Calculating frictional force on a block with different upward forces

[rockchalk1312]

A 2.50 kg block is initially at rest on a horizontal surface. A horizontal force F of magnitude 6.71 N and a vertical force P are then applied to the block. 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)7.00 N and (b)10.0 N. (The upward pull is insufficient to move the block vertically.)

relevant equations:
F=ma
Fk=μkFn
Fs,max=μsFnattempt at solution:
(2.50)(9.8)=24.5-7=17.5 net weight including upward pull of 7N
(.4)(17.5)=7N so force 6.71N would not move block
(.25)(17.5)=4.375N force of 6.71N would move block
6.71N-4.375N=2.335N

On my online homework it said this was the wrong answer.

Am I confusing coefficients I'm supposed to use at certain times? Thank you!

 

[haruspex]

(.4)(17.5)=7N so force 6.71N would not move block
(.25)(17.5)=4.375N force of 6.71N would move block

You've already established that the force will not move the block, so you do not have to consider dynamic friction.

6.71N-4.375N=2.335N

Even if it were appropriate to consider dynamic friction, what are you calculating there? It looks like it's the excess force over dynamic friction if the block were to move. That's not what the question asks for.
Remember that normal force * coefficient of static friction tells you the maximum static frictional force. If the force applied, F, is insufficient, what is the actual frictional force?

 

[rockchalk1312]

Remember that normal force * coefficient of static friction tells you the maximum static frictional force. If the force applied, F, is insufficient, what is the actual frictional force?

Wow I'm an idiot. Ok your emphasis on the word maximum helped me a lot. Just 6.71N. Thank you very much!