Explanation Given information: The magnitude of the force W is 144 lb. The length of the member BC is l = 15 in . . The spring is unstretched when θ = 30 ° . The spring constant is k = 20 lb/in . . Calculation: Show the free-body diagram of the arrangement as in Figure 1. Find the curved distance s using the relation. s = r ( θ − 30 ° × π 180 ° ) = r ( θ − π 6 ) Find the potential energy ( V ) using the relation; V = V s + V g = 1 2 k s 2 + W ( − l sin θ ) = 1 2 k r 2 ( θ − π 6 ) 2 − W l sin θ Differentiate the equation; d V d θ = k r 2 ( θ − π 6 ) − W l cos θ (1) Differentiate the Equation (1). d 2 V d θ 2 = k r 2 + W l sin θ (2) Equate the Equation (1) to zero for equilibrium. d V d θ = 0 k r 2 ( θ − π 6 ) − W l cos θ = 0 Substitute 20 lb/in

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ISBN: 9781259977121

Author: BEER

Publisher: MCG

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Chapter 10.2, Problem 10.76P

To determine

Find the value of θ corresponding to the equilibrium.

Check whether the equilibrium is stable.

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Chapter 10.2, Problem 10.75P

Chapter 10.2, Problem 10.77P

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Chapter 10 Solutions

VECTOR MECHANICS FOR ENGINEERS: STATICS

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Find the value of θ corresponding to the equilibrium. Check whether the equilibrium is stable.

Solution Summary: The author shows the free-body diagram of the arrangement as in Figure 1.