How Much Load Can a 300-kg Beam Support Before Exceeding Strut Safety Limits?

[CoogsDownFall]

Homework Statement

A uniform 300-kg, 6.0 M long, freely pivoted at P, as shown in the figure. The beam is supported in a horizontal position by a light strut, 5.0 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. The Maximum mass M of the load is closest to Pic

Homework Equations

$∑Fx=0$
$∑Fy=0$
$∑Ftorque=0$
$Torque = r * F$
$F = m * a$

The Attempt at a Solution

I started with first trying to find the forces of both the X and Y axis and then the force of the torque and setting this equal to 23,000 N. What is really messing me up is the beam that is at an angle. So for the forces acting on Y I set Fy = m * (9.8). I know there are forces acting on the X-axis, I am just struggling to be able to find them. Then for the Torque value I have Torque = m * (9.8) * (3.0). I figured because the large beam was cut in half that is the new radius. Any help would be great!

 

[haruspex]

I see no figure. Please supply a figure or more detailed description. E.g. where are P, Q, R and S in relation to the beam and to each other?

 

[CoogsDownFall]

There is a link to it if you click the word "pic"

 

[haruspex]

There is a link to it if you click the word "pic"

OK - doesn't stand out that well on my screen.
Yes, torque is the way to go, but which point are you taking torque about? What forces have a torque about that point? For each of those, what is the distance from its line of action to the point?