200 lb and 100 lb, respectivel. If a horiontal force of P = 200 lb is applied to the center of disk E, determine the normal reactions at the points of contact with the ground at,, and. 8. The smooth disks D and E have a weight of 200 lb and 100 lb, respectivel. Determine the largest horiontal force P that can be applied to the center of disk E without causing the disk D to move up the incline. 6 ft u k 3 ft 3 1. ft D 1 ft E P Prob. Probs. 7/8

120 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved.. TW- ND THREE-RE MEMERS man stands out at the end of the diving board, which is supported b two springs and, each having a stiffness of k = 1 kn>m. In the position shown the board is horiontal. If the man has a mass of 0 kg, determine the angle of tilt which the board makes with the horiontal after he jumps off. Neglect the weight of the board and assume it is rigid. 61. If spring is unstretched with u = 0 and the bell crank achieves its equilibrium position when u = 1, determine the force applied perpendicular to segment D and the horiontal and vertical components of reaction at pin. Spring remains in the horiontal postion at all times due to the roller at. k 2 kn/m 10 1 m 3 m 300 mm u D 00 mm Prob. 9 Prob. 61 * 60. The uniform rod has a length l and weight W. It is supported at one end b a smooth wall and the other end b a cord of length s which is attached to the wall as shown. Show that for equilibrium it is required that h = [(s 2 - l 2 )>3] 1> The thin rod of length l is supported b the smooth tube. Determine the distance a needed for equilibrium if the applied load is P. a 2r h s l P l Prob. 60 Prob. 62

121 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 236 HPTER EQ U I L I R I U M R IGID DY NEPTUL PRLEMS P. The tie rod is used to support this overhang at the entrance of a building. If it is pin connected to the building wall at and to the center of the overhang, determine if the force in the rod will increase, decrease, or remain the same if (a) the support at is moved to a lower position D, and (b) the support at is moved to the outer position. Eplain our answer with an equilibrium analsis, using dimensions and loads. ssume the overhang is pin supported from the building wall. P 7. Like all aircraft, this jet plane rests on three wheels. Wh not use an additional wheel at the tail for better support? (an ou think of an other reason for not including this wheel?) If there was a fourth tail wheel, draw a free-bod diagram of the plane from a side (2 D) view, and show wh one would not be able to determine all the wheel reactions using the equations of equilibrium. D P 7 P P 6. The man attempts to pull the four wheeler up the incline and onto the truck bed. rom the position shown, is it more effective to keep the rope attached at, or would it be better to attach it to the ale of the front wheels at? Draw a free-bod diagram and do an equilibrium analsis to eplain our answer. *P 8. Where is the best place to arrange most of the logs in the wheelbarrow so that it minimies the amount of force on the backbone of the person transporting the load? Do an equilibrium analsis to eplain our answer. P 6 P 8

122 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved.. REE-DY DIGRMS 237 EQUILIRIUM IN THREE DIMENSINS. ree-od Diagrams The first step in solving three-dimensional equilibrium problems, as in the case of two dimensions, is to draw a free-bod diagram. efore we can do this, however, it is first necessar to discuss the tpes of reactions that can occur at the supports. Support Reactions. The reactive forces and couple moments acting at various tpes of supports and connections, when the members are viewed in three dimensions, are listed in Table 2. It is important to recognie the smbols used to represent each of these supports and to understand clearl how the forces and couple moments are developed. s in the two-dimensional case: force is developed b a support that restricts the translation of its attached member. couple moment is developed when rotation of the attached member is prevented. or eample, in Table 2, item (), the ball-and-socket joint prevents an translation of the connecting member; therefore, a force must act on the member at the point of connection. This force has three components having unknown magnitudes,,,. Provided these components are known, one can obtain the magnitude of force, = , and the force s orientation defined b its coordinate direction angles a, b, g, Eqs. 2 7.* Since the connecting member is allowed to rotate freel about an ais, no couple moment is resisted b a ball-and-socket joint. It should be noted that the single bearing supports in items () and (7), the single pin (8), and the single hinge (9) are shown to resist both force and couple-moment components. If, however, these supports are used in conjunction with other bearings, pins, or hinges to hold a rigid bod in equilibrium and the supports are properl aligned when connected to the bod, then the force reactions at these supports alone are adequate for supporting the bod. In other words, the couple moments become redundant and are not shown on the free-bod diagram. The reason for this should become clear after studing the eamples which follow. * The three unknowns ma also be represented as an unknown force magnitude and two unknown coordinate direction angles. The third direction angle is obtained using the identit cos 2 a + cos 2 b + cos 2 g = 1, Eq. 2 8.

