# What is the Force of Kinetic Friction?

1. Oct 19, 2016

### JustynSC

1. The problem statement, all variables and given/known data
a 10 kg box is moving on a level floor. The coefficient of static friction between the box and floor is 0.3. What is the force of kinetic friction?

2. Relevant equations
fk = FN * μk

3. The attempt at a solution
I rearranged the formula so that everything was set equal to the normal force, but I'm convinced that there is information missing to solving this.

2. Oct 19, 2016

### Staff: Mentor

Looks to me that everything you need is in the problem statement. Use the formula that you quoted to calculate the friction force. What is the normal force in this situation?

3. Oct 19, 2016

### JustynSC

There is no normal force given. I have no knowledge on how many Newtons are being acted on the 10kg box. Is it required to have an FN given to solve that equation? Otherwise I only have one known variable, the 0.3 constant for static friction. I solved that the FN required to overcome static friction must be: FN>49N. But that is a separate problem that does not state that same force was used in this question. Besides if the the box were pushed at 98N would that change the Fk?

4. Oct 19, 2016

### Staff: Mentor

The normal force is easy to calculate from the given information. Hint: What forces act on the box?

5. Oct 19, 2016

### JustynSC

At rest the box is acted on by Earth's gravity (9.8 m/s2) multiplied by the mass of the box (10kg)and lastly the static friction constant between the two surfaces (0.3) . So the Normal force is in the perpendicular to the surface the box is on. FN= 49 Newtons. But that is fs max, a threshold that once overcome will cause the box to move. So an additional force pushing parallel to the horizontal surface could be as little as 49.00000000001N or les as long as it is greater than 49. Otherwise the box will remain at rest?

6. Oct 19, 2016

### Staff: Mentor

Right! That's a downward force acting on the box. The normal force is the upward force from the surface that supports the box. What must that normal force equal?

(Forget all the other stuff for the moment. Just get this problem solved.)

7. Oct 19, 2016

### JustynSC

(FN=ma=10kg * 9.8m/s2)
So then, FN =98N ?

P.S I am taking physics in college, but have had to teach myself because the professor does not do a great job at explain the reasoning behind the concepts. And his English is pretty limited, so there is a severe language barrier. So, sorry if I am missing something big here, or something obvious for that matter.

8. Oct 19, 2016

### Staff: Mentor

You got it.

The normal force must equal the weight of the box (otherwise the box would have a vertical acceleration). Thus FN = mg.

Now just apply the formula you already know for kinetic friction and you're done.

9. Oct 19, 2016

### JustynSC

Okay just to check my answer:
fk=FN * μk
Therefore: fk=98 N * 0.3μ

10. Oct 19, 2016

### Staff: Mentor

Looks good to me!

11. Oct 19, 2016

### JustynSC

Awesome. Thank you for working that through without just telling me the answer. I think I actually understand the concept of how friction effects an object now.

12. Oct 19, 2016

### Staff: Mentor

Excellent. That's what we try to do here.