Solving Tension & Pulley Problems: Acceleration & Tension Explained

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Homework Help Overview

The discussion revolves around a physics problem involving tension and acceleration in a system with two masses connected by a pulley and string. The original poster describes a scenario where mass m1 is on a frictionless table and mass m2 hangs off the edge, with a specific relationship between their masses (m1=2m2). The goal is to determine the acceleration and tension in the string.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate acceleration and tension using Newton's second law, but some participants question the application of tension to the different masses. There is a suggestion to analyze the forces acting on each mass separately to derive the equations needed for solving the problem.

Discussion Status

The discussion is ongoing, with participants providing guidance on how to approach the problem by applying Newton's laws to each mass. There is recognition of the need to clarify the role of tension in relation to both masses, and some participants have expressed confusion about the calculations and relationships involved.

Contextual Notes

Participants are working under the assumption that the pulley and table are frictionless, and there is a specific mass relationship stated (m1=2m2). The discussion includes attempts to reconcile differing interpretations of how tension affects the system.

amcavoy
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I have attached a picture of what I am describing below (Edit: also a link if the attachment didn't go through):

http://img368.imageshack.us/img368/949/mech036fig012zi.jpg

If a mass m1 sits on top of a frictionless table (assume the pulley is as well), attached to a massless string with a mass m2 on the other end hanging off the table, what is the accel. and tension in the string assuming that m1=2m2?

I figured that if it were attached to a wall (rather than m1), the tension would be |T|=m2g. However, since the mass m1 will move, this is what I did:

[tex]\vec{a}=\frac{F}{m}=\frac{m_2g}{m_2+2m_2}=\frac{g}{3}[/tex]

[tex]T=m\vec{a}=\frac{m_2g}{3}[/tex]

Am I correct?

Thanks for the help :smile:

Alex
 

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apmcavoy said:
However, since the mass m1 will move, this is what I did:
[tex]\vec{a}=\frac{F}{m}=\frac{m_2g}{m_2+2m_2}=\frac{g}{3}[/tex]
Good!
[tex]T=m\vec{a}=\frac{m_2g}{3}[/tex]
Careful: [tex]T = m_1 a[/tex]

Just for the exercise, I recommend doing the problem a second way: Apply Newton's 2nd law to each mass separately and combine the two equations. (You'll get the same answer, of course.)
 
The acceleration is correct, but T is the tension, providing acceleration, on m1, not m2.
 
Fermat said:
The acceleration is correct, but T is the tension, providing acceleration, on m1, not m2.
Hmm... Could you explain a bit more why it isn't m2? It seems like the hanging mass (m2) would determine the tension.

Thanks again,

Alex
 
Alright I tried your suggestion Doc Al:

[tex]F_1=\frac{2m_2g}{3}[/tex]

[tex]F_2=\frac{m_2g}{3}[/tex]

[tex]F_1+F_2=m_2g[/tex]

I'm not coming up with m1g like you suggested. Where did I go wrong?

Thank you.

Alex
 
apmcavoy said:
Alright I tried your suggestion Doc Al:
[tex]F_1=\frac{2m_2g}{3}[/tex]
[tex]F_2=\frac{m_2g}{3}[/tex]
[tex]F_1+F_2=m_2g[/tex]
I'm not coming up with m1g like you suggested. Where did I go wrong?
I suggested starting over and looking at each mass separately:
(1) What forces act on mass 1? Write Newton's 2nd law for mass 1.
(1) What forces act on mass 2? Write Newton's 2nd law for mass 2.​
When you do that, you'll get two equations. They will allow you to solve for the acceleration of the masses and the tension in the cord. (And you'll also see why [itex]T = m_1 g[/itex].)
 
Edit: Nevermind, I see what you're saying. The force on the mass m1 is equal to the tension, thus T=m1a. Thanks a lot for the help, it's appreciated.

Alex
 
Last edited:
You only need to consider the forces parallel to the direction of the acceleration:
on [itex]m_1[/itex]: The only horizontal force is the tension in the string
on [itex]m_2[/itex]: There two vertical forces, the tension in the string and the weight
 

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