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

[Blockade]

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?:
ΣF_{top blockx}: -T₁ + μ*2kg*g = [2kg*0] -> T₁ = μ*2kg*g
ΣF_{top blocky}: N_top = 2kg*g

ΣF_{bottom blockx}: T₂ - μ*9kg*g - T₁ + F = [9kg*0] -> T₂ = μ*9kg*g + ₁ - F
ΣF_{bottom blocky}: N_top = 9kg*g

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

 

[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?

 

[Blockade]

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?

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.

 

[haruspex]

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.

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

 

[Blockade]

So... therefore, "ΣF_{bottom block y}" N_bot should be subtracted to N_top as it is below where N_bot = 2kg*g ?
ΣFtop blockx: -T1 + μ*2kg*g = [2kg*0] -> T1 = μ*2kg*g
ΣFtop blocky: N_top = 2kg*g

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

 

[haruspex]

ΣFbottom blockx: T2 - μ*9kg*g ...

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

 

[Blockade]

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

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

 

[Blockade]

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

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".

 

[haruspex]

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".

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.

 

[Blockade]

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.

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

 

[haruspex]

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

Sure, but there's no friction there.

 

[Blockade]

Sure, but there's no friction there.

Hang on let me double check.

 

[Blockade]

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.

 

[haruspex]

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.

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

 

[Blockade]

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

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

since T2 = F it will just cancel out?

 

[haruspex]

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

since T2 = F it will just cancel out?

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.

 

[Blockade]

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.

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

 

[haruspex]

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

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.