123 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 238 HPTER EQ U I L I R I U M R IGID DY TLE 2 Supports for Rigid odies Subjected to Three-Dimensional orce Sstems Tpes of onnection Reaction Number of Unknowns (1) ne unknown. The reaction is a force which acts awa from the member in the known direction of the cable. cable (2) ne unknown. The reaction is a force which acts perpendicular to the surface at the point of contact. smooth surface support (3) roller ne unknown. The reaction is a force which acts perpendicular to the surface at the point of contact. () Three unknowns. The reactions are three rectangular force components. ball and socket () single journal bearing M M our unknowns. The reactions are two force and two couple-moment components which act perpendicular to the shaft. Note: The couple moments are generall not applied if the bod is supported elsewhere. See the eamples. continued

124 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved.. REE-DY DIGRMS 239 TLE 2 ontinued Tpes of onnection Reaction Number of Unknowns (6) M single journal bearing with square shaft M M ive unknowns. The reactions are two force and three couple-moment components. Note: The couple moments are generall not applied if the bod is supported elsewhere. See the eamples. (7) M single thrust bearing M ive unknowns. The reactions are three force and two couple-moment components. Note: The couple moments are generall not applied if the bod is supported elsewhere. See the eamples. (8) M single smooth pin M ive unknowns. The reactions are three force and two couple-moment components. Note: The couple moments are generall not applied if the bod is supported elsewhere. See the eamples. M (9) single hinge M ive unknowns. The reactions are three force and two couple-moment components. Note: The couple moments are generall not applied if the bod is supported elsewhere. See the eamples. (10) M M M Si unknowns. The reactions are three force and three couple-moment components. fied support

125 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 20 HPTER EQ U I L I R I U M R IGID DY Tpical eamples of actual supports that are referenced to Table 2 are shown in the following sequence of photos. This ball-and-socket joint provides a connection for the housing of an earth grader to its frame. () This journal bearing supports the end of the shaft. () This thrust bearing is used to support the drive shaft on a machine. (7) This pin is used to support the end of the strut used on a tractor. (8) ree-od Diagrams. The general procedure for establishing the free-bod diagram of a rigid bod has been outlined in Sec..2. Essentiall it requires first isolating the bod b drawing its outlined shape. This is followed b a careful labeling of all the forces and couple moments with reference to an established,, coordinate sstem. It is suggested to show the unknown components of reaction as acting on the free-bod diagram in the positive sense. In this wa, if an negative values are obtained, the will indicate that the components act in the negative coordinate directions.

126 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved.. REE-DY DIGRMS 21 EXMPLE.1 onsider the two rods and plate, along with their associated free-bod diagrams shown in ig. 23. The,, aes are established on the diagram and the unknown reaction components are indicated in the positive sense. The weight is neglected. N m N m SLUTIN 00 N 00 N Properl aligned journal bearings at,,. The force reactions developed b the bearings are sufficient for equilibrium since the prevent the shaft from rotating about each of the coordinate aes. M M 200 lb ft 200 lb ft T 300 lb Pin at and cable. 300 lb Moment components are developed b the pin on the rod to prevent rotation about the and aes. 00 lb 00 lb Properl aligned journal bearing at and hinge at. Roller at. ig. 23 nl force reactions are developed b the bearing and hinge on the plate to prevent rotation about each coordinate ais. No moments at the hinge are developed.

