Atwood's Machine: Solving for Force & Acceleration

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The discussion centers on solving a physics problem involving a movable pulley and its impact on force and acceleration. The initial assumption was that the acceleration of one mass (a1) would be half that of the other (a2) due to the force required by the movable pulley. However, it was clarified that the length of the rope remains constant, leading to the conclusion that if one mass moves up by 1 cm, the other mass moves up by 2 cm. This results in the correct relationship being established as a1 equals half of a2, confirming that option C is the correct answer. The explanation helped solidify the understanding of the mechanics involved in the pulley system.
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Homework Statement


http://[url%3Dhttps://postimg.org/image/pov8gzkr5/][ATTACH=full]200173[/ATTACH]

[ATTACH=full]200174[/ATTACH]
The question is shown above in the image.
[h2]Homework Equations[/h2]
F=ma

[h2]The Attempt at a Solution[/h2]
I personally think the answer should be option B. The movable pulley requires 1/2 of the force compare with the fixed pulley, therefore the a1 is 2 times a2. Can anyone confirm my logics please?
 

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KevinFan said:

Homework Statement


http://[url%3Dhttps://postimg.org/image/pov8gzkr5/][ATTACH=full]200175[/ATTACH]

[ATTACH=full]200176[/ATTACH]
The question is shown above in the image.
[h2]Homework Equations[/h2]
F=ma

[h2]The Attempt at a Solution[/h2]
I personally think the answer should be option B. The movable pulley requires 1/2 of the force compare with the fixed pulley, therefore the a1 is 2 times a2. Can anyone confirm my logics please?[/QUOTE]
No. The ratio is determined by fact that the length of the whole rope is constant.
 

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ehild said:
No. The ratio is determined by fact that the length of the whole rope is constant.
Could you elaborate a bit?
 
Imagine that m1 raises by 1 cm. How much does the piece of rope keeping the left pulley get shorter? The length of the whole rope does not change. So the right piece gets longer. What is the diplacement of m2?
 
ehild said:
Imagine that m1 raises by 1 cm. How much does the piece of rope keeping the left pulley get shorter? The length of the whole rope does not change. So the right piece gets longer. What is the diplacement of m2?
I think the displacement of m2 would be 1cm as well
 
Why? The right piece of string should be longer by the same as the left piece gets shorter. But the left piece consists of two pieces, and both of them gets shorter by 1 cm.

upload_2016-12-15_13-20-42.png
 
ehild said:
Why? The right piece of string should be longer by the same as the left piece gets shorter. But the left piece consists of two pieces, and both of them gets shorter by 1 cm.

View attachment 110436
oh... so the displacement of m2 should be 2 cm and a1=1cm/t^2, a2=2cm/t^2. abs(a1)=abs(1/2(a2))
Option C is the correct answer
 
Yes :smile:
 
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ehild said:
Yes :smile:
Thank you very much for your explanation
 
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