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