# Strength of wall regarding a system of hanging mass When we assume that mass m is directly at the center of L, and $$\alpha$$ is the angle between the rope and the horizontal, I am able to show that for equilibrium, $$\sin{\alpha}=\frac{m}{2M}$$ There is also a resultant reaction force from the roller due to the resultant of left side tension and weight of M (which are equal in case of equilibrium). However, let's say that there is a constraint as to how much of a load the wall anchor can hold. Assuming M is still kept the same, how do I proceed to calculate $$m_{max}$$ and $$\alpha$$ in that case?

• What's the "wall anchor"? The right side structure? – Bob D Nov 14 '20 at 22:03
• Yes, there is a roller to the left and a wall to the right. – Mark Nov 14 '20 at 22:03
• OK, so the structure to the right is the wall. Then how is the rope attached to the wall? – Bob D Nov 14 '20 at 22:08
• Well I assume its just mounted to the wall. The wall would provide a certain reaction force to the pull of the right side tension. As to the constraint I mentioned, would perhaps be the maximum reaction force the wall can provide (hence the load) – Mark Nov 14 '20 at 22:11
• When you say "let's say that there is a constraint as to how much of a load the wall anchor can hold", you need to say what you mean by "hold". Do you mean the maximum force before the said "anchor" pulls out of the wall"? If that's the case, it may depend as much on the type of anchor being used as the strength of the wall itself. – Bob D Nov 14 '20 at 22:20

It's not completely sure what you mean with the maximum load that the wall anchor can hold. But one way or the other that is either the tension force $$T$$ in the rope, or $$T\cos\alpha$$ or $$T\sin\alpha$$.
You equate this load to the maximum tension force. The unknown are $$m$$ and $$\alpha$$. Together with the equation for equilibrium that you already showed in the question, you end up with two equations with two unknown.