Question

Exercise 5.4.31 The collar A slides on the smooth vertical bar. Properties are given for the masses A = 23 kg and B = 12 kg, and the spring constant k = 360 N/m. When h=0.2 m, the spring is unstretched. Determine the value of h when the system is in equilibrium. + 0.25 m- the tolerance is +/-9%

Answer 1

Write down the given data: Mass of A, m = 23 kg Mass of B, mg = 12 kg Spring constant, k = 360 N/m Distance h= 0.2 m Height when system is in equilibrium, h 0.25 m lih = 0.2 m Find the value of L. L= V0.252 +0.22 L = 0.32 m Find the value of L'. L' = 70.252 +(H)2 Find the value of sin 0. sin e = sin 0 = = 10.252 + (H)2 Find the value of elongation in spring. S=L'-L 6 = 0.25 +(1) -(0.32)

Balancing the forces at A. T sine I sin =m48 = 23 x 9.81 FO.252 +(1) 225.63 x 70.252 +(W)? T=- ... (1) Balancing the forces at B. mag + ks T = m;8+ks T =12x9.81 + 360 x 5 T =117.72+360( 70.252 + (1 *- (0.32)) ... (2) Equate the equation (1) and (2) and find height when system is in equilibrium. 225.63xpo.25° +(1)° - 117.72 +360(0.252 + (H')? – (0.32) 225.634 10.2.52+(1)* = 117.721 +36061 40.252+(19 –(0.32)) h' = 0.620 m