One end of a uniform 4.80-m-long rod of weight Fg is supported by a cable at an

Question

One end of a uniform 4.80-m-long rod of weight Fg is supported by a cable at an angle of = 37° with the rod. The other end rests against the wall, where it is held by friction as shown in the figure below. The coefficient of static friction between the wall and the rod is s = 0.560. Determine the minimum distance x from point A at which an additional object, also with the same weight Fg, can be hung without causing the rod to slip at point A.

One end of a uniform 4.80-m-long rod of weight Fg is supported by a cable at an angle of = 370 with the rod. The other end rests against the wall, where it is held by friction as shown in the figure below. The coefficient of static friction between the wall and the rod is us 0.560. Determine the minimum distance>x from point A at which an additional object, also with the same weight Fg, can be hung without causing the rod to slip at point A.

Explanation / Answer

tan 37° = 0.6/0.8 = 3/4
Tx = Ty/tan 37° = Ty*4/3 (Ty and Tx being vertical and horizontal components of rope tension T)....therefore :
equilibriun on A :
Fg*X+Fg*4.8/2 = Ty*4.8 = Tx*3/4*4.8 = 3.6Tx
equilibrium on B
Tx*s*4.8 = Fg*2.4+Fg(4,8-X)

Tx = Fg(7.2-X)/(4.8*0.56)
Fg*(X+2.4)*3.6 = Fg(7.2-X)/(4.8*0.56)
3.6*Fg*X+8.64 = 2.678*Fg-0.4167*Fg*X
4.0167*Fg*X= 2.678Fg-8.64
X = (2.678Fg-8.64)/4.0167*Fg = 0.67-2.15/Fg

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