Atwood's Machine: Solving for Force & Acceleration

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Homework Help Overview

The discussion revolves around an Atwood's machine problem, focusing on the relationship between forces and accelerations in a system involving pulleys. Participants are analyzing the mechanics of the system as depicted in an image linked in the thread.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the implications of pulley types on force and acceleration ratios. There is a debate regarding the displacement of masses and how it relates to the lengths of the rope segments. Questions arise about the mechanics of the system and the reasoning behind the displacement calculations.

Discussion Status

Some participants have provided insights that challenge initial assumptions about the displacement of the masses. There is an ongoing exploration of the relationships between the forces and accelerations, with some participants expressing agreement on the revised understanding of the displacement ratios.

Contextual Notes

Participants are working within the constraints of the problem as presented in the image, and there is a focus on the constant length of the rope, which influences their reasoning. The discussion reflects a mix of interpretations and clarifications regarding the mechanics involved.

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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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Last edited by a moderator:
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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