Question

Physics 12 Equilibrium Question!

Answer 1

It looks like you did a lot to identify the situation, and that's the first step! :) Some people miss it.

Now, where to focus? There is always so much going on. Well, our limiting factor for the mass is the cable, which will break with too great a force. So, we work with the maximum tension. What mass can it afford to support? Well, you'll need to relate the cable's tension to the mass at the end.

We don't care about whatever chord holds the mass - I guess that stuff's invincible.

Forming a relation... Let's look at the forces. FT is what is limited. It's force is transferred through the beam to the mass. What forces matter to the beam? Let's look at that. And we'll start off thinking about how the net torque on the beam about it's fulcrum must be zero or it would be spinning. Torque is simple, the perpendicular force multiplied with radius from the fulcrum.

Perpendicular forces.
Well, for all three forces, we know the originating force. Either our maximum tension or a force of gravity. Tension is given and gravity is mass multiplied with gravitational acceleration. All are known. The rest comes down to trigonometry to find the components - we can do that since all we need is the angle, and the angle is known.

Radii.
These are all given, but the beam radii is tricky. Since it is uniform, we can imagine all of the weight to be at its center - so half of its length. The cable and mass are attached at given radii.

So, to the algebra, and beyond.
Tension must be 0:\[T=0\]Pertaining to our situation, that truth is:\[=TT+T_{beam}+T_{load}=0\]Breaking that down into the torque definitions (perpendicular force multiplied with radius):\[=FT_\perp\times rT+FG_{beam\perp}\times r_{beam}+FG_{load\perp}\times r_{load}=0\]
Now, you can solve

I guess I could've said, "hey, let's find the torques separately and plug them in." But I didn't. I made it messy. But I can't keep the mess up. Let's find the forces and radii separately, and then I'll leave it up to you to disagree with this solution or plug in the variables and values.

I'll have counterclockwise be positive force, and clockwise be negative.
\[FT_\perp=FT\sin(40^\circ)\]\(FG_{beam\perp}=-m_{beam}g\sin(40^\circ)\quad\) where \(\quad g>0\)
\(FG_{load\perp}=-m_{load}g\sin(40^\circ)\quad\) where \(\quad g>0\)

\[r_{beam}=\frac12\times 4.2[\rm{meter}]\]

So, when you formulate that beast of an equation, isolate the \(FT\) (that came from \(FT_\perp=FT\sin(40^\circ)\)). Or, pro tip, isolate \(FT_\perp\). Then isolating \(FT\) will be easier.

Good luck!

Answer 2

Thank you for your answer.