# 3 pulley and 2 masses,relation between their accelerations [closed]

I want to find acceleration of each mass in this system. lets take length of rope between three pulleys as $L$. then we can use this relations $$L=x_1-x_0+(x_1-x_0+x_0-x_2)+x_2-x_0+x_3-x_0=2x_1+x_3-3x_0$$

$$constant=L+3x_0=2x_1+x_3$$

So i think the relation between their accelerations is $2a_2=-a_1$

but I think i did something wrong and the relation is not true because this relation is for a simple case like this

So my question is:

what is the relation between accelerations of $m_1$ and $m_2$ and how we can find it(is there any other way than writing equations for rope length)?

UPDATE

i think I made a mistake at my equations the correct one will be $$L=x_1-x_0+(x_1-x_0+x_0-x_2)+x_2-x_0+x_3-x_2=2x_1+x_3-3x_0$$

$$constant=L+2x_0=2x_1+x_3-x_2$$

So i think the equation for the accelerations will be $2a_2+a_1-a_{P_1}=0$

$P_1$ is the pulley with the smallest circle in picture.If this is true, it means that we can't neglect the mass of this pulley. why this happens (a reason explaining with physics)? and how we should proceed to solve this because we just have one relation between accelerations?

2nd-UPDATE

taking the upward direction as $y+$, this will be our equations for forces:

$m_1$:$\sum f_y=T-m_1g=m_1a_1$

$P_{smallest}$:$\sum f_y=T-2T-m_{P_{smallest}}g=m_{P_{smallest}}a_{P_{smallest}}$

$P_{medium}$:$\sum f_y=2T-T'-m_{P_{medium}}g=m_{P_{medium}}a_{P_{medium}}$

$m_1$:$\sum f_y=T'-m_2g=m_2a_2$

$P_{largest}$:$\sum f_y=T''-2T-m_{P_{largest}}g=m_{P_{largest}}a_{P_{largest}}$

we can neglect $P_{largest}$ and $P_{medium}$ masses.So we have $$T''=2T$$ $$T'=2T$$

so we going to have four equations: $$\left\{ \begin{array}{c} -T-m_{P_s}g=m_{P_s}a_{P_s} \\ 2T-m_2g=m_2a_2\\ T-m_1g=m_1a_1\\ 2a_2+a_1-a_{P_s}=0 \end{array} \right.$$

Now what we should do to find $a_2$,$a_1$ and $a_{P_s}$?

## closed as off-topic by ACuriousMind♦, Danu, JamalS, Pranav Hosangadi, BMSDec 25 '14 at 23:02

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• You should be able to solve this system of equations you end up with. Perhaps use linear algebra? – Danu Dec 25 '14 at 14:01
• @Danu,whatever i try to do with equations, i can't find $a_1$ or $a_2$.for example i tried to find $a_1$ and got this $a_1=\frac{m_1g+m_{P_s}g+2m_{P_s}a_2}{-(m_{P_s}+m_1)}$ – user2838619 Dec 25 '14 at 14:29

I see your update but I don't understand the equality, more exactly the right hand side. I get there

$2(x_1 −x_0) +(x_3 −x_2 )$.

This equality is also directly obvious. From it one gets indeed your relation between accelerations

$2a_2 +a_1 −a_{P_{smallest}} =0$

• thanks for your answer. right now i want to know that how we should find the values of the accelerations. more exactly ,should we have value of $M_{P_smallest}$? also does there any other relation between accelerations? – user2838619 Dec 25 '14 at 13:02
• I have to think. I believe that the masses are needed, both of the smallest and medium pulley and $m_1$ and $m_2$. It's obvious to me that for certain masses all the system will be in equilibrium and the accelerations will vanish. – Sofia Dec 25 '14 at 13:20
• Let's label the medium pulley as $P_2$, and the smallest pulley as you named it $P_1$. On each one of these pulleys act three forces, through the ropes of the left and right of each pulley and one force due to gravity. – Sofia Dec 25 '14 at 13:31
• I updated the post about this. please take a look at it. – user2838619 Dec 25 '14 at 13:39
• Label the medium pulley $P_2$. We need its mass too. The masses have weights and they matter no less than the weights of $m_1$ and $m_2$. I also need to know what are the $T$ and $T'$ in your equations. Note that there are two tension forces that act on $P_1$ (besides the gravity force). – Sofia Dec 25 '14 at 13:46