4 1. If A, B, and D are given vectors, prove the distributive law for the vector cross product, i.e., A : (B + D) = (A : B) + (A : D). Consider the three vectors; with A vertical. Note obd is perpendicular to A. od = ƒ A * (B + D) ƒ = ƒ A ƒƒb + D ƒ sin u 3 ob = ƒ A * B ƒ = ƒ A ƒƒb ƒ sin u 1 bd = ƒ A * D ƒ = ƒ A ƒƒd ƒ sin u 2 Also, these three cross products all lie in the plane obd since they are all perpendicular to A. As noted the magnitude of each cross product is proportional to the length of each side of the triangle. The three vector cross products also form a closed triangle o b d which is similar to triangle obd. Thus from the figure, A * (B + D) = (A * B) + (A * D) (QED) Note also, A = A x i + A y j + A z k B = B x i + B y j + B z k D = D x i + D y j + D z k i j k A * (B + D) = 3 A x A y A z 3 B x + D x B y + D y B z + D z = [A y (B z + D z ) - A z (B y + D y )]i - [A x (B z + D z ) - A z (B x + D x )]j + [A x (B y + D y ) - A y (B x + D x )]k = [(A y B z - A z B y )i - (A x B z - A z B x )]j + (A x B y - A y B x )k + [(A y D z - A z D y )i - (A x D z - A z D x )j + (A x D y - A y D x )k i j k i j k = 3 A x A y A z 3 + 3 A x A y A z 3 B x B y B z D x D y D z = (A * B) + (A * D) (QED)
4 2. Prove the triple scalar product identity A # (B : C) = (A : B) # C. As shown in the figure Thus, But, Thus, Area = B(C sin u) = B * C Volume of parallelepiped is B * C h h = A # u(b * C) = ` A # a B * C B * C b ` Volume = A # (B * C) Since (A * B) # C represents this same volume then A # (B : C) = (A : B) # C (QED) Also, LHS = A # (B : C) i j k = (A x i + A y j + A z k) # 3 B x B y B z 3 C x C y C z = A x (B y C z - B z C y ) - A y (B x C z - B z C x ) + A z (B x C y - B y C x ) = A x B y C z - A x B z C y - A y B x C z + A y B z C x + A z B x C y - A z B y C x RHS = (A : B) # C i j k = 3 A x A y A z 3 # (Cx i + C y j + C z k) B x B y B z = C x (A y B z - A z B y ) - C y (A x B z - A z B x ) + C z (A x B y - A y B x ) = A x B y C z - A x B z C y - A y B x C z + A y B z C x + A z B x C y - A z B y C x Thus, LHS = RHS A # (B : C) = (A : B) # C (QED)
4 3. Given the three nonzero vectors A, B, and C, show that if A # (B : C) = 0, the three vectors must lie in the same plane. Consider, A # (B * C) = A B * C cos u = ( A cos u) B * C = h B * C = BC h sin f = volume of parallelepiped. If A # (B * C) = 0, then the volume equals zero, so that A, B, and C are coplanar.
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4 6. The crane can be adjusted for any angle 0 u 90 and any extension 0 x 5m. For a suspended mass of 120 kg, determine the moment developed at A as a function of x and u. What values of both x and u develop the maximum possible moment at A? Compute this moment. Neglect the size of the pulley at B. 1.5 m A 9m θ x B a+ M A = -12019.81217.5 + x2 cos u = 5-1177.2 cos u17.5 + x26 N # m = 51.18 cos u17.5 + x26 kn # m (Clockwise) The maximum moment at A occurs when u = 0 and x = 5m. a+ 1M A 2 max = 5-1177.2 cos 0 17.5 + 526 N # m = -14 715 N # m = 14.7 kn # m (Clockwise)
4 7. Determine the moment of each of the three forces about point A. F 1 250 N 30 A F 2 60 300 N 2 m 3 m 4 m The moment arm measured perpendicular to each force from point A is d 1 = 2 sin 60 = 1.732 m d 2 = 5 sin 60 = 4.330 m B 5 4 3 F 3 500 N d 3 = 2 sin 53.13 = 1.60 m Using each force where M A = Fd, we have a + 1M F1 2 A = -25011.7322 = -433 N # m = 433 N # m a + 1M F2 2 A = -30014.3302 (Clockwise) = -1299 N # m = 1.30 kn # m a + 1M F3 2 A = -50011.602 (Clockwise) = -800 N # m = 800 N # m (Clockwise)
4 8. Determine the moment of each of the three forces about point B. F 1 250 N 30 A 2 m 3 m F 2 300 N 60 The forces are resolved into horizontal and vertical component as shown in Fig. a. For F 1, a+ M B = 250 cos 30 (3) - 250 sin 30 (4) = 149.51 N # m = 150 N # m d For F 2, a + M B = 300 sin 60 (0) + 300 cos 60 (4) = 600 N # m d Since the line of action of F 3 passes through B, its moment arm about point B is zero. Thus M B = 0 4 m B 5 4 3 F 3 500 N
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4 11. The railway crossing gate consists of the 100-kg gate arm having a center of mass at G a and the 250-kg counterweight having a center of mass at G W. Determine the magnitude and directional sense of the resultant moment produced by the weights about point A. 2.5 m A G a 0.75 m B G W 0.5 m 1 m 0.25 m +(M R ) A = gfd; (M R ) A = 100(9.81)(2.5 + 0.25) - 250(9.81)(0.5-0.25) = 2084.625 N # m = 2.08 kn # m (Counterclockwise)
4 12. The railway crossing gate consists of the 100-kg gate arm having a center of mass at G a and the 250-kg counterweight having a center of mass at G W. Determine the magnitude and directional sense of the resultant moment produced by the weights about point B. 2.5 m A a+(m R ) B = gfd; (M R ) B = 100(9.81)(2.5) - 250(9.81)(0.5) G a 0.75 m B G W 0.5 m 1 m = 1226.25 N # m = 1.23 kn # m (Counterclockwise) 0.25 m
4 13. The force F acts on the end of the pipe at B. Determine (a) the moment of this force about point A, and (b) the magnitude and direction of a horizantal force, applied at C, which produces the same moment. Given: F a b c 70 N 0.9 m 0.3 m 0.7 m 60 deg Solution: c a (a) M A F sin F cos M A 73.9 Nm M A (b) F C ( a) M A F C F C 82.2 N a
4 14. The force F acts on the end of the pipe at B. Determine the angles ( 0 180 ) of the force that will produce maximum and minimum moments about point A. What are the magnitudes of these moments? Given: F a b c 70 N 0.9 m 0.3 m 0.7 m Solution: M A F sinc Fcosa For maximum moment d M A d cfcos afsin 0 max atan c max 37.9 deg a M Amax F sin max c Fcos max a M Amax 79.812 Nm For minimum moment M A F sinc Fcosa 0 a min 180 deg atan min 128 deg c M Amin F a Fcsin min ( ) cos min M Amin 0Nm
4 15. The Achilles tendon force of F t = 650 N is mobilized when the man tries to stand on his toes.as this is done, each of his feet is subjected to a reactive force of N f = 400 N. Determine the resultant moment of F t and N f about the ankle joint A. 5 F t A Referring to Fig. a, 200 mm a+(m R ) A = Fd; (M R ) A = 400(0.1) - 650(0.065) cos 5 =-2.09 N # m = 2.09 N # m (Clockwise) 65 mm 100 mm N f 400 N
4 16. The Achilles tendon force F t is mobilized when the man tries to stand on his toes. As this is done, each of his feet is subjected to a reactive force of N t = 400 N. If the resultant moment produced by forces F t and N f about the ankle joint A is required to be zero, determine the magnitude of. F f 5 F t A Referring to Fig. a, 200 mm a+(m R ) A = Fd; 0 = 400(0.1) - F cos 5 (0.065) F = 618 N 65 mm 100 mm N f 400 N
4 17. The total hip replacement is subjected to a force of F = 120 N. Determine the moment of this force about the neck at A and the stem at B. 120 N 15 Moment About Point A: The angle between the line of action of the load and the neck axis is 20-15 = 5. a+ M A = 120 sin 5 10.042 (Counterclockwise) Moment About Point B: The dimension l can be determined using the law of sines. Then, = 0.418 N # m l sin 150 = 55 l = 158.4 mm = 0.1584 m sin 10 A 150 10 B 40 mm 15 mm a+ M B = -120 sin 15 10.15842 = -4.92 N # m = 4.92 N # m (Clockwise)
4 18. The tower crane is used to hoist the 2-Mg load upward at constant velocity. The 1.5-Mg jib BD, 0.5-Mg jib BC, and 6-Mg counterweight C have centers of mass at G 1, G 2, and G 3, respectively. Determine the resultant moment produced by the load and the weights of the tower crane jibs about point A and about point B. G 3 C G 2 7.5 m 4 m B 9.5m 12.5 m G 1 D 23 m Since the moment arms of the weights and the load measured to points A and B are the same, the resultant moments produced by the load and the weight about points A and B are the same. A a + (M R ) A = (M R ) B = Fd; (M R ) A = (M R ) B = 6000(9.81)(7.5) + 500(9.81)(4) - 1500(9.81)(9.5) - 2000(9.81)(12.5) = 76 027.5 N # m = 76.0 kn # m (Counterclockwise)
4 19. The tower crane is used to hoist a 2-Mg load upward at constant velocity. The 1.5-Mg jib BD and 0.5-Mg jib BC have centers of mass at G 1 and G 2, respectively. Determine the required mass of the counterweight C so that the resultant moment produced by the load and the weight of the tower crane jibs about point A is zero. The center of mass for the counterweight is located at G 3. G 3 C G 2 7.5 m 4 m B 9.5m 12.5 m G 1 D 23 m A a+ (M R ) A = Fd; 0 = M C (9.81)(7.5) + 500(9.81)(4) - 1500(9.81)(9.5) - 2000(9.81)(12.5) M C = 4966.67 kg = 4.97 Mg
4 20. Determine the magnitude of the force F that should be applied at the end of the lever such that this force creates a clockwise moment M about point O. Given: M 15 N m 60 deg 30 deg a 50 mm b 300 mm Solution: M Fcos a bsin F sinb cos F M cosa bsin sinb cos F 77.6 N
4 21. Determine the angle (0 <= <= 90 deg) so that the force F develops a clockwise moment M about point O. Given: F M 100 N 20 Nm 60 deg a b Solution: 50 mm 300 mm Initial Guess 30 deg Given M Fcos a bsin F sinb cos Find 28.6 deg
4 22. The tool at A is used to hold a power lawnmower blade stationary while the nut is being loosened with the wrench. If a force of 50 N is applied to the wrench at B in the direction shown, determine the moment it creates about the nut at C. What is the magnitude of force F at A so that it creates the opposite moment about C? 400 mm A F 13 12 5 C 50 N c+ M A = 50 sin 60 (0.3) M A = 12.99 = 13.0 N # m a + M A = 0; -12.99 + Fa 12 13 b(0.4) = 0 F = 35.2 N 300 mm B 60
4 23. The towline exerts a force of P = 4kN at the end of the 20-m-long crane boom. If u = 30, determine the placement x of the hook at A so that this force creates a maximum moment about point O. What is this moment? B P 4kN O u 20 m 1.5 m A Maximum moment, OB BA x a+ (M O ) max =- 4kN(20) = 80 kn # m b 4 kn sin 60 (x) - 4 kn cos 60 (1.5) = 80 kn # m x = 24.0 m
4 24. The towline exerts a force of P = 4 k N at the end of the 20-m-long crane boom. If x = 25 m, determine the position u of the boom so that this force creates a maximum moment about point O. What is this moment? B P 4kN 20 m O u 1.5 m A Maximum moment, OB BA x c + (M O ) max = 4000(20) = 80 000 N # m = 80.0 kn # m 4000 sin f(25) - 4000 cos f(1.5) = 80 000 25 sin f - 1.5 cos f = 20 f = 56.43 u = 90-56.43 = 33.6 Also, (1.5) 2 + z 2 = y 2 2.25 + z 2 = y 2 Similar triangles 20 + y z = 25 + z y 20y + y 2 = 25z + z 2 20(22.25 + z 2 ) + 2.25 + z 2 = 25z + z 2 z = 2.260 m y = 2.712 m u = cos -1 a 2.260 2.712 b = 33.6
4 25. Determine the resultant moment of the forces about point A. Solve the problem first by considering each force as a whole, and then by using the principle of moments. Units Used: kn 10 3 N Given: F 1 250 N a 2m F 2 F 3 1 2 300 N b 3m 500 N c 4m 60 deg d 3 30 deg e 4 Solution Using Whole Forces: Geometry atan d d L a b e e c M A F 1 ( a)cos 2 F 2 a b ( ) sin 1 F 3 L e e 2 d 2 M A 2.532 knm Solution Using Principle of Moments: M A F 1 cos 2 a F 2 sin 1 ( a b) F 3 d c F 3 e ( a b) d 2 e 2 d 2 e 2 M A 2.532 10 3 Nm
4 26. If the resultant moment about point A is M clockwise, determine the magnitude of F 3. Units Used: kn 10 3 N Given: M 4.8 knm a 2m F 1 F 2 1 2 300 N b 3m 400 N c 4m 60 deg d 3 30 deg e 4 Solution: Initial Guess F 3 1N Given d e M F 1 cos 2 a F 2 sin 1 ( a b) F 3 d 2 e 2 F 3 d 2 e 2 F 3 Find F 3 F 3 1.593 kn c ( a b)
4 27. The connected bar BC is used to increase the lever arm of the crescent wrench as shown. If the applied force is F = 200 N and d = 300 mm, determine the moment produced by this force about the bolt at A. d C 15 F 30 300 mm B By resolving the 200-N force into components parallel and perpendicular to the box wrench BC, Fig. a, the moment can be obtained by adding algebraically the moments of these two components about point A in accordance with the principle of moments. A a+(m R ) A = Fd; M A = 200 sin 15 (0.3 sin 30 ) - 200 cos 15 (0.3 cos 30 + 0.3) = -100.38 N # m = 100 N # m (Clockwise)
