Why is the acceleration of M₂ twice that of M₁ in this pulley system?

[Gbox]

Homework Statement

Find the acceleration of the masses

Homework Equations

$F=Ma$

The Attempt at a Solution

We know that the system is moving so for $M_1$ we have $M_1gsin\theta$ on the other hand on $M_2$ we have ##M_1gsin\theta-M_2gsin\theta##

How can I conclude that the acceleration of $M_2$ is twice of $M_1$?[/B]

 

[haruspex]

How can I conclude that the acceleration of M2 is twice of M1?

I assume you meant the other way around.
It follows from the assumption that the string length does not change.

 

[Gbox]

I assume you meant the other way around.
It follows from the assumption that the string length does not change.

Sorry $M_1$ is twice the acceleration as $M_2$.

Yes the string length does not change

 

[haruspex]

Sorry $M_1$ is twice the acceleration as $M_2$.

Yes the string length does not change

Do you see how it follows from that?

 

[Gbox]

Do you see how it follows from that?

No I can't understand the forces map

 

[haruspex]

No I can't understand the forces map

It's not to do with the forces. Consider the lengths of the three sections of strings. They add up to a constant, and two are always the same as each other. That allows you to express two in terms of the third. Then see how they relate to the two accelerations.

 

[Crystal037]

It's not to do with the forces. Consider the lengths of the three sections of strings. They add up to a constant, and two are always the same as each other. That allows you to express two in terms of the third. Then see how they relate to the two accelerations.

Can you show me the free body diagram of the forces acting in the system

 

[haruspex]

Can you show me the free body diagram of the forces acting in the system

This thread was about the algebraic relationship between the two accelerations, which has nothing to do with the forces. It follows entirely from the constancy of the total string length.
If you want to discuss the free body diagrams for this problem, post it as a new thread with your own attempt.