force problem!? Find X!?

Question

One end of a uniform 4.20 m long rod of weight Fg is supported by a cable. The other end rests against the wall, where it is held by friction. The coefficient of static friction between the wall and the rod is µs = 0.495. Determine the minimum distance x from point A at which an additional weight Fg (the same as the weight of the rod) can be hung without causing the rod to slip at point A (where the rod intersects with the wall).

Answer 1

In order to compute the answer to this question some more data are needed.
First, is the rod horizontal? I will assume that it is
Second, you need to know the angle subtended by the cable and the rod. I will call it th.

With that info, here's the solution

Sum the torques about point A
0=T*sin(th)*L-Fg*L/2-Fg*X

Where T is the tension in the cable and L is the length of the rod

Next, sum the horizontal forces to compute the reaction force at the wall.
T*cos(th)
Next, sum the vertical forces. At the point where static friction breaks the following will be true
T*cos(th)*µs=2*Fg-T*sin(th)

Solve for T
T=2*Fg/(cos(th)*µs+sin(th))

Substitute above
0=2*Fg*sin(th)*L/(cos(th)*µs+
sin(th))-Fg*L/2-Fg*X

Note that Fg divides out
X=2*sin(th)*L/(cos(th)*µs+
sin(th))-L/2

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