Solving Acceleration of M1: Blocks and Pulleys

[Patta1667]

Homework Statement

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?

Homework Equations

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?

 

[Doc Al]

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

 

[Patta1667]

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

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.

 

[Doc Al]

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.

Looks right to me. Show exactly what you did.

 

[Doc Al]

THe blocks should have the same acceleration (one up, one down of course)...

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

 

[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?

 

[Doc Al]

$$M_1a = M_1 g - T_1$$
$$M_2 a = T_2 - M_2 g$$ (a_1 = a_2, so just replaced by a)

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

 

[Patta1667]

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

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?

 

[Patta1667]

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

 

[Doc Al]

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?

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.

[edit] $$a_1 = 2 a_2$$?

Almost. (Careful which is which.)

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