127 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 22 HPTER EQ U I L I R I U M R IGID DY.6 Equations of Equilibrium s stated in Sec..1, the conditions for equilibrium of a rigid bod subjected to a three-dimensional force sstem require that both the resultant force and resultant couple moment acting on the bod be equal to ero. Vector Equations of Equilibrium. The two conditions for equilibrium of a rigid bod ma be epressed mathematicall in vector form as = 0 M = 0 ( ) where is the vector sum of all the eternal forces acting on the bod and M is the sum of the couple moments and the moments of all the forces about an point located either on or off the bod. Scalar Equations of Equilibrium. If all the eternal forces and couple moments are epressed in artesian vector form and substituted into Eqs., we have = i + j + k = 0 M = M i + M j + M k = 0 Since the i, j, and k components are independent from one another, the above equations are satisfied provided = 0 = 0 = 0 ( 6a) and M = 0 M = 0 M = 0 ( 6b) These si scalar equilibrium equations ma be used to solve for at most si unknowns shown on the free-bod diagram. Equations 6a require the sum of the eternal force components acting in the,, and directions to be ero, and Eqs. 6b require the sum of the moment components about the,, and aes to be ero.

128 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..7 NSTRINTS ND STTIL DETERMINY 23.7 onstraints and Statical Determinac To ensure the equilibrium of a rigid bod, it is not onl necessar to satisf the equations of equilibrium, but the bod must also be properl held or constrained b its supports. Some bodies ma have more supports than are necessar for equilibrium, whereas others ma not have enough or the supports ma be arranged in a particular manner that could cause the bod to move. Each of these cases will now be discussed. Redundant onstraints. When a bod has redundant supports, that is, more supports than are necessar to hold it in equilibrium, it becomes staticall indeterminate. Staticall indeterminate means that there will be more unknown loadings on the bod than equations of equilibrium available for their solution. or eample, the beam in ig. 2a and the pipe assembl in ig. 2b, shown together with their free-bod diagrams, are both staticall indeterminate because of additional (or redundant) support reactions. or the beam there are five unknowns, M,,,, and, for which onl three equilibrium equations can be written ( = 0, = 0, and M = 0, Eqs. 2). The pipe assembl has eight unknowns, for which onl si equilibrium equations can be written, Eqs. 6. The additional equations needed to solve staticall indeterminate problems of the tpe shown in ig. 2 are generall obtained from the deformation conditions at the points of support. These equations involve the phsical properties of the bod which are studied in subjects dealing with the mechanics of deformation, such as mechanics of materials. * 2 kn m 2 kn m 00 N 00 N (a) ig. 2 M M M 00 N 00 N 200 N 200 N (b) * See R.. Hibbeler, Mechanics of Materials, 7th edition, Pearson Education/Prentice Hall, Inc.

129 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 2 HPTER EQ U I L I R I U M R IGID DY Improper onstraints. Having the same number of unknown reactive forces as available equations of equilibrium does not alwas guarantee that a bod will be stable when subjected to a particular loading. or eample, the pin support at and the roller support at for the beam in ig. 2a are placed in such a wa that the lines of action of the reactive forces are concurrent at point. onsequentl, the applied loading P will cause the beam to rotate slightl about, and so the beam is improperl constrained, M Z 0. In three dimensions, a bod will be improperl constrained if the lines of action of all the reactive forces intersect a common ais. or eample, the reactive forces at the ball-and-socket supports at and in ig. 2b all intersect the ais passing through and. Since the moments of these forces about and are all ero, then the loading P will rotate the member about the ais, M Z 0. P P (a) P P (b) ig. 2

130 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..7 NSTRINTS ND STTIL DETERMINY 2 P P (a) 100 N 100 N (b) ig. 26 nother wa in which improper constraining leads to instabilit occurs when the reactive forces are all parallel. Two- and threedimensional eamples of this are shown in ig. 26. In both cases, the summation of forces along the ais will not equal ero. In some cases, a bod ma have fewer reactive forces than equations of equilibrium that must be satisfied. The bod then becomes onl partiall constrained. or eample, consider member in ig. 27a with its corresponding free-bod diagram in ig. 27b. Here = 0 will not be satisfied for the loading conditions and therefore equilibrium will not be maintained. To summarie these points, a bod is considered improperl constrained if all the reactive forces intersect at a common point or pass through a common ais, or if all the reactive forces are parallel. In engineering practice, these situations should be avoided at all times since the will cause an unstable condition. 100 N (a) 100 N (b) ig. 27

