Meriam, Kraige & Bolton — Internal Forces and Moments: Problem 7_36

⚡ Mecademy AIENG정역학 · ch7  Problem Statement The figure shows the cross section of a uniform 60-kg ventilator door hinged along its upper horizontal edge at O. The door is controlled by the spring-loaded cable which passes over the small pulley at A. The spring has a stiffness of 160 N per meter of stretch and is undeformed when . Determine the angle for equilibrium. Problem 7/36 (a) Determination of the equilibrium angle 1. Formula: For a mechanical system in equilibrium, the first derivative of the total potential energy with respect to the generalized coordinate must be zero. The total potential energy is the sum of the gravitational potential energy and the elastic potential energy . Let be the length of the door and its mass. Taking the horizontal line as the datum for gravitational potential energy: From the geometry of triangle , where , the length of the cable segment is: Since the spring is undeformed at (where ), the stretch is . The elastic potential energy is: θ=0θ θ Vθ = dθ dV 0 V g V e V=V + g V e L=1.4 mm=60 kg OA V= g −m g sin θ( 2 L ) OABOA=OB=L AB s= = L+L−2Lcosθ 222 2Lsin ( 2 θ ) θ=0s=0x=s V = e kx = 2 1 2 k2Lsin = 2 1 ( 2 θ ) 2 2 kL sin = 22 ( 2 θ )kL(1− 2 cosθ) Combining these, the total potential energy is: 2. Substitution: Differentiate with respect to and set it to zero: Rearranging gives the equilibrium equation: Substitute the given values: , , , and . 3. Calculation: Perform the numerical calculation step-by-step: Numerator: Denominator: Ratio: 4.