How Is Acceleration Calculated in a Pulley System with Unequal Masses?

[dauto]

@dauto There is only one string attaching all the three masses.. I guess I won't be able to solve it on my own but thanks anyway

Don't give up just yet. The second string doesn't have to be a string. It might be a rod or something. The point is that the third mass must be connected to the axis of the pulley

 

[dauto]

I think he's on the right track. You're assuming T1 = T2 which turns out to be true for this problem, but this is not because they are the same string; rather it's because the pulley is assumed to have negligible mass.

Well, duh...

 

[paisiello2]

Well, duh...

So you admit you were wrong then.

 

[dauto]

So you admit you were wrong then.

No I wasn't wrong. I said the two tensions at opposite ends of the string in the problem are identical. That's correct.

 

[paisiello2]

Yeah, but you said it was because it was part of the same string. This isn't the real reason. And in fact the forces are not identical but rather the difference is considered negligible like the mass of the pulley.

 

[Dumbledore211]

So, the two tensions are identical because the third mass is connected to the axis of the pulley which means T1=T2 in this problem. So, we have four equations and we still need another one...

 

[paisiello2]

No, the two tensions are approximately equal because the assumption is (among the plethora of assumptions for this problem) that the pulley has negligible mass and negligible friction when it rotates. If you do a free body diagram of the pulley by itself you can show simply this is the case.

The other one is the standard pulley relationship between the various segments of the string: assume the total length of the string as the sum of it's assumed segment lengths and then take the 2nd derivative to get the relationship between a1, a2, and a3.

 

[Dumbledore211]

@Paisiello I have sent an attachment file and see if the overall free body diagram of the problem is correct. This problem is seriously flying over my head and one of the problems I have struggled with the most.. I still don't get how I can take the separate sums of each segment and establish a derivative relationship among a1, a2 and a3 and finally manage to get a result of one of the accelerations having 17 in the denominator

 

[paisiello2]

Your FBD is not correct.

The problem as interpreted by dauto, and I think he is correct, is that the string starts attached to m goes over the edge of the table, loops through the pulley, goes back up and over the table edge again, and connects back to mass 2m. Then mass 3m is attached to the center of the pulley. You have to visualize it in 3 dimensions.

To get the acceleration relationships you have to do what every pulley problem requires: setup an equation for the total length of the rope assuming some arbitrary lengths of each segment. Then take the 2nd derivative to get the relative accelerations of each segment of the rope.

 

[rl.bhat]

Same tension T is acting on m and 2m. So the displacement in m is greater then 2m.
The lengths of the vertical segments must be the same. Hence at any instant, the total length of the string L = x₁ + x₂ + 2x.Taking derivative, you can find the relation between the accelerations of m, 2m and 3m.
Now apply Newton's law for three masses and solve for T.
Using the value of T, you can find the acceleration of 3m.