How to find the tilted angle from a set of hanging masses between two pulleys?

Question

The problem is as follows:

The figure from below shows three masses tied to a central one which is hanging with respect of a floor. Find the angle so that the system remains in static equilibrium:

The alternatives are as follows:

$\begin{array}{ll} 1.&50^{\circ}\\ 2.&80^{\circ}\\ 3.&60^{\circ}\\ 4.&70^{\circ}\\ 5.&45^{\circ}\\ \end{array}$

What I've attempted to do was to equate what it is on the vertical components for the tension of the string and for the horizontal components.

What I obtained was as follows:

In the vertical:

$Mg\cos 50^{\circ}+mg\cos \alpha = mg$

In the horizontal:

$Mg\sin 50^{\circ} = mg\sin\alpha$

Therefore:

$M=\frac{m\sin\alpha}{\sin 50^{\circ}}$

If I do insert this in the equation for the vertical components. I'm obtaining the following:

$\frac{mg\sin\alpha}{\sin 50^{\circ}}\cos 50^{\circ}+mg\cos \alpha = mg$

Dividing by similar terms and multiplying by \sin 50^{\circ} to tall:

$\sin\alpha\cos 50^{\circ}+\cos \alpha\sin50^{\circ}=\sin 50$

This is reduced to:

$\sin(\alpha+50^{\circ})=\sin 50^{\circ}$

Therefore $\alpha =0$

But this doesn't make sense. Could it be that I'm missinterpreting something. Can somebody help me here?.

Answer 1

Hint: $\sin(\alpha+50^{\circ})=\sin 50^{\circ}$ has solutions other than $\alpha=0$.

Answer 2

From $\sin\alpha\cos 50^{\circ}+\cos \alpha\sin50^{\circ}=\sin 50^{\circ}$, square both sides:

$$\sin^2\alpha\cos^2 50^{\circ}+2\sin\alpha\cos 50^{\circ}\cos \alpha\sin50^{\circ}+\cos^2 \alpha\sin^250^{\circ}=\sin^2 50^{\circ}$$ Reconcile the right-most terms: $$\sin^2\alpha\cos^2 50^{\circ}+2\sin\alpha\cos 50^{\circ}\cos \alpha\sin50^{\circ}-\sin^2 \alpha\sin^250^{\circ}=0$$ Divide through by $\sin^2\alpha$, which is acceptable if we assume the obvious from the context that $\sin\alpha\neq0$: $$\cos^2 50^{\circ}+2\cos 50^{\circ}\cot \alpha\sin50^{\circ}-\sin^250^{\circ}=0$$ Now you can solve for $\cot\alpha$: $$\cot\alpha=\frac{\sin^250^{\circ}-\cos^250^{\circ}}{2\sin50^{\circ}\cos 50^{\circ}}$$

And use trig identities: $$\cot\alpha=\frac{-\cos100^{\circ}}{\sin100^{\circ}}=-\cot100^{\circ}=\cot(180^{\circ}-100^{\circ})=\cot(80^{\circ})$$

So $80^\circ$ is one solution. Add an integer multiple of $180^\circ$ and you preserve tangents and cotangents, so you can get other algebraic solutions that way. But only $80^\circ$ makes sense in context.