Announcements. Moment of a Force


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1 Announcements Test observations Units Significant figures Position vectors Moment of a Force Today s Objectives Understand and define Moment Determine moments of a force in 2D and 3D cases Moment of a force Class Activities Applications Moments in 2D Moments in 3D Examples Engr221 Chapter 4 1
2 Applications What is the net effect of the two forces on the wheel shaft? Applications  continued What is the effect of the 30 N force on the lug nut? Engr221 Chapter 4 2
3 Moments in 2D The moment of a force about a point provides a measure of the tendency for rotation (sometimes called torque). Moments in 2D  continued In the 2D case, the magnitude of the moment is: M o = Fd As shown, d is the perpendicular distance from point O to the line of action of the force F. In 2D, the direction of M O is either clockwise or counterclockwise depending on the tendency for rotation. Engr221 Chapter 4 3
4 Moments in 2D  continued b O d a F b O a F y F F x For example, M O = Fd and the direction is counterclockwise. Often it is easier to determine M O by using the components of F. Using this approach, M O = (F Y a) (F X b). Note the different signs on the terms. The typical sign convention for a moment in 2D is that counterclockwise is considered positive. We can determine the direction of rotation by imagining the body pinned at O and deciding which way the body would rotate due to the force. 3D Moments and Vector Formulation Moments in 3D can be calculated using a scalar (2D) approach but it can be difficult and time consuming. Thus, it is often easier to use a mathematical approach called the vector cross product. Using the vector cross product, M O = r F Here, r is a position vector from point O to any point on the line of action of force F. Engr221 Chapter 4 4
5 Cross Product In general, the cross product of two vectors A and B results in another vector C, i.e., C = A B. The magnitude and direction of the resulting vector can be written as C = A B = (AB sinθ) u C Here, u C is the unit vector perpendicular to both the A and B vectors as shown (or to the plane containing the A and B vectors). The right hand rule is useful for determining the direction of the vector resulting from a cross product. For example: i j = k Cross Product  continued Note that a vector crossed into itself is zero, e.g., i i = 0 Engr221 Chapter 4 5
6 Cross Product  continued Of even more utility, the cross product can be written as and each component can be determined using 2 2 determinants: Moments in 3D So, using the cross product, a moment can be expressed as By expanding the above equation using the 2 2 determinants (see Section 4.2 in your textbook), we get M O = (r y F z  r z F y ) i  (r x F z  r z F x ) j + (r x F y  r y F x ) k The physical meaning of the above equation becomes evident by considering the force components separately and using a 2D formulation. Units of moments are Nm or lbft. Engr221 Chapter 4 6
7 Example A Given: A 400 N force is applied to the frame and θ = 20 Find: The moment of the force at A Plan: 1) Resolve the force along the x and y axes 2) Determine M A using scalar analysis Example A  continued + F y = 400 sin 20 N + F x = 400 cos 20 N + M A = {(400 cos 20 )(2) + (400 sin 20 )(3)} N m = 1162 N m = 1.16 kn m Engr221 Chapter 4 7
8 Example B o Given:a = 3 in, b = 6 in, and c = 2 in Find: Moment of F about point O Plan: 1) Find r OA 2) Determine M O = r OA F Solution r OA = {3 i + 6 j 0 k} in i j k M O = = [{6(1) 0(2)} i {3(1) 0(3)} j {3(2) 6(3)} k] lb in = {6 i + 3 j 12 k} lb in Example C Given: a = 3 in, b = 6 in, and c = 2 in Find: Moment of F about point P Plan: 1) Find r PA 2) Determine M P = r PA x F Solution: r PA = { 3 i + 6 j  2 k } in M P = i j k = { 2 i  3 j  12 k } lb in Engr221 Chapter 4 8
