Explanation Given information : The maximum force supported by each cable is 800 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 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 | ] (IX) Here, the force in cable AD is F A D . Write the Equation of equilibrium condition. Σ F = 0 F A B + F A C + F A D + F = 0 (X) Calculations : Substitute 3 ft for x B , 0 ft for x A , (–2 ft) for y B , 0 ft for y A , (–6 ft) for z B , and 0 ft for z A in Equation (I). r A B = ( 3 − 0 ) i + ( − 2 − 0 ) j + ( − 6 − 0 ) k = 3 i − 2 j − 6 k Substitute 3 ft for x B , 0 ft for x A , (–2 ft) for y B , 0 ft for y A , (–6 ft) for z B , and 0 ft for z A in Equation (II). | r A B | = ( 3 − 0 ) 2 + ( − 2 − 0 ) 2 + ( − 6 − 0 ) 2 = 9 + 4 + 36 = 7 ft Substitute (–3 ft) for x C , 0 ft for x A , (–1 ft) for y C , 0 ft for y A , (–6 ft) for z C , and 0 ft for z A in Equation (III). r A C = ( − 3 − 0 ) i + ( − 1 − 0 ) j + ( − 6 − 0 ) k = − 3 i − 1 j − 6 k Substitute (–3 ft) for x C , 0 ft for x A , (–1 ft) for y C , 0 ft for y A , (–6 ft) for z C , and 0 ft for z A in Equation (IV). | r A C | = ( − 3 − 0 ) 2 + ( − 1 − 0 ) 2 + ( − 6 − 0 ) 2 = 9 + 1 + 36 = 46 ft Substitute (–2 ft) for x D , 0 ft for x A , (–1 ft) for y D , 0 ft for y A , (–6 ft) for z D , and 0 ft for z A in Equation (V). r A D = ( − 2 − 0 ) i + ( 3 − 0 ) j + ( − 6 − 0 ) k = − 2 i + 3 j − 6 k Substitute (–3 ft) for x C , 0 ft for x A , (–1 ft) for y D , 0 ft for y A , (–6 ft) for z D , and 0 ft for z A in Equation (VI)

15th Edition

ISBN: 9780137569830

Author: HIBBELER

Publisher: VST

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Chapter 3, Problem 51P

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The greatest force F applied to the ring.

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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