Why Do Accelerations Differ in These Atwood Machine Configurations?

[ron_jay]

Homework Statement

In the three figures given in the attachment consisting of three atwood machines with, the blocks A, B and C of mass m have accelerations a1, a2 and a3 respectively.F1 and F2 are external forces of magnitude 2mg and mg acting on the first and third diagrams respectively.

How do the accelerations of the block differ and why is it so?

Homework Equations

Basic Newton's Laws of Motion equations.

The Attempt at a Solution

Acceleration of masses in atwood machine is given by:

$$a = (\frac{m_{2}-m_{1}}{m_{2}+m_{1}})g$$

I really don't get it how the acceleration are not the same in each case as the forces add to the same?

 

[Doc Al]

Don't use a "canned" formula for Atwood's machine--that's only good in certain situations. Instead, derive the acceleration yourself using Newton's 2nd law. Hint: In the cases with two masses, analyze each mass separately. Then combine the resulting equations to solve for the acceleration.

 

[ron_jay]

For the second pulley we can straight forward apply the equation and with that we get the acceleration a2 = g/3

Don't use a "canned" formula for Atwood's machine--that's only good in certain situations. Instead, derive the acceleration yourself using Newton's 2nd law. Hint: In the cases with two masses, analyze each mass separately. Then combine the resulting equations to solve for the acceleration.
T 22:29

Yes, you were right. The acceleration have to be calculated individually.

For the first pulley let the tension in the string be T1 which will be equal to the force pulling F1=2mg.

for the block we can write: ma=T-mg or ma=mg or a=g...(this must be right) or a1=g

For the third pulley I can't figure out.

 

[ron_jay]

Lets see...F2=mg,

$$F_{2}+mg-T=ma ...(1)$$
$$T-mg=ma ...(2)$$

Adding both sides, we get

$$mg+mg-mg+T-T=2ma$$
$$mg=2ma$$
$$a=\frac{g}{2}$$
then $$a_{3}=\frac{g}{2}$$

Therefore, the correct option would be no. (b)
a1>a3>a2

 

[Doc Al]

For the third pulley: What forces act on the second mass? Apply Newton!

(Looks like you did it while I was typing. Good!)

 

[ron_jay]

Is this correct?

 

[Doc Al]

Yes. You got it.

 

[ron_jay]

Thank you very much for your help and support!