9 Example D Given: A 40 N force is applied to the wrench Find: The moment of the force at O Solution: + F y =  40 cos 20 N + F x =  40 sin 20 N Plan: 1) Resolve the force along the x and y axes 2) Determine M O using scalar analysis + M O = {(40 cos 20 )(200) + (40 sin 20 )(30)}N mm = N mm = N m Questions 1. The moment of force F about point O is defined as M O = A) r x F B) F x r C) r F D) r * F 2. What is the moment of the 10 N force about point A (M A )? A) 10 Nm B) 30 Nm C) 13 Nm D) (10/3) Nm E) 7 Nm A F = 10 N d = 3 m Engr221 Chapter 4 9
10 Question If a force of magnitude F can be applied in four different 2D configurations (P, Q, R, & S), select the cases resulting in the maximum and minimum torque values on the nut. A) (Q, P) B) (R, S) C) (P, R) D) (Q, S) P Q S R Example Problem Determine the direction θ for 0 θ 180 of the force F so that F produces (a) the maximum moment about point A and (b) the minimum moment about point A. Calculate the moment in each case. Maximum θ= 56.3º M A = 1.44KNM ccw Mimimum θ = 146º M A = 0 NM Engr221 Chapter 4 10
11 Example Problem The two boys push on the gate with forces of F A = 30 lb and F B = 50 lb as shown. Determine the moment of each force about point C. Which way will the gate rotate, clockwise or counterclockwise? Neglect the thickness of the gate. M CA = 162 lbft cw M CB = 260 lbft ccw Example Problem Strut AB of the 1m diameter hatch door exerts a force of 450 N on point B. Determine the moment of this force about point O. M O = {373i 100j +173k} NM Engr221 Chapter 4 11
12 Textbook Problem 4.40 The force F = {600i + 300j 600k} N acts at the end of the beam. Determine the moment of the force about point A. M O = {840i + 360j  660k} NM Summary Understand and define Moment Determine moments of a force in 2D and 3D cases Engr221 Chapter 4 12
13 Announcements Moment About an Axis Today s Objectives Be able to determine the moment of a force about an axis using Scalar analysis Vector analysis Class Activities Applications Scalar analysis Vector analysis Examples Engr221 Chapter 4 13
14 Applications With the force F, a person is creating the moment M A. What portion of M A is used in turning the socket? The force F is creating the moment M O. How much of M O acts to unscrew the pipe? Scalar Analysis Recall that the moment of a force about any point A is M A = Fd A where d A is the perpendicular distance from the point to the force s line of action. This concept can be extended to find the moment of a force about an axis. In the figure above, the moment about the yaxis would be M y = 20(0.3) = 6 Nm. This calculation is not always trivial and vector analysis is usually preferable. Engr221 Chapter 4 14
15 Vector Analysis Our goal is to find the moment of F (the tendency to rotate the body) about the axis a a. First, compute the moment of F about any arbitrary point O that lies on the a a axis using the cross product: M O = r OA F Now, find the component of M O along the axis a a using the dot product: M a a = u a M O Vector Analysis  continued M a a can also be obtained as The above equation is called the triple scalar product. In this equation, u a represents the unit vector along the a a axis r is the position vector from any point on the a a axis to any point on the line of action of the force (A) F is the force vector Engr221 Chapter 4 15
16 A B Example A Given:A force is applied to the tool to open a gas valve Find: The magnitude of the moment of this force about the z axis Plan: 1) Find M z = u (r F) 2) Note that u = 1k 3) The vector r is the position vector from A to B 4) Force F is already given in Cartesian vector form Example A  continued A B u = 1k r AB = {0.25 sin 30 i cos30 j} m = {0.125 i j} m F = {60 i + 20 j + 15 k} N M z = u (r AB F) M z = d with the Trial Edition of 0 Smart = 1{0.125(20) (60)} Nm = 15.5 Nm Engr221 Chapter 4 16
17 Example B Given: A force of 80 lb acts along the edge DB Find: The magnitude of the moment of this force about the axis AC Plan: 1) Find M AC = u AC (r AB F DB ) 2) Find u AC = r AC /r AC 3) Find F DB = 80(u DB ) lb = 80(r DB /r DB ) lb 4) Complete the triple scalar product Example B  continued r AB = { 20 j } ft r AC = { 13 i + 16 j } ft r DB = { 5 i + 10 j 15 k } ft u AC = (13 i + 16 j) / ( ) ½ ft = i j F DB = 80 {r DB / ( ) ½ } lb = { i j k } lb Engr221 Chapter 4 17
