Explanation Given Information: The distance d is 0.75 ft. The unstretched length of the spring is 1 ft. Assumptions: Consider the state of member as tension where the force is pulling the member and as compression where the force is pushing the member. Consider the force indicating right side as positive and left side as negative in horizontal components of forces. Consider the force indicating upside is taken as positive and downside as negative in vertical components of forces. Consider clockwise moment as negative and anti-clock wise moment as positive wherever applicable. Apply the Equation of equilibrium wherever applicable. Spring force: Determine the elongation of the spring, x using the equation. x = 2 d − Unstretched length (I) Determine the spring force using the equation. F s = k x (II) Here, the stiffness is k and the elongation is x . Conclusion: Substitute 0.75 ft for d and 1 ft for unstretched length in Equation (I). x = 2 ( 0.75 ) − 1 = 0.5 ft Substitute 150 lb/ft for k and 0.5 ft for x in Equation (II). F s = 150 × 0.5 F s = 75 lb Joint B : Show the free body diagram of the joint B as in Figure (1). Using Figure (1), Determine the value of θ : sin θ = Opposite side Hypotenuse (III) Along the vertical direction: Determine the force in the members AB and BC by resolving the vertical component of forces. ∑ F y = 0 F A B sin θ 1 + F B C sin θ 2 − F s = 0 (IV) Along the horizontal direction: Determine the force in the members AB and BC b resolving the horizontal component of forces. ∑ F x = 0 F A B cos θ 1 − F B C cos θ 2 = 0 (V) Conclusion: Substitute 0.75 ft for opposite side and 1 ft for hypotenuse in Equation (III). sin θ 1 = 0 .75 1 θ 1 = 48.59 ° Substitute 0.75 ft for opposite side and 1 ft for hypotenuse in Equation (III). sin θ 2 = 0 .75 1 θ 2 = 48.59 ° Substitute 48.59 ° for θ 1 , 48.59 ° for θ 2 , and 75 lb for F s in Equation (IV). F A B sin 48.59 ° + F B C sin 48.59 ° − 75 = 0 0.75 F A B + 0.75 F B C = 75 F A B + F B C = 100 (VI) Substitute 48.59 ° for θ 1 and 48.59 ° for θ 2 in Equation (V)

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 6.6, Problem 106P

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The force F required for equilibrium.

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Chapter 6.6, Problem 105P

Chapter 6.6, Problem 107P

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Chapter 6 Solutions

Engineering Mechanics: Statics & Dynamics (14th Edition)

Ch. 6.3 - In each case, calculate the support reactions and...Ch. 6.3 - Identify the zero-force members in each truss....Ch. 6.3 - Determine the force in each member of the truss....Ch. 6.3 - Determine the force in each member of the truss....Ch. 6.3 - Determine the force in each member of the truss....Ch. 6.3 - Determine the greatest load P that can be applied...Ch. 6.3 - Identify the zero-force members in the truss....Ch. 6.3 - Determine the force in each member of the truss....Ch. 6.3 - Determine the force in each member of the truss...Ch. 6.3 - Determine the force in each member of the truss...

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The force F required for equilibrium.

Solution Summary: The author explains how to determine the force F required for equilibrium. The distance d is 0.75 ft and the unstretched length of the spring is 1.

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The force F required for equilibrium.

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