Explanation Given information : The weight of the platform ( W ) is 3,500 lb. The given value of d is 4 ft. Show the free body diagram of the forces acting on the cables 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, 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 = ( ( y C − d ) i + y C j + z C k ) − ( x A i + y A j + z A k ) = ( ( y C − d ) − 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 d , 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 | = ( ( y C − d ) − 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 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) Conclusion : Substitute (4 ft) for x B , 0 ft for x A , (–3 ft) for y B , 0 ft for y A , (–10 ft) for z B , and 0 m for z A in Equation (I). r A B = ( 4 − 0 ) i + ( − 3 − 0 ) j + ( − 10 − 0 ) k = ( 4 i − 3 j − 10 k ) ft Substitute (4 ft) for x B , 0 ft for x A , (–3 ft) for y B , 0 ft for y A , (–10 ft) for z B , and 0 m for z A in Equation (II). | r A B | = ( 4 − 0 ) 2 + ( − 3 − 0 ) 2 + ( − 10 − 0 ) 2 = 16 + 9 + 100 = 11.18 ft Substitute 4 ft for x C , 4 ft for d , 0 ft for x A , (3 ft) for y C , 0 ft for y A , (–10 ft) for z C , 0 ft for z A in Equation (III). r A C = ( ( 4 − 4 ) − 0 ) i + ( 3 − 0 ) j + ( − 10 − 0 ) k = ( 0 i + 3 j − 10 k ) ft Substitute 4 ft for x C , 2 ft for d , 0 ft for x A , (3 ft) for y C , 0 ft for y A , (–10 ft) for z C , and 0 ft for z A in Equation (IV)

13th Edition

ISBN: 9780132915540

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Question

Chapter 3.4, Problem 57P

To determine

The force acting in each cable for equilibrium.

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Chapter 3.4, Problem 56P

Chapter 3.4, Problem 58P

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Chapter 3 Solutions

Engineering Mechanics: Statics

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