A uniform horizontal bar of length L = 2 m and weight 218 N is pinned to a verti

Question

A uniform horizontal bar of length L = 2 m and weight 218 N is pinned to a vertical wall and supported by a thin wire that makes an angle of theta = 39o with the horizontal. A mass M, with a weight of 304 N, can be moved anywhere along the bar. The wire can withstand a maximum tension of 542 N. What is the maximum possible distance from the wall at which mass M can be placed before the wire breaks?

AND

With M placed at this maximum distance what is the horizontal component of the force exerted on the bar by the pin at A?

Explanation / Answer

here the net torque about the point A = 0

torque

due to rope = +T*sin39*L

due to weight of the bar = -w1*(L/2)

due to weight of M = -W2*g*x

in equilibrium net torque = 0

T*sin39*L = W1*(L/2) + W2*x

542*0.629*2 = (218*1) + (304*x)

x = 1.32 m from A <------answer

++++++

paart B)

along horizantal Fnet = 0

Fx - T*cos39 = 0

Fx = 542*cos39 = 421.2 N <---answer

along vertical Fnet = 0

Fy + T*sin39 - W1 -W2 = 0

Fy = 218 + 304 - 542*sin39 = 181 N <---answer

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