In the figure, a 50.0kg uniform square sign, of edge length L=2.00m, is hung fro

Question

In the figure, a 50.0kg uniform square sign, of edge length L=2.00m, is hung from a horizontal rod of length dh = 3.00m and negligible mass. A cable is attached to the end of the rod and to a point on the wall at a distance dv = 4.00m above the point where the rod is hinged to the wall. (a) What Is the tension in the cable? What are the (b) magnitude and (c) direction (left or right) of the horizontal component of the force on the rod from the wall, and the (d) magnitude and (e) direction (up or down) of the vertical component of this force?

Explanation / Answer

mass of the uniform sign m = 50 kg length of the uniform sign L = 2 m length of the rod , dh = 3 m the distance between wall and rod , dv = 4 m from figure ,     tan = dv/dh             = (4 m) / (3 m) angle = tan-1(1.33)            = 53.130 from equilibrium condition , the net torque is      = (T sin)(dh) - (mg)(dh - L/2) = 0              (T sin)(dh) - (mg)(dh - L/2) = 0 therefore , tension in the cable is       T = [(mg)(dh - L/2)] / (dh) sin          = [(50 kg)(9.8 m/s2)(3 m - 1 m)] / {(3 m)sin(53.13)}          = 408.33 N ........................................................................ horizontal component of force:    Fx = T cos(53.13)         = 245 N direction : towards right ................................................................... vertical component of force :       Fy = mg - T sin(53.13)            = 163.33 N       Fy = mg - T sin(53.13)            = 163.33 N direction : upward direction

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