# Question: Solving for Accelerations in an Atwood Machine System

• Asterast
In summary, the conversation is about solving a physics question involving the accelerations of three masses connected by strings and a pulley. The conversation includes equations and discussions about the proper values for the accelerations of m2 and m3, as well as the tension in the strings and the motion of the pulley. Ultimately, the person seeking help realizes their mistake and asks for clarification on why the acceleration of m2 should be a'-a instead of a-a'.
Asterast
Summary:: I have solved the question but I'm getting answers wrong, some reaction equations seems to have trouble.

Question: Let m1 = 1 kg, m2 = 2 kg and m3 = 3 kg in figure. Find the accelerations of m1, m2 and m3. The string from the upper pulley to m1 is 20 cm when the system is released from rest. How long will it take before m1 strikes the pulley?

here's what I did:

Assuming acceleration of m1 be a
Assuming acceleration of m2 and m3 be a'
Tension in string between mSUB]2[/SUB] and m3 be T
so tension in upper string connected to m1 is 2T
So equations I made for the bodies:

equation for m1
2T-m1g = m1a

2T-g=a ...
equation 1

equation for m2

T-m2g = m2(a'-a)
Since m2 will be going up because of weight of m3 and system of mm2 and m3 will be going downwards.

T-2g=2(a-a') ...
equation 2

equation for m3

m3g-T = m3(a'+a)
since m3 will be going down simultaneously with m2 and m3

3g-T=3(a'+a) ...
equation 3
_________
Eliminating T from equation 1 and 2

3g = -3a+4a' ...
equation 4
_________
Eliminating T from equation 2 and 3

g = 5a+a' ...
equation 5

Solving equation 4 and 5 I get a' = 18g/23

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Asterast said:
Summary:: I have solved the question but I'm getting answers wrong
Solved ?
Assuming acceleration of m2 and m3 be a'
Do you really think they undergo the same acceleration ?

Asterast said:
so tension in upper string connected to m1 is 2T
Your conclusion must then be: lower pulley does not move

BvU said:
Solved ?
Do you really think they undergo the same acceleration ?
I assigned a' for the lower string system since motion of the rate of motion of m2 will be equal to m3
their final accelerations will be a'-a and a'+a for m2 and m3 respectively.

BvU said:
Your conclusion must then be: lower pulley does not move
How come? lower string has tension T and T on each side so upper string will have 2T.
Also check the equations of m2 and m3, their final accelerations are a-a' and a+a'

Asterast said:
I assigned a' for the lower string system since motion of the rate of motion of m2 will be equal to m3
their final accelerations will be a'-a and a'+a for m2 and m3 respectively.
Sounds much better than
Asterast said:
Assuming acceleration of m2 and m3 be a'

BvU said:
Your conclusion must then be: lower pulley does not move

Back to my drawing board

Asterast
Asterast said:
I assigned a' for the lower string system
And a' is different from a ? Still ambiguous

Asterast said:
motion of the rate of motion of m2 will be equal to m3
With respect to what ?

BvU said:
And a' is different from a ? Still ambiguous
Shouldn't it be different because m2 and m3 will be moving wrt to each other because m3>m2 with acceleration a'
The system of m3 and m2 will be moving wrt m1 with acceleration a.

Asterast said:
Shouldn't it be different because m2 and m3 will be moving wrt to each other because m3>m2 with acceleration a'
The system of m3 and m2 will be moving wrt m1 with acceleration a.
And their final accelerations are equated in eq1,eq2 and eq3

Asterast said:
equation for m1
2T-m1g = m1a

2T-g=a ...

equation for m2

T-m2g = m2(a'-a)*****
Since m2 will be going up because of weight of m3 and system of mm2 and m3 will be going downwards.

T-2g=2(a-a') ...*****
The equations with stars contradict to each other,
The first one is correct. The right side of the upper string goes downward, m2 goes up with acceleration a' with respect to the pulley, so its upward acceleration is a'-a.
Check also the equation for m3.

