# How to find coefficient of kinetic friction please?

1. Oct 3, 2006

### studentmom

how to find coefficient of kinetic friction... please?!!

I am lost on this question, hopefully someone can help:

a 300 kg crate is placed on an adjustable incline and as one end of the incline is raised the crate begins to move downward; if the crate slides down the plane at an acceleration of 0.7m/s2 when the incline angle is 25 degrees, what is the coefficient of kinetic friction between the ramp and the crate?

I drew a free-body diagram and that's about as far as I got... I know that the kinetic friction force is in the opposite direction that the crate is moving, but other than that i'm lost... please help!

Thanks!

2. Oct 3, 2006

do u have any lengths? Cause if you do you can just us tanθ and that would give you the answer...im doing a similar question and i figured out through other equations that tanθ is how u get it. So if the length of the incline was 4m and the height was 3 then ur answer would be 3/4.

3. Oct 3, 2006

### studentmom

no, the problem is stated as is, no lengths given...

4. Oct 3, 2006

### BishopUser

First thing i would do is draw a free body diagram, and create an coordinate system in which the x-axis is parallel to the plane. There are three forces acting on the box: gravity, friction, and the normal force. Friction = (coefficient of kinetic friction*normal force). On our coordinate system we know that the sum of the forces in the y direction must be 0 (the box isn't going to jump off the plane or anything like that). since friction does not have a component in the y-direction we can ignore it for now. The first step is to calculate the normal force.

Fy = -mgy + normal force
0 = -(300kg)(9.8m/s^s)(cos25) + Fn
Fn = 2664.5N

now that we have the normal force we can use F=ma in the X-direction. We have 2 forces with components in the x-direction: gravity, friction

Fnet = MA
Fgravity - Ffriction = MA
(9.8 m/s^2)(300kg)(sin25) - (2664.5N)(X) = (300kg)(.7m/s^2)
X = .39 (the coefficient of kinetic friction)

Last edited: Oct 3, 2006
5. Oct 3, 2006

### studentmom

Thank you so much for your help!

6. May 31, 2008

### Razza

I know this is an old thread and I know probably no ones gonna see it, but where did the force of friction (2664.5N) come from??? Seeing as 9.8*300 doesnt equal that, and 300*0.7 certainly doesnt...

7. Jun 1, 2008

### alphysicist

Hi Razza,

2664.5 is not the force of friction in that post; it is the normal force they calculate in the first three equations of post #4. In the equations after that they show that the magnitude of the force of friction is (2664.5)(X), where X is the coefficient of kinetic friction that they are solving for.

8. Jun 1, 2008

### Kushal

*edit* ooops alphysicist, i think we posted at the same time.

the force of friction is not 2664.5 N. It is (2664.5)(X) N, where 2664.5 N is the normal contact force and X is the coefficient of friction.

frictional force = normal contact force x coefficient of friction

9. Jun 1, 2008

### Razza

Thank you guys,

Haha its right above it how could I have missed. I think that was just because I was doing my own equation at the same time using my variables and thats where it got me lost.

Thank you for answering such an obvious question though. :rofl:

10. Jun 1, 2008

### Razza

Okayy wow..

With this, my equation works out to be:

F= -mgy+ Fn
0=-(0.012 kg)*(9.8 m/s squared) +Fn
-Fn = -0.1176 N

Then to sub it in

Fnet = MA
Fgravity - Ffriction = MA
(9.8 m/s^2)(0.012kg)(sin61) - (0.1176 N) (X) = (0.012kg)(15.24m/s^2) (as long as velocity is two times the average speed)

= 0.168

Now that sounds a lot better.

The answer I got given another formula, gave me 1.8, which seemed too high and gave my Force an answer of roughly 211 instead of 19 like it did for this.

F=0.168*117.6(my normal force)

Is that correct?

Last edited: Jun 1, 2008
11. Jun 1, 2008

### alphysicist

What is the statement of the problem? If it's the same type of problem as the original problem in this thread (object moving down an inclined plane) then I believe you are missing a trig factor here. As the angle of the incline increases, the normal force should decrease.

Here again, it's hard for me to be sure what's going on without the full problem. However, if this was just a normal incline problem with the only forces being from gravity and the incline, this acceleration seems too high. If the object was dropping freely, the acceleration would be 9.8 m/s^2. Here on an incline means the acceleration by gravity down the incline is not as much, and in addition you have friction to slow it down even more.