Question

The beam shown is supported by a pin at A and rope BDE as shown. Knowing that L = 1.2 m and that the mass suspended from C is 27 kg, determine the tension in rope BDE and the reactions at A. Enter the magnitude of the tension in rope BDE below, in N to the nearest whole N. (Take g = 9.81 m/s2) -2m-—2m-+

Answer 1

1413 70.8T V. 71 »Na Bi DGTE 2m-ly-2m-* L:12 M=27kg 1 wang igid diagram of The free body the de body diagram of the beam is shown in Fig. 2. Nx and Ny are the reactions at A in the horizontal and hertical direction respectively and T is the tension in the rope BDE. Efx=0 3 Nx = 0,6 T --- 0 fyro NyT+0.8T - W=0 9Ny-E1.8T+Mg) ---(2) Applying torque balance about point A, we get- 1. ETA=0 13) Til +0.87 (+1) - Wix(4+bx+1)=0 1 [ cow torque is assumed 3 T X 2 + 0.8T *(4) - 27x9.81512=0 ' ਮੈtv - J= [264.87 N] from (l), we get Nye - 0.6 x 244,87 = 158.922N7 from (2), we get » Ny = 29x9181 - 1.8 x 264,87) 73224435N1 = -211.896N Caus Reactions at at A, Nx = 158.922 N (o towards right) . Ny = 211, 896N (downward) Tersian in the rope BDE, T= 264.87 N.