Explanation Given information : The force vector ( F ) is { 300 i − 200 j + 150 k } N . 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 (I) 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 , and 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 modulus of position vector for point OA . | r O A | = ( x A − x O ) 2 + ( y A − y O ) 2 + ( z A − z O ) 2 (II) Determine the position vector for point AB . r A 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 (III) Here, the position vector for point B is r B , the position vector for point O is r O , 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 , the distance of point B from the origin along the z direction is z B . Unit vector: u O A = r O A | r O A | (IV) Determine the resultant moment of force F about x axis. M O A = u O A ⋅ ( r A B × F ) (V) Conclusion : Substitute 0.3 m for x A , 0 m for x O , 0.4 m for y A , 0 m for y O , 0 m for z A , and 0 m for z O in Equation (I). r O A = { ( 0.3 − 0 ) i + ( 0.4 − 0 ) j + ( 0 − 0 ) k } = ( 0.3 i + 0.4 j + 0 k ) m Substitute 0

Engineering Mechanics: Statics & Dynamics (14th Edition)

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ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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