# Four bodies connected through a sheave

1. Oct 30, 2014

### Chemist@

1. The problem statement, all variables and given/known data
Three bodies with masses m1 m2 and m3 connected with a string through a sheave with a body with mass m4. The system is moving right. Write the second Newton law for the movement of the objects. Ignore friction.

2. Relevant equations
2nd law for bodies:
m4a= m4g-T(the force of stretching the string)
m1a= T
m2a= T
m3a= T

3. The attempt at a solution
By summing up the last three equation I got:
a(m1+m2+m3)=3T but it is apparently incorrect. The book equation is:
a(m1+m2+m3)=T, how? Where am I wrong at?

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2. Oct 30, 2014

### BvU

You want to ask yourself if T is really the same in the various sections.

3. Oct 30, 2014

### Chemist@

4. Oct 30, 2014

### collinsmark

I believe what BvU is alluding to is that you are assuming that all sections of the string have the same tension.

While that is often true for a massless string when ignoring friction, it doesn't apply here because the string has masses attached to it. In other words, the tension of the string between bodies m1 and m2 is a different value than the tension between bodies m2 and m3 and so on.

You can still exploit the idea of tension though, just pick one of the tensions. The books seems to have picked the tension in the string at rightmost side, that passes through the sheave. Don't forget that the three leftmost bodies move together as a group.

5. Oct 30, 2014

### BvU

If the left block feels tension T1 and is accelerated with acceleration T1/m1, the next block feels tension T1 working to the left. If that block is also accelerated with acceleration a, the net force on block 2 must be m2a. So it feels a Tension T2 to the right for which m2a = T2-T1.

et cetera

6. Nov 1, 2014

### Chemist@

I see. When they all add up I get T3. Thank you very much.