Newton's laws of motion -- masses on pulleys...

  • Thread starter AbhinavJ
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  • #1
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Homework Statement


Wasnt able to upload file, therefore an external link

https://drive.google.com/file/d/0B0FIIKckKOcKWFcxcmtJRVRXZTA/view?usp=drivesdk

A force F= mg is applied on C, all pullies are frictionless and we need to find the reading of the spring balance, basically the tension in it.

Homework Equations


F=ma

The Attempt at a Solution


I assumed the pulley and block A are moving downwards with an acceleration a1, then used pseudo forces to calculate acceleration of B and C which comes out to be
g/2. Then I used the equation for A Tspring
- 2mg=2ma1.
I need another equation relating tension in spring to that in wire connecting B and C.
2Twire=Tspring?
Or i can consider the mass of pulley as that of the blocks combined which gives me 2Twire - Tspring=2ma1? The pulley is massless, though.
 
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Answers and Replies

  • #2
haruspex
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2Twire=Tspring?
i
Yes. The free body diagram for the pulley only has the massless pulley, the connection to the spring, and the wire running over it. It does not "know" anything about the blocks below.
 
  • #3
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i
Yes. The free body diagram for the pulley only has the massless pulley, the connection to the spring, and the wire running over it. It does not "know" anything about the blocks below.
But isnt the pulley accelerating downwards due to the tension in the wire being more than that in spring?
 
  • #4
haruspex
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But isnt the pulley accelerating downwards due to the tension in the wire being more than that in spring?
You are asked to find the reading on the spring,which implies it is a constant reading. That means the spring has reached a constant extension. If the spring balance has negligible mass, what does that tell you about the relationship between the tension above the spring and the tension below it?
 
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  • #5
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You are asked to find the reading on the spring,which implies it is a constant reading. That means the spring has reached a constant extension.
Got it, thanks :D
 
  • #6
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Is the reading in spring balance = (5/2)mg ?
 
  • #7
haruspex
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Is the reading in spring balance = (5/2)mg ?
From memory, that's not what I got, but it is in the ballpark.
Please post your working.
 
  • #8
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From memory, that's not what I got, but it is in the ballpark.
Please post your working.
In that case , I think I have misunderstood the setup .
A force F= mg is applied on C
I took the above statement as if a constant force F=mg is acting on block C .

If no external force acts on any block and the system is initially at rest , then none of the blocks should move .

@AbhinavJ , could you post the complete problem statement ?
 
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  • #9
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Is the reading in spring balance = (5/2)mg ?
Actually that is the answer, the spring balance shows a reading of 5 kg and and m=2kg, thereby 5mg/2 is correct.
 
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  • #10
haruspex
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From memory, that's not what I got, but it is in the ballpark.
Please post your working.
Edit:
I found my previous scribbles and confirm your answer. Sorry, I must have been thinking of a different pulley problem I replied to recently.... the answer there had 8/3 in it.
 

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