S in. S in 40 M s = (20.35)(30.0) M s = 611 in-lb clockwise = 2.12 m with a negative action. The moment about B is

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1 Problem 4.14 The moment eerted about point E b the weight is 299 in-lb. What moment does the weight eert about point S? S in. 12 in. E 40 The ke is the geometr rom trigonometr, cos 40 = d 2 13 in, cos 30 = d 1 12 in Thus d 1 = (12 in) cos 30 S in. d 1 =10.39 and d 2 = (13 in) cos in. E 40 d 2 =9.96 We are given that 299 in-lb = d 2 W =9.96 W Now, W =30.0 lb S d in W M s =(d 1 + d 2 )W 12 in 40 M s = (20.35)(30.0) M s = 611 in-lb clockwise E d 2 Problem 4.15 Three forces act on the square plate. etermine the sum of the moments of the forces (a) about, (b) about, (c) about. 3 m 200 N 200 N etermine the perpendicular distance between the points and the lines of action. etermine sign, and calculate moment. (a) The distances from point to the lines of action is ero, hence the moment about is M =0. (b) The perpendicular distances of the lines of action from are: 3 m for the ( force through, with a positive action, and for the force through, = 1 ) = 2.12 m with a negative action. The moment about is M = (3)(200) 2.12(200) = N-m (c) The distance of the force through from is 3 m, with a positive action, and the distance of the force through from is 3 m, with a positive action. The moment about is M = 2(3)(200) = 1200 N-m. 200 N 3 m 200 N 200 N 3 m 3 m 200 N

2 Problem 4.16 etermine the sum of moments of the three forces about (a) point, (b) point, (c) point. 100 lb 200 lb 100 lb 2 ft 2 ft 2 ft 2 ft (a) The sum of the moments about : M = (2)(100) + (4)(200) (6)(100) = lb 200 lb 100 lb (b) The sum of the moments about : 2 ft 2 ft 2 ft 2 ft M = +(2)(100) (2)(100) = 0 (c) The sum of the moments about : M = +(6)(100) (4)(200) + (2)(100) = 0. Problem 4.17 etermine the sum of the moments of the five forces acting on the Howe truss about point. 800 lb 400 lb 600 lb 600 lb E 400 lb 8 ft H I J K L 4 ft 4 ft 4 ft 4 ft 4 ft 4 ft G ll of the moments about are clockwise (negative). The equation for the sum of the moments about in units of ft-lb is given b: 800 lb M = 4(400) 8(600) 12(800) 16(600) 20(400) 600 lb 600 lb or M = 33,600 ft-lb. 400 lb 400 lb E H I J K L G 8 ft 4 ft 4 ft 4 ft 4 ft 4 ft 4 ft

3 Problem 4.22 The vector sum of the three forces is ero, and the sum of the moments of the three forces about is ero. (a) What are and? (b) What is the sum of the moments of the three forces about? The forces are: 80 N 900 mm 400 mm = (0i +1j), = (0i +1j), and = 80(0i 1j). The sum of the forces is: = + + =0, from which Y =( + 80)j =0. The sum of the moments: M = (0.9)(80) + (1.3)( )=0. (a) Solve these two equations to obtain: =55.4 N, and = 24.6 N (b) The moments about : M = (80)(0.4) (1.3) =0 Problem 4.23 The weights (in ounces) of fish,, and are 2.7, 8.1, and 2.1, respectivel. The sum of the moments due to the weights of the fish about the point where the mobile is attached to the ceiling is ero. What is the weight of fish? 12 in 3 in 6 in 7 in 2 in 2 in M Solving = (12)(2.7) 3(10.2+) =0.6 o 12 in 3 in 6 in 7 in 2 in 2 in ,7 ( ) = (10.2+)

