# Atwood's Machine: Understanding Acceleration and Tension

• FScheuer
In summary, the conversation discusses an Atwood machine with masses of m, 2m, and 3m, and the task is to find the acceleration of each mass. The conversation includes a discussion of the forces acting on each mass, including gravity and tension, and the need for free body diagrams and a constraint to solve the problem. Ultimately, an additional equation is provided to solve for the accelerations, resulting in -1/5mg for the first two masses and 3/5mg for the third.
FScheuer

## Homework Statement

There is an Atwood machine (image attached) with 3 masses, m, 2m, and 3m.

## Homework Equations

What is the acceleration of each mass?

## The Attempt at a Solution

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I’m not quite sure where to start here. I know that there is a gravitational force downward on each mass, mg, 2mg, and 3mg respectively. I also know that tension is pulling each of the masses upward. I think I would be able to begin solving the problem if I understood how the force of tension in the string was applied to each mass. Is the force of tension on each mass different?

#### Attachments

• IMG_0719.jpg
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If you can neglect friction and the pulleys are massless, the tension is the same in the entire string. What you need to do is to start making free body diagrams to your heart's delight. You will also need a constraint to keep the string length constant.

FScheuer
Orodruin said:
If you can neglect friction and the pulleys are massless, the tension is the same in the entire string. What you need to do is to start making free body diagrams to your heart's delight. You will also need a constraint to keep the string length constant.
I set it all up and got .5g upward for the acceleration of the first two masses and .5g downward for the acceleration of the third. Does that seem correct?

FScheuer said:
I set it all up and got .5g upward for the acceleration of the first two masses and .5g downward for the acceleration of the third. Does that seem correct?
With those numbers, the string is getting longer at 1g. The first three lengths each get shorter at .5g while the last gets longer at .5g.

FScheuer
haruspex said:
With those numbers, the string is getting longer at 1g. The first three lengths each get shorter at .5g while the last gets longer at .5g.
I set up three equations for the accelerations.

a1 = (mg + T)/m

a2 = (2mg + 2T)/2m

a3 = (3mg + T)/3m

I’m not quite sure how to use the constant string length as a constraint. Before I set a1 equal to a3 but I now realize that they are not necessarily equal.

Last edited by a moderator:
FScheuer said:
I’m not quite sure how to use the constant string length as a constraint.
Write the total string length as a function of the displacement of the pulleys and differentiate it twice.

FScheuer
Orodruin said:
Write the total string length as a function of the displacement of the pulleys and differentiate it twice.
What exactly do you mean by that? Isn’t the total string length constant?

FScheuer said:
What exactly do you mean by that? Isn’t the total string length constant?

Yes. But it can also be expressed in terms of the displacements. This is the entire point. What happens when you differentiate a constant? What do you get out when you differentiate the displacements twice?

Since you have only three equations and four variables (the accelerations and the tension), you need this additional condition to solve the system.

FScheuer
Express the length of string (aside from the parts in contact with the pulley) as ##L=L_1+2L_2+L_3##. Differentiate this twice with respect to time to get a relationship between the accelerations of the three masses.

FScheuer
Orodruin said:
Yes. But it can also be expressed in terms of the displacements. This is the entire point. What happens when you differentiate a constant? What do you get out when you differentiate the displacements twice?

Since you have only three equations and four variables (the accelerations and the tension), you need this additional condition to solve the system.
I added the equation and then solved to get -1/5mg for the first two masses, and 3/5mg for the third. This seems to match up with the answer I found online. Thanks.

FScheuer said:
I added the equation and then solved to get -1/5mg for the first two masses, and 3/5mg for the third. This seems to match up with the answer I found online. Thanks.
It is also the same result as I got (assuming that you define the positive direction to be down for all masses, which agrees with your sign for the gravitational term in post #5, but note that you have defined ##T## to be negative of the tension in the string as positive ##T## in those equations would lead to the tension accelerating the masses downwards too - of course this does not matter for the solution if you only solve for the accelerations, but it will matter if you are interested in the string tension).

## What is an Atwood's Machine?

An Atwood's Machine is a simple device used to demonstrate the relationship between acceleration, tension, and mass in a pulley system. It consists of two masses connected by a string or rope that passes over a pulley.

## How does an Atwood's Machine work?

When one mass is heavier than the other, the system will accelerate in the direction of the heavier mass. This acceleration is caused by the difference in tension between the two sides of the pulley. The heavier mass will experience a greater force of gravity, causing it to accelerate downward, while the lighter mass will accelerate upwards.

## What factors affect the acceleration in an Atwood's Machine?

The acceleration in an Atwood's Machine is affected by the difference in mass between the two masses, the force of gravity, and the amount of friction in the system. The larger the difference in mass, the greater the acceleration will be. The force of gravity also plays a role, as a stronger force will result in a greater acceleration. Additionally, friction in the pulley system will decrease the acceleration.

## What is the relationship between tension and acceleration in an Atwood's Machine?

In an Atwood's Machine, the tension in the string is directly related to the acceleration of the masses. As the masses accelerate, the tension in the string will increase. However, the tension will only be equal to the force of gravity acting on the heavier mass when the system is in equilibrium and not accelerating.

## How does an Atwood's Machine relate to Newton's Second Law of Motion?

An Atwood's Machine is an example of Newton's Second Law of Motion, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In the case of an Atwood's Machine, the net force is the difference between the forces of gravity acting on the two masses, and the acceleration is the resulting motion of the masses.

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