Finding the Frictional Force between Two Blocks

[ubiquinone]

Hi, I 'm just reviewing some forces problem. I have this one question that I was wondering if anyone could please clarify for me. Thank you.

Question: Block $$B$$ is initially at rest on block $$A$$, which is at rest on a smooth frictionless floor. There is friction between block $$A$$ and block $$B$$. The coefficient of static friction between the blocks is $$0.200$$ and the coefficient of kinetic friction between the blocks is $$0.150$$. $$m_A=6.00kg$$ and $$m_B=4.00kg$$. A horizontal force of $$F=23.0N$$ is applied to block $$B$$.
Calculate the magnitude of the frictional force on block $$B$$ showing wheteher $$B$$ slides on $$A$$ or not.

In order to find out whether block $$B$$ would slide on block $$A$$, I need to find out if $$F_{fmax}$$ is less than or greater than the horizontal force $$F$$ applied on block $$B$$
However, how can I find $$F_{fmax}$$? Do I just use the formula $$F_{fmax}=\mu_sm_Ag$$

I think it should be block A because the normal force of mass A is pushing block B up.

 

[Doc Al]

I think it should be block A because the normal force of mass A is pushing block B up.

The normal force is between A and B: Block B pushes down with the same force that Block A pushes up. What must that normal force be, considering that block B does not move vertically?

 

[ubiquinone]

Sorry Doc Al, but I don't understand what you mean. Can you please give me more explanation on how to solve for the normal force. Thanks.

 

[QuantumCrash]

The normal force is actually the reaction force due to block B's weight downwards, NIII law. Try it now.

 

[Doc Al]

The normal force is actually the reaction force due to block B's weight downwards, NIII law.

I know what you mean, but I wouldn't say it quite like that. (Taking the usual definition of weight as the gravitational force that the Earth exerts on an object, the reaction force to weight is the gravitational force that the object exerts on the earth.)

Since block B has no vertical acceleration, the net vertical force on it must be zero. And since only two vertical forces act on block B:

(1) gravity, down
(2) the normal force from A, up​

the normal force on B must equal B's weight.

Normal force is a contact force between two surfaces. The "reaction" force of A pushing up on B is the equal and opposite force of B pushing down on A.

 

[ubiquinone]

Thanks guys, I got it, so first check to see if $$m_B$$ will slide on block $$A$$
$$F_{fmax}=\mu_Sm_Bg=(0.200)(4.00kg)(9.8N/kg)=7.84N$$
Since $$F>F_{fmax}\Rightarrow$$ block $$B$$ will move
so frictional force on B would be kinetic friction $$F_{fk}=\mu_km_Bg=(0.15)(4.00kg)(9.8N/kg)=5.88N$$