4 Problem 4.28 ive forces act on a link in the gearshifting mechanism of a lawn mower. The vector sum of the five forces on the bar is ero. The sum of their moments about the point where the forces and act is ero. (a) etermine the forces,, and. (b) etermine the sum of the moments of the forces about the point where the force acts. The strateg is to resolve the forces into - and -components, determine the perpendicular distances from to the line of action, determine the sign of the action, and compute the moments. 650 mm 30 kn mm 350 mm 25 kn mm The angles are measured counterclockwise from the -ais. forces are The 2 = 30(i cos j sin 135 )= 21.21i j 1 = 25(i cos 20 + j sin 20 )=23.50i +8.55j. (a) The sum of the forces is = = mm 2 = 30 kn 45 1 = 25 kn mm Substituting: 650 mm 350 mm X =( X + X )i =0, and Y =( Y )j =0. Solve the second equation: Y = kn. The distances of the forces from are: the triangle has equal base and altitude, hence the angle is 45, so that the line of action of 1 passes through. The distance to the line of action of is 0.65 m, with a positive action. The distance to the line of action of the -component of 2 is ( ) = 1 m, and the action is positive. The distance to the line of action of the -component of 2 is ( ) = m, and the action is positive. The moment about is Solve: X = kn. Substitute into the force equation to obtain X =18.09 kn (b) The distance from to the line of action of the -component of 1 is m, and the action is negative. The distance from to the line of action of X is m and the action is negative. The distance from to the line of action of Y is 1 m and the action is positive. The distance from to the line of action of the -component of 2 is m and the action is negative. The sum of the moments about : M = (0.350)(21.21) (0.650)(18.09) + (1)(29.76) (0.450)(23.5) = 0 M =(8.55)(1) + (23.5)(0.2)+( X )(0.65)=0. Problem 4.29 ive forces act on a model truss built b a civil engineering student as part of a design project. The dimensions are b = 300 mm and h = 400 mm; = 100 N. The sum of the moments of the forces about the point where and act is ero. If the weight of the truss is negligible, what is the force? h b b b b b b The - and -components of the force are = (i cos 60 + j sin 60 )= (0.5i j) The distance from to the -component is h and the action is positive. The distances to the -component are 3b and 5b. The distance to is 6b. The sum of the moments about is h M =2 (0.5)(h) 3b (0.866) 5b (0.866) + 6b =0. Substitute and solve: = =93.2 N b b b b b b

5 Problem 4.36 The cable from to (the sailboat s foresta) eerts a 230-N force at. The cable from to (the backsta) eerts a 660-N force at. The bottom of the sailboat s mast is located at =4m, =0. What is the sum of the moments about the bottom of the mast due to the forces eerted at b the foresta and backsta? (4,13) m (0,1.2) m (9,1) m Triangle P tan α = ,α=18.73 Triangle Q (4,13) m tan β = 5 12,β= M +M = (13)(230) sin α (13)(660) sin β = 2340 N-m (4,13) (0,1.2) m (9,1) m 230 N 660 N 230 sin α 660 sin β α β α β (0,1.2) P 13 m (9,1) Q (4,0)

6 Problem 4.60 The direction cosines of the force are cos θ =0.818, cos θ =0.182, and cos θ = The support of the beam at will fail if the magnitude of the moment of about eceeds 100 kn-m. etermine the magnitude of the largest force that can safel be applied to the beam. The strateg is to determine the perpendicular distance from to the action line of, and to calculate the largest magnitude of from M =. The position vector from to the point of application of is r = 3i (m). Resolve the position vector into components parallel and normal to. The component parallel to is r P = (r e )e, where the unit vector e parallel to is e = i cos θ X + j cos θ Y + k cos θ Z = 0.818i j 0.545k. The dot product is r e = The parallel component is r P = 2.007i j k. The component normal to is r N = r r P = (3 2)i j k. The magnitude of the normal component is the perpendicular distance: = = m. The maimum moment allowed is M = = 100 kn-m, from which 3 m 3 m = 100 kn-m m =57.86 = 58 kn Problem 4.61 The force eerted on the grip of the eercise machine points in the direction of the unit vector e = 2 3 i 2 3 j+ 1 3k and its magnitude is 120 N. etermine the magnitude of the moment of about the origin. 150 mm force is The vector from to the point of application of the 200 mm r =0.25i +0.2j 0.15k m 250 mm and the force is = e or =80i 80j +40k N. The moment of about is 150 mm M = r = N-m 200 mm or M = 4i 22j 36k N-m 250 mm and M = N-m M =42.4N-m

