Explanation Given information : The give force ( F ) value is 800 N. Show the free body diagram of the forces acting on the rope 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 the x direction is x B , the distance of point B from the origin along the y direction is y B , the distance of point B from the origin along the z direction is z B , the distance of point A from the origin along the x direction is x A , the distance of point A from the origin along the y direction is y A , the distance of point A from the origin along the z direction is z A , the vector component along the x direction is i, vector component along the y direction is j, and vector component along the z direction is k. Determine the modulus of the 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 the x direction is x C , the distance of point C from the origin along the y direction is y C , and the distance of point C from the origin along the 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 the x direction is x D , the distance of point C from the origin along the y direction is y D , and the distance of point D from the origin along the 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 Rope AB : F A B = F A B [ r A B | r A B | ] (VII) Here, the force in rope AB is F A B . Force in Rope AC : F A C = F A C [ r A C | r A C | ] (VIII) Here, the force in rope AC is F A C . Force in Rope AD : F A D = F A D [ r A D | r A D | ] (IX) Here, the force in rope 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 (–1.5 m) for x B , 0 m for x A , (–2 m) for y B , 0 m for y A , (–6 m) for z B , and 0 m for z A in Equation (I). r A B = ( − 1.5 − 0 ) i + ( − 2 − 0 ) j + ( − 6 − 0 ) k = − 1.5 i − 2 j − 6 k Substitute (–1.5 m) for x B , 0 m for x A , (–2 m) for y B , 0 m for y A , (–6 m) for z B , and 0 m for z A in Equation (II). | r A B | = ( − 1.5 − 0 ) 2 + ( − 2 − 0 ) 2 + ( − 6 − 0 ) 2 = 42.25 = 6.5 m Substitute (2 m) for x C , 0 m for x A , (–3 m) for y C , 0 m for y A , (–6 m) for z C , and 0 m for z A in Equation (III)

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ISBN: 9780132915540

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 3.4, Problem 48P

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The tension developed in each rope; ropes AB, AC, and AD.

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Chapter 3.4, Problem 47P

Chapter 3.4, Problem 49P

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The tension developed in each rope; ropes AB , AC , and AD .

The tension developed in each rope; ropes AB, AC, and AD.

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