Explanation Given information : The torque about y axis is 50 N ⋅ m _ . Show the free body diagram of the forces acting on wrench as in Figure (1). Using Figure (1), Determine the position vector for point CB . r C B = r B − r C = ( x B i + y B j + z B k ) − ( x C i + y C j + z C k ) = ( x B − x C ) i + ( y B − y C ) j + ( z B − z C ) k (I) Here, the position vector for point B is r B , the position vector for point C is r C , the distance of point B from the origin along x direction is x B , the distance of point B from the origin along y direction is y B , the distance of point B from the origin along z direction is z B , the distance of point C from the origin along x direction is x C , the distance of point C from the origin along y direction is y C , the distance of point C from the origin along z direction is z C , the vector component along x direction is i, thevector component along y direction is j, and the vector component along z direction is k. Force vector at point B : F B = F B ( cos θ B i − sin θ B k ) (II) Here, the angle for the force at point B is θ B . Determine the moment produced by the force about y axis. M y = j ⋅ ( r C B × F B ) (III) Determine the position vector for point DA . r D A = r A − r D = ( x A i + y A j + z A k ) − ( x D i + y D j + z D k ) = ( x A − x D ) i + ( y A − y D ) j + ( z A − z D ) k (IV) Here, the position vector for point A is r A , the position vector for point D is r D , the distance of point A from the origin along x direction is x A , the distance of point A from the origin along y direction is y A , the distance of point A from the origin along z direction is z A , the distance of point D from the origin along x direction is x D , the distance of point D from the origin along y direction is y D , and the distance of point D from the origin along z direction is z D . Force vector at point A : F A = F A ( − cos θ A i + sin θ A k ) (V) Here, the angle for the force at point A is θ A . Determine the moment produced by the force about y axis. M y = j ⋅ ( r D A × F A ) (VI) Conclusion : Substitute ( − 300 sin 30 ° ) mm for x B , 0 mm for x C , 50 mm for y B , 50 mm for y C , ( − 300 cos 30 ° ) mm for z B , and 0 mm for z C in Equation (I). r C B = { ( ( − 300 sin 30 ° ) mm × 1 m 1 , 000 mm − 0 ) i + ( 50 − 50 ) j + ( ( − 300 cos 30 ° ) mm × 1 m 1 , 000 mm − 0 ) k } = ( − 0.15 i + 0 j − 0.2598 k ) m Substitute 60 ° for θ B in Equation (II). F B = F B ( cos 60 ° i − sin 60 ° k ) = { 0.5 F B i − 0.8660 F B k } N Substitute − 50 N ⋅ m for M y , ( − 0.15 i + 0 j − 0.2598 k ) m for r C B , and { 0

Engineering Mechanics: Statics and Study Pack (13th Edition)

13th Edition

ISBN: 9780133027990

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 4.5, Problem 65P

To determine

The magnitude of force FB in order to develop the given torque and the magnitude of force FA in order to counteract the given moment.

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Chapter 4.5, Problem 64P

Chapter 4.5, Problem 66P

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Chapter 4 Solutions

Engineering Mechanics: Statics and Study Pack (13th Edition)

Ch. 4.4 - F41. Determine the moment of the force about point...Ch. 4.4 - F42. Determine the moment of the force about point...Ch. 4.4 - F43. Determine the moment of the force about point...Ch. 4.4 - Neglect the thickness of the member.Ch. 4.4 - F45. Determine the moment of the force about point...Ch. 4.4 - F46. Determine the moment of the force about point...Ch. 4.4 - F47. Determine the resultant moment produced by...Ch. 4.4 - F48. Determine the resultant moment produced by...Ch. 4.4 - F49. Determine the resultant moment produced by...Ch. 4.4 - Express the result as a Cartesian vector.

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The magnitude of force F B in order to develop the given torque and the magnitude of force F A in order to counteract the given moment.

The magnitude of force FB in order to develop the given torque and the magnitude of force FA in order to counteract the given moment.

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Chapter 4 Solutions