Friction : BLOCKS AND A PULLEY

In summary: I'll be waiting for your response.Thank you for posting your attempt. However, it would be better if you could explain your thought process and any equations you used in solving the problem. This will help me understand where you might be going wrong and guide you in the right direction.
  • #1
judas_priest
174
0

Homework Statement



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?

Homework Equations





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 [tex] m3g - μ(m1+m2)g = μm1g[/tex] Correct me where I'm wrong.

minimum value of m3 for which block 1 will start to move relative to 2 is [tex]m3g>μm1g[/tex]
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
judas_priest said:
Force of friction exerted by block 2 on block 1 to keep them both moving without one sleeping over another is [tex] m3g - μ(m1+m2)g = μm1g[/tex]
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
haruspex said:
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?
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
judas_priest said:
Forces acting on 1 is friction between block 2 and block 1 towards left.
Towards the left on which block? Which way is block 1 accelerating?
It is accelerating with block 2, so friction gets stronger to maintain no slipping condition.
Why does the force need toi get stronger to accelerate at a constant rate?
The friction, since it is static, cannot be μN.
Static/kinetic refers to relative movement of the surfaces in contact. If 1 is not slipping on 2 it's static friction.
How do I get the value of friction force so that it moves with block 2?
You know the mass, you know the acceleration, so what's the force?
 
  • #5
I figured out the second part. Could you hint me with the third?
 
  • #6
Here's my attempt to the third part:
Since block 1 starts to move relative to block 2, it goes into kinetic friction. therefore:
[itex] m1a = mu*m1g [/itex]. Cancelling m1 would give me the minimum acceleration. How do I relate it to m3?
 
  • #7
After drawing FBD for block 2 and 3, and substituting the acceleration for in previous post, I et
[tex](2*mu*m2+mu*m1)/(1-mu)[/tex]

Is this correct?
 
  • #8
judas_priest said:
After drawing FBD for block 2 and 3, and substituting the acceleration for in previous post, I et
[tex](2*mu*m2+mu*m1)/(1-mu)[/tex]

Is this correct?

I get something slightly different, Pls post your working.
 
  • #9
For block 3:
[itex]m3g - T = m3a[/itex]
For block 2
[itex] T - mu*(m1 + m2)g = m2a [/itex]
From block 1, a = mu*g.

Substituted a in equations of block 2 and block 3.

Where am I going wrong?
 
  • #10
Need to learn Latex. Sorry for making it confusing.
 
  • #11
judas_priest said:
[itex] T - mu*(m1 + m2)g = m2a [/itex]
You've left out a force on m2.
 
  • #12
Which one? I don't see any more horizontal forces. Is it the reaction force of friction on block 1?
 
  • #13
Still don't get the answer. What did you get? How did you do it?
 
  • #14
judas_priest said:
Which one? I don't see any more horizontal forces. Is it the reaction force of friction on block 1?

Yes. What do you get when you include that?
 
  • #15
Okay, got it! Thanks!
 
  • #16
Why do not suppose that 1 "feeling" that 2 moves to the right generates a friction to the left to compensate m3g ?
 
  • #17
dodds said:
Why do not suppose that 1 "feeling" that 2 moves to the right generates a friction to the left to compensate m3g ?
I am unable to decipher your question.
 
  • #18
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
dodds said:
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 ?
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
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? :)
 
  • #21
hjkchorong said:
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? :)
As for a new thread, please post your attempt.
 
  • #22
haruspex said:
As for a new thread, please post your attempt.
Yes, I have now.
 

FAQ: Friction : BLOCKS AND A PULLEY

What is friction?

Friction is a force that resists motion between two surfaces that are in contact with each other. It is caused by the irregularities on the surfaces which create resistance when they come into contact.

How does friction affect blocks and a pulley?

Friction can affect blocks and a pulley in several ways. It can reduce the efficiency of the pulley system by creating resistance and causing energy loss. It can also cause wear and tear on the surfaces of the blocks, leading to decreased performance.

How can friction be reduced in a block and pulley system?

Friction can be reduced in a block and pulley system by using lubricants on the surfaces of the blocks and pulley, using smoother materials for the surfaces, and minimizing the contact area between the surfaces. Additionally, adjusting the tension on the pulley rope can also help reduce friction.

Can friction be beneficial in a block and pulley system?

Yes, friction can be beneficial in a block and pulley system. It helps to keep the blocks in place and prevents them from slipping or sliding. It also allows the pulley system to grip the rope and lift heavy objects without slipping.

How does the angle of the rope affect friction in a block and pulley system?

The angle of the rope can affect friction in a block and pulley system by increasing or decreasing the amount of contact between the surfaces. As the angle increases, the contact area decreases, resulting in less friction. However, if the angle is too steep, it can cause the rope to slip, leading to increased friction and decreased efficiency.

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