# How can i find the coefficient of kinetic friction?

1. Feb 18, 2015

### ciara0682

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

I know incline is 16 degrees
I know distance traveled is 2.85 meters
I know Vf is 2.71 meters/second

I calculated the meu static and got .287
2. Relevant equations

3. The attempt at a solution
Tried every thing so lost

2. Feb 18, 2015

### Staff: Mentor

Welcome to the PF.

Please fill out the Template a bit more. What is the exact problem statement? What are the relevant equations? Please show us your work where you say you got the static μ to be 0.287

3. Feb 18, 2015

### ciara0682

A flatbed truck slowly tilts its bed upwards to dispose of a crate. When the tilt angle is 16 degrees or above the create begins to slide. When the create reaches the bottom of the flatbed, it has slide 2.85 meters, its speed is 2.71 m/s.
I found meu static by summing the forces in both the x and y direction which all and all gave me the formula of meu static = tan of the angle. Tan(16)=.287 now I'm stuck on how to calculate meu kinetic because you need acceleration. But with no mass...I have two variables. I'm just very confused.

4. Feb 18, 2015

### Nathanael

Give it a try anyway. Sometimes (when the net force is proportional to the mass) you don't need to know the mass. For example, if I drop my computer right now, even though I have no idea what it's mass is, I can tell you it accelerates at 9.8 m/s/s

5. Feb 18, 2015

### ciara0682

But in the formula I reach...the masses just can't cancel? I'm sorry, I'm not quite understanding. I get your analogy, but still confused on how it pertains to a frictional problem.

6. Feb 18, 2015

### Staff: Mentor

Draw a free body diagram (FBD) for the crate sliding down the ramp. Show the component forces due to gravity and due to the sliding friction. Then use F=ma and sum the forces... Please show that work...

7. Feb 18, 2015

### ciara0682

I went back to the basics and started with the kinetic formula. Xf=x+Vot+1/2at^2...... I used Vf=Vo+at solved for t...plugged that into kinematic equation.
Now I have Xf=1/2a(Vf/a) ^2 solved for a. Then plugged that into.....meu k(mgcos(theta))-(mgsin(theta))=ma but I have two variables. This is where I'm lost. And who knows of I am even doing that part right.

8. Feb 18, 2015

### Nathanael

Have you actually tried it, or are you just guessing?

The analogy was not really supposed to give you any insight; the point of my post was to just try finding μk and see what happens!

If μk depends on mass like you assume it does, then you should at least determine exactly how it depends on mass.

9. Feb 18, 2015

### ciara0682

I provided you with the work above that I have done so far.

10. Feb 18, 2015

### Nathanael

What happens if you divide both sides by m?

11. Feb 18, 2015

### ciara0682

μk=a/g(cos(theta))-(sin(theta)) ???

12. Feb 18, 2015

### Nathanael

13. Feb 18, 2015

### ciara0682

μk=a/(gsintheta)-(gcostheta)......I solved for uk and I still got the incorrect answer. grrrrrrrrrr. sorry for growling.

14. Feb 18, 2015

### Nathanael

Your algebra is still wrong. Triple check it

As long as you don't bite we're okay.

15. Feb 18, 2015

### ciara0682

a+gsin(theta)/gcos(theta) ....

16. Feb 18, 2015

### Nathanael

From your equation, I get $\mu_k=\frac{a}{g\cos\theta}+\tan\theta$

Also when you plug in the value for a, remember which directions you chose to be positive and negative (you said that downhill was negative, so the acceleration will be negative)

17. Feb 18, 2015

### ciara0682

trig is not my strongest area. whoops. I realized my mistake in algebra... gracias mi amigo!

18. Feb 18, 2015