4 28. The connected bar BC is used to increase the lever arm of the crescent wrench as shown. If a clockwise moment of M A = 120 N # m is needed to tighten the bolt at A and the force F = 200 N, determine the required extension d in order to develop this moment. d C 30 15 F 300 mm B A By resolving the 200-N force into components parallel and perpendicular to the box wrench BC, Fig. a, the moment can be obtained by adding algebraically the moments of these two components about point A in accordance with the principle of moments. a+(m R ) A = Fd; -120 = 200 sin 15 (0.3 sin 30 ) - 200 cos 15 (0.3 cos 30 + d) d = 0.4016 m = 402 mm
4 29. The connected bar BC is used to increase the lever arm of the crescent wrench as shown. If a clockwise moment of M A = 120 N # m is needed to tighten the nut at A and the extension d = 300 mm, determine the required force F in order to develop this moment. d C 30 15 F 300 mm B By resolving force F into components parallel and perpendicular to the box wrench BC, Fig. a, the moment of F can be obtained by adding algebraically the moments of these two components about point A in accordance with the principle of moments. A a+(m R ) A = Fd; -120 = F sin 15 (0.3 sin 30 ) - F cos 15 (0.3 cos 30 + 0.3) F = 239 N
4 30. A force F having a magnitude of F = 100 N acts along the diagonal of the parallelepiped. Determine the moment of F about point A, using M A = r B : F and M A = r C : F. z C F -0.4 i + 0.6 j + 0.2 k F = 100 a b 0.7483 F x B r B 600 mm A 200 mm r C y 400 mm F = 5-53.5 i + 80.2 j + 26.7 k6 N M A = r B * F = 3 i j k 0-0.6 0 3 = 5-16.0 i - 32.1 k6 N # m - 53.5 80.2 26.7 Also, M A = r C * F = i j k -0.4 0 0.2-53.5 80.2 26.7 = -16.0 i - 32.1 k N # m
4 31. The force F = 5600i + 300j - 600k6 N acts at the end of the beam. Determine the moment of the force about point A. z A x O F r = {0.2i + 1.2j} m 1.2 m B 0.4 m i j k M O = r * F = 3 0.2 1.2 0 3 600 300-600 0.2 m y M O = {-720i + 120j - 660k} N# m
4 32. Determine the moment produced by force F B about point O. Express the result as a Cartesian vector. z 6m F C 420 N A F B 780 N Position Vector and Force Vectors: Either position vector r OA or r OB can be used to determine the moment of about point O. F B 2m C 2.5 m The force vector r OA = [6k] m F B is given by r OB = [2.5j] m 3m x O B y (0-0)i + (2.5-0)j + (0-6)k F B = F B u FB = 780B R = [300j - 720k] N 2(0-0) 2 + (2.5-0) 2 2 + (0-6) Vector Cross Product: The moment of F B about point O is given by i j k M O = r OA * F B = 3 0 0 6 0 300-720 3 = [-1800i] N# m = [-1.80i] kn # m or i j k M O = r OB * F B = 3 0 2.5 0 0 300-720 3 = [-1800i] N# m = [-1.80i] kn # m
4 33. Determine the moment produced by force F C about point O. Express. the result as a Cartesian vector z 6m F C 420 N A F B 780 N Position Vector and Force Vectors: Either position vector r OA or r OC can be used to determine the moment of about point O. F C 2m C 2.5 m r OA = {6k}m r OC = (2-0)i + (-3-0)j + (0-0)k = [2i - 3j] m 3m x O B y The force vector F C is given by (2-0)i + (-3-0)j + (0-6)k F C = F C u FC = 420B R = [120i - 180j - 360k] N 2(2-0) 2 + (-3-0) 2 2 + (0-6) Vector Cross Product: The moment of F C about point O is given by i j k M O = r OA * F C = 3 0 0 6 3 = [1080i + 720j] N# m 120-180 -360 or i j k M O = r OC * F C = 3 2-3 0 3 = [1080i + 720j] N# m 120-180 -360
4 34. Determine the resultant moment produced by forces F B and F C about point O. Express the result as a Cartesian vector. z 6m F C 420 N A F B 780 N Position Vector and Force Vectors: The position vector F C, Fig. a, must be determined first. r OA and force vectors F B and 2m C 2.5 m r OA = {6k} m (0-0)i + (2.5-0)j + (0-6)k F B = F B u FB = 780B R = [300j - 720k] N 2(0-0) 2 + (2.5-0) 2 2 + (0-6) 3m x O B y (2-0)i + (-3-0)j + (0-6)k F C = F C u FC = 420B R = [120i - 180j - 360k] N 2(2-0) 2 + (-3-0) 2 2 + (0-6) Resultant Moment: The resultant moment of and about point O is given by F B F C M O = r OA * F B + r OA * F C i j k i j k = 3 0 0 6 3 + 3 0 0 6 3 0 300-720 120-180 -360 = [-720i + 720j] N# m
4 35. Using a ring collar the 75-N force can act in the vertical plane at various angles u. Determine the magnitude of the moment it produces about point A, plot the result of M (ordinate) versus u (abscissa) for 0 u 180, and specify the angles that give the maximum and minimum moment. 2m z A 1.5 m i j k M A = 3 2 1.5 0 3 0 75 cos u 75 sin u x θ 75 N y = 112.5 sin u i - 150 sin u j + 150 cos u k M A = 21112.5 sin u2 2 + 1-150 sin u2 2 + 1150 cos u2 2 = 212 656.25 sin 2 u + 22 500 dm A du = 1 2 112 656.25 sin2 u + 22 5002-1 2 112 656.25212 sin u cos u2 = 0 sin u cos u = 0; u = 0, 90, 180 M max = 187.5 N # matu = 90 M min = 150 N # m at u = 0, 180
4 36. The curved rod lies in the x y plane and has a radius of 3 m. If a force of F = 80 N acts at its end as shown, determine the moment of this force about point O. z O y B 3m 45 3m r AC = {1i - 3j - 2k} m 1m A r AC = 2(1) 2 + (-3) 2 + (-2) 2 = 3.742 m F 80 N i j k M O = r OC * F = 3 4 0-2 3 1 3.742 (80) - 3 3.742 (80) - 2 3.742 (80) x 2m C M O = {-128i + 128j - 257k} N# m
4 37. The curved rod lies in the x y plane and has a radius of 3 m. If a force of F = 80 N acts at its end as shown, determine the moment of this force about point B. z O y B 3m 45 3m r AC = {1i - 3j - 2k} m r AC = 2(1) 2 + (-3) 2 + (-2) 2 = 3.742 m i j k M B = r BA * F = 3 3 cos 45 (3-3 sin 45 ) 0 3 1 3.742 (80) - 3 3.742 (80) - 2 3.742 (80) x 1m 2m C F 80 N A M B = {-37.6i + 90.7j - 155k} N# m
4 38. Force F acts perpendicular to the inclined plane. Determine the moment produced by F about point A. Express the result as a Cartesian vector. 3m z A F 400 N 3m x B 4m C y Force Vector: Since force F is perpendicular to the inclined plane, its unit vector is equal to the unit vector of the cross product, b = r AC * r BC, Fig. a. Here u F r AC = (0-0)i + (4-0)j + (0-3)k = [4j - 3k] m r BC = (0-3)i + (4-0)j + (0-0)k = [-3i + 4j] m Thus, i j k b = r CA * r CB = 3 0 4-3 3-3 4 0 = [12i + 9j + 12k] m 2 Then, u F = b b = 12i + 9j + 12k = 0.6247i + 0.4685j + 0.6247k 212 2 + 9 2 2 + 12 And finally F = Fu F = 400(0.6247i + 0.4685j + 0.6247k) = [249.88i + 187.41j + 249.88k] N Vector Cross Product: The moment of F about point A is i j k M A = r AC * F = 3 0 4-3 3 249.88 187.41 249.88 = [1.56i - 0.750j - 1.00k] kn # m
4 39. Solve 4 38 i f F 500 N. z 3m A F 500 N 3m x B 4m C y Force Vector: Since force F is perpendicular to the inclined plane, its unit vector is equal to the unit vector of the cross product, b = r AC * r BC, Fig. a. Here u F r AC = (0-0)i + (4-0)j + (0-3)k = [4j - 3k] m r BC = (0-3)i + (4-0)j + (0-0)k = [-3k + 4j] m Thus, i j k b = r CA * r CB = 3 0 4-3 3 = [12i + 9j + 12k] m 2-3 4 0 Then, u F = b b = 12i + 9j + 12k = 0.6247i + 0.4685j + 0.6247k 212 2 + 9 2 2 + 12 500 N And finally F = Fu F = 500(0.6247i + 0.4685j + 0.6247k) = [ 312. 35i + 234. 25j + 312. 35k] N Vector Cross Product: The moment of F about point B is i j k M B = r BC * F = 3-3 4 0 3 312. 35 234. 25 312. 35 = [1. 25i + 0. 937j - 1. 95k] kn # m
4 40. The pipe assembly is subjected to the 80-N force. Determine the moment of this force about point A. z A 400 mm B x 300 mm y Position Vector And Force Vector: r AC = {(0.55-0)i + (0.4-0)j + (-0.2-0)k} m = {0.55i + 0.4j - 0.2k} m F = 80(cos 30 sin 40 i + cos 30 cos 40 j - sin 30 k) N 200 mm C 30 200 mm 250 mm 40 = (44.53i + 53.07j - 40.0k} N F 80 N Moment of Force F About Point A: Applying Eq. 4 7, we have M A = r AC * F i j k = 3 0.55 0.4-0.2 3 44.53 53.07-40.0 = {-5.39i + 13.1j + 11.4k} N# m
4 41. The pipe assembly is subjected to the 80-N force. Determine the moment of this force about point B. z A 400 mm B x 300 mm y Position Vector And Force Vector: r BC = {(0.55-0)i + (0.4-0.4)j + (-0.2-0)k} m = {0.55i - 0.2k} m F = 80 (cos 30 sin 40 i + cos 30 cos 40 j - sin 30 k) N 200 mm C 30 200 mm 250 mm 40 = (44.53i + 53.07j - 40.0k} N F 80 N Moment of Force F About Point B: Applying Eq. 4 7, we have M B = r BC * F i j k = 3 0.55 0-0.2 3 44.53 53.07-40.0 = {10.6i + 13.1j + 29.2k} N# m
4 42. Strut AB of the 1-m-diameter hatch door exerts a force of 450 N on point B. Determine the moment of this force about point O. z B Position Vector And Force Vector: O 30 0.5 m F = 450 N 0.5 m A 30 y r OB = 510-02i + 11 cos 30-02j + 11 sin 30-02k6 m = 50.8660j + 0.5k6 m x r OA = 510.5 sin 30-02i + 10.5 + 0.5 cos 30-02j + 10-02k6 m = 50.250i + 0.9330j6 m 10-0.5 sin 30 2i + 31 cos 30-10.5 + 0.5 cos 30 24j + 11 sin 30-02k F = 450 210-0.5 sin 30 2 2 + 31 cos 30-10.5 + 0.5 cos 30 24 2 + 11 sin 30-02 N 2 = 5-199.82i - 53.54j + 399.63k6 N Moment of Force F About Point O: Applying Eq. 4 7, we have M O = r OB * F = 3 i j k 0 0.8660 0.5 3-199.82-53.54 399.63 = 5373i - 99.9j + 173k6 N # m Or M O = r OA * F = i j k 0.250 0.9330 0-199.82-53.54 399.63 = 373i - 99.9j + 173k N # m
. Determine the moment of the force at A about point O. Express the result as a cartesian vector. Units Used: 10 3 13 6 2.5 3 3 8 6 4 8 84 8 39
. Determine the moment of the force at A about point P. Express the result as a cartesian vector. Units Used: 10 3 13 6 2.5 3 3 8 6 4 8 116 16 135
4 45. A force of F = 56i - 2j + 1k6 kn produces a moment of M O = 54i + 5j - 14k6 kn # m about the origin of coordinates, point O. If the force acts at a point having an x coordinate of x = 1m, determine the y and z coordinates. M O z P F M O = r * F x d O y z 1m y i j k 4i + 5j - 14k = 3 1 y z 3 6-2 1 4 = y + 2z 5 = -1 + 6z -14 = -2-6y y = 2m z = 1m
4 46. The force F = 56i + 8j + 10k6 N creates a moment about point O of M O = 5-14i + 8j + 2k6 N # m. If the force passes through a point having an x coordinate of 1 m, determine the y and z coordinates of the point.also, realizing that M O = Fd, determine the perpendicular distance d from point O to the line of action of F. i j k -14i + 8j + 2k = 3 1 y z 3 6 8 10 M O x d z O y P z 1m F y -14 = 10y - 8z 8 = -10 + 6z 2 = 8-6y y = 1m z = 3m M O = 2(-14) 2 + (8) 2 + (2) 2 = 16.25 N # m F = 2(6) 2 + (8) 2 + (10) 2 = 14.14 N d = 16.25 14.14 = 1.15 m
4 47. Determine the magnitude of the moment of each of the three forces about the axis AB. Solve the problem (a) using a Cartesian vector approach and (b) using a scalar approach. z F 1 =60N a) Vector Analysis Position Vector and Force Vector: r 1 = 5-1.5j6 m r 2 = r 3 = 0 x B F 2 =85N 1.5 m 2m F 3 =45N A y F 1 = 5-60k6 N F 2 = 585i6 N F 3 = 545j6 N Unit Vector Along AB Axis: u AB = 12-02i + 10-1.52j = 0.8i - 0.6j 212-02 2 2 + 10-1.52 Moment of Each Force About AB Axis: Applying Eq. 4 11, we have 1M AB 2 1 = u AB # 1r1 * F 1 2 0.8-0.6 0 = 3 0-1.5 0 3 0 0-60 = 0.831-1.521-602 - 04-0 + 0 = 72.0 N # m 1M AB 2 2 = u AB # 1r2 * F 2 2 0.8-0.6 0 = 3 0 0 03 = 0 85 0 0 1M AB 2 3 = u AB # 1r3 * F 3 2 0.8-0.6 0 = 3 0 0 03 = 0 0 45 0 b) Scalar Analysis: Since moment arm from force and is equal to zero, F 2 F 3 1M AB 2 2 = 1M AB 2 3 = 0 Moment arm d from force to axis AB is d = 1.5 sin 53.13 = 1.20 m, F 1 M AB 1 = F 1 d = 60 1.20 = 72.0 N # m
4 48. Determine the moment produced by force F about the diagonal AF of the rectangular block. Express the result as a Cartesian vector. A z F {6i 3j 10k} N B C O D 1.5 m G 3 m y x 3 m F
4 49. Determine the moment produced by force F about the diagonal OD of the rectangular block. Express the result as a Cartesian vector. A z F {6i 3j 10k} N B C O D 1.5 m G 3 m y x 3 m F
4 50. Determine the magnitude of the moment of the force F about the base line CA of the tripod. Given: F 50 20 80 a 4m N b c d e f g 2.5 m 1m 0.5 m 2m 1.5 m 2m Solution: r CA g b g e r CA u CA r CD e M CA r CD Fu CA M CA 226 Nm r CA 0 a
