Explanation Given information : The force F is 20 lb. 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 (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 modulus of the position vector for point AB . | r A B | = ( r x i ) 2 + ( r y j ) 2 + ( r z k ) 2 (II) Express the force vector as follows : F = F ( r A B | r A B | ) (III) Determine the position vector for point OA . r O A = r A − r O = ( x O i + y O j + z O k ) − ( x A i + y A j + z A k ) = ( x O − x A ) i + ( y O − y A ) j + ( z O − z A ) k (IV) Here, the position vector for point O is r O , 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 , and the distance of point A from the origin along z direction is z A . Assumption : Take a point along y-axis as y ( 0 ft , y f t , 0 f t ) . Express the position vector along the axis O y as follows: r O y = i ( 0 − 0 ) ft + j ( y − 0 ) ft + k ( 0 − 0 ) ft = ( y j ) ft Express the unit vector that defines the direction of O y as follows: u O y = r O y | r O y | (V) Write the equation for the moment of force F about point A . ( M y ) C = u O y × r O A × F (VI) Show the calculation of the position vector, force vector, and moment calculation as follows: Conclusion : Substitute 2 ft for x B , –4.0 ft for x A , 2.0 ft for y B , 0 ft for y A , 4.0 for z B , and 0 ft for z A in Equation (I). r A B = ( 2.0 − 4.0 ) i + ( 2.0 − 0 ) j + ( 4.0 − 0 ) k = − 2

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The moment of the force F about the hinged axis y .

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