# Problem with Newtons Laws of Motion

1. Jul 28, 2011

### Ashu2912

1. The problem statement, all variables and given/known data
There are 2 blocks, one of mass m kept over other of mass 2m. Pls. refer to the diagram. The question is : the minimum value of u (coefficient of friction) between the 2 blocks for no slipping is :
(a) F/mg
(b) F/3mg
(c) 2F/3mg
(d) 4F/3mg
My answer is coming as (b) but the answer given is (d). Pls. help.....
Variables I used : N = normal reaction from ground
N = Normal reaction between the blocks
a (>0) = scalar coefficient of acc. of the system in x direction
2. Relevant equations
Newton's laws of motion (2nd and 3rd)
3. The attempt at a solution
Frame of reference: Ground
Chose coordinate axes as shown in the figure.
FOR PULLEY : F-T=0
FOR 2m BLOCK : T+uN=2ma and N=N+2mg
FOR m BLOCK : N=mg and F-uN=ma
Solving these, I get u = F/3mg

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2. Jul 28, 2011

### Staff: Mentor

What about the force exerted by the upper pulley?

3. Jul 28, 2011

### Ashu2912

What about it??? Will I get the answer with that. I don't recollect any such force. Any help will be appreciated. Thank you!!!

4. Jul 28, 2011

### Staff: Mentor

Why would I bring it up otherwise?
You want to consider all horizontal forces on the 2m block. The upper pulley attaches to that block, so its force must be accounted for. (To find that force, analyze forces on the upper pulley.)

Even simpler is to consider the entire bunch of stuff--both blocks and both pulleys--as a single system. That will also give you a second equation. (Do it both ways!)

5. Jul 28, 2011

### Ashu2912

I've already tried by taking both blocks+pulley+strings as a single system, but didn't get the answer. In face the answer I was getting was negative, which is simply not possible. Anyways, here's what I did :
In x dir : F=3ma
In y dir : N=3mg
Replacing a in T+uN=2ma and T=F, I'm getting u=-F/3mg....

6. Jul 28, 2011

### Staff: Mentor

This equation, for forces on the 2m block, is not correct. That's why I flagged it.

You only show two forces acting. What about that force from the upper pulley?

7. Jul 28, 2011

### Ashu2912

Thanks a lot DocAl..... Seriously.....
I replaced a=F/3m in N=mg and F-uN=ma and got the answer. So, my question is done.
However, I still don't understand what I missed in the equation for the 2m block. Can you please tell me now??? Thanks a lot once again!!!!

8. Jul 28, 2011

### Staff: Mentor

OK, I assume you used the second method--considering the entire bunch as a single system?
How about you tell me. There are three things touching the 2m block that exert a horizontal force:
- The 1m block, via friction: This force is μmg, acting to the right
- The bottom pulley, via its attachment: This force is T, acting to the right
- The upper pulley, via its attachment: You tell me

You left out that last force. Figure out what it must be, then correct your equation for the 2m block.

9. Jul 28, 2011

### Ashu2912

OK. Let me give it a try. The pulley is attached to the 2m block. So, is the force - the normal reaction between them?

10. Jul 28, 2011

### Staff: Mentor

You can think of it that way. What force must the attachment exert on the upper pulley?

11. Jul 28, 2011

### Ashu2912

Is it some kind or normal reaction only. I had once heard my teacher mentioning some hinge reaction, but it wasn't in our course. is it that only???

12. Jul 28, 2011

### Staff: Mentor

No need to give it a name. What must it equal? What's the net force on that upper pulley? What other forces act on it?

13. Jul 28, 2011

### Ashu2912

Two tension forces equal to F in magnitude one in negative x and other in negative y....

14. Jul 28, 2011

### Staff: Mentor

Good. Those are the other forces acting on the pulley. So what force must the attachment exert on the upper pulley to give zero net force? Then apply Newton's 3rd law to find the force that the pulley exerts on the 2m block.

15. Jul 29, 2011

### Ashu2912

Since the force F is non-zero, there must be 2 forces of magnitude F in positive x and positive y directions.....

16. Jul 29, 2011

### Staff: Mentor

Yes. Go on. What's that third horizontal force on the 2m block?

17. Jul 29, 2011

### Ashu2912

Is it a force of magnitude F, i.e. F(i cap + j cap) will be the force on the pulley, so F(-i cap - j cap) should be the force on the block due to the upper pulley. Is it correct now?? However, I still don't understand the origin of this force..... as in how is the block exerting force on the pulley. Can you help me out with this please??

18. Jul 31, 2011

### Staff: Mentor

Yes. So rewrite your equation for horizontal forces on the 2m block.
The block and the pulley are attached. The block is supporting the pulley--they exert forces on each other.