# Homework Help: Torque equilbrium involving support rod?

1. Oct 2, 2012

### genessis42

1. The problem statement, all variables and given/known data

A uniform 300-kg beam, 6 m long, is freely pivoted at P. The beam is supported in a horizontal position by a light strut, 5 m long, which is freely pivoted at Q and is loosely pinned to the beam at R. A load of mass is suspended from the end of the beam at S. A maximum compression of 23,000 N in the strut is permitted, due to safety. In Fig. 11.1, the maximum mass M of the load is closest to:

2. Relevant equations

Net Torque=0
Net Force=0

3. The attempt at a solution

I tried adding up all of the torque moments and then setting them equal to 23,000N.

300(9.8)(3m)+0(Pivot arm)+6m(mg for the mass)=23,000

Then I set that net torque equal to 23,000, isolated force of the mass, and divided by 9.8 m/s^2 in order to find the maximum mass. It's still wrong, and I don't know where to go from here
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 2, 2012

### voko

You should start from writing down the equations of equilibrium. Don't plug in any numbers just yet, use letters for everything.

3. Oct 2, 2012

### genessis42

Just figured it out.

I have to take the 23,000 maximum force into account that is exerted upwards.

So Net Torque=

0(0)(at the Pivot P)+ (-300kg*9.8)(3m) + -x(6m) + 23,000(4/5)

Its 4/5*23,000 because that is how much is being exerted upwards after looking at the triangle.

We solve for x, and divide by 9.8, to get 788 Kg