7 Problem 4.62 The force in Problem 4.61 points in the direction of the unit vector e = 2 3 i 2 3 j + 1 3k. The support at will safel support a moment of 560 N-m magnitude. (a) ased on this criterion, what is the largest safe magnitude of? (b) If the force ma be eerted in an direction, what is its largest safe magnitude? See the figure of Problem The moment in Problem 4.61 can be written as M = where = M =( i j 0.3k) If we set M = 560 N-m, we can solve for ma 560 = ma ma = 1586 N (b) If can be in an direction, then the worst case is when r. The moment in this case is M = r worst nd the magnitude of M is M =( ) M =0.353 r = = m 560 = (0.3536) WRST worst = 1584 N Problem 4.63 n engineer estimates that under the most adverse epected weather conditions, the total force on the highwa sign will be = ±1.4i 2.0j (kn). What moment does this force eert about the base? TUN The coordinates of the point of application of the force are: (0, 8, 8). The position vector is r =8j +8k. The cross product is r = =16i ( 1.4)(8)j +( 1.4)(8)k ± M =16i ± 11.2j 11.2k (N-m) heck: Use perpendicular distances to forces: M X =8(2)=16, M Y =8(±1.4) = ±11.2, M Z = 8(±1.4) = 11.2.

8 Problem 4.70 onsider the 70-m tower in Problem Suppose that the tension in cable is 4 kn, and ou want to adjust the tensions in cables and so that the sum of the moments about the origin due to the forces eerted b the cables at point is ero. etermine the tensions. rom Varignon s theorem, the moment is ero onl if the resultant of the forces normal to the vector r is ero. rom Problem 4.69 the unit vectors are: The tensions are T = 4e, T = T e, and T = T e. The components normal to r are e e e = r r = i j = i j k = r r = i j k = i j k = r r = i 70 j +0k =0.4963i j +0k 80.6 X =( T T )i =0 Z =( T T )k =0. The HP-28S calculator was used to solve these equations: T =2.23 kn, T =2.43 kn Problem 4.71 The tension in cable is 150 N. The tension in cable is 100 N. etermine the sum of the moments about due to the forces eerted on the wall b the cables. The coordinates of the points,, are (8, 0, 0), (0, 4, 5), (0, 8, 5), (0, 0, 5). The point is the intersection of the lines of action of the forces. The position vector is 5 m 5 m 4 m r =8i +0j 5k. The position vectors and are r = 8i +4j 5k, r = = m. r = 8i +8j +5k, r = = m. The unit vectors parallel to the cables are: e = i j k, e = i j k. The tensions are T = 150e = i j 73.19k, T = 100e = 64.68i j 40.42k. M = =( )(5)i ((8)(+32.77) ( 5)(181.79))j + (8)( )k M = 616.2i j 985.9k (N-m) (Note: n alternate method of solution is to epress the moment in terms of the sum: M =(r T +(r T ).) 5 m The sum of the forces eerted b the wall on is 5 m 4 m T = i j 32.77k. The force eerted on the wall b the cables is T. The moment about is M = r T,

9 Problem 4.98 The tension in cable is 80 lb. What is the moment about the line due to the force eerted b the cable on the wall at? 8 ft 6 ft (6, 0, 10) ft The strateg is to find the moment about the point eerted b the force at, and then to find the component of that moment acting along the line. The coordinates of the points,, are (8, 6, 0), (3, 6, 0), (3, 0, 0). The position vectors are: r = 8i +6j, r = 3i +6j, r = 3i. The vector parallel to is r = r r = 6j. The unit vector parallel to is e = 1j. The vector from point to is r = r r =5i. The position vector of is r =6i +10k. The vector parallel to is r = r r = 2i 6j +10k. The magnitude is r = ft. The unit vector parallel to is e = i j k. The tension acting at is T =80e = 13.52i 40.57j k. The magnitude of the moment about due to the tension acting at is M = e (r T )= = (ft lb). The moment about is M = 338.1e = 338.1j (ft lb). The sense of the moment is along the curled fingers of the right hand when the thumb is parallel to, pointing toward. 8 ft 6 ft (6, 0, 10)

