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.