Friction : BLOCKS AND A PULLEY

1. Jun 21, 2013

judas_priest

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

A block of mass m1 is on top of a block of mass m2. Block 2 is connected by an ideal rope passing through a pulley to a block of unknown mass m3 as shown. The pulley is massless and frictionless. There is friction between block 1 and 2 and between the horizontal surface and block 2. Assume that the coefficient of kinetic friction between block 2 and the surface, μ, is equal to the coefficient of static friction between blocks 1 and 2.

Q1 : The mass of block 3 has been changed such that block 1 and block 2 are moving together with a given acceleration of magnitude a. What is the magnitude and the direction of the force of friction exerted by block 2 on block 1?

Q2: What is the minimum value of m3 for which block 1 will start to move relative to block 2?

2. Relevant equations

3. The attempt at a solution

Force of friction exerted by block 2 on block 1 to keep them both moving without one sleeping over another is $$m3g - μ(m1+m2)g = μm1g$$ Correct me where I'm wrong.

minimum value of m3 for which block 1 will start to move relative to 2 is $$m3g>μm1g$$
Therefore, m3>μm1

Both my answers are wrong. Not able to go any where with this. Someone please explaing, and I'll solve.

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2. Jun 21, 2013

haruspex

Take it in smaller steps. You are given that the acceleration is a. What are the forces acting on block 1? If it's accelerating at rate a, what does that tell you about the frictional force between 1 and 2?

3. Jun 21, 2013

judas_priest

Forces acting on 1 is friction between block 2 and block 1 towards left. It is accelerating with block 2, so friction gets stronger to maintain no slipping condition. The friction, since it is static, cannot be μN.
How do I get the value of friction force so that it moves with block 2?

4. Jun 21, 2013

haruspex

Towards the left on which block? Which way is block 1 accelerating?
Why does the force need toi get stronger to accelerate at a constant rate?
Static/kinetic refers to relative movement of the surfaces in contact. If 1 is not slipping on 2 it's static friction.
You know the mass, you know the acceleration, so what's the force?

5. Jun 22, 2013

judas_priest

I figured out the second part. Could you hint me with the third?

6. Jun 22, 2013

judas_priest

Here's my attempt to the third part:
Since block 1 starts to move relative to block 2, it goes into kinetic friction. therefore:
$m1a = mu*m1g$. Cancelling m1 would give me the minimum acceleration. How do I relate it to m3?

7. Jun 22, 2013

judas_priest

After drawing FBD for block 2 and 3, and substituting the acceleration for in previous post, I et
$$(2*mu*m2+mu*m1)/(1-mu)$$

Is this correct?

8. Jun 22, 2013

haruspex

I get something slightly different, Pls post your working.

9. Jun 22, 2013

judas_priest

For block 3:
$m3g - T = m3a$
For block 2
$T - mu*(m1 + m2)g = m2a$
From block 1, a = mu*g.

Substituted a in equations of block 2 and block 3.

Where am I going wrong?

10. Jun 22, 2013

judas_priest

Need to learn Latex. Sorry for making it confusing.

11. Jun 22, 2013

haruspex

You've left out a force on m2.

12. Jun 22, 2013

judas_priest

Which one? I don't see any more horizontal forces. Is it the reaction force of friction on block 1?

13. Jun 22, 2013

judas_priest

Still don't get the answer. What did you get? How did you do it?

14. Jun 22, 2013

haruspex

Yes. What do you get when you include that?

15. Jun 22, 2013

judas_priest

Okay, got it! Thanks!

16. Sep 21, 2013

dodds

Why do not suppose that 1 "feeling" that 2 moves to the right generates a friction to the left to compensate m3g ?

17. Sep 21, 2013

haruspex

I am unable to decipher your question.

18. Sep 21, 2013

dodds

Instead of asserting that 2 generates a friction to prevent 1 to stay steady, why don't you assert that it's 1 which generates a friction to prevent 2 to move ?

19. Sep 21, 2013

haruspex

We are not concerned (in the last part of the question) with whether 2 moves. The critical point is whether 1 moves with 2 or lags behind. Besides, the friction between 1 and 2 could never prevent 2 from moving.

20. Jan 25, 2015

hjkchorong

Hello. I've been working on this problem as well and I have not been able to get my head around both of these questions. Could you please help me? :)