# Magnitude of frictional force

1. Oct 25, 2008

### hansel13

1. The problem statement, all variables and given/known data
A 2.5 kg block is initially at rest on a horizontal surface. A horizontal force F of magnitude 6.0 N and a vertical force P are then applied to the block. The coefficients of friction for the block and surface are us = 0.40 and uk = 0.25. Determine the magnitude of the frictional force acting on the block if the magnitude of P is
(a) 8 N
(b) 10 N
(c) 12 N

2. Relevant equations
fk = UkN
N = mg

3. The attempt at a solution

part (a)
N = 2.5kg*9.8m/s2 = 24.5 N

So, fk = 0.25*24.5 N = 6.125

But the answer to part (a) was 6.0, not 6.125... It feels like I'm quite a bit off track. Could someone give me a little step forward on part (a)?

2. Oct 25, 2008

### phyzmatix

You are assuming that the block is actually moving when the horizontal and vertical forces are applied.

Remember, the static frictional force increases as the applied force on your object increases (but in the opposite direction) until it reaches a maximum $$f_{smax}=\mu_s N$$.

In other words, until the applied force reaches a value equal to that of $$f_{smax}$$ there will be no motion AND the magnitude of the static frictional force will be equal to the magnitude of the applied force.

For (a) in your example (taking the negative direction of an imaginary y-axis as positive) when calculating the resultant "normal" force $$N_{res}$$ on your block (y-axis only) we get

$$N_{res} = F_{mg} + P = (2.5kg \times 9.8 m/s^2) \\- \\8N = 16.5N$$

You know the coefficient of static friction is $$\mu_s=0.40$$ and therefore

$$f_{smax} = 16.5N \times 0.40 = 6.6N$$

This is clearly larger than the horizontal force of 6.0N so we can conclude that there is no horizontal motion when a vertical force of 8N is applied (since $$f_{smax}$$ opposes the horizontal force until the horizontal force reaches this maximum value).

Therefore the magnitude of the frictional force at this point must still equal to that of the horizontal force, i.e. frictional force = 6.0N.

Now you should just apply the same principles to the remaining parts of your problem and you should be ok.

Hope that helps

3. Oct 25, 2008

### hansel13

Got it, thanks.

4. Oct 26, 2008

Good stuff!