Finding the Forces on a 7.5m Aluminum Pole

[pb23me]

Homework Statement

A 7.5 m long uniform aluminum pole with a mass of 12.0 kg supports a 21.5 kg traffic light. The aluminum pole is attached to a vertical pole by a pivot near the base. A massless horizontal rope supports the aluminum pole and is connected to the vertical pole 3.80 m above the pivot. The rope makes an angle of 37º where it connects to the aluminum pole. The traffic light hangs at the end of the aluminum pole. Find the tension in the massless rope and the horizontal and vertical components of the force exerted by the pivot on the aluminum pole.

Homework Equations

$\tau$_net=0
F_net=0

The Attempt at a Solution

i set the pivot point where the two poles meet then T(r)-mgpole(r)-mglight(r)=0 so T=404.25 vert tenision i then used that information to solve for horizontal tension cos(37)*672=536N ... i got 672 by concluding that if vert tension=404.25 then sin(37)*T=404.25 then solve for total tension. so the horizontal force from the vertical pole=-536N and the vertical force =42N

 

[BigJDubs]

The correct answer is T = 672 N, Fhorizontal = 536 N, and Fvertical = 136 N. You made a mistake in setting up the equation for the net torque about the pivot point. The net torque should be equal to zero, so you should have:τnet=T⋅rpole -mgpole⋅rpole -mglight⋅rpole=0 Where T is the tension in the rope, and rpole is the distance from the pivot point to the center of mass of the aluminum pole. Once you set this up correctly, you should be able to solve for the tension in the rope, which is what you need to solve for the other two forces.