4 51. Determine the magnitude of the moment produced by the force of F = 200 N about the hinged axis (the x axis) of the door. 0.5 m z B F 200 N 2 m 15 y A 2.5 m x 1 m
4 52. The lug nut on the wheel of the automobile is to be removed using the wrench and applying the vertical force F at A. Determine if this force is adequate, provided a torque M about the x axis is initially required to turn the nut. If the force F can be applied at A in any other direction, will it be possible to turn the nut? Given: F M a b c d 30 N 14 Nm 0.25 m 0.3 m 0.5 m 0.1 m Solution: M x F c 2 b 2 M x 12 Nm M x M No For M xmax, apply force perpendicular to the handle and the x-axis. M xmax Fc M xmax 15 Nm M xmax M Yes
4 53. The lug nut on the wheel of the automobile is to be removed using the wrench and applying the vertical force F. Assume that the cheater pipe AB is slipped over the handle of the wrench and the F force can be applied at any point and in any direction on the assembly. Determine if this force is adequate, provided a torque M about the x axis is initially required to turn the nut. Given: F 1 30 N M 14 Nm a 0.25 m b 0.3 m c 0.5 m d 0.1 m Solution: a c M x F 1 c c 2 b 2 M x 18 Nm M x M Yes M xmax occurs when force is applied perpendicular to both the handle and the x-axis. M xmax F 1 ( a c) M xmax 22.5 Nm M xmax M Yes
4 54. The board is used to hold the end of a four-way lug wrench in the position shown when the man applies a force of F = 100 N. Determine the magnitude of the moment produced by this force about the x axis. Force F lies in a vertical plane. z F 60 250 mm y Vector Analysis Moment About the x Axis: The position vector r AB,Fig.a, will be used to determine the moment of F about the x axis. x 250 mm r AB = (0.25-0.25)i + (0.25-0)j + (0-0)k = {0.25j} m The force vector F,Fig.a, can be written as F = 100( cos 60 j - sin 60 k) = {50j - 86.60k} N Knowing that the unit vector of the x axis is i, the magnitude of the moment of F about the x axis is given by 1 0 0 M x = i # rab * F = 3 0 0.25 0 3 0 50-86.60 = 1[0.25(-86.60) - 50(0)] + 0 + 0 = -21.7 N # m The negative sign indicates that M x is directed towards the negative x axis. Scalar Analysis This problem can be solved by summing the moment about the x axis M x = M x ; M x = -100 sin 60 (0.25) + 100 cos 60 (0) = -21.7 N # m
4 55. The board is used to hold the end of a four-way lug wrench in position. If a torque of 30 N # m about the x axis is required to tighten the nut, determine the required magnitude of the force F that the man s foot must apply on the end of the wrench in order to turn it. Force F lies in a vertical plane. z F 60 250 mm y x 250 mm Vector Analysis Moment About the x Axis: The position vector r AB,Fig.a, will be used to determine the moment of F about the x axis. r AB = (0.25-0.25)i + (0.25-0)j + (0-0)k = {0.25j} m The force vector F,Fig.a, can be written as F = F(cos 60 j - sin 60 k) = 0.5Fj - 0.8660Fk Knowing that the unit vector of the x axis is i, the magnitude of the moment of F about the x axis is given by 1 0 0 M x = i # rab * F = 30 0.25 0 3 0 0.5F - 0.8660F = 1[0.25(-0.8660F) - 0.5F(0)] + 0 + 0 = -0.2165F The negative sign indicates that M x is directed towards the negative x axis. The magnitude of F required to produce M x = 30 N # m can be determined from 30 = 0.2165F F = 139 N Scalar Analysis This problem can be solved by summing the moment about the x axis M x = M x ; -30 = -F sin 60 (0.25) + F cos 60 (0) F = 139 N
4 56. The cutting tool on the lathe exerts a force F on the shaft as shown. Determine the moment of this force about the y axis of the shaft. z F {6i 4j 7k}kN 40 0 1 0 M y = u y # (r * F) = 3 30 cos 40 0 30 sin 40 3 6-4 -7 30 mm y M y = 276.57 N # mm = 0.277 N # m x
4 57. The cutting tool on the lathe exerts a force F on the shaft as shown. Determine the moment of this force about the x and z axes. z F {6i 4j 7k} kn Moment About x and y Axes: Position vectors r x and r z shown in Fig. a can be conveniently used in computing the moment of F about x and z axes respectively. x 40 30 mm y r x = {0.03 sin 40 k} m r z = {0.03 cos 40 i} m Knowing that the unit vectors for x and z axes are i and k respectively. Thus, the magnitudes of moment of F about x and z axes are given by Thus, 1 0 0 M x = i # rx * F = 3 0 0 0.03 sin 40 3 6-4 -7 = 1[0(-07) - (-4)(0.03 sin 40 )] - 0 + 0 = 0.07713 kn # m = 77.1 N # m 0 0 1 M z = k # rz * F = 3 0.03 cos 40 0 0 3 6-4 7 = 0-0 + 1[0.03 cos 40 (-4) - 6(0)] = - 0.09193 kn # m =-91.9 N # m Mx = M x i = {77.1i} N # m M z = M z k = {-91.9 k} N # m
4 58. If F = 450 N, determine the magnitude of the moment produced by this force about the x axis. z 45 F 60 A B 60 100 mm x 300 mm 150 mm y
4 59. The friction at sleeve A can provide a maximum resisting moment of 125 N # m about the x axis. Determine the largest magnitude of force F that can be applied to the bracket so that the bracket will not turn. z 45 F 60 A B 60 100 mm x 300 mm 150 mm y
4 60. A vertical force of F = 60 N is applied to the handle of the pipe wrench. Determine the moment that this force exerts along the axis AB (x axis) of the pipe assembly. Both the wrench and pipe assembly ABC lie in the x-y plane. Suggestion: Use a scalar analysis. 500 mm z A y F 150 mm B x 45 200 mm C.
4 61. Determine the magnitude of the vertical force F acting on the handle of the wrench so that this force produces a component of moment along the AB axis (x axis) of the pipe assembly of (M A ) x = 5-5i6 N # m. Both the pipe assembly ABC and the wrench lie in the x-y plane. Suggestion: Use a scalar analysis. 500 mm z A y F 150 mm B x 45 200 mm C.
4 62. Determine the magnitude of the moments of the force F about the x, y, and z axes. Solve the problem (a) using a Cartesian vector approach and (b) using a scalar approach. A z 4 m m m y x 3 m C 2 m kn # m. B F {4i 12j 3k} kn kn # m. kn # m kn # m kn # m. kn # m...
4 63. Determine the moment of the force F about an axis extending between A and C. Express the result as a Cartesian vector. A z 4 m y x 3 m C 2 m B F {4i 12j 3k} kn m m m kn. kn # m kn # m kn # m.
4 64. The wrench A is used to hold the pipe in a stationary position while wrench B is used to tighten the elbow fitting. If F B = 150 N, determine the magnitude of the moment produced by this force about the y axis. Also, what is the magnitude of force F A in order to counteract this moment? z 50 mm 50 mm y Vector Analysis Moment of F B About the y Axis: The position vector r CB,Fig.a, will be used to determine the moment of F B about the y axis. r CB = (-0.15-0)j + (0.05-0.05)j + (-0.2598-0)k = {-0.15i - 0.2598k} m Referring to Fig. a, the force vector F B can be written as x 300 mm F A 135 A 30 300 mm 30 120 B F B F B = 150(cos 60 i - sin 60 k) = {75i - 129.90k} N Knowing that the unit vector of the y axis is j, the magnitude of the moment of F B about the y axis is given by M y = j # rcb * F B = 3 0 1 0-0.15 0-0.2598 3 75 0-129.90 = 0-1[-0.15(-129.90) - 75(-0.2598)] + 0 = -38.97 N # m = 39.0 N # m The negative sign indicates that M y is directed towards the negative y axis. Moment of F A About the y Axis: The position vector r DA,Fig.a, will be used to determine the moment of F A about the y axis. r DA = (0.15-0)i + [-0.05 - (-0.05)]j + (-0.2598-0)k = {0.15i - 0.2598k} m Referring to Fig. a, the force vector F A can be written as F A = F A (-cos 15 i + sin 15 k) = -0.9659F A i + 0.2588F A k Since the moment of F A about the y axis is required to counter that of F B about the same axis, F A must produce a moment of equal magnitude but in the opposite sense to that of F A. M x = j # rda * F B + 0.38.97 = 3 0 1 0 0.15 0-0.2598 3-0.9659 F A 0 0.2588 F A + 0.38.97 = 0-1[0.15(0.2588F A ) - (-0.9659F A )(-0.2598)] + 0 F A = 184 N Scalar Analysis This problem can be solved by first taking the moments of F B and then F A about the y axis. For F B we can write M y = M y ; M y = -150 cos 60 (0.3 cos 30 ) - 150 sin 60 (0.3 sin 30 ) = -38.97 N # m The moment of F A, about the y axis also must be equal in magnitude but opposite in sense to that of F B about the same axis M y = M y ; 38.97 = F A cos 15 (0.3 cos 30 ) - F A sin 15 (0.3 sin 30 ) F A = 184 N
4 65. The wrench A is used to hold the pipe in a stationary position while wrench B is used to tighten the elbow fitting. Determine the magnitude of force F B in order to develop a torque of 50 N # m about the y axis. Also, what is the required magnitude of force in order to counteract this moment? F A z 50 mm 50 mm y Vector Analysis Moment of F B About the y Axis: The position vector determine the moment of about the y axis. F B, Fig. a, will be used to r CB = (-0.15-0)i + (0.05-0.05)j + (-0.2598-0)k = {-0.15i - 0.2598k} m r CB x 300 mm F A 135 A 30 300 mm 30 120 B F B Referring to Fig. a, the force vector F B can be written as F B = F B (cos 60 i - sin 60 k) = 0.5F B i - 0.8660F B k Knowing that the unit vector of the y axis is j, the moment of F B about the y axis is required to be equal to -50 N # m, which is given by M y = j # rcb * F B -50i = -50 = 0-1[-0.15(-0.8660F B ) - 0.5F B (-0.2598)] + 0 F B = 192 N 0 1 0-0.15 0-0.2598 0.5F B 0-0.8660F B Moment of F A About the y Axis: The position vector used to determine the moment of about the y axis. F A r DA, Fig. a, will be r DA = (0.15-0)i + [-0.05 - (-0.05)]j + (-0.2598-0)k = {-0.15i - 0.2598k} m Referring to Fig. a, the force vector F A can be written as F A = F A (-cos 15 i + sin 15 k) = -0.9659F A i + 0.2588F A k Since the moment of F A about the y axis is required to produce a countermoment of 50 N # m about the y axis, we can write M y = j # rda * F A 50 = 0 1 0 0.15 0-0.2598-0.9659F A 0 0.2588F A 50 = 0-1[0.15(0.2588F A ) - (-0.9659F A )(-0.2598)] + 0 F A = 236 N # m Scalar Analysis This problem can be solved by first taking the moments of F B and then F A about the y axis. For F B we can write M y = M y ; -50 = -F B cos 60 (0.3 cos 30 ) - F B sin 60 (0.3 sin 30 ) F B = 192 N For F A, we can write M y = M y ; 50 = F A cos 15 (0.3 cos 30 ) - F A sin 15 (0.3 sin 30 ) F A = 236 N
4 66. The wooden shaft is held in a lathe. The cutting tool exerts force F on the shaft in the direction shown. Determine the moment of this force about the x axis of the shaft. Express the result as a Cartesian vector. The distance OA is a. Given: a 25 mm 30 deg F 5 3 8 N Solution: r 0 ( a)cos ( a)sin i 1 0.211 0 M x r F ii M x 0 0 0 Nm
4 67. A twist of 4N# m is applied to the handle of the screwdriver. Resolve this couple moment into a pair of couple forces F exerted on the handle and P exerted on the blade. P F 30 mm 4 N m For the handle P 5mm F M C = M x ; F10.032 = 4 F = 133 N For the blade, M C = M x ; P10.0052 = 4 P = 800 N
4 68. The ends of the triangular plate are subjected to three couples. Determine the plate dimension d so that the resultant couple is 350 N # m clockwise. 100 N 600 N d 100 N 30 600 N a+ M R = M A ; -350 = 2001d cos 30 2-6001d sin 30 2-100d 200 N 200 N d = 1.54 m
4 69. The caster wheel is subjected to the two couples. Determine the forces F that the bearings exert on the shaft so that the resultant couple moment on the caster is zero. F A B 500 N F 40 mm a + M A = 0; 500(50) - F(40) = 0 45 mm 100 mm F = 625 N 50 mm 500 N
4 70. If u = 30, determine the magnitude of force F so that the resultant couple moment is 100 N # m, clockwise. The disk has a radius of 300 mm. 300 mm u 300 N 15 F 30 30 F 15 u 300 N
4 71. If F = 200 N, determine the required angle u so that the resultant couple moment is zero. The dis khas a radius of 300 mm.
