Explanation Given information : The stiffness k of the spring is 400 N / m . The mass ( m ) of the disk is 20 kg. The un-stretched length ( l 0 ) of the spring is 1 m. Show the free body diagram of the disk supported on the smooth cylindrical surface by a spring as in Figure (1). Use Figure (1). Along the vertical direction : Assumption: Upward is positive. Apply the equation of equilibrium in the y -axis. ∑ F y = 0 R sin θ − m g = 0 (I) Here, the resultant force is R , the resolving angle is θ , and the acceleration due to gravity is g . Along the horizontal direction : Assumption: Towards right is positive. Apply the equation of equilibrium in the x -axis. ∑ F x = 0 R cos θ − F = 0 (II) Determine the spring force using the formula. F s p = k ( Stretched length − unstretched length ) (III) Conclusion : Substitute 20 kg for m and 9.81 m / s 2 for g in Equation (I). R sin θ − ( 20 × 9.81 ) = 0 R sin θ = 196.2 (IV) Equation (II) can be written as follows: R cos θ = F (V) Substitute 400 N / m for k , 2.2 m for stretched length, and 1.0 m for unstretched length in Equation (III). F s p = 400 ( 2.2 cos θ − 1.0 ) Here, the resolving angle of stretched length is θ . Divide Equation (III) by (IV) and it can be written as follows: R sin θ R cos θ = 196.2 F tan θ = 196.2 F (VI) Substitute 400 ( 2.2 cos θ − 1.0 ) for F s p in Equation (VI)

13th Edition

ISBN: 9780132915540

Author: Russell C. Hibbeler

Publisher: Prentice Hall

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Chapter 5.4, Problem 56P

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The angle for equilibrium of the roller.

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Chapter 5 Solutions

Engineering Mechanics: Statics

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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