Explanation Given information : The mass of the pot ( m ) is 50 kg. The force in cables AD and AC is one–half as great as the force in cable AB ( F A C = F A D = F A B 2 ) . Show the free body diagram of the forces acting on the cables as in Figure (1). 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 (I) Here, the position vector for point C is r C , the position vector for point A is r A , 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 , 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 AC . | r A C | = ( x C − x A ) 2 + ( y C − y A ) 2 + ( z C − z A ) 2 (II) 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 (III) 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 (IV) Force in Cable AB : F A B = ( F A B ) x i + ( F A B ) z k Force in Cable AC : F A C = F A C [ r A C | r A C | ] (V) Force in Cable AD : F A D = F A D [ r A D | r A D | ] (VI) Determine the weight of the pot. F = m g (VII) Here, the acceleration due to gravity is g . Write the Equation of equilibrium condition. Σ F = 0 F A B + F A C + F A D + F = 0 (VIII) Determine the height of the cable ( d ). d = x B tan θ (IX) Conclusion : Substitute (–6 m) for x C , 0 m for x A , (–2 m) for y C , 0 m for y A , (3 m) for z C , and 0 m for z A in Equation (I). r A C = ( − 6 − 0 ) i + ( − 2 − 0 ) j + ( 3 − 0 ) k = ( − 6 i − 2 j + 3 k ) m Substitute (–6 m) for x C , 0 m for x A , (–2 m) for y C , 0 m for y A , (3 m) for z C , and 0 m for z A in Equation (II). | r A C | = ( − 6 − 0 ) 2 + ( − 2 − 0 ) 2 + ( 3 − 0 ) 2 = 36 + 4 + 9 = 7 m Substitute (–6 m) for x D , 0 m for x A , (2 m) for y D , 0 m for y A , (3 m) for z D , and 0 m for z A in Equation (III). r A D = ( − 6 − 0 ) i + ( 2 − 0 ) j + ( 3 − 0 ) k = ( − 6 i + 2 j + 3 k ) m Substitute (–6 m) for x D , 0 m for x A , (2 m) for y D , 0 m for y A , (3 m) for z D , and 0 m for z A in Equation (IV). | r A D | = ( − 6 − 0 ) 2 + ( 2 − 0 ) 2 + ( 3 − 0 ) 2 = 49 = 7 m Substitute F A B 2 for F A C , ( − 6 i − 2 j + 3 k ) m for r A C , and 7 m for | r A C | in Equation (V). F A C = F A B 2 [ − 6 i − 2 j + 3 k 7 ] Substitute F A B 2 for F A D , ( − 6 i + 2 j + 3 k ) m for r A D , and 7 m for | r A D | in Equation (VI). F A D = F A B 2 [ − 6 i + 2 j + 3 k 7 ] Substitute 50 kg for m and 9

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ISBN: 9780132915540

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 3.4, Problem 53P

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The height of the cable AB (d) and the force in each cable.

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Chapter 3 Solutions

Engineering Mechanics: Statics

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The height of the cable AB ( d ) and the force in each cable.

The height of the cable AB (d) and the force in each cable.

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