# Homework Help: Atwood Machine Proof.

1. Sep 24, 2010

### vandersmissen

1. The problem statement, all variables and given/known data
It appears that the subscript is not working properly, please take m1 to means mass 1 and m2 to mean mass2

Atwood's machine consists of two masses connected by a string that passes over a pulley, as show in the figure. Consider the pulley to be massless and frictionless. Show that, if released for rest, m2 takes a time t=$$\sqrt{}2h(m2+m1)/g(m2+m1)$$ to reach the floor.
[PLAIN]http://rawrspace.com/atwood.jpg [Broken]

2. Relevant equations
I believe,
T-m1g=m1a
m2g-T=m2a

3. The attempt at a solution
So from the relevant equations I solve for T and set them equal. I get
m1a+m1g=m2g-m2a

I solved the equation so that the a's were on one side and the g's were on the other and factored.

a(m2+m1)=g(m2-m1)

Then I divided both sides by (m2-m1)

a(m2+m1)/(m2-m1)=g

Now I am kind of close I think but I am not sure where to go. I know that acceleration (a) is the distance travelled over time squared. So would I replace a with h/t2 and then solve the equation to get

t=$$\sqrt{}h(m2+m1)/g(m2-m1)$$

That is where I have hit a brick wall because they have it as 2h , I know that both masses move h distance, how did that get incorporated in however ? I also know that there is an equation 1/2(g)t2 that may be the way it was introduced, but I do not know how to relate them. Any help would be greatly appreciated.

Last edited by a moderator: May 4, 2017
2. Sep 24, 2010

### vela

Staff Emeritus
If you solve for the acceleration, you get a=g(m2-m1)/(m2+m1), which is constant. That means you can apply the equations you have for constant acceleration.

3. Sep 24, 2010

### vandersmissen

Alright so
We know that for mass 1 , T-m1g=m1a and for mass 2 , m2g-T=m2a
The reason I set this up is because I was trying to make sure the force is positive and since m1 is moving upward, T is greater and on the otherside, since m2 is moving downward, it is greater than T , is that the correct reasoning for how I came up with the formulas above ?

Once I have those formulas I rewrote it to

a(m2+m1)=g(m2-m1)

then solving for a you get the following
for a=g(m2-m1)/(m2+m1)

then using
x=x_0+v_0t+(1/2)at^2

you plugin and get
h=(1/2)[g(m2-m1)/(m2+m1)]t2

and finally rewriting you get
t=$$\sqrt{}2h(m2+m1)/g(m2+m1)$$

Is that the correct reasoning ? I just want to make sure I understand the process.

4. Sep 24, 2010

### vela

Staff Emeritus
Yup, that's correct. Good work.

5. Sep 24, 2010

### vandersmissen

Thank you very much for the help.

6. Sep 24, 2010

### vandersmissen

Let M be the mass of the pulley in the previous question. If the effect of friction on the pulley is considered , how long does it take m2 to reach the floor? The mass is again released from rest and height h.

To solve this would I follow the same general format as the last except now in this case I will have a frictional force acting on m2 since it is pulling downward. Or is this where something like torque would come in.

I believe for the equations I would have

T=m1a+m1g and for mass 2 , T=m2g-m2a-f

Then setting them equal and solving I get

(m2-m1)g-f=(m2+m1)a

Solving for a I get
a=[(m2-m1)g-f]/(m2+m1)

Plugging into the constant acceleration equation, I get

t=$$\sqrt{}2h(m2+m1)/g(m2-m1)-f$$

Does that seem correct ? I think it is because it is taking the external forces and subtracting frictional force from it, thus reducing the force that is doing the work.

Last edited: Sep 24, 2010