# Moving Pulleys

1. Jun 17, 2009

### Patta1667

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

Blocks of mass M1 and m2 (as shown in attachment) are connected by a system of frictionless pulleys by massless strings. What is the acceleration of M1?

2. Relevant equations

3. The attempt at a solution

Let T1 be the tension in the first (higher) string, and T2 be the tension in the second string. Then the total force on block M1 (assuming it falls) is $$F_{M_1} = M_1 g - T_1 = M_1 a$$ and the force on block M2 is $$F_{M_2} = T_2 - M_2 g = M_2 a$$. THe blocks should have the same acceleration (one up, one down of course) so the relavent equations may be set equal to each other and solved for $$T_1$$ as a function of $$T_2$$ with no accelerations involved. Unfortunately I get stuck here, as I cannot see the relationship between the tensions and how to eliminate them in the acceleration of the block $$M_1$$. Any help?

#### Attached Files:

• ###### Pulleys.jpg
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2. Jun 17, 2009

### Staff: Mentor

To relate the two tensions, analyze forces on pulley 2.

3. Jun 17, 2009

### Patta1667

Pulley 2 has the force $$T_1$$ upwards, and twice the force $$T_2$$ downwards, correct? This would lead to $$T_1 = 2 T_2$$, which I believe was the first technique I tried and failed with.

4. Jun 17, 2009

### Staff: Mentor

Looks right to me. Show exactly what you did.

5. Jun 17, 2009

### Staff: Mentor

There's the problem: The blocks do not have the same acceleration. Fix your equations for M1 and M2.

6. Jun 17, 2009

### Patta1667

$$M_1a = M_1 g - T_1$$
$$M_2 a = T_2 - M_2 g$$ (a_1 = a_2, so just replaced by a)
$$a = g - \frac{T_1}{M_1} = \frac{T_2}{M_2} - g$$
$$2g = \frac{T_1}{M_1} + \frac{T_2}{M_2} = \frac{2T_2}{M_1} + \frac{T_2}{M_2} = T_2 \left( \frac{2}{M_1} + \frac{1}{M_2} \right)$$

Since $$T_1 = 2T_2$$ we get:

$$T_2 = \frac{2g}{\frac{2}{M_1} + \frac{1}{M_2}} \implies T_1 = \frac{4g}{\frac{2}{M_1} + \frac{1}{M_2}}$$

Substituting this into the equation for $$a = g - \frac{T_1}{M_1}$$, we get:

$$a = g - \frac{4g}{\frac{2}{M_1} + \frac{1}{M_2}} \frac{1}{M_1} = g - \frac{4g M_1 M_2}{2M_2 + M_1} \frac{1}{M_1} = g - \frac{4 g M_2}{2M_2 + M_1}$$

The text's "hint" says let M_1 = M_2, then $$a = \frac{1}{5} g$$. Doing so with my equation,

$$a = g - \frac{4g M_1}{2M_1 + M_1} = g - \frac{4}{3} g = \frac{-1}{3} g$$.

Was my derivation incorrect?

7. Jun 17, 2009

### Staff: Mentor

Your assumption that a_1 = a_2 is incorrect. Hint: When pulley 2 moves up 1m, how far does M2 move up?

8. Jun 17, 2009

### Patta1667

Ah, okay.

$$a_1 = g - \frac{2 t_2}{M_1}$$
$$a_2 = \frac{T_2}{M_2} - g$$

I can eliminate T_2 between the equations, using T_1 = 2 T_2, and this leads to :

$$T_2 = M_2 ( a_2 + g) = \frac{M_1}{2} (g - a_1)$$

Now I can relate a_1 to a_2 simply, but I cannot find a way to absolutely express a_1 without tensions or a_2 in the equation. Am I missing an equation to help me solve this?

Last edited: Jun 17, 2009
9. Jun 17, 2009

### Patta1667

Oops, I means $$2 a_1 = a_2$$. This leads to the right answer, thanks Doc!

10. Jun 17, 2009

### Staff: Mentor

You are missing the constraint equation that relates the two accelerations. (They have a fixed relationship because of how the pulleys are connected.) See my hint in the last post to figure it out.
Almost. (Careful which is which.)

Edit: Ah... I see you got it. Good!