Problem with Newtons Laws of Motion

[Ashu2912]

Homework Statement

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

Homework Equations

Newton's laws of motion (2nd and 3rd)

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

 

[Doc Al]

FOR 2m BLOCK : T+uN`=2ma and N=N`+2mg

What about the force exerted by the upper pulley?

 

[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!

 

[Doc Al]

What about it? Will I get the answer with that.

Why would I bring it up otherwise?

I don't recollect any such force. Any help will be appreciated.

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!)

 

[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...
Thanks for your help, but I'm still not getting the answer...

 

[Doc Al]

Replacing a in T+uN`=2ma

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?

 

[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!

 

[Doc Al]

I replaced a=F/3m in N`=mg and F-uN`=ma and got the answer. So, my question is done.

OK, I assume you used the second method--considering the entire bunch as a single system?

However, I still don't understand what I missed in the equation for the 2m block. Can you please tell me now?

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.

 

[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?

 

[Doc Al]

The pulley is attached to the 2m block. So, is the force - the normal reaction between them?

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

 

[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?

 

[Doc Al]

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?

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?

 

[Ashu2912]

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

 

[Doc Al]

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

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.

 

[Ashu2912]

So what force must the attachment exert on the upper pulley to give zero net force?

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

 

[Doc Al]

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

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

 

[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??

 

[Doc Al]

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??

Yes. So rewrite your equation for horizontal forces on the 2m block.

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??

The block and the pulley are attached. The block is supporting the pulley--they exert forces on each other.