Explanation Given information : The force value ( F ) is 500 N. Explanation : Scalar Analysis: Determine the force along the x -axis. F x = sin θ x ( F cos θ x ) (I) Here, the angle of the force F along the x -axis is θ x and the distance of the force F along the x -axis is d x . Determine the force along the y -axis. F y = cos θ y ( F cos θ y ) (II) Here, the angle of the force F along the y -axis is θ y and the distance of the force F along the y -axis is d y . Determine the force along the z -axis. F z = F sin θ z (III) Here, the angle of the force F along the z -axis is θ z . Determine the moment of force F about the x -axis. M x = − F y d z + F z d y (IV) Here, the distance of the force F y about the x -axis is d z and the distance of the force F z about the x -axis is d y . Determine the moment of force F about the y- axis. M y = − F x d z − F z d x (V) Here, the distance of the force F x about the y -axis is d z and the distance of the force F z about the y -axis is d x . Determine the moment of force F about the z- axis. M z = F x d y − F y d x (VI) Here, the distance of the force F x about the z -axis is d y and the distance of the force F y about the z- axis is d x . Vector Analysis: Determine the position vector for the point OA . r O A = r A − r O = ( x A i + y A j + z A k ) − ( x O i + y O j + z O k ) = ( x A − x O ) i + ( y A − y O ) j + ( z A − z O ) k (VII) Here, the position vector for the point A is r A , the position vector for the point O is r O , the distance of the point A from the origin along the x direction is x A , the distance of the point A from the origin along the y direction is y A , and the distance of the point A from the origin along the z direction is z A , the distance of the point O from the origin along the x direction is x O , the distance of the point O from the origin along the y direction is y O , the distance of the point O from the origin along the z direction is z O , the vector component along the x direction is i, thevector component along the y direction is j, and vector component along the z direction is k. Determine the moment of force F about the x -axis. M x = i ⋅ ( r O A × F ) (VIII) Determine the moment of force F about the y -axis. M y = j ⋅ ( r O A × F ) (IX) Determine the moment of force F about the z -axis. M z = k ⋅ ( r O A × F ) (X) Conclusion : Substitute ( 3 5 ) for sin θ x , 500 N for F , and ( 4 5 ) for cos θ x in Equation (I). F x = ( 3 5 ) ( 500 ( 4 5 ) ) = 240 N Substitute ( 4 5 ) for cos θ y and 500 N for F in Equation (II). F y = ( 4 5 ) ( 500 ( 4 5 ) ) = 320 N Substitute 500 N for F and ( 3 5 ) for sin θ z in Equation (III). F z = 500 ( 3 5 ) = 300 N Substitute 320 N for F y , 3 m for d z , 300 N for F z , and 2 m for d y in Equation (IV). M x = − 320 × 3 + 300 × 2 = − 360 N ⋅ m Thus, the magnitude of the moment of force about the x -axis ( M x ) using the scalar analysis is − 360 N ⋅ m _ . Substitute 240 N for F x , 3 m for d z , 300 N for F z , and –2 m for d x in Equation (V). M y = − 240 × 3 − 300 × ( − 2 ) = − 120 N ⋅ m Thus, the magnitude of the moment of force about the y -axis ( M y ) using the scalar analysis is − 120 N ⋅ m _ . Substitute 240 N for F x , 2 m for d y , 320 N for F y , and 2 m for d x in Equation (VI)

13th Edition

ISBN: 9780132915540

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 4.5, Problem 18FP

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The moment of the force about the x, y, and z axes.

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The moment of the force about the x , y , and z axes.

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