Explanation Given information : The weight of the crate ( W ) is 300 lb. Show the free body diagram of the forces acting on the cable system as in 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, the vector component along y direction is j , and the 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 the position vector for point AC . | r A C | = ( x C − x A ) 2 + ( y C − y A ) 2 + ( z C − z A ) 2 (IV) Determine the position vector for point AD . r A D = r D − r A = ( x D i + y D j + z D k ) − ( x A i + y A j + z A k ) = ( x D − x A ) i + ( y D − y A ) j + ( z D − z A ) k (V) Here, the position vector for point D is r D , the distance of point D from the origin along x direction is x D , the distance of point C from the origin along y direction is y D , and the distance of point D from the origin along z direction is z D . Determine the modulus of the position vector for point AD . | r A D | = ( x D − x A ) 2 + ( y D − y A ) 2 + ( z D − z A ) 2 (VI) Force in Cable AB : F A B = F A B [ r A B | r A B | ] (VII) Here, the force in cable AB is F A B . Force in Cable AC : F A C = F A C [ r A C | r A C | ] (VIII) Here, the force in cable AC is F A C . Force in Cable AD : F A D = F A D [ r A D | r A D | ] X (IX) Here, the force in cable AD is F A D . Write the weight of the crate in vector format. W = − 300 k lb Write the Equation of equilibrium condition. Σ F = 0 F A B + F A C + F A D + W = 0 (X) Calculations : Substitute –3 ft for x B , 0 ft for x A , (–6 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 = ( − 3 − 0 ) i + ( − 6 − 0 ) j + ( 2 − 0 ) k = − 3 i − 6 j + 2 k Substitute –3 ft for x B , 0 ft for x A , (–6 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 | = ( − 3 − 0 ) 2 + ( − 6 − 0 ) 2 + ( 2 − 0 ) 2 = 49 = 7 ft Substitute (2 ft) for x C , 0 ft for x A , (–6 ft) for y C , 0 ft for y A , (3 ft) for z C , and 0 ft for z A in Equation (III)

15th Edition

ISBN: 9780137569830

Author: HIBBELER

Publisher: VST

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Chapter 3, Problem 54P

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The tension developed in each cable for equilibrium.

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The tension developed in each cable for equilibrium.

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

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The tension developed in each cable for equilibrium.

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