Explanation Given information : The weight of the crate ( W ) is 150 lb. Show the free body diagram of the forces acting on the wire system as in Figure 1. Using Figure (1), 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, vector component along y direction is j, and vector component along z direction is k. Determine the modulus of position vector for point AB . | r A B | = ( x B − x A ) 2 + ( y B − y A ) 2 + ( z B − z A ) 2 (II) Determine the position vector for point AC . r A 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 (III) 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 modulus of position vector for point AC . | r A C | = ( x C − x A ) 2 + ( y C − y A ) 2 + ( z C − z A ) 2 (IV) Force in Wire AB : F A B = F A B [ r A B | r A B | ] (V) Here, the force in wire AB is F A B . Force in Wire AC : F A C = F A C [ r A C | r A C | ] (VI) Here, the force in wire AC is F A C . Force in Wire AD : F A D = F A D i Here, the force in wire AD is F A D . Write the weight of the crate in vector format. W = − 150 k lb Write the Equation of equilibrium condition. Σ F = 0 F A B + F A C + F A D + W = 0 (VII) Conclusion : Substitute –6 ft for x B , 0 ft for x A , (3 ft) for y B , 0 ft for y A , (2 ft) for z B , and 0 ft for z A in Equation (I). r A B = ( − 6 − 0 ) i + ( 3 − 0 ) j + ( 2 − 0 ) k = − 6 i + 3 j + 2 k Substitute –6 ft for x B , 0 ft for x A , (3 ft) for y B , 0 ft for y A , (2 ft) for z B , and 0 ft for z A in Equation (II). | r A B | = ( − 6 − 0 ) 2 + ( 3 − 0 ) 2 + ( 2 − 0 ) 2 = 49 = 7 ft Substitute (–6 ft) for x C , 0 ft for x A , (–2 ft) for y C , 0 ft for y A , (3 ft) for z D , and 0 ft for z A in Equation (III). r A C = ( − 6 − 0 ) i + ( − 2 − 0 ) j + ( 3 − 0 ) k = − 6 i − 2 j+3k Substitute (–6 ft) for x C , 0 ft for x A , (–2 ft) for y C , 0 ft for y A , (3 ft) for z D , and 0 ft for z A in Equation (IV)

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ISBN: 9780133941357

Author: HIBBELER

Publisher: PEARSON CO

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Chapter 3.4, Problem 11FP

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The tension developed in wires AB, AC and AD.

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The tension developed in wires AB , AC and AD .

Solution Summary: The author shows the free body diagram of the forces acting on the wire system as in Figure 1.

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The tension developed in wires AB, AC and AD.

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