4 72. Friction on the concrete surface creates a couple moment of M O = 100 N # m on the blades of the trowel. Determine the magnitude of the couple forces so that the resultant couple moment on the trowel is zero. The forces lie in the horizontal plane and act perpendicular to the handle of the trowel. F 750 mm F M O Couple Moment: The couple moment of F about the vertical axis is M C = F(0.75) = 0.75F. Since the resultant couple moment about the vertical axis is required to be zero, we can write 1.25 mm (M c ) R = M z; 0 = 100-0.75F F = 133 N
4 73. The man tries to open the valve by applying the couple forces of F = 75 N to the wheel. Determine the couple moment produced. 150 mm 150 mm F a+m c = M; M c = -75(0.15 + 0.15) = -22.5 N # m = 22.5 N # m b F
4 74. If the valve can be opened with a couple moment of 25 N # m, determine 150 mm 150 mm the required magnitude of each couple force which must be applied to the wheel. F a+m c = M; -25 = -F(0.15 + 0.15) F = 83.3 N F
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4 76. Three couple moments act on the pipe assembly. Determine the magnitude of M 3 and the bend angle so that the resultant couple moment is zero. Given: 1 45 deg M 1 900 Nm M 2 500 Nm Solution: Initial guesses: 10 deg M 3 10 Nm Given M x = 0; M 1 M 3 cos M 2 cos 1 0 M y = 0; M 2 sin 1 M 3 sin 0 Find M 3 32.9 deg M 3 651 Nm M 3
4 77. The cord passing over the two small pegs A and B 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 C and D so that the resultant couple produced by the two couples is 15 N # m acting clockwise. Take u = 15. P u C 300 mm B 30 100 N 300 mm 100 N 30 A 45 D u P.
4 78. The cord passing over the two small pegs A and B 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 C and D, so that the resultant couple moment produced by the two cords is 20 N # m, clockwise. P u C 300 mm B 30 100 N 300 mm 100 N. 30 A 45 D u P
4 79. Express the moment of the couple acting on the pipe in Cartesian vector form. What is the magnitude of the couple moment? Given: F a b c d 125 N 150 mm 150 mm 200 mm 600 mm Solution: M c a b 0 0 0 F 37.5 M 25 Nm M 45.1 Nm 0
4 80. If the couple moment acting on the pipe has a magnitude M, determine the magnitude F of the forces applied to the wrenches. Given: M a 300 Nm 150 mm b c 150 mm 200 mm d 600 mm Solution: Initial guess: F 1N Given c a b 0 0 0 F M F Find( F) F 832.1 N
4 81. Two couples act on the cantilever beam. If F = 6kN, determine the resultant couple moment. 3m 5kN 4 5 3 B 3m 0.5 m 30 F A 30 F 0.5 m 4 3 5 5kN a) By resolving the 6-kN and 5-kN couples into their x and y components, Fig. a, the couple moments (M c ) 1 and (M c ) 2 produced by the 6-kN and 5-kN couples, respectively, are given by a+(m C ) 1 = 6 sin 30 (3) - 6 cos 30 (0.5 + 0.5) = 3.804 kn # m a+(m C ) 2 = 5a 3 5 b(0.5 + 0.5) - 5a 4 5 b(3) = -9kN # m Thus, the resultant couple moment can be determined from (M C ) R = (M C ) 1 + (M C ) 2 = 3.804-9 = -5.196 kn # m = 5.20 kn # m (Clockwise) b) By resolving the 6-kN and 5-kN couples into their x and y components, Fig. a,and summing the moments of these force components about point A, we can write a + (M C ) R = M A ; (M C ) R = 5a 3 5 b(0.5) + 5a 4 b(3) - 6 cos 30 (0.5) - 6 sin 30 (3) 5 + 6 sin 30 (6) - 6 cos 30 (0.5) + 5a 3 5 b(0.5) - 5a 4 5 b(6) = -5.196 kn # m = 5.20 kn # m (Clockwise)
4 82. Determine the required magnitude of force F, if the resultant couple moment on the beam is to be zero. 3m 5kN 4 5 3 B 3m 0.5 m 30 F By resolving F and the 5-kN couple into their x and y components, Fig. a, the couple moments (M c ) 1 and (M c ) 2 produced by F and the 5-kN couple, respectively, are given by A 30 F 0.5 m 4 3 5 5kN a+(m C ) 1 = F sin 30 (3) - F cos 30 (1) = 0.6340F a +(M C ) 2 = 5a 3 5 b(1) - 5a 4 b(3) = -9kN 5 The resultant couple moment acting on the beam is required to be zero. Thus, (M C ) R = (M C ) 1 + (M C ) 2 0 = 0.6340F - 9 F = 14.2 kn
4 83. Express the moment of the couple acting on the pipe assembly in Cartesian vector form. Solve the problem (a) using Eq. 4 13, and (b) summing the moment of each force about point O. Take F = 525k6 N. 300 mm O z 200 mm y (a) M C = r AB * (25k) 150 mm B F = 3 i j k -0.35-0.2 0 3 0 0 25 M C = {-5i + 8.75j} N# m x 400 mm F A 200 mm (b) M C = r OB * (25k) + r OA * (-25k) i j k i j k = 3 0.3 0.2 0 3 + 3 0.65 0.4 0 3 0 0 25 0 0-25 M C = (5-10)i + (-7.5 + 16.25)j M C = {-5i + 8.75j} N# m
4 84. If the couple moment acting on the pipe has a magnitude of z 400 N # m, determine the magnitude F of the vertical force applied to each wrench. 300 mm O 200 mm y M C = r AB * (Fk) 150 mm B F = 3 i j k -0.35-0.2 0 3 0 0 F M C = {-0.2Fi + 0.35Fj} N# m x 400 mm F A 200 mm M C = 2(-0.2F) 2 + (0.35F) 2 = 400 F = 400 2(-0.2) 2 + (0.35) 2 = 992 N
4 85. The gear reducer is subjected to the couple moments shown. Determine the resultant couple moment and specify its magnitude and coordinate direction angles. z M 2 =60N m Express Each Couple Moment as a Cartesian Vector: M 1 = 550j6 N # m M 2 = 601cos 30 i + sin 30 k2 N # m = 551.96i + 30.0k6 N # m Resultant Couple Moment: x 30 M 1 =50N m y M R = M; M R = M 1 + M 2 = 551.96i + 50.0j + 30.0k6 N # m = 552.0i + 50j + 30k6 N # m The magnitude of the resultant couple moment is M R = 251.96 2 + 50.0 2 + 30.0 2 = 78.102 N # m = 78.1 N # m The coordinate direction angles are a = cos -1 a 51.96 78.102 b = 48.3 b = cos -1 a 50.0 78.102 b = 50.2 g = cos -1 30.0 78.102 = 67.4
4 86. The meshed gears are subjected to the couple moments shown. Determine the magnitude of the resultant couple moment and specify its coordinate direction angles. M 2 =20N m 20 z 30 M 1 = 550k6 N # m y M 2 = 201-cos 20 sin 30 i - cos 20 cos 30 j + sin 20 k2 N # m = 5-9.397i - 16.276j + 6.840k6 N # m x Resultant Couple Moment: M 1 =50N m M R = M; M R = M 1 + M 2 = 5-9.397i - 16.276j + 150 + 6.8402k6 N # m = 5-9.397i - 16.276j + 56.840k6 N # m The magnitude of the resultant couple moment is M R = 21-9.3972 2 + 1-16.2762 2 + 156.8402 2 = 59.867 N # m = 59.9 N # m The coordinate direction angles are a = cos -1 a -9.397 59.867 b = 99.0 b = cos -1 a -16.276 59.867 b = 106 g = cos -1 56.840 59.867 = 18.3
4 87. If the resultant couple of the two couples acting on the fire hydrant is M R = { 15i + 30j} Nm, determine the force magnitude P. Given: a b 0.2 m 0.150 m M 15 30 0 Nm F 75 N Solution: Initial guess P 1N Given F a M Pb P Find( P) P 200 N 0
4 88. A couple acts on each of the handles of the minidual valve. Determine the magnitude and coordinate direction angles of the resultant couple moment. z 35 N 60 25 N M x = -35(0.35) - 25(0.35) cos 60 = -16.625 M y = -25(0.35) sin 60 = -7.5777 N # m M = 2(-16.625) 2 + (-7.5777) 2 = 18.2705 = 18.3 N # m 175 mm x 35 N 25 N 175 mm y a = cos -1 a -16.625 18.2705 b = 155 b = cos -1 a -7.5777 18.2705 b = 115 g = cos -1 0 a 18.2705 b = 90
4 89. Determine the resultant couple moment of the two couples that act on the pipe assembly. The distance from A to B is d = 400 mm. Express the result as a Cartesian vector. { 50i} N 250 mm C {35k} N B d 30 { 35k} N z Vector Analysis Position Vector: x A {50i} N 350 mm y r AB = {(0.35-0.35)i + (-0.4 cos 30-0)j + (0.4 sin 30-0)k} m = {-0.3464j + 0.20k} m Couple Moments: With F 1 = {35k} Nand F 2 = {-50i} N, applying Eq. 4 15, we have (M C ) 1 = r AB * F 1 i j k = 3 0-0.3464 0.20 3 = {-12.12i} N# m 0 0 35 (M C ) 2 = r AB * F 2 i j k = 3 0-0.3464 0.20 3 = {-10.0j - 17.32k} N# m -50 0 0 Resultant Couple Moment: M R = M; M R = (M C ) 1 + (M C ) 2 = {-12.1i - 10.0j - 17.3k}N # m Scalar Analysis: Summing moments about x, y, and z axes, we have (M R ) x = M x ; (M R ) x = -35(0.4 cos 30 ) = -12.12 N # m (M R ) y = M y ; (M R ) y = -50(0.4 sin 30 ) = -10.0 N # m (M R ) z = M z ; (M R ) z = -50(0.4 cos 30 ) = -17.32 N # m Express M R as a Cartesian vector, we have M R = {-12.1i - 10.0j - 17.3k} N# m
4 90. Determine the distance d between A and B so that the resultant couple moment has a magnitude of. M R = 20 N # m {35k} N z { 50i} N 250 mm C B 30 d { 35k} N Position Vector: r AB = {(0.35-0.35)i + (-d cos 30-0)j + (d sin 30-0)k} m x A {50i} N 350 mm y = {-0.8660d j + 0.50d k} m Couple Moments: With F 1 = {35k} Nand F 2 = {-50i} N, applying Eq. 4 15, we have (M C ) 1 = r AB * F 1 i j k = 3 0-0.8660d 0.50d 3 = {-30.31d i} N# m 0 0 35 (M C ) 2 = r AB * F 2 i j k = 3 0-0.8660d 0.50d 3 = {-25.0d j - 43.30d k} N# m -50 0 0 Resultant Couple Moment: M R = M; M R = (M C ) 1 + (M C ) 2 = {-30.31d i - 25.0d j - 43.30d k} N# m The magnitude of MR is 20 N # m, thus 20 = 2(-30.31d) 2 + (-25.0d) 2 + (43.30d) 2 d = 0.3421 m = 342 mm
4 91. If F = 80 N, determine the magnitude and coordinate direction angles of the couple moment. The pipe assembly lies in the x y plane. z F 300 mm x 300 mm F It is easiest to find the couple moment of F by taking the moment of F or F about point A or B, respectively, Fig. a. Here the position vectors r AB and r BA must be determined first. 200 mm 200 mm 300 mm y r AB = (0.3-0.2)i + (0.8-0.3)j + (0-0)k = [0.1i + 0.5j] m r BA = (0.2-0.3)i + (0.3-0.8)j + (0-0)k = [-0.1i - 0.5j] m The force vectors F and F can be written as F = {80 k} Nand - F = [-80 k] N Thus, the couple moment of F can be determined from M c = r AB * F = 3 i j k 0.1 0.5 0 3 = [40i - 8j] N # m 0 0 80 or M c = r BA * -F = 3 i j k -0.1-0.5 0 3 = [40i - 8j] N # m 0 0-80 The magnitude of M c is given by M c = 2M x 2 + M y 2 + M z 2 = 240 2 + (-8) 2 + 0 2 = 40.79 N # m = 40.8 N # m The coordinate angles of M c are a = cos - 1 M x 40 = cos M 40.79 = 11.3 b = cos - 1 M y -8 = cos M 40.79 = 101 g = cos - 1 M z M = cos 0 40.79 = 90