131 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 26 HPTER EQ U I L I R I U M R IGID DY Important Points lwas draw the free-bod diagram first when solving an equilibrium problem. If a support prevents translation of a bod in a specific direction, then the support eerts a force on the bod in that direction. If a support prevents rotation about an ais, then the support eerts a couple moment on the bod about the ais. If a bod is subjected to more unknown reactions than available equations of equilibrium, then the problem is staticall indeterminate. stable bod requires that the lines of action of the reactive forces do not intersect a common ais and are not parallel to one another. Procedure for nalsis Three-dimensional equilibrium problems for a rigid bod can be solved using the following procedure. ree-od Diagram. Draw an outlined shape of the bod. Show all the forces and couple moments acting on the bod. Establish the origin of the,, aes at a convenient point and orient the aes so that the are parallel to as man of the eternal forces and moments as possible. Label all the loadings and specif their directions. In general, show all the unknown components having a positive sense along the,, aes. Indicate the dimensions of the bod necessar for computing the moments of forces. Equations of Equilibrium. If the,, force and moment components seem eas to determine, then appl the si scalar equations of equilibrium; otherwise use the vector equations. It is not necessar that the set of aes chosen for force summation coincide with the set of aes chosen for moment summation. ctuall, an ais in an arbitrar direction ma be chosen for summing forces and moments. hoose the direction of an ais for moment summation such that it intersects the lines of action of as man unknown forces as possible. Realie that the moments of forces passing through points on this ais and the moments of forces which are parallel to the ais will then be ero. If the solution of the equilibrium equations ields a negative scalar for a force or couple moment magnitude, it indicates that the sense is opposite to that assumed on the free-bod diagram.

132 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..7 NSTRINTS ND STTIL DETERMINY 27 EXMPLE.1 The homogeneous plate shown in ig. 28a has a mass of 100 kg and is subjected to a force and couple moment along its edges. If it is supported in the horiontal plane b a roller at, a ball-and-socket joint at, and a cord at, determine the components of reaction at these supports. SLUTIN (SLR NLYSIS) 1. m 2 m 300 N 200 N m 3 m ree-od Diagram. There are five unknown reactions acting on the plate, as shown in ig. 28b. Each of these reactions is assumed to act in a positive coordinate direction. (a) Equations of Equilibrium. Since the three-dimensional geometr is rather simple, a scalar analsis provides a direct solution to this problem. force summation along each ais ields 300 N 200 N m 981 N T = 0; = 0 ns. = 0; = 0 ns. = 0; + + T N N = 0 (1) Recall that the moment of a force about an ais is equal to the product of the force magnitude and the perpendicular distance (moment arm) from the line of action of the force to the ais. lso, forces that are parallel to an ais or pass through it create no moment about the ais. Hence, summing moments about the positive and aes, we have 1 m 1 m 1. m (b) ig m M = 0; M = 0; T 12 m2-981 N11 m m2 = N11. m N11. m2-13 m2-13 m2-200 N# m = 0 (2) (3) The components of the force at can be eliminated if moments are summed about the and aes. We obtain M = 0; M = 0; 981 N11 m N12 m2-12 m2 = N11. m2-981 N11. m2-200 N# m + T 13 m2 = 0 () () Solving Eqs. 1 through 3 or the more convenient Eqs. 1,, and ields = 790 N = -217 N T = 707 N ns. The negative sign indicates that acts downward. NTE: The solution of this problem does not require a summation of moments about the ais. The plate is partiall constrained since the supports cannot prevent it from turning about the ais if a force is applied to it in the plane.

133 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 28 HPTER EQ U I L I R I U M R IGID DY EXMPLE.16 Determine the components of reaction that the ball-and-socket joint at, the smooth journal bearing at, and the roller support at eert on the rod assembl in ig. 29a. 900 N 900 N 0. m 0. m 0. m D 0. m 0.6 m 0. m 0. m 0. m 0. m 0.6 m (a) ig. 29 SLUTIN ree-od Diagram. s shown on the free-bod diagram, ig. 29b, the reactive forces of the supports will prevent the assembl from rotating about each coordinate ais, and so the journal bearing at onl eerts reactive forces on the member. Equations of Equilibrium. direct solution for can be obtained b summing forces along the ais. = 0; = 0 ns. The force can be determined directl b summing moments about the ais. M = 0; (0.6 m) N(0. m) = 0 = 600 N ns. Using this result, can be determined b summing moments about the ais. M = 0; (0.8 m) N(1.2 m) N(0. m) = 0 = -0 N ns. The negative sign indicates that acts downward. The force can be found b summing moments about the ais. M = 0; Thus, = 0; inall, using the results of and. = 0; - (0.8 m) = 0 = = 0 = 0 (b) + (-0 N) N N = 0 = 70 N ns. ns. ns.