10 Problem our forces and a couple act on the beam. The vector sum of the forces is ero, and the sum of the moments about the left end of the beam is ero. What are the forces,, and? 200 N-m 800 N The sum of the forces about the -ais is X = Y + 800=0. The sum of the forces about the -ais is X = X =0. 4 m 4 m 3 m The sum of the moments about the left end of the beam is ML =11 8(800) 200= N rom the moments: 200 N-m = = 600 N. Substitute into the forces balance equation to obtain: Y = = 200 N 4 m 4 m 3 m Problem The force =40i +24j +12k (N). (a) What is the moment of the couple? (b) etermine the perpendicular distance between the lines of action of the two forces. (6, 3, 2) m (10, 0, 1) m (a) The moment of the couple is given M = r M =( 4i +3j +1k) (40i +24j +12k) M =12i +88j 216k (N-m) (6, 3, 2) m (b) M = d sin 90 = =48.2N M = M 2 + M 2 + M 2 = N d = perpendicular distance d = M / d =4.85 m (10, 0, 1) m

11 Problem etermine the sum of the moments eerted on the plate b the three couples. (The 80-lb forces are contained in the - plane.) 20 lb 20 lb 40 lb 8 ft The moments of two of the couples can be determined from inspection: 40 lb lb lb M 1 M 2 = (3)(20)k = 60k ft lb. = (8)(40)j = 320j ft lb The forces in the 3rd couple are resolved: = (80)(i sin 60 + k cos 60 )=69.282i +40k 20 lb 20 lb 8 ft 40 lb The two forces in the third couple are separated b the vector 40 lb r 3 =(6i +8k) (8k) =6i lb lb The moment is M 3 = r 3 3 = = 240j. The sum of the moments due to the couples: M = 60k + 320j 240j =80j 60k ft lb Problem What is the magnitude of the sum of the moments eerted on the T -shaped structure b the two couples? 50j (lb) 50i + 20j 10k (lb) inspection: The moment of the 50 lb couple can be determined b M 1 = (50)(3)k = 150k ft lb. 50j (lb) The vector separating the other two force is r =6k. The moment is 50i 20j + 10k (lb) M 2 = r = = 120i + 300j The sum of the moments is M = 120i + 300j 150k. 50j (lb) 50j (lb) The magnitude is M = = lb

12 Problem In Problem 4.125, the forces =2i +6j +3k (kn), = i 2j +2k (kn), and the couple M = M j + M k (kn-m). etermine the values for M and M, so that the sum of the moments of the two forces and the couple about is ero. rom the solution to Problem 4.125, the sum of the moments of the two forces about is M orces =0i 7j +2k (kn-m). The required moment, M, must be the negative of this sum. Thus M =7(kN-m), and M = 2 (kn-m). Problem Two wrenches are used to tighten an elbow fitting. The force = 10k (lb) on the right wrench is applied at (6, 5, 3) in., and the force on the left wrench is applied at (4, 5, 3) in. (a) etermine the moment about the ais due to the force eerted on the right wrench. (b) etermine the moment of the couple formed b the forces eerted on the two wrenches. (c) ased on the results of (a) and (b), eplain wh two wrenches are used. The position vector of the force on the right wrench is r R =6i 5j 3k. The magnitude of the moment about the -ais is M R = e X (r R ) = = 50 in lb from which M XL =50i in lb, which is opposite in direction and equal in magnitude to the moment eerted on the -ais b the right wrench. The left wrench force is applied 2 in nearer the origin than the right wrench force, hence the moment must be absorbed b the space between, where it is wanted. (a) The moment about the -ais is M R = M R e X = 50i (in lb). (b) The moment of the couple is M =(r R r L ) R = = 20j in lb (c) The objective is to appl a moment to the elbow relative to connecting pipe, and ero resultant moment to the pipe itself. resultant moment about the -ais will affect the joint at the origin. However the use of two wrenches results in a net ero moment about the -ais the moment is absorbed at the juncture of the elbow and the pipe. This is demonstrated b calculating the moment about the -ais due to the left wrench: M X = e X (r L L )= =50in lb