Torque equilbrium involving support rod?

[genessis42]

Homework Statement

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:

Homework Equations

Net Torque=0
Net Force=0

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

 

[voko]

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

 

[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