134 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..7 NSTRINTS ND STTIL DETERMINY 29 EXMPLE.17 The boom is used to support the 7-lb flowerpot in ig. 30a. Determine the tension developed in wires and. SLUTIN ree-od Diagram. The free-bod diagram of the boom is shown in ig. 30b. 2 ft 2 ft Equations of Equilibrium. We will use a vector analsis. = a r 2i - 6j + 3k6 ft b = r a 2 (2 ft) 2 + (-6 ft) 2 + (3 ft) b 2 3 ft = 2 7 i j k = a r -2i-6j + 3k6 ft b = r a 2 (-2 ft) 2 + (-6 ft) 2 + (3 ft) b 2 = i j k We can eliminate the force reaction at b writing the moment equation of equilibrium about point. (a) ig ft M = 0; r * ( + + W) = 0 (6j) * ca 2 7 i j kb + a- 2 7 i j kb + (-7k) d = 0 a bi + a bk = 0 M = 0; M = 0; = 0 0 = 0 (1) 2 ft 2 ft M = 0; = 0 Solving Eqs. (1) and (2) simultaneousl, = = 87. lb (2) ns. 3 ft r W 7 lb 6 ft (b)

135 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 20 HPTER EQ U I L I R I U M R IGID DY EXMPLE.18 Rod shown in ig. 31a is subjected to the 200-N force. Determine the reactions at the ball-and-socket joint and the tension in the cables D and E. SLUTIN (VETR NLYSIS) 1. m ree-od Diagram. ig. 31b. 2 m 200 N 2 m E 1. m (a) 1 m D Equations of Equilibrium. Representing each force on the free-bod diagram in artesian vector form, we have ppling the force equation of equilibrium. = 0; = 0; = 0; = 0; = i + j + k T E = T E i T D = T D j = -200k6 N + T E + T D + = T E 2i T D 2j k = 0 + T E = 0 + T D = = 0 (1) (2) (3) Summing moments about point ields M = 0; r * + r * 1T E + T D 2 = 0 r Since r = 1 2 r, then 10.i + 1j - 1k2 * 1-200k2 + 11i + 2j - 2k2 * 1T E i + T D j2 = 0 Epanding and rearranging terms gives 200 N T E r (b) ig. 31 T D 12T D i + 1-2T E j + 1T D - 2T E 2k = 0 M = 0; 2T D = 0 M = 0; -2T E = 0 M = 0; T D - 2T E = 0 Solving Eqs. 1 through, we get T D = 100 N T E = 0 N = -0 N = -100 N = 200 N () () (6) ns. ns. ns. ns. ns. NTE: The negative sign indicates that and have a sense which is opposite to that shown on the free-bod diagram, ig. 31b.

136 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..7 NSTRINTS ND STTIL DETERMINY 21 EXMPLE.19 The bent rod in ig. 32a is supported at b a journal bearing, at D b a ball-and-socket joint, and at b means of cable. Using onl one equilibrium equation, obtain a direct solution for the tension in cable. The bearing at is capable of eerting force components onl in the and directions since it is properl aligned on the shaft. 1 m SLUTIN (VETR NLYSIS) ree-od Diagram. s shown in ig. 32b, there are si unknowns. Equations of Equilibrium. The cable tension ma be obtained directl b summing moments about an ais that passes through points D and. Wh? The direction of this ais is defined b the unit vector u, where u = r D = - 1 r D 22 i j = i j T 0. m E 0. m (a) D 100 kg Hence, the sum of the moments about this ais is ero provided M D = u # 1r * 2 = 0 Here r represents a position vector drawn from an point on the ais D to an point on the line of action of force (see Eq. 11). With reference to ig. 32b, we can therefore write u # 1r * T + r E * W2 = 0 T r u i j2 # 1-1j2 * 1T k j2 * 1-981k2D = i j2 # [1-T i] = 0 0. m W 981 N 0. m r E D D D D T = 0 (b) T = 90. N ns. ig. 32 Since the moment arms from the ais to and W are eas to obtain, we can also determine this result using a scalar analsis. s shown in ig. 32b, T M D = 0; T (1 m sin ) N(0. m sin ) = 0 T = 90. N ns.