4 92. If the magnitude of the couple moment acting on the pipe assembly is 50 N # m, determine the magnitude of the couple forces applied to each wrench.the pipe assembly lies in the x y plane. z F 300 mm x 300 mm F It is easiest to find the couple moment of F by taking the moment of either F or F about point A or B, respectively, Fig. a. Here the position vectors r AB and r BA must be determined first. 200 mm 200 mm 300 mm y r AB = (0.3-0.2)i + (0.8-0.3)j + (0-0)k = [0.1i + 0.5j] m r BA = (0.2-0.3)i + (0.3-0.8)j + (0-0)k = [-0.1i - 0.5j] m The force vectors F and F can be written as F = {Fk} Nand -F = [-Fk]N Thus, the couple moment of F can be determined from i j k M c = r AB * F = 3 0.1 0.5 0 3 = 0.5Fi - 0.1Fj 0 0 F The magnitude of M c is given by M c = 2M x 2 + M y 2 + M z 2 = 2(0.5F) 2 + (0.1F) 2 + 0 2 = 0.5099F Since Mc is required to equal 50 N # m, 50 = 0.5099F F = 98.1 N
4 93. If F = 5100k6 N, determine the couple moment that acts on the assembly. Express the result as a Cartesian vector. Member BA lies in the x y plane. z O B f = tan -1 a 2 b - 30 = 3.69 3 r 1 = {-360.6 sin 3.69 i + 360.6 cos 3.69 j} = {-23.21i + 359.8j} mm x 300 mm 60 150 mm F 200 mm 200 mm A F y u = tan -1 a 2 b + 30 = 53.96 4.5 r 2 = {492.4 sin 53.96 i + 492.4 cos 53.96 j} = {398.2i + 289.7j} mm M c = (r 1 - r 2 ) * F = 3 i j k - 421.4 70.10 0 3 0 0 100 M c = {7.01i + 42.1j} N# m
4 94. If the magnitude of the resultant couple moment is 15 N # m, determine the magnitude F of the forces applied to the wrenches. z O B f = tan -1 a 2 b - 30 = 3.69 3 r 1 = {-360.6 sin 3.69 i + 360.6 cos 3.69 j} = {-23.21i + 359.8j} mm x 300 mm 60 150 mm F 200 mm 200 mm A F y u = tan -1 a 2 b + 30 = 53.96 4.5 r 2 = {492.4 sin 53.96 i + 492.4 cos 53.96 j} = {398.2i + 289.7j} mm M c = (r 1 - r 2 ) * F = 3 i j k - 421.4 70.10 0 3 0 0 F M c = {0.0701Fi + 0.421Fj} N# m M c = 2(0.0701F) 2 + (0.421F) 2 = 15 F = 15 2(0.0701) 2 + (0.421) 2 = 35.1 N Also, align y axis along BA. M c = -F(0.15)i +F(0.4)j 15 = 2(F(-0.15)) 2 + (F(0.4)) 2 F = 35.1 N
4 95. If F 1 = 100 N, F 2 = 120 N and F 3 = 80 N, determine the magnitude and coordinate direction angles of the resultant couple moment. z F 4 [150 k] N 0.3 m 30 x 0.2 m F 1 0.3 m 0.2 m F 4 [150 k] N 0.2 m 0.2 m F 1 y Couple Moment: The position vectors r 1, r 2, r 3, and r 4,Fig.a, must be determined first. F 2 0.2 m F 2 r 1 = {0.2i} m r 2 = {0.2j} m From the geometry of Figs. b and c, we obtain r 3 = {0.2j} m F 3 0.2 m F 3 r 4 = 0.3 cos 30 cos 45 i + 0.3 cos 30 sin 45 j - 0.3 sin 30 k = {0.1837i + 0.1837j - 0.15k} m The force vectors F 1, F 2, and F 3 are given by F 1 = {100k} N Thus, F 2 = {120k} N F 3 = {80i} N M 1 = r 1 * F 1 = (0.2i) * (100k) = {-20j} N # m M 2 = r 2 * F 2 = (0.2j) * (120k) = {24i} N # m M 3 = r 3 * F 3 = (0.2j) * (80i) = {-16k} N # m M 4 = r 4 * F 4 = (0.1837i + 0.1837j - 0.15k) * (150k) = {27.56i - 27.56j} N # m Resultant Moment: The resultant couple moment is given by (M c ) R = M c ; (M c ) R = M 1 + M 2 + M 3 + M 4 = (-20j) + (24i) + (-16k) + (27.56i-27.56j) = {51.56i - 47.56j - 16k} N # m The magnitude of the couple moment is (M c ) R = 2[(M c ) R ] x 2 + [(M c ) R ] y 2 + [(M c ) R ] z 2 = 2(51.56) 2 + (-47.56) 2 + (-16) 2 = 71.94 N # m = 71.9 N # m The coordinate angles of (M c ) R are a = cos -1 a [(M c) R ] x (M c ) R b = cos a 51.56 71.94 b = 44.2 b = cos -1 a [(M c) R ] y (M c ) R b = cos a -47.56 71.94 b = 131 g = cos -1 a [(M c) R ] z (M c ) R b = cos a -16 71.94 b = 103
4 96. Determine the required magnitude of F 1, F 2, and F 3 so that the resultant couple moment is (M c ) R = [50 i - 45 j - 20 k] N # m. z F 4 [150 k] N 0.3 m 30 Couple Moment: The position vectors r 1, r 2, r 3, and r 4,Fig.a, must be determined first. r 1 = {0.2i} m r 2 = {0.2j} m r 3 = {0.2j} m x F 2 0.2 m 0.2 m F 2 F 1 0.3 m 0.2 m F 4 [150 k] N 0.2 m 0.2 m F 1 y From the geometry of Figs. b and c, we obtain r 4 = 0.3 cos 30 cos 45 i + 0.3 cos 30 sin 45 j - 0.3 sin 30 k = {0.1837i + 0.1837j - 0.15k} m The force vectors F 1, F 2, and F 3 are given by F 3 0.2 m F 3 F 1 = F 1 k Thus, F 2 = F 2 k F 3 = F 3 i M 1 = r 1 * F 1 = (0.2i) * (F 1 k) = -0.2 F 1 j M 2 = r 2 * F 2 = (0.2j) * (F 2 k) = 0.2 F 2 i M 3 = r 3 * F 3 = (0.2j) * (F 3 i) = -0.2 F 3 k M 4 = r 4 * F 4 = (0.1837i + 0.1837j - 0.15k) * (150k) = {27.56i - 27.56j} N # m Resultant Moment: The resultant couple moment required to equal (M c ) R = {50i - 45j - 20k} N # m. Thus, (M c ) R = M c ; (M c ) R = M 1 + M 2 + M 3 + M 4 50i - 45j - 20k = (-0.2F 1 j) + (0.2F 2 i) + (-0.2F 3 k) + (27.56i - 27.56j) 50i - 45j - 20k = (0.2F 2 + 27.56)i + (-0.2F 1-27.56)j - 0.2F 3 k Equating the i, j, and k components yields 50 = 0.2F 2 + 27.56 F 2 = 112 N -45 = -0.2F 1-27.56 F 1 = 87.2 N -20 = -0.2F 3 F 3 = 100 N
4 97. Replace the force and couple system by an equivalent force and couple moment at point O. 8kN 5m y 3m m O 6kN 4m 13 12 5 3m 4kN 60 P x : + F Rx = F x ; F Rx = 6a 5 b - 4 cos 60 13 = 0.30769 kn + c F Ry = F y ; F Ry = 6a 12 b - 4 sin 60 13 = 2.0744 kn A 4m F R = 2(0.30769) 2 + (2.0744) 2 = 2.10 kn u = tan -1 c 2.0744 a 0.30769 d = 81.6 a + M O = M O ; M O = 8-6a 12 13 b(4) + 6a 5 b(5) - 4 cos 60 (4) 13 M O = -10.62 kn # m = 10.6 kn # mb
4 98. Replace the force and couple system by an equivalent force and couple moment at point P. 8kN 5m y 3m m O 6kN 4m 13 12 5 3m 4kN 60 P x : + F Rx = F x ; F Rx = 6a 5 b - 4 cos 60 13 A 4m = 0.30769 kn + c F Ry = F y ; F Ry = 6a 12 b - 4 sin 60 13 = 2.0744 kn F R = 2(0.30769) 2 + (2.0744) 2 = 2.10 kn u = tan -1 c 2.0744 a 0.30769 d = 81.6 a + M P = M P ; M P = 8-6a 12 13 b(7) + 6a 5 b(5) - 4 cos 60 (4) + 4 sin 60 (3) 13 M P = -16.8 kn # m = 16.8 kn # mb
4 99. Replace the force system acting on the beam by an equivalent force and couple moment at point A. 2.5 kn 1.5 kn 30 5 3 4 3 kn A B 2 m 4 m 2 m : + F Rx = F x ; F Rx = 1.5 sin 30-2.5a 4 5 b = -1.25 kn = 1.25 kn ; + cf Ry = F y ; F Ry = -1.5 cos 30-2.5a 3 5 b - 3 = -5.799 kn = 5.799 kn T Thus, F R = 2F 2 R x + F 2 R y = 21.25 2 + 5.799 2 = 5.93 kn and u = tan - 1 F R y F Rx = tan - 1 a 5.799 1.25 b = 77.8 d a +M RA = M A ; M RA = -2.5a 3 b(2) - 1.5 cos 30 (6) - 3(8) 5 = -34.8 kn # m = 34.8 kn # m (Clockwise)
4 100. Replace the force system acting on the beam by an equivalent force and couple moment at point B. 2.5 kn 1.5 kn 30 5 3 4 3 kn A B 2 m 4 m 2 m : + F Rx = F x ; F Rx = 1.5 sin 30-2.5a 4 5 b = -1.25 kn = 1.25 kn ; + cf Ry = F y ; F Ry = -1.5 cos 30-2.5a 3 5 b - 3 = -5.799 kn = 5.799 kn T Thus, F R = 2F 2 R x + F 2 R y = 21.25 2 + 5.799 2 = 5.93 kn and u = tan - 1 F R y F Rx = tan - 1 a 5.799 1.25 b = 77.8 d a+ M RB = M RB ; M B = 1.5cos 30 (2) + 2.5a 3 5 b(6) = 11.6 kn # m (Counterclockwise)
4 101. Replace the loading system acting on the beam by an equivalent resultant force and couple moment at point O. Given: F 1 200 N F 2 M 450 N 200 Nm a b c d 0.2 m 1.5 m 2m 1.5 m 30 deg Solution: 0 F R F 1 1 0 F 2 sin cos 0 225 F R 190 N F R 294 N 0 b c M O a 0 0 M O 0 39.6 Nm 0 F 1 1 0 b a 0 F 2 sin cos 0 M 0 0 1
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4 104. The system of four forces acts on the roof truss. Determine the equivalent resultant force and specify its location along AB, measured from point A. 1.375 275 kn lb 14 mft 1.5 300 knlb 14 mft 0.75 150 kn lb 41 ft m 200 1 knlb 30 B A 30 +F Rx = ΣF x ; F Rx = 1 sin 30 = 0.5 kn +F Ry = ΣFy; FRy = 0.75 + 1.5 + 1.375 + 1 cos 30 = 4.491 kn F Ry = 4.491 kn 2 2 F R = (0.5) + (4.491) = 4.52 kn 0.5 = tan 1 4.491 = 6.35 = 30 6.35 = 23.6 + M RA = ΣM A ; 4.491 (d) = 1 (1.5) + 2 (1.375) + 3 cos 30 (1) F Rx = 0.5 kn d = 1.52 m F Rx = 0.5 kn F Ry = 4.491 kn
4 105. Replace the force system acting on the frame by a resultant force and couple moment at point A. 5 kn 3 5 4 C 4m 2 kn 1 m 1 m B 3 kn 13 12 5 D 5 m Equivalent Resultant Force: Resolving F 1, F 2, and F 3 into their x and y components, Fig. a, and summing these force components algebraically along the x and y axes, we have A : + (F R) x = F x ; + c (F R) y = F y ; (F R ) x = 5a 4 5 b - 3a 5 b = 2.846 kn : 13 (F R ) y = -5a 3 5 12 b - 2-3a b = -7.769 kn = 7.769 kn T 13 The magnitude of the resultant force F R is given by F R = 2(F R ) x 2 + (F R ) y 2 = 22.846 2 + 7.769 2 = 8.274 kn = 8.27 kn The angle u of F R is u = tan -1 c (F R) y d = tan -1 c 7.769 d = 69.88 = 69.9 (F R ) x 2.846 c Equivalent Couple Moment: Applying the principle of moments and summing the moments of the force components algebraically about point A, we can write a + (M R) A = M A ; (M R ) A = 5a 3 5 b(4) - 5a 4 5 12 5 b(5) - 2(1) - 3a b(2) + 3a 13 13 b(5) = -9.768 kn # m = 9.77 kn # m (Clockwise)
4 106. Replace the force system acting on the bracket by a resultant force and couple moment at point A. 600 N 30 B 450 N 45 Equivalent Resultant Force: Forces F 1 and are resolved into their x and y components, Fig. a. Summing these force components algebraically along the x and y axes, we have F 2 A 0.6 m 0.3 m : + (F R ) x = F x ; + c(f R ) y = F y ; (F R ) x = 450 cos 45-600 cos 30 = -201.42 N = 201.42 N ; (F R ) y = 450 sin 45 + 600 sin 30 = 618.20 N c The magnitude of the resultant force F R is given by F R = 2(F R ) x 2 + (F R ) y 2 = 2201.4 2 + 618.20 2 = 650.18 kn = 650 N F R The angle u of is u = tan - 1 c (F R) y d = tan - 1 c 618.20 d = 71.95 = 72.0 (F R ) x 201.4 b Equivalent Resultant Couple Moment: Applying the principle of moments, Figs. a and b, and summing the moments of the force components algebraically about point A, we can write a+ (M R ) A = M A ; (M R ) A = 600 sin 30 (0.6) + 600 cos 30 (0.3) + 450 sin 45 (0.6) - 450 cos 45 (0.3) = 431.36 N # m = 431 N # m (Counterclockwise)
4 107. A biomechanical model of the lumbar region of the human trunk is shown. The forces acting in the four muscle groups consist of F R = 35 N for the rectus, F O = 45 N for the oblique, F L = 23 N for the lumbar latissimus dorsi, and F E = 32 N for the erector spinae. These loadings are symmetric with respect to the y z plane. Replace this system of parallel forces by an equivalent force and couple moment acting at the spine, point O. Express the results in Cartesian vector form. F R F O z F E F L O F R F E F L F O F R = F z ; F R = {2(35 + 45 + 23 + 32)k } = {270k}N x 75 mm 15 mm 45 mm 50 mm 40 mm30 mm y M ROx = M Ox ; M RO = [-2(35)(0.075) + 2(32)(0.015) + 2(23)(0.045)]i M RO = {-2.22i}N# m
