Explanation Given information : The force F 1 is ( − 300 i + 150 j + 200 k ) N . The force F 2 is ( − 450 k ) N . Determine the resultant force using the equation of equilibrium. F R = ∑ F = F 1 + F 2 (I) Determine the position vector for 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 (II) Here, the position vector for point A is r A , the position vector for point O is r O , the distance of point A from the origin along the x direction is x A , the distance of point A from the origin along the y direction is y A , the distance of point A from the origin along the z direction is z A , the distance of point O from the origin along the x direction is x O , the distance of point O from the origin along the y direction is y O , the distance of point O from the origin along the z direction is z O , the vector component along the x direction is i, the vector component along the y direction is j, and the vector component along the z direction is k. Determine the position vector for point OB . r O B = r B − r O = ( x B i + y B j + z B k ) − ( x O i + y O j + z O k ) = ( x B − x O ) i + ( y B − y O ) j + ( z B − z O ) k (III) Here, the position vector for point B is r B , the distance of point B from the origin along the x direction is x B , the distance of point B from the origin along the y direction is y B , and the distance of point B from the origin along the z direction is z B . Express the force 1 as follows: F 1 = ( − 300 i + 150 j + 200 k ) N (IV) Express the force 1 as follows: F 2 = ( − 450 k ) N (V) Write the equation for moment of force about point O . ( M R ) o = ∑ M ( M R ) o = r O B × F 1 + r O A × F 2 (VI) Conclusion : Substitute ( − 300 i + 150 j + 200 k ) N for F 1 and ( − 450 k ) N for F 2 in Equation (I). F R = ( − 300 i + 150 j + 200 k ) + ( − 450 k ) = ( − 300 i + 150 j − 250 k ) N Substitute 0 m for x B , 0 m for x O , 2.0 m for y B , 0 m for y O , 0 m for z B , and 0 m for z O in Equation (II)

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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The area of science and mathematics concerned the motion of the objects, the relation between the force matter, and motion if specified. Forces exerted over the objects result in displacements or changes in the position of an object concerning its enviro…

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