In the figure, a 53.6 kg uniform square sign, of edge length L = 2.01 m, is hung

Question

In the figure, a 53.6 kg uniform square sign, of edge length L = 2.01 m, is hung from a horizontal rod of length dh = 3.01 m and negligible mass. A cable is attached to the end of the rod and to a point on the wall at distance dv 4.63 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 Cable dy Hinge Rod H. Perex DENTIST UnitsI

Explanation / Answer

angle made by the cable with the rod , theta = tan^-1(dv/dh) = 57 degrees

In equilibrium net torque = 0

T*sintheta*dh - m*g*(dh-L/2) = 0

T*sin57*3.01 = 53.6*9.8*(3.01-2.01/2)

T = 417.2 N

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(b)

along horizontal Fnet = 0

Fh - T*costheta= 0

Fh = T*costheta

Fh = 417.2*cos57 = 227.2 N

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(c)

along vertical

Fnet = 0

Fv + T*sintheta - mg = 0

Fv = mg - T*sintheta

Fv = (53.6*9.8) - (417.2*sin57)

Fv = 175.4 N

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