4 108. Replace the two forces acting on the post by a resultant force and couple moment at point O. Express the results in Cartesian vector form. z A C F B 5 kn F D 7 kn Equivalent Resultant Force: The forc es F B and F D,Fig. a,expressed in Cartesian vector form can be written as (0-0)i + (6-0)j + (0-8)k F B = F B u AB = 5C (0-0) 2 + (6-0) 2 S = [3j - 4k] kn 2 + (0-8) x 6 m 2 m D 3m 6 m O B 8 m y (2-0)i + (-3-0)j + (0-6)k F D = F D u CD = 7C (2-0) 2 + (-3-0) 2 S = [2i - 3j - 6k] kn 2 + (0-6) The resultant force F R is given by F R =πf; F R = F B + F D = (3j - 4k) + (2i - 3j - 6k) = [2i - 10k] kn Equivalent Resultant Force: The position vectors r OB and r OC are r OB = {6j} m r OC = [6k] m Thus, the resultant couple moment about point O is given by (M R ) O = M O ; (M R ) O = r OB * F B + r OC * F D i j k i j k = 3 0 6 0 3 + 3 0 0 6 3 0 3-4 2-3 -6 = [-6i + 12j] kn # m
4 109. Replace the force system by an equivalent force and couple moment at point A. z F 2 {100i 100j 50k} N F 1 {300i 400j 100k} N 4 m F R = F; F R = F 1 + F 2 + F 3 F 3 { 500k} N 1 m 8 m = 1300 + 1002i + 1400-1002j + 1-100 - 50-5002k = 5400i + 300j - 650k6 N The position vectors are r AB = 512k6 m and r AE = 5-1j6 m. M RA = M A ; M RA = r AB * F 1 + r AB * F 2 + r AE * F 3 i j k i j k = 3 0 0 12 3 + 3 0 0 12 3 300 400-100 100-100 -50 + i j k 0-1 0 0 0-500 = -3100i + 4800j N # m
4 110. The belt passing over thepulley issubjected to forces F 1 and F 2, each having a magnitude of 40 N. F 1 acts in the -k direction. Replace these forces by an equivalent force and couple moment at point A. Express the result in Cartesian vector form. Set u = 0 so that acts in the -j direction. F 2 z r 80 mm y 300 mm F R = F 1 + F 2 A x F R = {-40j - 40k}N F 2 M RA = (r * F) F 1 i j k i j k = 3-0.3 0 0.08 3 + 3-0.3 0.08 0 3 0-40 0 0 0-40 M RA = {-12j + 12k}N# m
4 111. The belt passing over the pulley is subjected to two forces and F 2, each having a magnitude of 40 N. F 1 acts in the -k direction. Replace these forces by an equivalent force and couple moment at point A. Express the result in Cartesian vector form. Take u = 45. F 1 z r 80 mm y 300 mm F R = F 1 + F 2 A x = -40 cos 45 j + (-40-40 sin 45 )k F 2 Also, F R = {-28.3j - 68.3k} N r AF1 = {-0.3i + 0.08j} m r AF2 = -0.3i - 0.08 sin 45 j + 0.08 cos 45 k = {-0.3i - 0.0566j + 0.0566k} m M RA = (r AF1 * F 1 ) + (r AF2 * F 2 ) = 3 i j k i j k -0.3 0.08 0 3 + 3-0.3-0.0566 0.0566 3 0 0-40 0-40 cos 45-40 sin 45 M RA = {-20.5j + 8.49k} N # m F 1 M RAx = M Ax M RAx = 28.28(0.0566) + 28.28(0.0566) - 40(0.08) M RAx = 0 M RAy = M Ay M RAy = -28.28(0.3) - 40(0.3) M RAy = -20.5 N # m M RAz = M Az M RAz = 28.28(0.3) M RAz = 8.49 N # m M RA = {-20.5j + 8.49k} N # m
4 112. Handle forces F 1 and F 2 are applied to the electric drill. Replace this force system by an equivalent resultant force and couple moment acting at point O. Express the results in Cartesian vector form. F 2 {2j 4k} N z 0.15 m F 1 {6i 3j 10k} N 0.25 m 0.3 m O F R = F; F R = 6i - 3j - 10k + 2j - 4k x y = {6i - 1j - 14k} N M RO = M O ; i j k i j k M RO = 3 0.15 0 0.3 3 + 3 0-0.25 0.3 3 6-3 -10 0 2-4 = 0.9i + 3.30j - 0.450k + 0.4i = {1.30i + 3.30j - 0.450k} N # m Note that F Rz = -14 N pushes the drill bit down into the stock. (MRO ) x = 1.30 N # m and (M RO) y = 3.30 N # m cause the drill bit to bend. (M RO ) z = -0.450 N # m causes the drill case and the spinning drill bit to rotate about the z-axis.
4 113. Replace the force at A by an equivalent force and couple moment at point O. Given: F a b c d 375 N 2m 4m 2m 1m 30 deg Solution: F v sin 187.5 F cos F v 324.76 0 0 N M O a 0 b F v M O 0 0 100.481 Nm
4 114. Replace the force at A by an equivalent force and couple moment at point P. Given: F a b c d 375 N 2m 4m 2m 1m 30 deg Solution: F v M P sin 187.5 F cos F v 324.76 0 0 a c 0 b d F v M P 0 0 736.538 N Nm
4 115. Determine the magnitude and direction of force F and its placement d on the beam so that the loading system is equivalent to a resultant force F R acting vertically downward at point A and a clockwise couple moment M. Units Used: kn Given: 10 3 N F 1 F 2 F R M Solution: 5kN a 3m 3kN b 4m 12 kn c 6m 50 knm e 7 f 24 Initial guesses: F 1kN 30 deg d 2m Given e e 2 f 2 f e 2 f 2 F 1 F 1 F cos 0 F sin F 2 F R f e 2 f 2 F 1 a a Fsin ( b d) F 2 ( a b) M F d Find F d F 4.427 kn 71.565 deg d 3.524 m
4 116. Determine the magnitude and direction of force F and its placement d on the beam so that the loading system is equivalent to a resultant force F R acting vertically downward at point A and a clockwise couple moment M. Units Used: kn Given: 10 3 N F 1 F 2 F R M Solution: 5kN a 3m 3kN b 4m 10 kn c 6m 45 knm e 7 f 24 Initial guesses: F 1kN 30 deg d 1m Given e e 2 f 2 f e 2 f 2 F 1 F 1 F cos 0 F sin F 2 F R f e 2 f 2 F 1 a a Fsin ( b d) F 2 ( a b) M F d Find F d F 2.608 kn 57.529 deg d 2.636 m
4 117. Replace the loading acting on the beam by a single resultant force. Specify where the force acts, measured from end A. A 60 450 N 700 N 300 N 30 B 2 m 4 m 3 m 3000 N m : + F Rx = F x ; F Rx = 450 cos 60-700 sin 30 = -125 N = 125 N ; + cf Ry = F y ; F Ry = -450 sin 60-700 cos 30-300 = -1296 N = 1296 N T F = 2(-125) 2 + (-1296) 2 = 1302 N u = tan -1 a 1296 125 b = 84.5 d c+ M RA = M A ; 1296(x) = 450 sin 60 (2) + 300(6) + 700 cos 30 (9) + 3000 x = 8.51 m
4 118. Replace the loading acting on the beam by a single resultant force. Specify where the force acts, measured from B. A 60 450 N 700 N 300 N 30 B 2 m 4 m 3 m 3000 N m : + F Rx = F x ; F Rx = 450 cos 60-700 sin 30 = -125 N = 125 N ; + cf Ry = F y ; F Ry = -450 sin 60-700 cos 30-300 = -1296 N = 1296 N T F = 2(-125) 2 + (-1296) 2 = 1302 N u = tan -1 a 1296 125 b = 84.5 d c+ M RB = M B ; 1296(x) = -450 sin 60 (4) + 700 cos 30 (3) + 3000 x = 2.52 m (to the right)
4 119. The system of parallel forces acts on the top of the Warren truss. Determine the equivalent resultant force of the system and specify its location measured from point A. 1 kn 500 N 500 N 500 N A 2 kn 1 m 1 m 1 m 1 m..
4 120. Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member AB, measured from A. C 300 N 1m 2m 3m B 250 N 5 3 4 D 400 N m 2m : + F x = F Rx ; F Rx = -250a 4 b - 500(cos 60 ) = -450 N = 450 N ; 5 60 500 N 3m A + c F y = F y ; F Ry = -300-250a 3 b - 500 sin 60 = -883.0127 N = 883.0127 N T 5 F R = 2(-450) 2 + (-883.0127) 2 = 991 N u = tan -1 a 883.0127 b = 63.0 d 450 a+ M RA = M A ; 450y = 400 + (500 cos 60 )(3) + 250a 4 5 b(5) - 300(2) - 250a 3 5 b(5) y = 800 450 = 1.78 m
4 121. Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member CD, measured from end C. C 300 N 1m 2m 3m B 250 N 5 3 4 D 400 N m 2m : + F x = F Rx ; F Rx = -250a 4 b - 500(cos 60 ) = -450 N = 450 N ; 5 60 500 N 3m + c F y = F y ; F Ry = -300-250a 3 b - 500 sin 60 = -883.0127 N = 883.0127 N T 5 A F R = 2(-450) 2 + (-883.0127) 2 = 991 N u = tan -1 a 883.0127 b = 63.0 d 450 c + M RA = M C ; 883.0127x = -400 + 300(3) + 250a 3 b(6) + 500 cos 60 (2) + (500 sin 60 )(1) 5 x = 2333 883.0127 = 2.64 m
4 122. Replace the force and couple system by an equivalent force and couple moment at point O. Units Used : kn 10 3 N Given: M a b c d 30kNm 60deg 3m f 12 3m g 5 4m F 1 8kN 4m F 2 6kN e 6m Solution: F 1 F R f 2 g 2 g f 0 F 2 cos( ) sin( ) 0 0.077 F R 2.188 kn F R 2.19 kn 0.000 0 M O 0 M c e 0 F 1 f 2 g 2 g f 0 0 d 0 F 2 cos( ) sin( ) 0 0.00 M O 0.00 6.9 2 knm
4 123. Replace the force and couple system by an equivalent force and couple moment at point P. Units Used : kn 10 3 N Given: M a b c d 30kNm 60deg 3m f 12 3m g 5 4m F 1 8kN 4m F 2 6kN e 6m Solution: F 1 F R f 2 g 2 g f 0 F 2 cos( ) sin( ) 0 0.077 F R 2.188 kn F R 2.19 kn 0.000 0 M P 0 M c b e 0 F 1 f 2 g 2 g f 0 b d 0 F 2 cos( ) sin( ) 0 0.000 M P 0.000 0.358 knm
4 124. Replace the force system acting on the post by a resultant force, and specify where its line of action intersects the post AB measured from point A. B 0.5 m 1 m 500 N 30 0.2 m 3 4 5 250 N 1 m 300 N Equivalent Resultant Force: Forces F 1 and F 2 are resolved into their x and y components, Fig. a. Summing these force components algebraically along the x and y axes, 1 m A : + (F R ) x = F x ; (F R ) x = 250a 4 b - 500cos 30-300 = -533.01 N = 533.01 N ; 5 + c(f R ) y = F y ; (F R ) y = 500 sin 30-250a 3 5 b = 100 N c The magnitude of the resultant force F R is given by F R = 2(F R ) x 2 + (F R ) y 2 = 2533.01 2 + 100 2 = 542.31 N = 542 N The angle u of F R is u = tan - 1 B (F R) y R = tan - 1 100 c d = 10.63 = 10.6 b (F R ) x 533.01 Location of the Resultant Force: Applying the principle of moments, Figs. a and b, and summing the moments of the force components algebraically about point A, a +(M R ) A = M A ; 533.01(d) = 500 cos 30 (2) - 500 sin 30 (0.2) - 250a 3 5 b(0.5) - 250a 4 b(3) + 300(1) 5 d = 0.8274 mm = 827 mm
4 125. Replace the force system acting on the post by a resultant force, and specify where its line of action intersects the post AB measured from point B. B 0.5 m 1m 500 N 30 0.2 m 3 4 5 250 N 1m 300 N Equivalent Resultant Force: Forces F 1 and F 2 are resolved into their x and y components, Fig. a. Summing these force components algebraically along the x and y axes, 1m A : + (F R ) x = F x ; (F R ) x = 250a 4 b - 500 cos 30-300 = -533.01N = 533.01 N ; 5 + c(f R ) y = F y ; (F R ) y = 500 sin 30-250a 3 5 b = 100 N c The magnitude of the resultant force F R is given by F R = 2(F R ) x 2 + (F R ) y 2 = 2533.01 2 + 100 2 = 542.31 N = 542 N The angle u of is F R u = tan - 1 B (F R) y R = tan - 1 100 c d = 10.63 = 10.6 b (F R ) x 533.01 Location of the Resultant Force: Applying the principle of moments, Figs. a and b, and summing the moments of the force components algebraically about point B, a +(M R ) B = M b ; -533.01(d) = -500cos 30 (1) - 500sin 30 (0.2) - 250a 3 b(0.5) - 300(2) 5 d = 2.17 m
4 126. Replace the force and couple moment system acting on the overhang beam by a resultant force, and specify its location along AB measured from point A. A 0.3 m 30 30 kn 45 knm 26 kn 12 13 5 B 0.3 m 2 m 1 m 1 m 2 m
