Explanation Given information: The system is in equilibrium when θ 1 = θ 2 = 0 . Calculation: Draw the free-body diagram of the arrangement as in Figure (1). Consider the movement of spring at left end is from a to b , and the right end is from a ′ to b ′ . Find the elongation of the spring ( s ) using the relation. s = a ′ b ′ − a b = r θ 1 − r θ 2 = r ( θ 1 − θ 2 ) Find the potential energy ( V ) using the relation. V = V g + V s = 1 2 k s 2 + P l cos θ 1 − W l cos θ 2 = 1 2 k r 2 ( θ 1 − θ 2 ) 2 + P l cos θ 1 − W l cos θ 2 (1) Here, the spring constant is k . Differentiate the Equation (1) with respect to θ 1 . ∂ V ∂ θ 1 = 2 × 1 2 k r 2 ( θ 1 − θ 2 ) − P l sin θ 1 − 0 = k r 2 ( θ 1 − θ 2 ) − P l sin θ 1 (2) Differentiate the Equation (2) with respect to θ 1 . ∂ 2 V ∂ θ 1 2 = k r 2 − P l cos θ 1 Differentiate the equation (2) with θ 2 to find the derivative of ∂ 2 V ∂ θ 1 ∂ θ 2 . ∂ 2 V ∂ θ 1 ∂ θ 2 = − k r 2 Differentiate the Equation (1) with respect to θ 2 . ∂ V ∂ θ 2 = − 2 × 1 2 k r 2 ( θ 1 − θ 2 ) + 0 − W l sin θ 2 = − k r 2 ( θ 1 − θ 2 ) − W l sin θ 2 (3) Differentiate the Equation (3) with respect to ∂ θ 2 . ∂ 2 V ∂ θ 2 2 = k r 2 + W l cos θ 2 Condition 1: When the equilibrium is stable, θ 1 = θ 2 = 0 . Substitute 0 for θ 1 and 0 for θ 2 in Equation (2). ∂ V ∂ θ 1 = k r 2 ( 0 − 0 ) − P l sin 0 = 0 Substitute 0 for θ 1 and 0 for θ 2 in Equation (3). ∂ V ∂ θ 2 = − k r 2 ( 0 − 0 ) − W l sin 0 = 0 ∂ V ∂ θ 1 = ∂ V ∂ θ 2 = 0 The condition is satisfied

Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9780073398242

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Free undamped vibrations

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Chapter 10.2, Problem 10.99P

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The range of values of P for which the equilibrium is stable.

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Chapter 10.2, Problem 10.98P

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Vector Mechanics for Engineers: Statics and Dynamics

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