Forces in Equilibrium - Tension in Strings

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travi5
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Homework Statement



[PLAIN]http://img62.imageshack.us/img62/2120/derppz.jpg
Need to find T1 + T2I can do it when the two masses are connected to each other, I was just wondering how I'd do it now there is another 5kg in the middle?

Also if the acceleration is already given as 2m/s to the left, do I just plug that into F=ma for both Tensions?

Cheers
 
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Actually, the acceleration of the center block is to the left, the acceleration of the hanging block is down, and the acceleration of the block on the incline is up parallel to the incline. The magnitude of each block's acceleration is the same. You must draw free body diagrams of each block to identify the forces acting on each block, and then use Newton's 2nd law on each block. Is this a homework problem or a take home exam?
 
PhanthomJay said:
Actually, the acceleration of the center block is to the left, the acceleration of the hanging block is down, and the acceleration of the block on the incline is up parallel to the incline. The magnitude of each block's acceleration is the same. You must draw free body diagrams of each block to identify the forces acting on each block, and then use Newton's 2nd law on each block. Is this a homework problem or a take home exam?

A past paper question
It's driving me crazy

If I attempt a and b, and post my workings would you be able to tell me where I'm going wrong?
 
Is this how the diagrams should look?

[PLAIN]http://img805.imageshack.us/img805/4020/diagrams.jpg
 
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The first diagram looks good. In the 2nd diagram, you are missing the friction force (uR) and its direction. In the third diagram, the weight always acts vertically down, not at an angle. This is the hardest of the diagrams...you have to break the weight up into its components parallel and perpendicular to the incline, then find the normal force and friction force ( as a function of the Normal force). Hint: the magnitude of the Normal force, R, in this diagram is equal to the component of the weight perpendicular to the incline.