A uniform horizontal bar of length L = 4 m and weight 226 N is pinned to a verti

Question

A uniform horizontal bar of length L = 4 m and weight 226 N is pinned to a vertical wall and supported by a thin wire that makes an angle of theta = 29o with the horizontal. A mass M, with a weight of 312 N, can be moved anywhere along the bar. The wire can withstand a maximum tension of 501 N. What is the maximum possible distance from the wall at which mass M can be placed before the wire breaks?


I keep getting that the equation would be

x= (T*l*sin(theta)-(mg*(L/2))/Mg ..... the answer I keep getting when I plug in is -1.13079. can anyone see what I am doing wrong

Explanation / Answer

Let's set the fulcrum to be at the point where the wall and the bar connect.

The bar's weight Wb is creating a clockwise torque

Wb*(L/2) since the center of mass is at L/2

The tension of the wire is creating a counter clockwise torque

T*sin(29o)*L

Now you want to place a mass on the bar.

Assume that the mass is x meter away from the bar

The weight of mass Wm is creating a clockwise torque

Wm*x

Since the bar is not moving and is horizontal,

clockwise torque = counter clockwise torque

Wb*(L/2) + Wm*x = T*sin(29o)*L

Let's plug in the value. from the quesiton we know Wb = 226, Wm = 312, L = 4

226*(4/2) + 312*x = T*sin(29o)*4

Since the question is asking for maximum distance, we want x to be max.

Thus, we need to set T to be max to get maximum x.

Tmax = 501 N

Now the equation become

226*2 + 312*x = 501*sin(29o)*4

x = [ T*sin(29o)*L - Wb*(L/2) ]/Wm = [ 501*sin(29o)*4 - 226*2 ] / 312 = 1.665 (m)

Looks like your equation is correct... I'll say you could check whether you are using degree or radian in sin fuction.

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