Pulley system on a big block

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1. May 23, 2017

Buffu

1. The problem statement, all variables and given/known data

Find the accelaration of $M_1$ in the given system if $F = 0$.

2. Relevant equations

3. The attempt at a solution

$x_3 -x_1 = k \iff \ddot x_3 = \ddot x_1$

and $h - y_3 + x_3 - x_2 = l \iff \ddot y_3 + \ddot x_2 = \ddot x_3 \qquad (*)$

h is the height of block $M_1$ and $l$ is the length of string between $M_2$ and $M_3$.

Now from the Free body diagram,

$-M\ddot x_1 = N^{\prime \prime \prime}$

$T = M_2 \ddot x_2$

$N^{\prime \prime \prime} = -M_3\ddot x_3$

$\therefore -M\ddot x_1 = M_2 \ddot x_2 - M_3\ddot x_3 \qquad (1)$

Now from vertical force on $M_3$,

$M_3 - T = -M_3 \ddot y_3$

$-M_3 g + M_2 \ddot x_2 = M_3 \ddot y_3$

Substituting for $\ddot y_3$ in $(*)$

$\ddot x_3 = x_2 + \dfrac{-M_3 g + M_2 \ddot x_2 }{M_3}$

Solving for $\ddot x_2$

$x_2 = \dfrac {M_3(\ddot x_3 + g) }{M_3 + M_2}$

Substituting this in $(1)$

$-M_1 \ddot x_1 = \dfrac {M_3M_2(\ddot x_3 + g) }{M_3 + M_2} - M_3\ddot x_3$

Since $\ddot x_1 = \ddot x_3$

$-M_1 \ddot x_1 = \dfrac {M_3M_2(\ddot x_1 + g) }{M_3 + M_2} - M_3\ddot x_1$

Solving for $\ddot x_1$

$\ddot x_1 = \dfrac{-g(M_2M_3)}{M_1M_2 + M_3M_1 - M_3^2}$

Which is incorrect as the given answer is $\ddot x_1 = \dfrac{-g(M_2M_3)}{M_1M_2 + M_3M_1 \color{red}{ + 2M_2M_3 +} M_3^2}$.

What is the problem ?

2. May 23, 2017

kuruman

This equation doesn't look right.

3. May 23, 2017

Buffu

Oh sorry that was a pure typo.
That equation should be,
$M_3g - T = -M\ddot y_3$.

Last edited: May 23, 2017
4. May 23, 2017

kuruman

Still not right. On the left side (Fnet) "down" is positive, but on the right side (mass x acceleration) "down" is negative.

Also, check equation (1). The center of mass does not accelerate because the only external force to the three mass system is a vertical force. So $M_1 \ddot{x}_1+M_2 \ddot{x}_2+M_3 \ddot{x}_3=0$

5. May 23, 2017

Buffu

Yes I got the correct answer, but I think $M_3g - T = -M\ddot y_3$ is correct because $y_3$ is pointing upwards and so is $\ddot y_3$ and therefore I put a negative sign, since $M_3g - T$ is downwards as you said. Am I wrong ?

6. May 23, 2017

kuruman

No, it's OK. I fooled myself because I didn't notice $y_3$ in the diagram.