In the figure, a 45.2 kg uniform square sign, of edge length L = 1.51 m, is hung

Question

In the figure, a 45.2 kg uniform square sign, of edge length L = 1.51 m, is hung from a horizontal rod of length dh = 2.95 m and negligible mass. A cable is attached to the end of the rod and to a point on the wall at distance dv = 3.51 m above the point where the rod is hinged to the wall. (a) What is the tension in the cable? (b) What is the horizontal component of the force on the rod from the wall? Take the positive direction to be to the right. (c) What is the vertical component of this force? Take the positive direction to be upward.

Explanation / Answer

a)the tension in the cable is

T = m x g

where m = 45.2 kg and g = 9.8 m/s^2

b)the horizontal component of force is

Fx = T x cos(theta)

where theta = tan^-1(3.51/2.95)

c)the vertical component of force is

Fy = T x sin(theta)

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