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...
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...