A mass of 4.70 kg is suspended from the end of a thin, uniform, horizontal rod w

Question

A mass of 4.70 kg is suspended from the end of a thin, uniform, horizontal rod with a mass of 1.50 kg. As shown below, one end of the rod is in contact with a wall and is supported by a thin wire attached to the wall. Friction between the wall and rod keeps the rod from slipping.

Calculate the tension in the cable.
Note: the grid spacing in the figure is 10 cm, in both horizontal and vertical directions.

Calculate the minimum value of the coefficient of static friction between the wall and the rod which is required to keep the rod from slipping.

Explanation / Answer

given,

mass = 4.7 kg

mass of the rod = 1.5 kg

equating the vertical forces

mass * g = tension * sin(theta)

theta = tan^-1(5/10)

theta= 26.57 degree

4.7 * 9.8 = tension * sin(26.57)

tension = 102.976 N

for rod normal force = tension * cos(26.57)

normal force on rod = 102.976 * cos(26.57)

normal force on rod = 92.1006 N

for preventing rod from slipping

mass of rod * g = coefficient of friction * normal force

1.5 * 9.8 = coefficient of friction * 92.1006

coefficient of friction = 0.15961

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