Explanation Given information : Mass of the lamp ( m ) is 15 kg. Show the free body diagram of the forces acting on the cables as in Figure (1). Using Figure (1), Determine the position vector for point AO . r A O = r O − r A = ( x O i + y O j + z O k ) − ( x A i + y A j + z A k ) = ( x O − x A ) i + ( y O − y A ) j + ( z O − z A ) k (I) Here, the position vector for point O is r O , the position vector for point A is r A , the distance of point O from the origin along x direction is x O , the distance of point O from the origin along y direction is y O , the distance of point O from the origin along z direction is z O , 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 AO . | r A O | = ( x O − x A ) 2 + ( y O − y A ) 2 + ( z O − z A ) 2 (II) 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 (III) Here, the position vector for point B is r B , 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 , and the distance of point B from the origin along z direction is z B . 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 (IV) 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 (V) 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 (VI) Force in Cable AO : F A O = F A O [ r A O | r A O | ] (VII) Here, the force in cable AO is F A O . Force in Cable AB : F A B = F A B [ r A B | r A B | ] (VIII) 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 | ] (IX) Here, the force in cable AC is F A C . Determine the weight of the lamp. W = m g (X) Here, the acceleration due to gravity is g . Write the Equation of equilibrium condition. Σ F = 0 F A O + F A B + F A C + W = 0 (XI) Conclusion : Substitute (2 m) for x O , 0 m for x A , (–1.5 m) for y O , 0 m for y A , (6 m) for z O , and 0 m for z A in Equation (I). r A O = ( 2 − 0 ) i + ( − 1.5 − 0 ) j + ( 6 − 0 ) k = ( 2 i − 1.5 j + 6 k ) m Substitute (2 m) for x O , 0 m for x A , (–1.5 m) for y O , 0 m for y A , (6 m) for z O , and 0 m for z A in Equation (II). | r A O | = ( 2 − 0 ) 2 + ( − 1.5 − 0 ) 2 + ( 6 − 0 ) 2 = 42.25 = 6.5 m Substitute (–6 m) for x B , 0 m for x A , (3 m) for y B , 0 m for y A , (–6 m) for z B , and 0 m for z A in Equation (III). r A B = ( − 6 − 0 ) i + ( 3 − 0 ) j + ( − 6 − 0 ) k = ( − 6 i + 3 j − 6 k ) m Substitute (–6 m) for x B , 0 m for x A , (3 m) for y B , 0 m for y A , (–6 m) for z B , and 0 m for z A in Equation (IV)

13th Edition

ISBN: 9780132915540

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 3.4, Problem 50P

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The forces in cables AO, AB, and AC.

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The forces in cables AO , AB , and AC .

The forces in cables AO, AB, and AC.

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