Problem with Newtons Laws of Motion

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Homework Help Overview

The discussion revolves around a problem involving Newton's Laws of Motion, specifically focusing on two blocks with different masses and the forces acting between them. The original poster is trying to determine the minimum coefficient of friction required to prevent slipping between the blocks while considering various forces, including tension and normal reactions.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore different equations of motion for the blocks and question the inclusion of forces from the pulleys. The original poster attempts to analyze the system by considering both blocks and pulleys as a single unit, but encounters difficulties leading to negative values for the coefficient of friction.

Discussion Status

Some participants have provided guidance on considering additional forces acting on the blocks, particularly from the pulleys. The original poster has made progress in understanding the forces involved but still seeks clarification on specific aspects of the equations used.

Contextual Notes

There is an ongoing discussion about the forces acting on the blocks, including the need to account for the upper pulley and its effects. The original poster has expressed confusion regarding the nature of these forces and their origins, which remains a point of exploration in the discussion.

Ashu2912
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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
 

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Ashu2912 said:
FOR 2m BLOCK : T+uN`=2ma and N=N`+2mg
What about the force exerted by the upper pulley?
 
What about it? Will I get the answer with that. I don't recollect any such force. Any help will be appreciated. Thank you!
 
Ashu2912 said:
What about it? Will I get the answer with that.
Why would I bring it up otherwise? :smile:
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!)
 
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...
 
Ashu2912 said:
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?
 
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!
 
Ashu2912 said:
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.
 
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
Ashu2912 said:
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?
 
  • #11
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
Ashu2912 said:
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?
 
  • #13
Two tension forces equal to F in magnitude one in negative x and other in negative y...
 
  • #14
Ashu2912 said:
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.
 
  • #15
Doc Al said:
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...
 
  • #16
Ashu2912 said:
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?
 
  • #17
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
Ashu2912 said:
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.
 

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