137 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 22 HPTER EQ U I L I R I U M R IGID DY UNDMENTL PRLEMS ll problem solutions must include an D. 7. The uniform plate has a weight of 00 lb. Determine the tension in each of the supporting cables. 3 ft lb 8. Determine the reactions at the roller support, the ball-and-socket joint D, and the tension in cable for the plate. 9. The rod is supported b smooth journal bearings at, and and is subjected to the two forces. Determine the reactions at these supports. 2 ft 2 ft 0.2 m 900 N 600 N 0. m 0. m 0. m 0.3 m D 0.1 m Determine the support reactions at the smooth journal bearings,, and of the pipe assembl. 0.6 m 0. m 0.6 m 0.6 m 11. Determine the force developed in cords D, E, and and the reactions of the ball-and-socket joint on the block N 12. Determine the components of reaction that the thrust bearing and cable eert on the bar. 3 m D 6 kn 11 9 kn E 1. m m 80 lb 0.6 m 600N 00N D 0.6 m 0.6 m 0. m 6 ft D 1. ft 1. ft 9 12

138 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..7 NSTRINTS ND STTIL DETERMINY 23 PRLEMS ll problem solutions must include an D. 63. The cart supports the uniform crate having a mass of 8 kg. Determine the vertical reactions on the three casters at,, and. The caster at is not shown. Neglect the mass of the cart. 0.1 m 0. m 0.2 m 0.2 m 0. m 0.3 m 0.6 m 0.3 m 6. If P = 6 kn, = 0.7 m and = 1 m, determine the tension developed in cables, D, and E. Neglect the weight of the plate. 66. Determine the location and of the point of application of force P so that the tension developed in cables, D, and E is the same. Neglect the weight of the plate. Prob. 63 P D * 6. The pole for a power line is subjected to the two cable forces of 60 lb, each force ling in a plane parallel to the - plane. If the tension in the gu wire is 80 lb, determine the,, components of reaction at the fied base of the pole,. E 2 m Probs. 6/66 2 m 1 ft ft 60 lb 67. Due to an unequal distribution of fuel in the wing tanks, the centers of gravit for the airplane fuselage and wings and are located as shown. If these components have weights W = 000 lb, W = 8000 lb, and W = 6000 lb, determine the normal reactions of the wheels D, E, and on the ground. 60 lb 80 lb 10 ft D E 3 ft 8 ft 6 ft 8 ft 6 ft 20 ft 3 ft ft Prob. 6 Prob. 67

139 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 2 HPTER EQ U I L I R I U M R IGID DY * 68. Determine the magnitude of force that must be eerted on the handle at to hold the 7-kg crate in the position shown. lso, determine the components of reaction at the thrust bearing and smooth journal bearing. 70. Determine the tension in cables D and D and the,, components of reaction at the ball-and-socket joint at. 0.1 m 0.6 m 0. m D 300 N 3 m 0.2 m 0.1 m 1. m Prob m 1 m Prob The shaft is supported b three smooth journal bearings at,, and. Determine the components of reaction at these bearings. 71. The rod assembl is used to support the 20-lb clinder. Determine the components of reaction at the ball-andsocket joint, the smooth journal bearing E, and the force developed along rod D. The connections at and D are ball-and-socket joints. 900 N 600 N 0 N 0.6 m 0.6 m 0.9 m 0.9 m 0.9 m 00 N 0.9 m 0.9 m 1 ft 1 ft 1 ft D 1 ft E 1. ft Prob. 69 Prob. 71

