# Friction between 3 blocks conceptual problem

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1. Dec 26, 2014

### PhysicsKid703

1. The problem statement, all variables and given/known data
Find the accelerations of each of the following three blocks:
2kg block, stacked on top of a 3kg block, stacked on top of a 7kg block
When a force of 10N is applied on the 2kg block.
Given:
Coefficient of friction between 2kg and 3kg block is 0.2, coefficient of friction between 3kg and 7kg block is 0.3, and surface between 7kg block and ground is smooth.

2. Relevant equations
None needed or given, taking g=10

3. The attempt at a solution
It's a pretty straightforward question but I suddenly had a conceptual problem. First off, I check the maximum friction between the 2kg and 3kg block (uN), and I get that to be 4N. So obviously the the resultant force on the 2kg block is 6N, and by Newton's second law, acceleration of he 2kg block is 3m/s2.
Next up, the 3kg block. So we know that by Newton's third law, an equal and opposite force ( I.e 4N) will act on this 3kg block because of the friction between it and the 2kg. Upon checking the maximum frictional force between the 3kg and 7kg block, we find that it is a whopping 18N, so there will be no motion, (and another 4N will act on the 7kg block, quite obvious). THIS IS WHERE I HAVE A PROBLEM. According to my textbook, this means that the 3kg block will be at rest "with respect to the 7kg block", and hence acceleration of both 7kg and 3kg block will be the same, i.e 0.4 m/s2. Why do they say that it is at rest only w.r.t to the 7kg block? Since the net force acting on it is zero, shouldn't it just be at rest, and not move at all? And therefore shouldn't the 7kg block move with an acceleration of 4/7 m/s2? I'm thoroughly confused. By their logic, wouldn't that also mean that the 2kg block is accelerating with an acceleration of 3m/s2 w.r.t to 3kg block (not w.r.t the ground)? A detailed explanation would be appreciated, thanks all!

2. Dec 26, 2014

### jbriggs444

Your reasoning for the first block is correct. Your reasoning for the 4 N force by the first block on the second is correct.

The "obvious" argument that the force between second and third block will be 4 N is suspect. What law of motion would allow you to conclude this?

3. Dec 26, 2014

### PhysicsKid703

The force due to the first block on the second block is 4N, so won't the friction between the second and the third try to oppose this, and be able to do so since the max it can give is 18? Won't it self adjust to give a 4N frictional force?

4. Dec 26, 2014

### jbriggs444

The friction between the second and third blocks will self-adjust to oppose what exactly?

5. Dec 26, 2014

### PhysicsKid703

Agh I'm confused. Could you explain the concept? If we have two blocks on top of each other, and we exert a force less than uN on the top block, won't the friction adjust itself the the value of the applied force and hence leave the top block at rest while the bottom accelerates? I find this counterintuitive but it seems right equation wise.

6. Dec 26, 2014

### ehild

The 2 kg block slides on the 3 kg one, so it is kinetic friction between them. Kinetic friction has a definite value, 4 N.
The force of kinetic friction between the 3 kg block and the 7 kg block would be 18 N backwards. The 4 N force exerted by the upper block is not enough to overcome it. So the 3 kg block can not slide on the 7 kg block. They interact with static friction, which is less then 18 N.
The 3 kg block and 7 kg blocks move together as a single body on the frictionless surface. What forces act on that body? What is its acceleration?
If you know the common acceleration both the 3 kg block and the 7 kg block you can find the force acting between them.

Last edited: Dec 26, 2014
7. Dec 26, 2014

### PhysicsKid703

Ohhhkay, so you're saying that
a) 4N kinetic friction on 3kg block isn't enough to overcome the 18N, therefore
b) static friction acts between 7kg and 3kg
c) this static friction is something less than 18N. One last thing, why isn't the static friction exactly equal to the 4N acting in the block, since that it what static friction does- adjust itself upto the max value?

8. Dec 26, 2014

### jbriggs444

Static friction self-adjusts to avoid relative movement between the surfaces that are in contact.

9. Dec 26, 2014

### PhysicsKid703

But on a simple table top, if max friction is, let's say, 100N, and we exert a force of 10N on a body, isn't the force of static friction adjusted to 10N there?

10. Dec 26, 2014

### ehild

It is be true if the body does not move. But those boxes will accelerate with respect to the ground. See the picture attached to my previous post.

11. Dec 26, 2014

### PhysicsKid703

I see. That diagram makes things quite clear.
How will I know, in future questions, whether the boxes will accelerate?

12. Dec 26, 2014

### ehild

Just in the same way. You assume that they accelerate with respect to each other, and if that proves to be false you stick the two boxes together and treat them as a single body.

13. Dec 26, 2014

### PhysicsKid703

Like in this sum? How does it prove false?

14. Dec 26, 2014

### ehild

You considered the forces acting on the middle block, assuming it slides on the 7kg one. There was the 4 N forward force from the interaction with the 2kg block. Sliding friction on the top of the 7 kg block would mean 18 N backward force, what the 4 N force can not overcome. So there is no sliding, the 3 kg block moves together with the 7 kg one.

15. Dec 26, 2014

### PhysicsKid703

Gotcha. Thanks for everything man!

16. Dec 26, 2014

### ehild

You are welcome :)

17. Jun 12, 2015

### Mana_69

How is the max frictional force between 3 Kg and 7 Kg block 18N

18. Jun 12, 2015

### ehild

Good catch, it is 15 N.