18 Example B  continued Now find the triple product, M AC = u AC ( r AB F DB ) M AC = ft lb M AC = {20 (64.14) (0 0)} lb ft = 809 lb ft IMPORTANT: The negative sign indicates that the sense of M AC is opposite to that of u AC Questions 1. For finding the moment of the force F about the xaxis, the position vector in the triple scalar product should be A) r AC B) r BA C) r AB D) r BC 2. If r = {1i + 2j} m and F = {10i + 20j + 30k} N, then the moment of F about the yaxis is m A) 10 B) 30 C) 40 D) None of the above Engr221 Chapter 4 18
19 Questions 1. When determining the moment of a force about a specified axis, the axis must be along A) the x axis B) the y axis C) the z axis D) any line in 3D space E) any line in the xy plane 2. The triple scalar product u ( r F ) results in: A) a scalar quantity ( + or  ) B) a vector quantity C) zero D) a unit vector E) an imaginary number Question The force F is acting along DC. Using the triple product to determine the moment of F about the bar BA, you could use any of the following position vectors except A) r BC B) r AD C) r AC D) r DB E) r BD Engr221 Chapter 4 19
20 Summary Be able to determine the moment of a force about an axis using Scalar analysis Vector analysis Announcements Engr221 Chapter 4 20
21 Moment of a Couple Today s Objectives Define a couple Determine the moment of a couple Class activities Applications Moment of a Couple Examples Applications A torque or moment of 12 Nm is required to rotate the wheel. Which one of the two grips of the wheel above will require less force to rotate the wheel? Engr221 Chapter 4 21
22 Applications  continued The crossbar lug wrench is being used to loosen a lug net. What is the effect of changing dimensions a, b, or c on the force that must be applied? Moment of a Couple A couple is defined as two parallel forces with the same magnitude but opposite in direction, separated by a perpendicular distance d. The moment of a couple is defined as: M = Fd (using a scalar analysis) or as M = r F (using a vector analysis). Here, r is any position vector from the line of action of F to the line of action of F. Engr221 Chapter 4 22
23 Moment of a Couple  continued The net external effect of a couple is that the net force equals zero and the magnitude of the net moment equals Fd. Since the moment of a couple depends only on the distance between the forces, the moment of a couple is a free vector. It can be moved anywhere on the body and have the same external effect on the body. Moments due to couples can be added using the same rules as adding any vectors. Example A Scalar Approach Given: Two couples act on the beam and d = 8 ft Find: The resultant couple Plan: 1) Resolve the forces in x and y directions so they can be treated easily as couples 2) Determine the net moment due to the two couples Engr221 Chapter 4 23
24 The net moment is: Example A  continued + ΣM = (48)(4) + (40)(cos 30º)(8) lb ft = = 85.1 lb ft The x and y components of the top 60 lb force are: (4/5)(60 lb) = 48 lb vertically up (3/5)(60 lb) = 36 lb to the left Similarly, for the top 40 lb force: (40 lb) (sin 30 ) up (40 lb) (cos 30 ) to the left Example B Vector Approach Given: A force couple acting on the rod A B Find: Plan: The couple moment acting on the rod in Cartesian vector notation 1) Use M = r F to find the couple moment 2) Set r = r AB and F = {14i 8j 6k} N 3) Calculate the cross product to find M Engr221 Chapter 4 24
25 Example B  continued A r AB = {0.8i + 1.5j 1k} m F = {14i 8j 6k} N B M = r AB F = i j k the 0.8 Trial Edition 1.51of S Nm = {i (9 (8)) j (4.8 (14)) + k ( (1.5))} Nm = {17 i 9.2 j 27.4 k} Nm Example C Scalar Approach Given: Two couples act on the beam. The resultant couple is zero. Find: Plan: The magnitudes of the forces P and F and the distance d. 1) Use the definition of a couple to find P and F 2) Resolve the 300N force in x and y directions 3) Determine the net moment 4) Equate the net moment to zero to find d Engr221 Chapter 4 25
