Finding coefficient of kinetic friction

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?

Astronuc

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.

F_gx = mg sin 35° = 363 sin 35° = 208.2 N
F_gy = mg cos 35° =363 cos 35° = 297.4 N

Now the normal force of the box on the incline produces friction according to [itex]\mu[/itex]F_gy

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

So the friction force is [itex]\mu[/itex]297.4 N

However, a better approach is to think about the net force acting down the incline: F_gx-F_frict = mg sin 35° - [itex]\mu[/itex]mg cos 35°.

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

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


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

missashley

Sorry for posting twice and thanks for the help!

Astronuc

You are very welcome!

If one of one's posts goes missing, one can click on one's username and select 'Find More Posts by username', then select the post.