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

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SUMMARY

The discussion focuses on solving a simple pulley problem involving two connected blocks with masses of 2kg and 3kg. The key equations used are F = m*a and Q = m*g, where Q1 and Q2 represent the gravitational forces acting on the masses. Two methods for finding the tension (T) in the cord and the acceleration (a) of the masses are presented: a quick method using resultant forces and a more rigorous method applying Newton's second law separately to each mass. The second method is recommended for consistent results in similar problems.

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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.

24waglf.jpg
 
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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:

906rsy.jpg
 
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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