26 Example C  continued From the definition of a couple: P = 500 N and F = 300 N Resolve the 300 N force into vertical and horizontal components. The vertical component is (300 cos 30º) N and the horizontal component is (300 sin 30º) N It was given that the net moment equals zero. So + ΣM =  (500)(2) + (300 cos 30º)(d)  (300 sin 30º)(0.2) = 0 Solving this equation for d yields d = ( sin 30º) / (300 cos 30º) = 3.96 m Example D Vector Approach Given: F = {25 k} N and  F = { 25 k} N Find: The couple moment acting on the pipe assembly, using Cartesian vector notation. Plan: 1) Use M = r F to find the couple moment 2) Set r = r AB and F = {25 k} N 3) Calculate the cross product to find M Engr221 Chapter 4 26
27 Example D  continued r AB = { i 200 j } mm = { i 0.2 j } m F = {25 k} N M = r AB F = i j k Nm = { i (5 0 ) j ( ) + k (0) } Nm = { 5 i j } Nm Questions 1. A couple is applied to the beam as shown. Its moment equals Nm A) 50 B) 60 C) 80 D) You can determine the couple moment as M = r F If F = {20k} lb, then r is A) r BC B) r AB C) r CB D) r AC 1m 50 N 2m Engr221 Chapter 4 27
28 Questions 1. In statics, a couple is defined as separated by a perpendicular distance. A) two forces in the same direction B) two forces of equal magnitude C) two forces of equal magnitude acting in the same direction D) two forces of equal magnitude acting in opposite directions 2. The moment of a couple is called a vector. A) free B) spin C) romantic D) sliding Questions 1. F 1 and F 2 form a couple. The moment of the couple is given by A) r 1 F 1 B) r 2 F 1 C) F 2 r 1 D) r 2 F 2 r1 F1 F2 r2 2. If three couples act on a body, the overall result is that A) the net force is not equal to 0 B) the net force and net moment are equal to 0 C) the net moment equals 0 but the net force is not necessarily equal to 0 D) the net force equals 0 but the net moment is not necessarily equal to 0 Engr221 Chapter 4 28
29 Textbook Problem 479 Express the moment of the couple acting on the pipe assembly in Cartesian vector form. Solve the problem (a) using Eq. 413, and (b) summing the moment of each force about point O. Take F = {25k} N. M C = (5i j) Nm Summary Define a couple Determine the moment of a couple Engr221 Chapter 4 29
30 Announcements Equivalent Force and Couple Systems Today s Objectives Determine the effect of moving a force Find an equivalent forcecouple system for a system of forces and couples Class Activities Applications Equivalent systems System reduction Examples Engr221 Chapter 4 30
31 Applications What is the resultant effect on the person s hand when the force is applied in four different ways? Applications  continued Several forces and a couple moment are acting on this vertical section of an Ibeam.?? Can you replace them with just one force and one couple moment at point O that will have the same external effect? If yes, how will you do that? Engr221 Chapter 4 31
32 An Equivalent Force System = When a number of forces and couple moments are acting on a body, it is easier to understand their overall effect on the body if they are combined into a single force and couple moment having the same external effect. The two force and couple systems are called equivalent systems since they have the same external effect on the body. Moving a Force On its Line of Action Moving a force from A to O, when both points are on the vectors line of action, does not change the external effect. Hence, a force vector is called a sliding vector. (But the internal effect of the force on the body does depend on where the force is applied). Engr221 Chapter 4 32
33 Moving a Force Off its Line of Action Moving a force from point A to O (as shown above) requires creating an additional couple moment. Since this new couple moment is a free vector, it can be applied at any point P on the body. Resultant of a Force and Couple System When several forces and couple moments act on a body, you can move each force and its associated couple moment to a common point O. Now you can add all the forces and couple moments together and find one resultant forcecouple moment pair. Engr221 Chapter 4 33
