Explanation Given information : The magnitude of the force ( F ) is 100 N. Draw the free body diagram of the forces acting on the parallelepiped as in Figure (1). Determine the position vector for point B . r B = r B − r A = ( x B i + y B j + z B k ) − ( x A i + y A j + z A k ) = ( x B − x A ) i + ( y B − y A ) j + ( z B − z A ) k (I) Here, the position vector for point B is r B , the position vector for point A is r A , 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 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 vector component along x direction is i, the vector component along y direction is j, and the vector component along z direction is k. Determine the position vector for point C . r C = r C − r A = ( x C i + y C j + z C k ) − ( x A i + y A j + z A k ) = ( x C − x A ) i + ( y C − y A ) j + ( z C − z A ) k (II) Here, the position vector for point C is r C , 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 , and the distance of point C from the origin along z direction is z C . Determine the position vector for point BC . r B C = r C − r B = ( x C i + y C j + z C k ) − ( x B i + y B j + z B k ) = ( x C − x B ) i + ( y C − y B ) j + ( z C − z B ) k (III) Determine the modulus of position vector for point BC . | r B C | = ( x C − x B ) 2 + ( y C − y B ) 2 + ( z C − z B ) 2 (IV) Force vector: F = F ( r B C | r B C | ) (V) Write the equation for moment of force F about point A . M A = r B × F (VI) Write the equation for moment of force F about point A . M A = r C × F (VII) Conclusion : Substitute 0 m for x B , 0 m for x A , –600 mm for y B , 0 m for y A , 0 m for z B , and 0 m for z A in Equation (I). r B = { 0 i + ( − 600 mm × 1 m 1 , 000 mm − 0 ) j + 0 k } = ( 0 i − 0.6 j+ 0 k ) m Substitute –400 mm for x C , 0 m for x A , 0 m for y C , 0 m for y A , 200 mm for z C , and 0 m for z A in Equation (II). r C = { ( − 400 mm × 1 m 1 , 000 mm − 0 ) i + ( 0 − 0 ) j + ( 200 mm × 1 m 1 , 000 mm − 0 ) k } = ( − 0.4 i + 0 j + 0.2 k ) m Substitute –400 mm for x C , 0 m for x B , 600 mm for y C , 0 m for y B , 200 mm for z C , and 0 m for z B in Equation (III)

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 4.4, Problem 47P

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The moment of the force about point A.

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Chapter 4.4, Problem 48P

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