Solving Pulley on a Pulley Problem using Force Analysis

[pmr]

I'm currently reading Kleppner and Kolenkow's "Introduction to Mechanics", and I'm working through the problem sets.

I'm only on chapter 2, but I happen to know Lagrangian mechanics. When I get stuck on a problem I occasionally take the Lagrangian approach to find the solution, and then I try to reverse-engineer the solution using old-style force diagrams and force analysis.

I'm currently looking at this problem here:
https://dl.dropboxusercontent.com/u/1818234/pulley-on-pulley/lagrangian-solution.pdf

The PDF shows problem from the textbook, and also Lagrangian solution that I was able to work out. However, I'm 100% baffled as to how one would obtain that solution using only force diagrams and force analysis. This is a chapter 2 problem, so the authors presumably expect you to solve it using more primitive methods, but I'm completely stuck.

Could anyone perhaps shed some light on how one might approach this without Lagrangian mechanics?

 

[Orodruin]

This should not be worse than drawing some free body diagrams and noting that the free pulley is massless (thus the tension in the top string must be twice that in the lower).

 

[pmr]

That the one tension would be twice the other makes sense. But having the relative tensions doesn't produce an answer. To find an absolute acceleration we need an absolute tension, which we don't yet have.

 

[Orodruin]

Did you use the geometrical constraints to relate the accelerations?

 

[pmr]

Yep, it's all in the PDF I linked to. The relation between the accelerations is on the last line of the first page; the acceleration of the second block is twice that of the first (and in the opposite direction).

 

[haruspex]

Yep, it's all in the PDF I linked to. The relation between the accelerations is on the last line of the first page; the acceleration of the second block is twice that of the first (and in the opposite direction).

Right, so if you let the lower tension be T and acceleration of M1 be a then you can write two equations with those two unknowns.

 

[pmr]

I take back my previous statement. The problem is indeed solvable once you know that one tension is twice the other. After doing some algebra I see that we can find the accelerations based on only just the ratio between the two tensions.

However, I now have a different confusion. It no longer makes sense to me that one tension is twice the other. It makes perfect sense that the net acceleration of the second mass is twice that of the first, but I don't see how we go from this simple statement about accelerations to any a claim about tensions.

I know that the tensions are indeed in the same ratio as the accelerations, because my Lagrangian approach tells me so. Yet, I can't figure out how I would explain that using simple intuition or logic. I want to solve this problem in a way that uses only the means provided in chapter 2 of this book, and I currently can't see how to do that.

Just for the sake of argument, how would I explain to a classroom of fresh students that relating the accelerations allows you to likewise relate the tensions?

Should I open up a new post of this more general question?

 

[ehild]

Draw the free-body diagram also for the moving pulley. If it is massless, the net force is zero, otherwise you would get infinite acceleration.

ehild

 

[haruspex]

I know that the tensions are indeed in the same ratio as the accelerations, because my Lagrangian approach tells me so. Yet, I can't figure out how I would explain that using simple intuition or logic.

In general, it does not follow that the ratios are the same. If the pulleys had mass, there would be a different relationship between the tensions, but the accelerations would still be 2:1. Similarly, a different arrangement might keep the tensions 2:1 but not the accelerations. E.g., instead of tying the lower rope to the floor, tie both of its ends to M2.
In the arrangement as given, the ratios are the same (or, more correctly, inverses of each other) because of the principles of mechanical advantage. In a small movement, M2 travels twice as far as M1, exerting effectively half the force on the pulley (the other half coming from the tied end).
To see that it is true from the basic laws of mechanics, consider the free body diagram for the lower pulley. It has no mass, therefore no net force on it (or the acceleration would be infinite).

 

[Orodruin]

drawing some free body diagrams and noting that the free pulley is massless (thus the tension in the top string must be twice that in the lower).

As stated ... The mass of the lower pulley would hardly be of interest for any free body diagram than for that of they pulley itself ;)

All being said, I do prefer the Lagrangian approach here.