Explanation Given: The magnitude of the component force, F 2 = 150 lb . The magnitude of the component force, F 3 = 52 lb . The magnitude of the component force, F 1 = 80 lb . The angle between the magnitude of the component force F 1 and u is 25 ° . The angle between the magnitude of the component force F 2 and u axes, θ = 55 ° . Show the magnitude of component forces into x and y axis components as in Figure (1). Show the magnitude of resultant force F R into x and y axis components as in Figure (2). Conclusion: Using Figure (1), the values of cos θ and sin θ have been measured. cos θ = 5 13 sin θ = 12 13 Express the magnitude of resultant force measured clockwise from the positive x axis using Figure (1). + → ( F R ) x = ∑ F x ( F R ) x = F 1 + F 3 cos θ + F 2 cos ( θ + 25 ° ) (I) Substitute 80 lb for F 1 , 5 13 for cos θ , 52 lb for F 3 , and 55 ° for θ in Equation (I). ( F R ) x = 80 + [ ( 52 ) 5 13 ] + 150 cos ( 55 ° + 25 ° ) = 80 + [ ( 52 ) 5 13 ] + 150 cos 80 ° = 80 + 20 + 26.05 = 126.05 lb → Express the magnitude of resultant directed measured clockwise from the positive x axis using Figure (1). + ↑ ( F R ) y = ∑ F y ( F R ) y = F 3 sin θ − F 2 sin ( θ + 25 ° ) (II) Substitute 52 lb for F 3 , 12 13 for sin θ , and 55 ° for θ in Equation (II)

13th Edition

ISBN: 9780132915540

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 2.4, Problem 57P

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The magnitude and its direction θ of the resultant force FR measured clockwise from the positive x axis.

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Chapter 2.4, Problem 56P

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Chapter 2 Solutions

Engineering Mechanics: Statics

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The magnitude and its direction θ of the resultant force F R measured clockwise from the positive x axis.

The magnitude and its direction θ of the resultant force FR measured clockwise from the positive x axis.

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