Asterast
ehild said:
The equations with stars contradict to each other,
The first one is correct. The right side of the upper string goes downward, m2 goes up with acceleration a' with respect to the pulley, so its upward acceleration is a'-a.
Check also the equation for m3.
I'm sorry must have been typing mistake. Let me fix it and check the solution I've solved on paper real quick.

ehild said:
The equations with stars contradict to each other,
The first one is correct. The right side of the upper string goes downward, m2 goes up with acceleration a' with respect to the pulley, so its upward acceleration is a'-a.
Check also the equation for m3.
Hello
So the first starred equation was a typo for me. Actually I didn't know that acceleration for m2 should be a'-a, I have been trying to solve the problem by taking the acceleration a-a'; because m2 will be going down even though it is moving upwards hence I took a-a' can you please explain me why it shouldn't be a-a'

Asterast said:
Hello
So the first starred equation was a typo for me. Actually I didn't know that acceleration for m2 should be a'-a, I have been trying to solve the problem by taking the acceleration a-a'; because m2 will be going down even though it is moving upwards hence I took a-a' can you please explain me why it shouldn't be a-a'
Ser picture

It is sure that m1 accelerates upward, with a. The moving pulley accelerates downward with a. m3 accelerates downward with a' with respect to the moving pulley , as it is heavier than m2, and . m2 accelerates upward with a' with respect to the moving pulley. Taking upward the positive direction relative to the ground, acceleration of m2 is a'-a , and that of m3 is -(a'+a).
You can not know beforehand if m2 moves up or down.

Last edited:
Asterast and BvU
ehild said:
Ser picture
View attachment 258338
It is sure that m1 accelerates upward, with a. The moving pulley accelerates downward with a. m3 accelerates downward with a' with respect to the moving pulley , as it is heavier than m2, and . m2 accelerates upward with a' with respect to the moving pulley. Taking upward the positive direction relative to the ground, acceleration of m2 is a'-a , and that of m3 is -(a'+a).
You can not know beforehand if m2 moves up or down.
Isn't m2 actually going downwards overall, even though it's moving upwards wrt m3.
So direction of acceleration of m2 should be downwards??

Asterast said:
Isn't m2 actually going downwards overall, even though it's moving upwards wrt m3.'
So direction of acceleration of m2 should be downwards??
You can not know. and it does not matter. The acceleration of m2 with respect to the ground is a'-a., it goes upward if a'>a and goes downward if a'<a. In this case you can say that m2 accelerates downward with a-a'. You can not decide with logic alone. Solve for a and a' and you will see.
You have to fit the sign of the forces to the direction of acceleration.

Asterast
ehild said:
You can not know. and it does not matter. The acceleration of m2 with respect to the ground is a'-a., it goes upward if a'>a and goes downward if a'<a. In this case you can say that m2 accelerates downward with a-a'. You can not decide with logic alone. Solve for a and a' and you will see.
You have to fit the sign of the forces to the direction of acceleration.
Thank you so much I got it.

## 1. What is an Atwood machine?

An Atwood machine is a simple mechanical device consisting of two masses connected by a string or rope that runs over a pulley. It is used to demonstrate the principles of acceleration and force in physics.

## 2. How does an Atwood machine work?

The two masses in an Atwood machine are typically unequal, causing a net force to be applied to the system. This causes the masses to accelerate towards each other, with the heavier mass moving downward and the lighter mass moving upward.

## 3. What is the equation for an Atwood machine?

The equation for an Atwood machine is F = (m1 - m2)g, where F is the net force applied to the system, m1 and m2 are the masses of the two objects, and g is the acceleration due to gravity (9.8 m/s^2).

## 4. How can I calculate the acceleration of an Atwood machine?

The acceleration of an Atwood machine can be calculated using the equation a = (m1 - m2)g / (m1 + m2), where a is the acceleration of the system, m1 and m2 are the masses of the two objects, and g is the acceleration due to gravity.

## 5. What factors affect the acceleration of an Atwood machine?

The acceleration of an Atwood machine is affected by the difference in masses between the two objects, the length of the string or rope, and the friction between the string and the pulley. It is also affected by external factors such as air resistance and the strength of the gravitational field.

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