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

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Homework Help Overview

The problem involves two blocks with masses connected by a cord over a massless and frictionless pulley. The original poster seeks to find the tension in the cord and the acceleration of the masses after releasing the system from rest.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the equality of tension in the cord and the relationship between the masses and their acceleration. Some suggest applying the resultant force equations for each mass separately, while others mention different methods of analysis, including a simpler approach versus a more rigorous one.

Discussion Status

There is ongoing exploration of different methods to analyze the problem. Some participants have provided guidance on how to set up the equations, while others are questioning the assumptions made about the forces acting on each mass.

Contextual Notes

Participants are considering the implications of the system being released from rest and the assumptions regarding the massless string and frictionless pulley. There is also a mention of the need to clarify the resultant forces acting on each mass.

travikk
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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.
 

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