What is the minimum distance x from point A at which another weight Fg can be held without the rod slipping?

Question

The rod is 4m long, with a weight Fg. One end, Point B, is supported by a cable from the wall. The other end, Point A, rests against a wall with μ s = 0.5. The angle between the cable and Point B is 37 degrees. Basically, the question is asking how far out from the wall can an extra weight, Fg, same as the weight of the rod, be placed without causing the rod to slip at Point A. (Diagram looks smiliar to a triangle). No actual weights are given.

Answer 1

μTcos37 + Tsin37 = 2Fg
cFg + xFg = LTsin37
xFg = LTsin37 - cFg
x = (LT/Fg)sin37 - c
T(μcos37 + sin37) = 2Fg
T = 2Fg/(μcos37 + sin37)
x = (2LFg/Fg(μcos37 + sin37))sin37 - c
x = (2L/(μcos37 + sin37))sin37 - c
x = (2*4/(0.5cos37 + sin37))sin37 - 2
x ≈ 2.809 m