# Can someone tell me how many forces of friction are in here?

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1. Apr 3, 2016

Let's assume that the force of friction is only present between the two blocks and that the pulley is massless. On top of that, force F pulling the two blocks at a constant velocity; as a result, the acceleration = 0m/s2.
May someone tell me how many forces of friction are going on between the "2kg" and "7kg" blocks? I guessed two since the top block is moving left and the bottom box is moving right. Am I right?

And is it true that the force of gravity that applies on the "7kg" block is actually (2kg+7kg)*g = 9kg*g?

Also may you please check if my sum of all forces are corrrect?:
ΣFtop blockx: -T1 + μ*2kg*g = [2kg*0] -> T1 = μ*2kg*g
ΣFtop blocky: Ntop = 2kg*g

ΣFbottom blockx: T2 - μ*9kg*g - T1 + F = [9kg*0] -> T2 = μ*9kg*g + 1 - F
ΣFbottom blocky: Ntop = 9kg*g

note: sorry but I don't know if I should still multiply kg * gravity or just leave it as kg.

2. Apr 3, 2016

### haruspex

Action and reaction are equal and opposite. If the normal force experienced by the 2kg block from the 7kg block is 2kg * g then what is the normal force the 7kg block experiences from the 2kg block?

3. Apr 3, 2016

Hello again :). The normal force the 7kg block experiences from the 2kg block is 2kg * g because of Newton's 3rd law as you mentioned above.

4. Apr 3, 2016

### haruspex

Right, so what is the frictional force on the 7kg block?

5. Apr 3, 2016

So... therefore, "ΣFbottom block y" Nbot should be subtracted to Ntop as it is below where Nbot = 2kg*g ?
ΣFtop blockx: -T1 + μ*2kg*g = [2kg*0] -> T1 = μ*2kg*g
ΣFtop blocky: Ntop = 2kg*g

ΣFbottom blockx: T2 - μ*9kg*g - T1 + F = [9kg*0] -> T2 = μ*9kg*g + 1 - F
ΣFbottom blocky: Ntop - Nbot = 9kg*g

6. Apr 3, 2016

### haruspex

Where does that μ*9kg*g term come from?

7. Apr 3, 2016

The force of friction on the 7kg block should feel μ*2kg*g.

8. Apr 3, 2016

Well I thought that since the 2kg block is on top of the 7kg block, then the force of gravity for the 7kg block would be (2kg+7kg)*g "the sum of both masses times gravity".

9. Apr 3, 2016

### haruspex

Sure, but where will that total weight be felt? You already accepted in post #3 that the normal force at the boundary between the blocks is only 2kg * g.

10. Apr 3, 2016

I figured the "9kg * g" force would be felt from block 7kg to the ground and back to up at it again through the Nbot force.

11. Apr 3, 2016

### haruspex

Sure, but there's no friction there.

12. Apr 3, 2016

Hang on let me double check.

13. Apr 3, 2016

ΣFbottom blockx: T2 - μ*2kg*g - T1 + F = [2kg*0] -> T2 = μ*2kg*g + 1 - F
ΣFbottom blocky: Ntop - Nbot = 7kg*g

I think I've fixed it.

14. Apr 3, 2016

### haruspex

You have T1 and T2. There's only one tension acting on the bottom block (other than F).

15. Apr 3, 2016

So I don't need T2 since there's already a force F pulling on it?

since T2 = F it will just cancel out?

16. Apr 3, 2016

### haruspex

It can't cancel. Either take F as acting directly on the block, or take T2 as acting on the block and set T2=F. It comes to the same thing.

17. Apr 3, 2016

Alright. I just thought that force F was acting on the rope since it's pulling it but thank you, if the bottom equation looks correct. You've been a great help.

ΣFbottom blockx: μ*2kg*g - T1 + F = [2kg*0] -> F = μ*2kg*g + 1 + T1
ΣFbottom blocky: Ntop - Nbot = 7kg*g

18. Apr 3, 2016

### haruspex

Yes, that's right.
If you want to think of F as a force someone is applying to a rope, the tension in the rope is F. The block feels the tension F, but it knows nothing about the person pulling on the other end. When constructing the FBD for a body, it's important to think in terms of the forces that act directly on the body, and not get mixed up with more distant origins of the forces.