Atwood's Machine: Solving for Force & Acceleration

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In summary, the conversation discusses a physics problem involving a movable and fixed pulley. The question is about the displacement of m2 and the ratio between a1 and a2. It is determined that the displacement of m2 should be 2 cm and the correct answer is option C, with a1=1cm/t^2 and a2=2cm/t^2. The ratio is determined by the fact that the length of the whole rope is constant.
  • #1
KevinFan
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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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  • #2
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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  • #3
ehild said:
No. The ratio is determined by fact that the length of the whole rope is constant.
Could you elaborate a bit?
 
  • #4
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?
 
  • #5
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
 
  • #6
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
 
  • #7
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
 
  • #8
Yes :smile:
 
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  • #9
ehild said:
Yes :smile:
Thank you very much for your explanation
 

Related to Atwood's Machine: Solving for Force & Acceleration

What is an Atwood's Machine?

An Atwood's Machine is a simple mechanical device consisting of two masses connected by a string or rope that passes over a pulley. It is used to demonstrate the principles of force, mass, and acceleration in physics.

How do you solve for force and acceleration in an Atwood's Machine?

To solve for force and acceleration in an Atwood's Machine, you can use Newton's second law of motion, which states that the sum of all forces acting on an object is equal to the mass of the object multiplied by its acceleration. You can also use the formula F = ma, where F represents force, m represents mass, and a represents acceleration.

What are the variables involved in solving for force and acceleration in an Atwood's Machine?

The variables involved in solving for force and acceleration in an Atwood's Machine are the masses of the two objects, the acceleration of the system, and the force of gravity acting on the objects. Other factors such as friction and air resistance may also need to be taken into account.

What is the difference between the force of tension and the force of gravity in an Atwood's Machine?

The force of tension is the force transmitted through a string or rope, while the force of gravity is the force of attraction between two objects with mass. In an Atwood's Machine, the force of tension is equal and opposite on each side of the pulley, while the force of gravity is exerted by each object on the other.

How does changing the mass ratio affect the force and acceleration in an Atwood's Machine?

The mass ratio, or the ratio of the two masses in an Atwood's Machine, directly affects the force and acceleration of the system. As the mass ratio increases, the force of tension increases, resulting in a greater acceleration. Conversely, a lower mass ratio will result in a lower force of tension and a slower acceleration.

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