Question

A 325 kg uniform square sign, 2/00 m on a side, is hung from a 3.00 m rod of negligible mass. A cable is attached to the end of the rod to a point on the wall 4.00 m above the point where the rod is fixed to the wall (see the figure). What is the tension in the cable? What is the horizontal component of the force exerted by the wall on the rod? What is the vertical compound of the force exerted by the wall on the rod?

Answer 1

length of the cable = r= sqrt(y^2+x^2)= sqrt(4^2+3^2) = 5 m

\let theta be the angle made with vertical

tantheta = x/y

theta = 37

In equilibrium net torque = 0

torque due to cable = T*cos37*x

torque due to sign board = -325*2

net torque = 0

T*cos37*x - 325*2 = 0

T*cos37*3 - 325*2 = 0

T = 271.3 N

+++++++++++++++++++++++++

along horizantal Fnet = 0

T*sin37 = Fh

Fh = 163.3 N

______________

along vertical

Fnet = 0

T*cos37 = mg

271.3*cos37 + Fv = 325*9.8

Fv = 2968.3 N