Simple pulley problem with weights

AI Thread Summary
The problem involves two 0.25 kg blocks in equilibrium connected by a thread over a frictionless pulley, with a third 0.20 kg block added to one of the blocks. The addition of the 0.20 kg block disrupts the equilibrium, prompting a discussion on how to calculate its acceleration. The approach suggested involves applying Newton's second law (F=ma) to both sides of the pulley system to find the forces acting on each mass. The tension in the thread and gravitational forces must be accounted for in the equations to solve for acceleration accurately. The final solution requires careful consideration of the masses and their respective forces.
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Homework Statement


Two 0.25 kg blocks are connected by a thread that passes over a frictionless pulley. The thread has no mass, and no friction. The two blocks are in balance. If a third block that weighs 0.20 kg is put on one of the blocks, then what is the acceleration of the 0.20 kg block?


Homework Equations





The Attempt at a Solution



I was thinking that since the 0.20 kg block just disrupts the equilibrium of the 0.25 kg blocks, then wouldn't the 0.20kg block just accelerate as if it were just dropped, at 9.8 m/s^2?? That seems like too obvious of an answer though...
 
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Something accelerates at 9.8 m/s^2 only if there are no forces on it other than gravity--but that's not the case here, since it's in contact with another block.

How would you figure out the acceleration using Newton's laws?
 
Well.. I could try using F=ma, and then solve for acceleration. But if I were to use this equation, how would I figure out what F was?
 
Why not treat it as a standard pulley problem? Apply F=ma to both sides, then combine the equations. What forces are acting on each mass?
 
Ok, so how about this. If we do T-m1g=ma, and then T-m2g=-ma, then combine both equations and solve for a?
 
djokoman95 said:
Ok, so how about this. If we do T-m1g=ma, and then T-m2g=-ma, then combine both equations and solve for a?
Great! But make sure that the masses in the 'ma' terms have the proper subscripts also. For example: T - m1g = m1a.
 
Oh yes, that's important. Thanks for your help!
 

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