Calculating Acceleration in Pulley System with m_1 = 5.0 kg and m_2 = 4.0 kg

[dekoi]

If $$m_1 = 5.0 \ kg \ and \ m_2 = 4.0 \ kg$$, what is the acceleration of $$m_2$$?

I think the following equations describing the situation are correct:
$$m_1a_1=T_1 - m_1g$$
$$m_2a_2=T_2 - m_2g$$

But I don't know how I can go about solving for $$a_2$$.
Is there a relationship between the two tensions, and the two accelerations?

 

[HappMatt]

well in a pulley system the Tension on m1=m2g I am moslty sure but after my test today I am not sure what i know

 

[DaMastaofFisix]

Okay, take a deep breath. You are off to the right start, what with the Newton equations and all. Seems like you have a simple atwood pulley, where you have two hanging masses around a frictionless, massless pulley.

First thing you got to do is hit up the free body diagram. By correctly identifying that each mass experiences two forces, its weight and the corresponding tension of the rope, you can set them into their respective net force equations.

Here's where it gets tricky and easy, both at the same time. As the system moves (ya know, the masses move around the pulley) they seem to share something in common, and its not the tension. seems like both have the same sort of "movement" of you will. IN fact, they share a common acceleration! Not only that, but in this simple special case, they do indeed share a common tension as well! So whatyou have is two equations with two unknowns; a pretty simple mathematical procedure.

Here's the catch though. In your equations, you have the force of gravity to be negative for both of the masses. While gravity indeeds points downward, in this case, you must DEFINE a direction of rotation (clockwise or CC, it's totally up to you) as either positive or negative. By doing this, your free body diagrams will correctly label what is a "positve" force and a "negative" force.

I know this was a lot to handle, but try to rework it out and tell us what your final result for the acceleration of the system was. Good luck!