140 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..7 NSTRINTS ND STTIL DETERMINY 2 * 72. Determine the components of reaction acting at the smooth journal bearings,, and. 7. If the load has a weight of 200 lb, determine the,, components of reaction at the ball-and-socket joint and the tension in each of the wires. 300 N m 0 N 0. m 0.6 m D 2 ft 2 ft ft 0.8 m 0. m 3 ft Prob ft 2 ft G 2 ft E Prob Determine the force components acting on the balland-socket at, the reaction at the roller and the tension on the cord D needed for equilibrium of the quarter circular plate. 7. If the cable can be subjected to a maimum tension of 300 lb, determine the maimum force which ma be applied to the plate. ompute the,, components of reaction at the hinge for this loading. 1 m 30 N 2 m D N 3 m 200 N 3 ft 2 ft 9 ft 1 ft 3 ft Prob. 73 Prob. 7

141 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 26 HPTER EQ U I L I R I U M R IGID DY * 76. The member is supported b a pin at and a cable. If the load at D is 300 lb, determine the,, components of reaction at the pin and the tension in cable. 79. The boom is supported b a ball-and-socket joint at and a gu wire at. If the -kn loads lie in a plane which is parallel to the plane, determine the,, components of reaction at and the tension in the cable at. 1 ft kn 2 ft 2 ft kn 3 m 2 ft 6 ft 2 ft 2 m D 1. m Prob The plate has a weight of W with center of gravit at G. Determine the distance d along line GH where the vertical force P = 0.7W will cause the tension in wire D to become ero. 78. The plate has a weight of W with center of gravit at G. Determine the tension developed in wires, D, and E if the force P = 0.7W is applied at d = L/2. Prob. 79 * 80. The circular door has a weight of lb and a center of gravit at G. Determine the,, components of reaction at the hinge and the force acting along strut needed to hold the door in equilibrium. Set u =. 81. The circular door has a weight of lb and a center of gravit at G. Determine the,, components of reaction at the hinge and the force acting along strut needed to hold the door in equilibrium. Set u = 90. H L 2 L 2 L 2 L 2 P d G E D 3 ft G 3 ft u Probs. 77/78 Probs. 80/81

142 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved..7 NSTRINTS ND STTIL DETERMINY Member is supported at b a cable and at b a smooth fied square rod which fits loosel through the square hole of the collar. If = 20i - 0j - 7k6 lb, determine the,, components of reaction at and the tension in the cable. 83. Member is supported at b a cable and at b a smooth fied square rod which fits loosel through the square hole of the collar. Determine the tension in cable if the force = -k6 lb. 8. The circular plate has a weight W and center of gravit at its center. If it is supported b three vertical cords tied to its edge, determine the largest distance d from the center to where an vertical force P can be applied so as not to cause the force in an one of the cables to become ero. 86. Solve Prob. 8 if the plate s weight W is neglected. 8 ft 6 ft P r 120 d 12 ft 10 ft ft Probs. 82/83 * 8. Determine the largest weight of the oil drum that the floor crane can support without overturning. lso, what are the vertical reactions at the smooth wheels,, and for this case. The floor crane has a weight of 300 lb, with its center of gravit located at G. Probs. 8/ uniform square table having a weight W and sides a is supported b three vertical legs. Determine the smallest vertical force P that can be applied to its top that will cause it to tip over. a/2 a/2 3 ft G 1.ft a 2. ft 2. ft ft 1 ft 2 ft Prob. 8 Prob. 87

143 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 28 HPTER EQ U I L I R I U M R IGID DY HPTER REVIEW Equilibrium bod in equilibrium does not rotate but can translate with constant velocit, or it does not move at all. = 0 M = Two Dimensions efore analing the equilibrium of a bod, it is first necessar to draw its free-bod diagram. This is an outlined shape of the bod, which shows all the forces and couple moments that act on it. ouple moments can be placed anwhere on a free-bod diagram since the are free vectors. orces can act at an point along their line of action since the are sliding vectors. ngles used to resolve forces, and dimensions used to take moments of the forces, should also be shown on the free-bod diagram. Some common tpes of supports and their reactions are shown below in two dimensions. 2 m 1 m 2 m 1 m 00 N m 00 N m T Remember that a support will eert a force on the bod in a particular direction if it prevents translation of the bod in that direction, and it will eert a couple moment on the bod if it prevents rotation. u u u M roller smooth pin or hinge fied support The three scalar equations of equilibrium can be applied when solving problems in two dimensions, since the geometr is eas to visualie. = 0 = 0 M = 0

