# Beam Torque Problem

1. Apr 11, 2006

### Mr. Snookums

There is an 80kg beam attached to a hinge on a wall and supported by a cord. The angle between the cord and the beam is 25 degrees. The beam is 4.5m long., and the cord is attached to the beam a metre from the beam's end. There is a 20kg beam hanging from the very tip of the beam. Find the cord tension and the force exerted by the hinge.

CW Torque=(80kg)(9.8m/s^2)(2.25m)+(20kg)(9.8m/s^2)(4.50m)
=2646Nm

Since the torques must balance:

2646Nm=F(3.5m)
F=756N

Plug this into the right triangle created by the cord and the beam:

Tension in the cord=cos65/756N=1790N

Now here's where I have the trouble. How do I get the force exerted by the hinge? I found the x xomponent, which is 1790sin65. But when I use this answer to find the force vector of the hinge, I don't get the right answer.

(1790sin65)(sin25)=687N. The answer is 1600N. Where have I gone wrong?

2. Apr 12, 2006

### dav2008

The hinge will excert an x and y component force.

You're right that the x-component will be 1790 cos(25) [or as you put it, sin(65)].

Now you have to find the y-component.

3. Apr 12, 2006

### Mr. Snookums

The y-component is the force from the hinge that would balance all the other forces along the beam? Because the torque is already balanced.

4. Apr 12, 2006

### dav2008

Don't think of it as "balancing" torques. Think of it as setting all of the torques set to 0.

You're right in that if you set all of the vertical forces set to 0 then you can find the y-component of the reaction force at the hinge. The thing to realize though is that if you take the moments (torques) about any point on the beam, they should add to zero. (Well actually if you take the moments about any point in space they should add to zero.)

Last edited: Apr 12, 2006