Explanation Given information : The mass of the crate ( m ) is 130 kg. Show the free body diagram of the forces acting on the cable system as in Figure 1. Determine the position vector for point DA . r D A = r A − r D = ( x A i + y A j + z A k ) − ( x D i + y D j + z D k ) = ( x A − x D ) i + ( y A − y D ) j + ( z A − z D ) k (I) Here, the position vector for point A is r A , the position vector for point D is r D , 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 distance of point D from the origin along the x direction is x D , the distance of point D from the origin along the y direction is y D , the distance of point D from the origin along the z direction is z D , 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 position vector for point DA . | r D A | = ( x A − x D ) 2 + ( y A − y D ) 2 + ( z A − z D ) 2 (II) Determine the position vector for point DB . r D B = r B − r D = ( x B i + y B j + z B k ) − ( x D i + y D j + z D k ) = ( x B − x D ) i + ( y B − y D ) j + ( z B − z D ) k (III) Here, the position vector for point B is r B , 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 , and the distance of point B from the origin along the z direction is z B . Determine the modulus of position vector for point DB . | r D B | = ( x B − x D ) 2 + ( y B − y D ) 2 + ( z B − z D ) 2 (IV) Determine the position vector for point DC . r D C = r C − r D = ( x C i + y C j + z C k ) − ( x D i + y D j + z D k ) = ( x C − x D ) i + ( y C − y D ) j + ( z C − z D ) k (V) 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 position vector for point DC . | r D C | = ( x C − x D ) 2 + ( y C − y D ) 2 + ( z C − z D ) 2 (VI) Force in Cable DA : F D A = F D A [ r D A | r D A | ] (VII) Here, the force in cable DA is F D A . Force in Cable DB : F D B = F D B [ r D B | r D B | ] (VIII) Here, the force in cable DB is F D B . Force in Cable DC : F D C = F D C [ r D C | r D C | ] (IX) Here, the force in cable DC is F D C . Determine the weight of the crate in vector form. W = ( − m g ) k (X) Here, the mass of crate is m and the acceleration due to gravity is g . Write the Equation of equilibrium. Σ F = 0 F A B + F A C + F A D + W = 0 (XI) Conclusion : Substitute 2 m for x A , 0 m for x D , (–1 m) for y A , 0 m for y D , 1 m for z A , and 0 m for z D in Equation (I). r D A = ( 2 − 0 ) i + ( − 1 − 0 ) j + ( 1 − 0 ) k = 2 i − 1 j+1k Substitute 2 m for x A , 0 m for x D , (–1 m) for y A , 0 m for y D , 1 m for z A , and 0 m for z D in Equation (II). | r D A | = ( 2 − 0 ) 2 + ( − 1 − 0 ) 2 + ( 1 − 0 ) 2 = 6 = 2.45 m Substitute –2 m for x B , 0 m for x D , (–1 m) for y B , 0 m for y D , 1 m for z B , and 0 m for z D in Equation (III). r D B = ( − 2 − 0 ) i + ( − 1 − 0 ) j + ( 1 − 0 ) k = − 2 i − 1 j+1k Substitute –2 m for x B , 0 m for x D , (–1 m) for y B , 0 m for y D , 1 m for z B , and 0 m for z D in Equation (IV)

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 3.4, Problem 63P

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The tension developed in each cable for equilibrium.

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Chapter 3.4, Problem 62P

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Chapter 3 Solutions

Engineering Mechanics: Statics & Dynamics (14th Edition)

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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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Chapter 3 Solutions