# Homework Help: Moving pulley question

1. Jan 11, 2013

### ShizukaSm

1. The problem statement, all variables and given/known data
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Find acceleration of block 1, knowing that the mass of block 1 is 2kg and of block 2 is 4kg. You may ignore friction and pulley mass.

2. Relevant equations

Fr = ma

3. The attempt at a solution

I Drew free body diagrams to block 1, 2 and the pulley, then made the equations:
$\\ \sum F_{r2} = m_2a_2\\ P_2 - T_2 = m_2a_2\\ \\ \sum F_{rpulley} = m_{pulley}*a\\ 2T_2 = T_1\\ \\ \sum F_{1}=m_1a_1\\ T_1=m_1a_1\rightarrow T_2 = \frac{m_1a_1}{2} \\\\a_1=2a_2 \rightarrow T_2=\frac{m_1(2a_2)}{2} = m_1a_2 \\(Substituting\ in \ first\ equation)\\ P_2-(m_1a_2)= m_2a_2\\ a_2(m_2+m_1) = m_2*g\\\\ a_2=\frac{m_2g}{m_2+m_1} = 6.53\frac{m}{s^2}\rightarrow a_1 = 13.06\frac{m}{s^2}$
The answer, however, is supposed to be 4.4 m/s²

2. Jan 11, 2013

### Staff: Mentor

Is $a_1 = 2 a_2$ true?

3. Jan 11, 2013

### ShizukaSm

Well, I think it is, isn't it?

If 1 moves a distance "d", 2 will move only d/2 because the rope has to run on both sides of the pulley.

4. Jan 11, 2013

### Staff: Mentor

On the other hand, if 1 moves a distance "d", the pulley "takes" a length "d" of rope from both sides of the pulley. This rope can only come from the free end holding m2.

#### Attached Files:

• ###### Fig1.gif
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5. Jan 11, 2013

### ShizukaSm

Wow, that... that makes a lot of sense, so in fact a1 = 2a2

Thanks a lot, gneill!