4 127. The tube supports the four parallel forces. Determine the magnitudes of forces F C and F D acting at C and D so that the equivalent resultant force of the force system acts through the midpoint O of the tube. 600 N D F D z A F C 400 mm O 500 N C Since the resultant force passes through point O, the resultant moment components about x and y axes are both zero. x 400 mm zb 200 mm 200 mm y M x = 0; F D (0.4) + 600(0.4) - F C (0.4) - 500(0.4) = 0 F C - F D = 100 (1) M y = 0; 500(0.2) + 600(0.2) - F C (0.2) - F D (0.2) = 0 F C + F D = 1100 (2) Solving Eqs. (1) and (2) yields: F C = 600 N F D = 500 N
4 128. Replace the force and couple-moment system by an equivalent resultant force and couple moment at point P. Express the results in Cartesian vector form. Units Used: kn 10 3 N Given: F 8 6 8 kn M 20 70 20 knm a 3m b c d 3m e 5m 4m f 6m 6m g 5m Solution: f 8 46 F R F M R M e F F R 6 kn M R 66 d g 8 56 knm
4 129. Replace the force and couple-moment system by an equivalent resultant force and couple moment at point Q. Express the results in Cartesian vector form. Units Used: kn 10 3 N Given: F M a b c d 8 6 8 3m 20 70 20 kn knm 3m e 5m 4m f 6m 6m g 5m Solution: F R 0 8 10 F M R M e F F R 6 kn M R 30 g 8 20 knm
4 130. The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location (x, y) on the slab. Take F 1 = 30 kn, F 2 = 40 kn. 20 kn z 50 kn F 1 F 2 x 3m 8m 6m 4m y + cf R = F z ; F R = -20-50 - 30-40 = -140 kn = 140 kn T 2m (M R ) x = M x ; -140y = -50(3) - 30(11) - 40(13) y = 7.14 m (M R ) y = M y ; 140x = 50(4)+ 20(10)+ 40(10) x = 5.71 m
4 131. The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location (x, y) on the slab. Take F 1 = 20 kn, F 2 = 50 kn. 20 kn z 50 kn F 1 F 2 +TF R = F z ; F R = 20 + 50 + 20 + 50 = 140 kn x 3m 8m 2m 6m 4m y M Ry = M y ; 140(x) = (50)(4) + 20(10) + 50(10) x = 6.43 m M Rx = M x ; -140(y) = -(50)(3) - 20(11) - 50(13) y = 7.29 m
4 132. If F A = 40 kn and F B = 35 kn, determine the magnitude of the resultant force and specify the location of its point of application (x, y) on the slab. z 30 kn F B 2.5 m 0.75 m 0.75 m 90 kn 2.5 m 20 kn Equivalent Resultant Force: By equating the sum of the forces along the z axis to the resultant force, Fig. b, F R 0.75 m x 3m 3m 0.75 m F A y + cf R = F z ; -F R = -30-20 - 90-35 - 40 F R = 215 kn Point of Application: By equating the moment of the forces and y axes, F R, about the x and (M R ) x = M x ; -215(y) = -35(0.75) - 30(0.75) - 90(3.75) - 20(6.75) - 40(6.75) y = 3.68 m (M R ) y = M y ; 215(x) = 30(0.75) + 20(0.75) + 90(3.25) + 35(5.75) + 40(5.75) x = 3.54 m
4 133. If the resultant force is required to act at the center of the slab, determine the magnitude of the column loadings F A and and the magnitude of the resultant force. F B z 30 kn F B 2.5 m 0.75 m 0.75 m 90 kn 2.5 m 20 kn 0.75 m x 3m 3m F A y Equivalent Resultant Force: By equating the sum of the forces along the z axis to the resultant force F R, 0.75 m + cf R = F z ; -F R = -30-20 - 90 - F A - F B F R = 140 + F A + F B (1) Point of Application: By equating the moment of the forces and y axes, F R, about the x and (M R ) x = M x ; -F R (3.75) = -F B (0.75) - 30(0.75) - 90(3.75) - 20(6.75) - F A (6.75) F R = 0.2F B + 1.8F A + 132 (2) (M R ) y = M y ; F R (3.25) = 30(0.75) + 20(0.75) + 90(3.25) + F A (5.75) + F B (5.75) F R = 1.769F A + 1.769F B + 101.54 (3) Solving Eqs.(1) through (3) yields F A = 30kN F B = 20 kn F R = 190kN
4 134. Replace the two wrenches and the force, acting on the pipe assembly, by an equivalent resultant force and couple moment at point O. z 100N m 300 N Force And Moment Vectors: F 1 = 5300k6 N F 3 = 5100j6 N x O 0.5 m A B 0.6 m 0.8 m 45 200 N C 100 N y F 2 = 2005cos 45 i - sin 45 k6 N 180N m = 5141.42i - 141.42k6 N M 1 = 5100k6 N # m M 2 = 1805cos 45 i - sin 45 k6 N # m = 5127.28i - 127.28k6 N # m Equivalent Force and Couple Moment At Point O: F R = F; F R = F 1 + F 2 + F 3 = 141.42i + 100.0j + 1300-141.422k = 5141i + 100j + 159k6 N The position vectors are r 1 = 50.5j6 m and r 2 = 51.1j6 m. M RO = M O ; M RO = r 1 * F 1 + r 2 * F 2 + M 1 + M 2 i j k = 3 0 0.5 0 3 0 0 300 + i j k 0 1.1 0 141.42 0-141.42 + 100k + 127.28i - 127.28k = 122i - 183k N # m
4 135. Replace the parallel force system acting on the plate by a resultant force and specify its location on the x z plane. 1 m 0.5 m z 2 kn 5 kn 1 m 1 m y 0.5 m x 3 kn
4 136. The pipe assembly is subjected to the action of a wrench at B and a couple at A. Determine the magnitude F of the couple forces so that the system can be simplified to a wrench acting at point C. Given: a 0.6 m b 0.8 m c 0.25 m d e f g h P Q 0.7 m 0.3 m 0.3 m 0.5 m 0.25 m 60 N 40 N Solution: Initial Guess F 1N M C 1Nm Given M C 0 0 P( c h) 0 0 0 0 F( e f) a b 0 Q 0 0 F M C Find F M C M C 30 Nm F 53.3 N
4 137. Replace the three forces acting on the plate by a wrench. Specify the magnitude of the force and couple moment for the wrench and the point P(x, y) where its line of action intersects the plate. A z F A F B {500i} N {800k} N y P x B x 4m C 6m y F R = {500i + 300j + 800k} N F C {300j} N F R = 2(500) 2 + (300) 2 + (800) 2 = 990 N u FR = {0.5051i + 0.3030j + 0.8081k} M Rx = M x ; M Rx = 800(4 - y) M Ry = M y ; M Ry = 800x M Rz = M z ; M Rz = 500y + 300(6 - x) Since M R also acts in the direction of u FR, M R (0.5051) = 800(4 - y) M R (0.3030) = 800x M R (0.8081) = 500y + 300(6 - x) M R = 3.07 kn # m x = 1.16 m y = 2.06 m
. The loading on the bookshelf is distributed as shown. Determine the magnitude of the equivalent resultant location, measured from point O. 30 50 1 1.5 0.5 1 1 = 2 2 = ( ) 70 0.107
4 139. Replace the distributed loading with an equivalent resultant force, and specify its location on the beam measured from point O. 3 kn/m O Loading: The distributed loading can be divided into two parts as shown in Fig. a. Equations of Equilibrium: Equating the forces along the y axis of Figs. a and b, we have 3 m 1.5 m +T F R = F ; F R = 1 2 (3)(3) + 1 (3)(1.5) = 6.75 kn T 2 If we equate the moment of F R,Fig.b, to the sum of the moment of the forces in Fig. a about point O, we have a + (M R) O = M O ; 1-6.75(x) = - 2 (3)(3)(2) - 1 2 (3)(1.5)(3.5) x = 2.5 m
4 140. Replace the loading by an equivalent force and couple moment acting at point O. 200 N/m O 4m 3m Equivalent Force and Couple Moment At Point O: + c F R = F y ; F R = -800-300 = -1100 N = 1.10 kn T a+ M RO = M O ; M RO = -800122-300152 = -3100 N # m = 3.10 kn # m (Clockwise)
4 141. Replace the loading by an equivalent resultant force and couple moment acting at point A. Units Used: kn Given: 10 3 N w 1 600 N m w 2 600 N m a b 2.5 m 2.5 m Solution: F R w 1 a w 2 b F R 0N M RA w 1 a a b M RA 3.75 knm 2
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4 143. The masonry support creates the loading distribution acting on the end of the beam. Simplify this load to a single resultant force and specify its location measured from point O. O 1kN/m 0.3 m 2.5 kn/m Equivalent Resultant Force: + c F R = F y ; F R = 1(0.3) + 1 (2.5-1)(0.3) 2 = 0.525 kn c Location of Equivalent Resultant Force: a + 1M R 2 O = M O ; 0.5251d2 = 0.30010.152 + 0.22510.22 d = 0.171 m
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4 145. Replace the distributed loading by an equivalent resultant force, and specify its location on the beam, measured from the pin at C. Units Used: kn 10 3 N Given: w 8000 N m a 5m b 5m 30deg Solution : wb F R wa F 2 R 60 kn F R x = a wa 2 wb 2 a b 3 a wb b wa a 2 2 3 x x 3.89 m F R
4 146. Replace the distributed loading with an equivalent resultant force, and specify its location on the beam measured from point A. w 0 w 0 A B L 2 L 2 Loading: The distributed loading can be divided into two parts as shown in Fig. a. The magnitude and location of the resultant force of each part acting on the beam are also shown in Fig. a. Resultants: Equating the sum of the forces along the y axis of Figs. a and b, +TF R = F; F R = 1 2 w 0a L 2 b + 1 2 w 0a L 2 b = 1 2 w 0L T If we equate the moments of F R, Fig. b, to the sum of the moment of the forces in Fig. a about point A, a+(m R ) A = M A ; - 1 1 w0l(x) = - w0a L 2 2 2 bal 6 b - 1 w0a L 2 2 ba2 3 Lb x = 5 12 L
4 147. The beam supports the distributed load caused by the sandbags. Determine the resultant force on the beam and specify its location measured from point A. Units Used: kn 10 3 N Given: w 1 1.5 kn a 3m m w 2 1 kn b 3m m w 3 2.5 kn c 1.5 m m Solution: F R w 1 a w 2 b w 3 c F R 11.25 kn M A w 1 a a b c w 2 ba w 3 ca b 2 2 2 M A 45.563 knm d M A d 4.05 m F R
4 148. If the soil exerts a trapezoidal distribution of load on the bottom of the footing, determine the intensities w 1 and w 2 of this distribution needed to support the column loadings. 80 kn 1 m 60 kn 50 kn 2.5 m 3.5 m 1 m Loading: The trapezoidal reactive distributed load can be divided into two parts as shown on the free-body diagram of the footing, Fig. a. The magnitude and location measured from point A of the resultant force of each part are also indicated in Fig. a. Equations of Equilibrium: Writing the moment equation of equilibrium about point B, we have w 1 w 2 a+ M B = 0; w 2 (8) 4-8 3 + 60 8 3-1 - 80 3.5-8 3-50 7-8 3 = 0 w 2 = 17.1875 kn >m = 17.2 kn>m Using the result of w 2 and writing the force equation of equilibrium along the y axis, we obtain + c F y = 0; 1 2 (w 1-17.1875)8 + 17.1875(8) - 60-80 - 50 = 0 w 1 = 30.3125 kn>m = 30.3 kn>m
4 149. Determine the length b of the triangular load and its position a on the beam such that the equivalent resultant force is zero and the resultant couple moment is M clockwise. Units Used: kn Given: 10 3 N w 1 4 kn w 2 2.5 kn m m M 8kNm c 9m Solution: Initial Guesses: a 1m b 1m Given 1 2 w 1 1 b 2 w 2 c 0 1 2 w 1 ba 2b 3 1 2 w 2 c 2c 3 M a b Find( a b) a b 1.539 m 5.625
