# Newton's Second law with Friction

1. Feb 21, 2015

### SnakeDoc

1. The problem statement, all variables and given/known data

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.)
2. Relevant equations
f=μN
F=MA
N=mg-P

3. The attempt at a solution
So first I wrote the x and y forces separately
Fx=max
max=Fk

Fy=may
may=-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 μs 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 Fkk*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.

2. Feb 21, 2015

### 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?

3. Feb 21, 2015

### SnakeDoc

That they are in a state of equilibrium?

4. Feb 21, 2015

### SnakeDoc

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
max=F-ƒ so
0=F
ƒ=F so a) is 7.39

5. Feb 21, 2015

### PeroK

Yes, you got there on your own.