Simple Pulley Problem: Finding Tension and Acceleration for Two Connected Blocks

AI Thread Summary
The discussion revolves around solving a physics problem involving two connected blocks with masses of 2kg and 3kg over a frictionless pulley. Participants emphasize the importance of understanding that tension is uniform throughout the string and that both masses share the same acceleration due to the inextensibility of the cord. Two methods for calculating tension and acceleration are presented: a quick approach using resultant forces and a more rigorous method applying Newton's second law separately to each mass. The second method is recommended for its reliability in various scenarios, while the first is noted for its simplicity in straightforward cases. Ultimately, both methods yield the same results, reinforcing the principles of dynamics in connected systems.
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Homework Statement


Two blocks with masses m = 2kg and M = 3kg are connected by a cord that passes over a
massless and frictionless pulley. The system is released from rest. Find the tension T in the cord and acceleration of the masses.


Homework Equations


F = m*a
Q = m*g


The Attempt at a Solution


I actually made a sketch of a system but I am not sure, If equations are correct. Could you check and show me the correct way to handle this simple problem? Thanks.

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If it's a light string and a frictionless pulley, then T1 = T2 = T
The tension is the same throughout the string.
You need to apply Resultant force = mass x acceleration for the two masses separately.
Both masses also have the same acceleration magnitude (=a) as the string is not extensible. (I guess!)
For the left mass, the resultant force is T-Mg
The mass is M and the acceleration is a
Do the same for the right mass.
 
Thanks for your answer!
Is it this way then:

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There are 2 ways of looking at this. The simple (quick) way, and the longer, more rigorous way.

The quick way will work here as it is a simple problem.
In this case, you can say that the resultant force is Q1 - Q2(Q1 is greater than Q2 and the 3kg mass will accelerate downwards, and the 2kg upwards.)
This resultant force accelerates the combined mass M+m
So your equation becomes
Q1-Q2 = (M+m)a

The more rigorous method is to apply F=ma separately to the two masses, as I said in my first post. (F is resultant force)

On the left
Q1 -T1 = Ma
On the right
T2-Q2 = ma

If you combine these two equations to eliminate T, you will get the same result as in the first method.
This 2nd method always works. The 1st method only works in simple cases such as this.
I recommend always approaching such problems using the 2nd method.
 
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