Newton's Second law with Friction

[SnakeDoc]

Homework Statement

A 2.60 kg block is initially at rest on a horizontal surface. A horizontal force F^→ of magnitude 7.39 N and a vertical force P^→ are then applied to the block (see the figure). The coefficients of friction for the block and surface are μs = 0.4 and μk = 0.25. Determine the magnitude of the frictional force acting on the block if the magnitude of P^→ is(a)6.00 N and (b)9.00 N. (The upward pull is insufficient to move the block vertically.)

Homework Equations

f=μN
F=MA
N=mg-P

The Attempt at a Solution

So first I wrote the x and y forces separately
F_x=ma_x
ma_x=F^→-ƒ_k

F_y=ma_y
ma_y=-mg+N+P the y acceleration is equal to zero
0=-mg+N+P
N=mg-P

Then I found the maximum possible values of μ_s for both values of P
μ*N=.4*(2.6*9.81-6)=7.8024
μ*N=.4*(2.6*9.81-9)=6.6024

I was able to get b as 4.31 by F_k=μ_k*N
but I'm not sure what to do for part a because the maximum value for P=6N is higher than the F^→ force and I tried doing the same as part b but it says the answer is wrong.

 

[PeroK]

You seem to have done the hard part! In the second case, the block moves and friction is kinetic.

In the first case, the block does not move. What can you say about the forces on a block that isn;t moving?

 

[SnakeDoc]

You seem to have done the hard part! In the second case, the block moves and friction is kinetic.

In the first case, the block does not move. What can you say about the forces on a block that isn;t moving?

That they are in a state of equilibrium?

 

[SnakeDoc]

You seem to have done the hard part! In the second case, the block moves and friction is kinetic.

In the first case, the block does not move. What can you say about the forces on a block that isn;t moving?

I figured it out thank you. I don't know why it took me so long to realize that I had already answered my own question. Since they are in a state of equilibrium then the acceleration is equal to zero so
ma_x=F^→-ƒ so
0=F^→-ƒ
ƒ=F^→ so a) is 7.39

 

[PeroK]

I figured it out thank you. I don't know why it took me so long to realize that I had already answered my own question. Since they are in a state of equilibrium then the acceleration is equal to zero so
ma_x=F^→-ƒ so
0=F^→-ƒ
ƒ=F^→ so a) is 7.39

Yes, you got there on your own.