# Finding coefficient of kinetic friction

1. Nov 2, 2007

### missashley

A 37 kg box slides down a 35 degree ramp with an acceleration of 1.35 m/s^2. The acceleration of gravity is 9.81 m/s^2.

Find the coefficient of kinetic friction between the box and the ramp.

Ff= MkFn

37 * 1.35 = 49.95 N
Fgx = 49.95 sin 35 = 28.650143
Fgy = 49.95 cos 32 = 40.91664461

Ax = (1/m)(Ff - Fgx)

Ax= ((1/m)(Ff)) - ((1/m)(Fgx))
Ax + ((1/m)(Fgx)) = ((1/m)(Ff))
1.35 + ((1/37)(28.650143)) = ((1/37)(Ff))
2.124328189 = ((1/37)(Ff))
2.124328189 * 37 = Ff
78.600143 = Ff

Mk = 78.600143 / 40.916
Mk = 1.921012391

I think I did something wrong because the coefficient is greater than one

Please check what I did wrong and how can I fix it?

2. Nov 3, 2007

### Staff: Mentor

This approach is not correct.

The acceleration of 1.35 m/s2 is the consequence of friction, and has no effect on the normal force.

Start with the weight components normal and parallel with the plane of the incline.

mg = 37 kg * 9.81 m/s2 = 363 N.

Fgx = mg sin 35° = 363 sin 35° = 208.2 N
Fgy = mg cos 35° =363 cos 35° = 297.4 N

Now the normal force of the box on the incline produces friction according to $\mu$Fgy

http://hyperphysics.phy-astr.gsu.edu/hbase/mincl.html#c2

So the friction force is $\mu$297.4 N

However, a better approach is to think about the net force acting down the incline: Fgx-Ffrict = mg sin 35° - $\mu$mg cos 35°.

Dividing the force by the mass being accelerated gives the acceleration, so

a = g sin 35° - $\mu$g cos 35° = 0.574 g - $\mu$ 0.819 g = 1.35 m/s2

BTW, this appears to be a homework problem, so please post in the HW forum, Introductory Physics.

3. Nov 4, 2007

### missashley

Sorry for posting twice and thanks for the help!

4. Nov 4, 2007

### Staff: Mentor

You are very welcome!

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