Pulley Problem, 1 crate hanging, 1 crate on an incline

[cd80187]

Here is a question that I am not sure where to start.
A block of mass m1 = 6.07 kg on a frictionless plane inclined at angle = 33.3° is connected by a cord over a massless, frictionless pulley to a second block of mass m2 = 2.46 kg hanging vertically (Fig. 5-54). (a) What is the acceleration of the hanging block (choose the positive direction up)? (b) What is the tension in the cord?

So far, I have drawn a free body diagram. I will call the crate on the incline m1, and the crate that is hanging m2. So to start, I have drawn a free body diagram. For m2, there is the weight of the crate (M x g), and the tension in the cord that is suspending the crate. Therefore, the tension should be equal to the weight of the box, correct? Next for m1, I altered the axes to corrsepond with the incline. The weight is still acting straight downwards, while the force that is sliding the crate down the incline is equal to (Weight x sin 33.3 degrees). There is also in the tension acting towards the edge of the incline which is opposite of the force acting down the incline. (?). So I have all of these forces, so to find the acceleration, do I simply add the weight of m1 and m2, and also add the tension of wire from the crate that is dangling, then use the mass of m2 to find the acceleration? And for the second part of the question, I just unsure what to do, if i just add up the tension on the wire from crates m1 and m2? Thank you for the help

 

[Eds]

Tension is less than weight of crate. You have 2 equations involving T and a:
Mg – T = Ma (for crate) &
T – mgsin(33.3°) = ma (mass on slope).
Now all you have to do is solve these 2 simultaneous equations for T & a. Don't you have any notes on this?