One end of a uniform 4.40-m-long rod of weight F g is supported by a cable at an
ID: 1504351 • Letter: O
Question
One end of a uniform 4.40-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 = 0.530. 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 4.40-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 = 0.530. 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
solution:
L = length of the rod = 4.4 m
Fg = weight of the rod = weight of the ball
= angle of the cable = 37°
µ = coefficient of friction = 0.53
X = distance to find = ?
Find the tension in the cable.
Fg×X + Fg×L/2 = T×L×sin()
T = (Fg×(X + L/2)) / (L×sin())
Find the horizontal reaction force at point A.
Rh = T×sin()
Rh = Fg×(X + L/2) / L
Rh = Fg×(X + 2.2) / 4.4
Find the vertical reaction force at point A.
Fg×L/2 + Fg×(L - X) = Rv×L
Rv = [Fg×L/2 + Fg×(L - X)] / L
Rv = Fg×(1.5 - X/L)
Rv = Fg×(1.5 - X/4.4)
Now, you can say:
µ×Rh = Rv
so
0.57×Fg×(X + 2.2) / 4.4 = Fg×(1.5 - X/4.4)
0.57(X + 2.2) /4.4 = 1.5 - X/4.4
0.57×X + 1.254 = 6.6 - X
1.57×X = 5.346
X = 5.346 / 1.57
X = 3.405 m ans
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