144 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. HPTER REVIEW 29 or the most direct solution, tr to sum forces along an ais that will eliminate as man unknown forces as possible. Sum moments about a point that passes through the line of action of as man unknown forces as possible. = 0; - P 2 = 0 = P 2 M = 0; P 2 d 2 + d - P 1 d 1 = 0 P 1 d 1 d P 2 d 2 = P 1d 1 - P 2 d 2 d Three Dimensions Some common tpes of supports and their reactions are shown here in three dimensions. M roller ball and socket fied support M M In three dimensions, it is often advantageous to use a artesian vector analsis when appling the equations of equilibrium. To do this, first epress each known and unknown force and couple moment shown on the free-bod diagram as a artesian vector. Then set the force summation equal to ero. Take moments about a point that lies on the line of action of as man unknown force components as possible. rom point direct position vectors to each force, and then use the cross product to determine the moment of each force. The si scalar equations of equilibrium are established b setting the respective i, j, and k components of these force and moment summations equal to ero. = 0 M = 0 = 0 = 0 = 0 M = 0 M = 0 M = 0 Determinac and Stabilit If a bod is supported b a minimum number of constraints to ensure equilibrium, then it is staticall determinate. If it has more constraints than required, then it is staticall indeterminate. To properl constrain the bod, the reactions must not all be parallel to one another or concurrent. 2 kn m 00 N Staticall indeterminate, five reactions, three equilibrium equations 600 N 200 N 100 N Proper constraint, staticall determinate

145 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. 260 HPTER EQ U I L I R I U M R IGID DY REVIEW PRLEMS * 88. Determine the horiontal and vertical components of reaction at the pin and the force in the cable. Neglect the thickness of the members. 91. Determine the normal reaction at the roller and horiontal and vertical components at pin for equilibrium of the member. 10 kn 0.6 m 0.6 m 3 m 200 N/m 0.8 m 6 kn m 100 N 0. m 60. m Prob. 91 Prob Determine the horiontal and vertical components of reaction at the pin and the reaction at the roller required to support the truss. Set = 600 N. 90. If the roller at can sustain a maimum load of 3 kn, determine the largest magnitude of each of the three forces that can be supported b the truss. * 92. The shaft assembl is supported b two smooth journal bearings and and a short link D. If a couple moment is applied to the shaft as shown, determine the components of force reaction at the journal bearings and the force in the link. The link lies in a plane parallel to the plane and the bearings are properl aligned on the shaft. 20 D 2 m 120 mm 20 mm 300 mm 2 m 2 m 2 m 20 N m 00 mm Probs. 89/90 Prob. 92

146 2010 Pearson Education, Inc. Upper Saddle River, NJ. ll Rights Reserved. REVIEW PRLEMS Determine the reactions at the supports and of the frame. 10 kip kip 7 kip 2 kip 8 ft 6 ft 6 ft 9. vertical force of 80 lb acts on the crankshaft. Determine the horiontal equilibrium force P that must be applied to the handle and the,, components of force at the smooth journal bearing and the thrust bearing. The bearings are properl aligned and eert onl force reactions on the shaft. 80 lb 10 in. 8 ft 9. skeletal diagram of the lower leg is shown in the lower figure. Here it can be noted that this portion of the leg is lifted b the quadriceps muscle attached to the hip at and to the patella bone at. This bone slides freel over cartilage at the knee joint. The quadriceps is further etended and attached to the tibia at. Using the mechanical sstem shown in the upper figure to model the lower leg, determine the tension in the quadriceps at and the magnitude of the resultant force at the femur (pin), D, in order to hold the lower leg in the position shown. The lower leg has a mass of 3.2 kg and a mass center at G 1 ; the foot has a mass of 1.6 kg and a mass center at G 2. 6 ft Prob mm 0. kip 8 in. in. P 6 in. Prob. 9 1 in. * 96. The smmetrical shelf is subjected to a uniform load of kpa. Support is provided b a bolt (or pin) located at each end and and b the smmetrical brace arms, which bear against the smooth wall on both sides at and. Determine the force resisted b each bolt at the wall and the normal force at for equilibrium. 1 in. 2 mm D 7 30 mm 300 mm G 1 G 2 kpa D 0.1 m 1. m 0.2 m Prob. 9 Prob. 96