34 Resultant of a Force and Couple System If the force system lies in the xy plane (the 2D case), then the reduced equivalent system can be obtained using the following three scalar equations: ForceMoment Reduction = = If F R and M RO are perpendicular to each other, then the system can be further reduced to a single force, F R, by simply moving F R from O to P. In three special cases, concurrent, coplanar, and parallel systems of forces, the system can always be reduced to a single force. Engr221 Chapter 4 34
35 Example A Given: A 2D force and couple system as shown. Find: The equivalent resultant force and couple moment acting at A, and then the equivalent single force location along the beam AB. Plan: 1) Sum all the x and y components of the forces to find F RA 2) Find and sum all the moments resulting from moving each force to A 3) Shift F RA to a distance d such that d = M RA /F Ry Example A  continued + ΣF Rx = sin 30 = 42.5 lb + ΣF Ry = cos 30 = lb F R + M RA = 35 cos30 (2) + 20(6) 25(3) = lb ft The equivalent single force F R can be located on the beam AB at a distance d measured from point A. d = M RA /F Ry = 105.6/50.31 = 2.10 ft F R = ( ) 1/2 = 65.9 lb θ = tan 1 (50.31/42.5) = 49.8 Engr221 Chapter 4 35
36 1) Find F RO = F i = F Rzo k Example B 2) Find M RO = (r i F i ) = M RxO i + M RyO j Given: The building slab has four columns. F 1 and F 2 = 0. Find: The equivalent resultant force and couple moment at the origin O. Also find the location (x,y) of the single equivalent resultant force. Plan: 3) The location of the single equivalent resultant force is given as x = M RyO /F RzO and y = M RxO /F RzO Example B  continued F RO = {50 k 20 k} = {70 k} kn M RO = (10 i) (20 k) + (4 i + 3 j) (50 k) = {200 j j 150 i} kn m = {150 i j } kn m The location of the single equivalent resultant force is given as: x = M Ryo /F Rzo = 400/(70) = 5.71 m y = M Rxo /F rzo = (150)/(70) = 2.14 m Engr221 Chapter 4 36
37 Example C Given:A 2D force and couple system as shown Find: The equivalent resultant force and couple moment acting at A Plan: 1) Sum all the x and y components of the forces to find F RA 2) Find and sum all the moments resulting from moving each force to A and add them to the 500 lbft free moment to find the resultant M RA Example C  continued Summing the force components: + ΣF x = (4/5) 150 lb + 50 lb sin 30 = 145 lb + ΣF y = (3/5) 150 lb + 50 lb cos 30 = 133 lb Now find the magnitude and direction of the resultant: F RA = ( ) 1/2 = 197 lb and θ = tan 1 (133.3/145) = M RA = { (4/5)(150)(2) 50 cos30 (3) + 50 sin30 (6) } = 760 lb ft Engr221 Chapter 4 37
38 Example D Given: Handle forces F 1 and F 2 are applied to the electric drill Find: Plan: An equivalent resultant force and couple moment at point O a) Find F RO = Σ F i b) Find M RO = Σ ( r i F i ) F i are the individual forces in Cartesian vector form r i are the position vectors from the point O to any point on the line of action of F i Example D  continued F 1 F 2 = {6 i 3 j 10 k} N = {0 i + 2 j 4 k} N F RO = {6 i 1 j 14 k} N r 1 r 2 = {0.15 i k} m = {0.25 j k} m M RO = r 1 F 1 + r 2 F 2 M RO = { i j k i j k } Nm = {0.9 i j 0.45 k i + 0 j + 0 k} Nm = {1.3 i j 0.45 k} Nm Engr221 Chapter 4 38
39 Questions 1. A general system of forces and couple moments acting on a rigid body can be reduced to a A) single force B) single moment C) single force and two moments D) single force and a single moment 2. The original force and couple system and an equivalent forcecouple system have the same effect on a body. A) internal B) external C) internal and external D) microscopic Questions 1. The forces on the pole can be reduced to a single force and a single moment at point A) P B) Q C) R D) S E) Any of these points 2. Consider two couples acting on a body. The simplest possible equivalent system at any arbitrary point on the body will have: A) one force and one couple moment B) one force C) one couple moment D) two couple moments X Z S R Q P Y Engr221 Chapter 4 39
40 Summary Determine the effect of moving a force Find an equivalent forcecouple system for a system of forces and couples Announcements Engr221 Chapter 4 40
