Explanation Given information : The weight ( W ) of the crate is 80 lb. Show the free body diagram of the boom as in Figure (1). Use Figure (1). Determine the position vector for point CE : r C E = r E − r C = ( x E i + y E j + z E k ) − ( x C i + y C j + z C k ) = ( x E − x C ) i + ( y E − y C ) j + ( z E − z C ) k (I) Here, the position vector for point C is r C , the position vector for point E is r E , the distance of point E from the origin along x direction is x E , the distance of point E from the origin along y direction is y E , the distance of point E from the origin along z direction is z E , 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 , the distance of point C from the origin along z direction is z C , the vector component along x direction is i, the vector component along y direction is j, and the vector component along z direction is k. Determine the modulus of position vector for point CE . | r C E | = ( x E − x C ) 2 + ( y E − y C ) 2 + ( z E − z C ) 2 (II) Express the force F C E as follows: F C E = F C E [ r C E | r C E | ] (III) Determine the position vector for point CD . r C D = r D − r C = ( x D i + y D j + z D k ) − ( x C i + y C j + z C k ) = ( x D − x C ) i + ( y D − y C ) j + ( z D − z C ) k (IV) 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 D 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 CD . | r C D | = ( x D − x C ) 2 + ( y D − y C ) 2 + ( z D − z C ) 2 (V) Express the force F C D as follows: F C D = F B D C [ r C D | r C D | ] (VI) Determine the position vector for point BD . r B D = r D − r B = ( x D i + y D j + z D k ) − ( x B i + y B j + z B k ) = ( x D − x B ) i + ( y D − y B ) j + ( z D − z B ) k (VII) 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 BD . | r B D | = ( x D − x B ) 2 + ( y D − y B ) 2 + ( z D − z B ) 2 (VIII) Express the force F B D as follows: F B D = F B D C [ r B D | r B D | ] (IX) Take moment about point A . Apply the Equation of equilibrium in the x axis. ∑ M x = 0 0.6247 F B D C ( a ) + 0.4364 F C E ( b ) + 0.3077 F B D C ( b ) − W ( b ) = 0 (X) Here, the horizontal distance between F B D C , F C E and W and point A is a and b . Apply the Equation of equilibrium in the z axis. ∑ M z = 0 0.4685 F B D C ( a ) + 0.2308 F B D C ( b ) − 0.2182 F C E ( b ) = 0 (XI) Here, the horizontal distance between F B D C and F C E and point A is a and b . Along the horizontal direction : Apply the Equation of equilibrium in the x axis. ∑ F x = 0 A x + 0.2182 F C E − 0.2308 F B D C − 0.4685 F B D C = 0 (XII) Along the vertical direction : Apply the Equation of equilibrium in the y axis. ∑ F y = 0 A y − 0.8729 F C E − 0.9231 F B D C − 0.6247 F B D C = 0 (XIII) Along the z direction : Apply the Equation of equilibrium in the z axis. ∑ F z = 0 A z + 0.4364 F C E + 0.3077 F B D C + 0.6247 F B D C − W = 0 (XIV) Conclusion : Substitute 3.0 m for x E , 0 m for x C , –12.0 m for y E , 0 m for y C , 6.0 m for z E , and 0 m for z C in Equation (I). r C E = ( 3.0 − 0 ) i + ( − 12 − 0 ) j + ( 6 − 0 ) k = ( 3i − 12 j + 6 k ) m Substitute 3.0 m for x E , 0 m for x C , –12.0 m for y E , 0 m for y C , 6.0 m for z E , and 0 m for z C in Equation (II). | r C E | = ( 3.0 − 0 ) 2 + ( − 12 − 0 ) 2 + ( 6.0 − 0 ) 2 = 189 m Substitute ( 3i − 12 j + 6 k ) m for r C E and 189 m for | r C E | in Equation (III). F C E = F C E [ ( 3i − 12 j + 6 k ) 189 ] = 3 F C E i 189 − 12 F C E j 189 + 6 F C E k 189 = 0

Engineering Mechanics: Statics and Study Pack (13th Edition)

13th Edition

ISBN: 9780133027990

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 5.7, Problem 76P

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The tension in the cables and the x, y, z components of reaction at A.

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Chapter 5 Solutions

Engineering Mechanics: Statics and Study Pack (13th Edition)

Ch. 5.2 - Prob. 1P Ch. 5.2 - Prob. 2P Ch. 5.2 - Prob. 3P Ch. 5.2 - Prob. 4P Ch. 5.2 - Prob. 5P Ch. 5.2 - Prob. 6P Ch. 5.2 - Prob. 7P Ch. 5.2 - Prob. 8P Ch. 5.2 - Prob. 9P Ch. 5.4 - Determine the horizontal and vertical components...

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The tension in the cables and the x , y , z components of reaction at A .

The tension in the cables and the x, y, z components of reaction at A.

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Chapter 5 Solutions