Explanation Given information : The force F 1 is 860 N. The force F 2 is 450 N. Show the free body diagram of the pole as in Figure (1). Using Figure (1), Express the force vector F A of A as follows: F A = A x i + A y j + A z k (I) Here, 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. Express the force F 1 in vector form as follows: F 1 = F 1 ( cos θ i − sin θ k ) (II) Here, the vertical and horizontal resolving angles of F 1 are θ . Express the force F 2 in vector form as follows: F 2 = F 2 ( − cos θ cos θ 1 i + cos θ sin θ 1 j − sin θ k ) (III) Here, the vertical and horizontal resolving angles of F 2 along each direction are θ and θ 1 . Determine the position vector for point BC . r B C = r C − r B = ( x C i + y C j + z C k ) − ( x B i + y B j + z B k ) = ( x C − x B ) i + ( y C − y B ) j + ( z C − z B ) k (IV) Here, the position vector for point A is r A , the position vector for point B is r B , 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 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 BC . | r B C | = ( x C − x B ) 2 + ( y C − y B ) 2 + ( z C − z B ) 2 (V) Express the tension in cable BC as follows: F B C = F B C ( r B C | r B C | ) (VI) Determine the position vector for point ED . r E D = r D − r E = ( x D i + y D j + z D k ) − ( x E i + y E j + z E k ) = ( x D − x E ) i + ( y D − y E ) j + ( z D − z E ) k (VII) Here, the position vector for point D is r D , the position vector for point E is r E , 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 , the distance of point D from the origin along z direction is z D , 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 , and the distance of point E from the origin along z direction is z E . Determine the modulus of position vector for point ED . | r D E | = ( x D − x E ) 2 + ( y D − y E ) 2 + ( z D − z E ) 2 (VIII) Express the tension in cable ED as follows: F E D = F E D ( r E D | r E D | ) (IX) Express the force equation of equilibrium as follows: ∑ F = 0 F A + F 1 + F 2 + F E D + F B C = 0 (X) Conclusion : Substitute 860 N for F 1 and 45 ° for θ in Equation (II). F 1 = 860 ( cos 45 ° i − sin 45 ° k ) = { 608.11 i − 608.11 k } N Substitute 450 N for F 2 , 20 ° for θ , and 30 ° for θ 1 in Equation (III). F 2 = 450 ( − cos 20 ° cos 30 ° i + cos 20 ° sin 30 ° j − sin 20 ° k ) = { − 366.21 i + 211.43 j − 153.91 k } N Substitute 6.0 m for x C , 0 m for x B , –4.5 m for y C , 0 m for y B , 0 m for z B , and –4.0 m for z B in Equation (IV). r B C = ( 6 − 0 ) i + ( ( − 4.5 ) − 0 ) j + ( 0 − 4.0 ) k = ( 6 i − 4.5 j − 4.0 k ) m Substitute 6.0 m for x C , 0 m for x B , –4.5 m for y C , 0 m for y B , 0 m for z B , and –4.0 m for z B in Equation (V). | r B C | = ( 6 − 0 ) 2 + ( − 4.5 − 0 ) 2 + ( 0 − 4 ) 2 = 8.5 m Substitute ( 6 i − 4.5 j − 4.0 k ) m for r B C and 8.5 m for | r B C | in Equation (VI). F B C = F B C ( 6 i − 4.5 j − 4.0 k 8.5 ) = ( 6 8.5 F B C i − 4.5 8.5 F B C j − 4 8.5 F B C k ) N Substitute –6.0 m for x D , 0 m for x E , –3.0 m for y D , 0 m for y E , 0 m for z D , and 6.0 m for z E in Equation (VII). r E D = ( − 6 − 0 ) i + ( − 3 − 0 ) j + ( 0 − 6 ) k = ( − 6.0 i − 3.0 j − 6.0 k ) m Substitute –6.0 m for x D , 0 m for x E , –3.0 m for y D , 0 m for y E , 0 m for z D , and 6.0 m for z E in Equation (VIII). | r E D | = ( − 6 − 0 ) 2 + ( − 3 − 0 ) 2 + ( 0 − 6 ) 2 = 9 m Substitute ( − 6.0 i − 3.0 j − 6.0 k ) m for r E D and 9 m for | r E D | in Equation (IX). F E D = F E D ( − 6 i − 3.0 j − 6.0 k 9 ) = − 6 9 F E D i − 3 9 F E D j − 6 9 F E D k = ( − 2 3 F E D i − 1 3 F E D j − 2 3 F E D k ) N Substitute A x i + A y j + A z k for F A , { 608.11 i − 608.11 k } N for F 1 , { − 366.21 i + 211.43 j − 153.91 k } N for F 2 , ( − 2 3 F E D i − 1 3 F E D j − 2 3 F E D k ) N for F E D , and ( 6 8.5 F B C i − 4.5 8.5 F B C j − 4 8.5 F B C k ) N for F B C in Equation (X). { ( A x i + A y j + A z k ) + ( 608.11 i − 608.11 k ) + ( − 366.21 i + 211.43 j − 153.91 k ) + ( − 2 3 F E D i − 1 3 F E D j − 2 3 F E D k ) + ( 6 8

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 72P

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The components of reaction of A and the tension in each of the guy wires BC and ED.

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Chapter 5.7, Problem 71P

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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 components of reaction of A and the tension in each of the guy wires BC and ED .

The components of reaction of A and the tension in each of the guy wires BC and ED.

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Chapter 5 Solutions