# Inclined pulley. Finding the $\frac{M_1}{M_2}$ [closed]

Problem

As you can see in the picture, $M_1$ and $M_2$ are attached by a massless rope over a pulley. the height difference of $M_1$ and $M_2$ is $h$. The masses start to move and after $τ$ seconds, $M_1$ and $M_2$ will be on same height. If the coefficient of kinetic friction is $k$, find the value of $\frac{M_1}{M_2}$ .

My ideas and work

when the both objects have same height, $M_2$ will be up $h_f$ meters, and $M_1$ will be down $h_f$ meters. So, we have these equations $$\sin\alpha.x_t=h_f$$ $$\sin\beta.x_t=h-h_f$$

where $x_t$ is the distance $M_1$ moved after $τ$ seconds.

we can get that $x_t=\frac{h}{\sin\alpha + \sin\beta}$

We also know that $x_t=x_0+v_0t+\frac{1}{2}at^2$ so $x_t=\frac{1}{2}aτ^2$ and finally $a$ will be $$a=\frac{2h}{τ^2(\sin\alpha + \sin\beta)}$$

For $M_1$:$\left\{ \begin{array}{c} \sum f_x=-T-f_{k_1}+M_1g\sin\alpha=M_1a \\ \sum f_y=N_1-M_1g\cos\alpha=0 \end{array} \right.$

For $M_2$:$\left\{ \begin{array}{c} \sum f_x=T-f_{k_2}-M_2g\sin\beta=M_2a \\ \sum f_y=N_2-M_2g\cos\beta=0 \end{array} \right.$

finding $a$ from this equations:$$a=\frac{M_1g(\sin\alpha-(\cos\alpha)k)-M_2g((\cos\beta)k+\sin\beta)}{M_1+M_2}$$

Now I amstuck as what to do next. I think i should go for $\frac{M_1a}{M_2a}$ and finding $\frac{M_1}{M_2}$ that way but i think it will be a lot of work because i should find $T$ and then doing other things. any better way for finding $\frac{M_1}{M_2}$? Also I think the question has gotten us many information and some of them may not be useful very much as I can't see any connection between $h$, $τ$ and kinematics between dynamic part of problem.

Update

I did this $$\frac{2h}{τ^2(\sin\alpha + \sin\beta)}=\frac{M_1g(\sin\alpha-(\cos\alpha)k)-M_2g((\cos\beta)k+\sin\beta)}{M_1+M_2}$$

And I finally got this $$\frac{M_1}{M_2}=\frac{2h+gτ^2(\sin\alpha + \sin\beta)(\sin\beta+k\cos\beta)}{-2h+gτ^2(\sin\alpha + \sin\beta)(\sin\alpha-k\cos\alpha)}$$

I hope I did everything right. If somebody thinks the result is wrong, I would be pleased if he/she explains it in comments or as an answer.

## closed as off-topic by John Rennie, Kyle Kanos, ACuriousMind♦, Danu, JimDec 22 '14 at 17:05

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