# Homework Help: Block Interactions

1. Apr 18, 2015

### Jazzman

1. The problem statement, all variables and given/known data

Analyze the motion of the system in an inertial frame of reference fixed to the table on which M1 slides. You can't analyze the system with Newton's Laws in the frame of M1 because M1 will accelerate, and that frame will not be inertial. When you do this analysis, you're also going to need to find constraints relating the accelerations of the various masses. Try to determine these constraints in a way similar to what I did in class with the block on the wedge. It may help to note that the since M2 and M3 are attached by a rope, their motions are related.

2. The attempt at a solution

I am completely lost. I spent pretty much all day working on this problem, to no avail. I know what the final answer should be, but I am having trouble finding how to get there.

Here is what the final answer should be:
m2m3g/(m1m2 + m2m3 + m1m3 + m2m3 + m3m3)

Last edited: Apr 18, 2015
2. Apr 18, 2015

### Orodruin

Staff Emeritus
If you spent most of a day working on the problem you must have had some thoughts. Why don't you share them with us? It will help us identify where you are going wrong and allow us to help you better. (It is also required by the PF homework rules.)

3. Apr 19, 2015

### Jazzman

Thanks for letting me know about the homework rules Orodruin, I was unaware of those.

I am pretty confident that the tension force will cause block 2 to move to the right. As a consequence, block 3 will move downwards, since it is attached to block 2. I found this tension force to be equal to m3g, which makes block 2's acceleration to be equal to m3g/m2.

Since blocks 2 and 3 are attached by a rope, the x position of block 2 and the y position of block 3 should equal a constant. I made that into an equation, and after taking two derivatives of both sides with respect to time, I got an equation relating the two blocks' accelerations, and I learned that block 3's acceleration in the y direction is equal to -m3g/m2.

However, all of this information doesn't really tell me anything about block 1's acceleration, considering the fact that there is no friction.

I drew a free body diagram for block 1 and it looks like the only forces acting on it are the downward forces of block 2 and gravity, and the upward normal force of the ground. These don't affect the y-position of block 1 because the net force is 0. I don't see any forces acting on the block in the x-direction, so I'm inclined to think that block 1's acceleration is 0. However, I know that that is not the correct answer; as noted above I know that the correct value of block 1's acceleration should be nonzero.

Last edited: Apr 19, 2015
4. Apr 19, 2015

### Orodruin

Staff Emeritus
There is no friction, so there is no external force acting on the system in the horizontal direction. What does this tell you about the system's total momentum in the horizontal direction?

5. Apr 19, 2015

### Jazzman

Yes, someone has pointed that out to me. Since there is no external force on the system as a whole, the system should have a net acceleration of 0. Since block 2 is moving in the positive x-direction, block 1 should move in the negative x-direction to balance out this force.

HOWEVER, we haven't learned about systems in our physics class yet. So we aren't supposed to solve the problem in that manner... That's what I'm having trouble with--figuring out how to solve the problem with the basic physics knowledge I've learned in class so far.

6. Apr 19, 2015

### Orodruin

Staff Emeritus
You do not need to consider much of system properties. I assume you have discussed some conservation laws ...

7. Apr 19, 2015

### Jazzman

I only learned about conservation laws from others who are in higher level physics classes. So far in my class, conservation laws weren't even mentioned yet

8. Apr 19, 2015

### Orodruin

Staff Emeritus
Well, you can do it by drawing free body diagram for all of the components, but it is nothing I would recommend ... You will also need some additional constraints on the movements of the blocks.

9. Apr 19, 2015

### Jazzman

That's what other higher level physics students have told me... but free body diagrams are the only tools I have. Maybe that's why I'm having so much trouble with the problem

10. Apr 19, 2015

### Orodruin

Staff Emeritus
So why don't you tell us what you've got for each of the blocks so far? What relations do you have?

11. Apr 19, 2015

### Jazzman

12. Apr 19, 2015

### Orodruin

Staff Emeritus
You are missing several forces, namely the forces that are acting:
1. Horizontally on the mass $m_1$.
2. Horizontally on the mass $m_3$.
You are also missing a constraint based on the length of the string being constant.

Edit: Think back on Newton's third law. For every action, there is an equal and opposite reaction. Do all your force pairs consist of equal, but opposite forces?

13. Apr 19, 2015

### Jazzman

Are you assuming that m1 and m3 are touching? I've always assumed that they weren't.

14. Apr 19, 2015

### Orodruin

Staff Emeritus
If they are not, then the string will not be vertical. You need the contact force in order to accelerate $m_3$ in the x-direction. But that is not the only thing that is missing.

15. Apr 19, 2015

### Jazzman

Hmm ok, thanks for pointing that out!

Would you mind explaining what else is missing? Or do you think I should figure this out on my own

16. Apr 19, 2015

### Orodruin

Staff Emeritus
I think you can figure this out if you only think a bit and ponder on Newton's third law. For each force you draw in your free body diagrams, there needs to be an equal and opposite counter force. For which forces that you drew is this not true?

17. Apr 19, 2015

### Jazzman

All of the forces are accounted for, except for the forces exerted onto the ground and the forces exerted onto the rope. I assumed that these are irrelevant, but maybe they are actually relevant.

I could be missing something though

18. Apr 19, 2015

### Orodruin

Staff Emeritus
The forces in the vertical direction are all irrelevant (except the one acting on $m_3$) as they will be countered by normal forces transmitted to the ground.
The horizontal force on the pulley from the rope is definitely not irrelevant. The pulley is a part of $m_1$!

19. Apr 19, 2015

### SammyS

Staff Emeritus
The rope also exerts force on M1 via the pulley.

20. Apr 19, 2015

### Orodruin

Staff Emeritus
Yes, I have been hinting at this for 20 minutes and wanted him to figure this out for himself ...

21. Apr 19, 2015

### Jazzman

Can you explain why the rope applies a force onto the pulley in that specific direction? I guess I'm really confused about how rope forces work.

22. Apr 19, 2015

### Orodruin

Staff Emeritus
Try drawing a free body diagram for any pulley attached to some ropes. Note that the horizontal force is not the only force acting on the pulley. There is also the vertical force which is the force pair with the tension force acting on $m_3$, but that one is going to be countered by a normal force from the ground.