4 150. Replace the loading by an equivalent force and couple moment acting at point O. 6kN/m 15 kn 500 kn m O + cf R = F y ; F R = -22.5-13.5-15.0 7.5 m 4.5 m = -51.0 kn = 51.0 knt a+ M Ro = M o ; M Ro = -500-22.5(5) - 13.5(9) - 15(12) = -914 kn # m = 914 kn # m (Clockwise)
4 151. Replace the loading by a single resultant force, and specify the location of the force measured from point O. 6kN/m 15 kn 500 kn m O Equivalent Resultant Force: 7.5 m 4.5 m + c F R = F y ; -F R = -22.5-13.5-15 F R = 51.0 kn T Location of Equivalent Resultant Force: a+ (M R ) O = M O ; -51.0(d) = -500-22.5(5) - 13.5(9) - 15(12) d = 17.9 m
4 152. Replace the loading by an equivalent resultant force and couple moment at point A. 50 1 kn/m lb/ft 150 kn/m lb/ft B 1.2 4 ft m 6 ft 1.8 m 100 2 kn/m lb/ft A 60 F 3 = 1.2 kn F 1 = 0.9 kn F 2 = 1.8 kn 0.9 m 0.6 m F Rx = 2.338 kn F 1 = 1 2 (1.8) (1) = 0.9 kn F 2 = (1.8) (1) = 1.8 kn F 3 = (1.2) (1) = 1.2 kn + F Rx = ΣF x ; F Rx = 0.9 sin 60 + 1.8 sin 60 = 2.338 kn (1.8 cos 60 + 0.6) m F Ry = 2.55 kn + F Ry = ΣF y ; F Ry = 0.9 cos 60 + 1.8 cos 60 + 1.2 = 2.55 kn 2 2 F R = (2.338) + (2.55) = 3.460 kn 2.55 = tan 1 2.338 = 47.5 + M RA = ΣM A ; M RA = 0.9 (0.6) + 1.8 (0.9) + 1.2 (1.8 cos 60 + 0.6) = 3.96 kn m
4 153. Replace the loading by an equivalent resultant force and couple moment acting at point B. 150 kn/m lb/ft 150 kn/m lb/ft B 1.2 4 ftm 6 ft 1.8 m 100 2 kn/m lb/ft A 60 F 1 = 1 2 (1.8) (1) = 0.9 kn F 2 = (1.8) (1) = 1.8 kn F 3 = (1.2) (1) = 1.2 kn + F Rx = ΣF x ; + F Ry = ΣF y ; F Rx = 0.9 sin 60 + 1.8 sin 60 = 2.338 kn F Ry = 0.9 cos 60 + 1.8 cos 60 + 1.2 = 2.55 kn 2 2 F R = (2.338) + (2.55) = 3.460 kn 2.55 = tan 1 2.338 = 47.5 + M RB = ΣM B ; M RB = 0.9 cos 60 (1.2 cos 60 + 1.2) + 0.9 sin 60 (1.2 sin 60 ) + 1.8 cos 60 (0.9 cos 60 + 1.2) + 1.8 sin 60 (0.9 sin 60 ) + 1.2 (0.6) M RB = 5.04 kn m
4 154. Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects member AB, measured from A. 200 N/m B 100 N/m C ; + F Rx = F x ; F Rx = 1000 N +TF Ry = F y ; F R = 2(1000) 2 + (900) 2 = 1345 N F R = 1.35 kn u = tan -1 c 900 1000 d = 42.0 d F Ry = 900 N a+ M RA = M A ; 1000y = 1000(2.5) - 300(2) - 600(3) y = 0.1 m 200 N/m A 5m 6m
4 155. Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects member BC, measured from C. 200 N/m B 100 N/m C ; + F Rx = F x ; F Rx = 1000 N +TF Ry = F y ; F R = 2(1000) 2 + (900) 2 = 1345 N F R = 1.35 kn F Ry = 900 N u = tan -1 c 900 1000 d = 42.0 d a+ M RC = M C ; 900x = 600(3) + 300(4) - 1000(2.5) x = 0.556 m 200 N/m A 5m 6m
4 156. Replace the distributed loading with an equivalent resultant force, and specify its location on the beam measured from point A. w w w 0 sin ( p 2Lx) w 0 A x Resultant: The magnitude of the differential force df R is equal to the area of the element shown shaded in Fig. a.thus, L df R = w dx = w 0 sin p 2L x dx Integrating df R over the entire length of the beam gives the resultant force. F R L +T F R = df R = w 0 sin L L 0 p 2L x dx = - 2w 0L p cos p 2L x ` L 0 = 2w 0L p T Location: The location of df R on the beam is x c = x measured from point A.Thus, the location x of measured from point A is given by F R x c df R x = L = df R L L0 L x w 0 sin p 2L x dx 2w 0 L p = 4w 0 L 2 p 2 2w 0 L p = 2L p
4 157. Replace the distributed loading with an equivalent resultant force, and specify its location on the beam measured from point A. 10 kn/m w w 1 6 (x 2 4x 60) kn/m A B x Resultant: The magnitude of the differential force df R is equal to the area of the element shown shaded in Fig. a.thus, 6 m df R = w dx = 1 6 (- x2-4x + 60)dx Integrating df R over the entire length of the beam gives the resultant force F R. 6 m 1 +T F R = df R = L L 0 6 (-x2-4x + 60)dx = 1 6 B - x 3 6m 3-2x2 + 60xR ` 0 = 36 kn T Location: The location of df R on the beam is x c = x, measured from point A. Thus the location x of measured from point A is F R x c df R x = L = df R L = 2.17 m L0 6 m xb 1 6 (- x2-4x + 60)Rdx 36 = L 0 6m 1 6 (-x3-4x 2 + 60x)dx 36 = 1 6 - x4 4-4x3 6m 3 + 30x2 ` 0 36
4 158. Determine the equivalent resultant force and couple moment at point O. Units Used: 2 kn Given: 10 3 N a 3m w O 3 kn m 2 x wx ( ) w O a Solution: a F R wx ( ) dx F R 3kN 0 a M O wx ( )( a x) dx M O 2.25 knm 0
4 159. Wet concrete exerts a pressure distribution along the wall of the form. Determine the resultant force of this distribution and specify 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 5 m. p 4m p (4 z 1 /2 ) kpa Equivalent Resultant Force: : + z F R = F x ; -F R = - L da = - wdz L 4 m F R = a20z 1 2 b A103 B dz L = 106.67A10 3 B N = 107 kn ; Location of Equivalent Resultant Force: 0 0 h z 8 kpa z = LA zda da LA = L zwdz 0 = L 0 = L 0 z z wdz L0 4 m 4 m 4 m zc A20z 1 2 B(10 3 ) ddz 4 m A20z 1 2 B(10 3 )dz L0 c A20z 3 2 B(10 3 ) ddz A20z 1 2 B(10 3 )dz L0 = 2.40 m Thus, h = 4 - z = 4-2.40 = 1.60 m
4 160. Replace the loading by an equivalent force and couple moment acting at point O. w 1 w = (200x 2 )N/m 600 N/m O x 9m Equivalent Resultant Force And Moment At Point O: + c F R = F y ; x F R = - da = - wdx LA L 9 m F R = - A200x 1 2 Bdx L 0 0 = -3600 N = 3.60 kn T a + M RO = M O ; M RO x = - xwdx L 0 9 m = - xa200x 1 2 Bdx L 0 9 m = - A200x 3 2 Bdx L 0 = -19 440 N # m = 19.4 kn # m (Clockwise)
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4 162. Determine the equivalent resultant force of the distributed loading and its location, measured from point A. Evaluate the integral using Simpson s rule. w w 5x (16 x 2 ) 1/2 kn/m 5.07 kn/m 2 kn/m A B x 3 m 1 m 4 F R = wdx = 45x + (16 + x 2 ) 1 2 dx L L 0 F R = 14.9 kn 4 4 x df = (x) 45x + (16x + x 2 ) 1 2 dx L0 L 0 = 33.74 kn # m x = 33.74 14.9 = 2.27 m
4 163. Determine the resultant couple moment of the two couples that act on the assembly. Member OB lies in the x-z plane. z A y For the 400-N forces: M C1 = r AB * 1400i2 150 N O 600 mm 45 500 mm x 400 N i j k = 3 0.6 cos 45-0.5-0.6 sin 45 3 400 0 0 = -169.7j + 200k 150 N C B 400 mm 400 N For the 150-N forces: M C2 = r OB * 1150j2 i j k = 3 0.6 cos 45 0-0.6 sin 45 3 0 150 0 = 63.6i + 63.6k M CR = M C1 + M C2 M CR = 63.6i - 170j + 264k N # m
4 164. The horizontal 30-N force acts on the handle of the wrench. What is the magnitude of the moment of this force about the z axis? B z 200 mm A 30 N 45 45 50 mm 10 mm Position Vector And Force Vectors: x O y r BA = {-0.01i + 0.2j} m r OA = [(-0.01-0)i + (0.2-0)j + (0.05-0)k} m = {-0.01i + 0.2j + 0.05k} m F = 30(sin 45 i - cos 45 j) N = [21.213i - 21.213j] N Moment of Force F About z Axis: The unit vector along the z axis is k. Applying Eq. 4 11, we have M z = k # (rba * F) 0 0 1 = 3-0.01 0.2 0 3 21.213-21.213 0 = 0-0 + 1[(-0.01)(-21.213) - 21.213(0.2)] = -4.03 N # m Or M z = k # (roa * F) 0 0 1 = 3-0.01 0.2 0.05 3 21.213-21.213 0 = 0-0 + 1[(-0.01)(-21.213) - 21.213(0.2)] = -4.03 N # m The negative sign indicates that M z, is directed along the negative z axis.
4 165. The horizontal 30-N force acts on the handleofthewrench. Determinethemomentofthis force about point O. Specify the coordinate direction angles a, b, g of the moment axis. B z 200 mm A 30 N 45 45 50 mm 10 mm Position Vector And Force Vectors: x O y r OA = {(-0.01-0)i + (0.2-0)j + (0.05-0)k} m = {-0.01i + 0.2j + 0.05k} m F = 30(sin 45 i - cos 45 j)n = {21.213i - 21.213j} N Moment of Force F About Point O: Applying Eq. 4 7, we have M O = r OA * F i j k = 3-0.01 0.2 0.05 3 21.213-21.213 0 = {1.061i + 1.061j - 4.031k} N # m = {1.06i + 1.06j - 4.03k} N # m The magnitude of M O is M O = 21.061 2 + 1.061 2 + (-4.031) 2 = 4.301 N# m The coordinate direction angles for M O are a = cos - 1 a 1.061 4.301 b = 75.7 b = cos - 1 a 1.061 4.301 b = 75.7 g = cos - 1 a -4.301 4.301 b = 160
4 166. The forces and couple moments that are exerted on the toe and heel plates of a snow ski are Ft = 5-50i + 80j - 158k6N, M t = 5-6i + 4j + 2k6N # m, and Fh = 5-20i + 60j - 250k6 N, M h = 5-20i + 8j + 3k6 N # m, respectively. Replace this system by an equivalent force and couple moment acting at point P. Express the results in Cartesian vector form. Ft F h Mt z O M h P 800 mm 120 mm y x F R = F t + F h = {-70i + 140j - 408k} N i j k i j k M RP = 3 0.8 0 0 3 + 3 0.92 0 0 3 + (-6i + 4j + 2k) + (-20i + 8j + 3k) -20 60-250 -50 80-158 M RP = (200j + 48k) + (145.36j + 73.6k) + (-6i + 4j + 2k) + (-20i + 8j + 3k) M RP = {-26i + 357.36j + 126.6k} N# m M RP = {-26i + 357j + 127k} N# m
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4 170. Determine the moment of the force F c about the door hinge at A. Express the result as a Cartesian vector. z C 2.5 m 1.5 m 250 N F C a Position Vector And Force Vector: 30 A r AB = {[-0.5 - (-0.5)]i + [0 - (-1)]j + (0-0)k} m= {1j} m F C = 250 [-0.5 - (-2.5)]i + {0 - [ - (1 + 1.5 cos 30 )]}j + (0-1.5 sin 30 )k [ -0.5 - (-2.5)] 2 + B{0 - [ - (1 + 1.5 cos 30 )]} 2 + (0-1.5 sin 30 ) 2 N = [159.33i + 183.15j - 59.75k]N x a 1m B 0.5 m y Moment of Force F c About Point A: Applying Eq. 4 7, we have M A = r AB * F i j k = 3 0 1 0 3 159.33 183.15-59.75 = {-59.7i - 159k}N # m
4 171. Determine the magnitude of the moment of the force about the hinged axis aa of the door. F c z C 2.5 m 1.5 m 250 N F C a Position Vector And Force Vectors: 30 A r AB = {[-0.5 - (-0.5)]i + [0 - (-1)]j + (0-0)k} m= {1j} m F C = 250 {-0.5 - (-2.5)]i + {0 - [ - (1 + 1.5 cos 30 )]}j + (0-1.5 sin 30 )k [-0.5 - (-2.5)] 2 + B{0 - [ - (1 + 1.5 cos 30 )]} 2 + (0-1.5 sin 30 ) 2 N = [159.33i + 183.15j - 59.75k] N x a 1m B 0.5 m y Moment of Force F c About a - a Axis: The unit vector along the a a axis is i. Applying Eq. 4 11, we have M a - a = i # (rab * F C ) 1 0 0 = 3 0 1 0 3 159.33 183.15-59.75 = 1[1(-59.75) - (183.15)(0)] - 0 + 0 = -59.7 N # m The negative sign indicates that M a - a is directed toward the negative x axis. M a - a = 59.7 N # m
4 172. The ends of the triangular plate are subjected to three couples. Determine the magnitude of the force F so that the resultant couple moment is M clockwise. Given: F 1 F 2 600 N 250 N a 1m 40 deg M 400 Nm Solution: Initial Guess F 1N a Given F 1 2 cos F 2 a F a 2 cos M F Find( F) F 830 N