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

Jun 2017
399
6
Lima, Peru
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?.
 

romsek

Math Team
Sep 2015
2,974
1,682
USA
look at the behavior of $\sin(x)$

could it be that there is another solution to $\sin(\alpha + 50^\circ) = \sin(50^\circ)$ other than $0^\circ$ ?

One that's even on your list of choices!

On the plus side you did all the setting up of the problem correctly!