41 Distributed Loading Reduction Today s Objective Determine an equivalent force for a distributed load Class Activities Applications Equivalent force Examples Applications A distributed load on the beam exists due to the weight of the lumber. Is it possible to reduce this force system to a single force that will have the same external effect? If yes, how? Engr221 Chapter 4 41
42 Applications  continued The sandbags on the beam create a distributed load. How can we determine a single equivalent resultant force and its location? Distributed Loading In many situations a surface area of a body is subjected to a distributed load. Such forces are caused by winds, fluids, or the weight of items on the body s surface. We will analyze the most common case of a distributed pressure loading. This is a uniform load along one axis of a flat rectangular body. In such cases, w is a function of x and has units of force per length. Engr221 Chapter 4 42
43 Magnitude of Resultant Force Consider an element of length dx. The force magnitude df acting on it is given as df = w(x) dx The net force on the beam is given by + F R = L df = L w(x) dx = A Here, A is the area under the loading curve w(x). Location of the Resultant Force The force df will produce a moment of (x)(df) about point O. The total moment about point O is given as + M RO = L x df = L x w(x) dx Assuming that F R acts at x, it will produce the moment about point O as + M RO = x (F R ) = L x w(x) dx Engr221 Chapter 4 43
44 Location of the Resultant Force  continued Comparing the last two equations, we get In other words, F R acts through a point C, which is called the geometric center or centroid of the area under the loading curve w(x). Examples We will usually consider only rectangular and triangular loading diagrams whose centroids are well defined and shown on the inside back cover of your textbook. In the rectangular loading, F R = = 4,000 lb, x = 5 ft In the triangular loading, F R = (0.5) (600) (6) = 1,800N and x = 6 (1/3) 6 = 4 m Note that the centroid in a right triangle is at a distance one third the width of the triangle as measured from its base. Engr221 Chapter 4 44
45 Example A Given: The loading on the beam as shown Find: The equivalent force and its location from point A Plan: 1) Consider the trapezoidal loading as two separate loads (one rectangular and one triangular) 2) Find F R and x for each of these two distributed loads 3) Determine the overall F R and for the three point loadings x Example A  continued For the rectangular loading of height 0.5 kn/m and width 3m, F R1 = 0.5 kn/m 3 m = 1.5 kn x 1= 1.5m from A For the triangular loading of height 2 kn/m and width 3m, F R2 = (0.5) (2 kn/m) (3 m) = 3 kn and its line of action is at 1m from A For the combined loading of the three forces, F R = 1.5 kn + 3 kn kn = 6 kn + M RA = (1.5) (1.5) + 3 (1) + (1.5) 4 = knm M RA = F R (x) x Hence, x = x(11.25) / (6) = 1.88m from A Engr221 Chapter 4 45
46 Questions 100 N/m F R 12 m x 1. F R = A) 12 N B) 100 N C) 600 N D) 1200 N 2. x = A) 3 m B) 4 m C) 6 m D) 8 m w Distributed load curve F R Questions x 1. The resultant force (F R ) due to a distributed load is equivalent to the under the distributed loading curve, w = w(x). A) centroid B) arc length C) area D) volume 2. The line of action of the distributed load s equivalent force passes through the of the distributed load. A) centroid B) midpoint C) left edge D) right edge Engr221 Chapter 4 46
47 Questions A B A d F R B 1. What is the location of F R, i.e., the distance d? A) 2m B) 3m C) 4m 3 m 3 m D) 5m E) 6m x 2 F 1 F 2 x F R 2. If F 1 = 1N, x 1 = 1m, F 2 = 2N and x 2 = 2m, what is the location of F R, i.e., the distance x? x 1 A) 1m B) 1.33m C) 1.5m D) 1.67m E) 2m Textbook Problem Simplify the distributed loading to an equivalent resultant force and specify the magnitude and location of the force, measured from A. F R = 1.25 kn x A = 8.00 m Engr221 Chapter 4 47
48 Summary Determine an equivalent force for a distributed load Engr221 Chapter 4 48
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