What Is the Coefficient of Kinetic Friction for a Box Sliding Down a Ramp?

[cavery4]

A box of canned goods slides down a ramp from street level into the basement of a grocery store with acceleration 0.65 m/s2 directed down the ramp. The ramp makes an angle of 35 degrees with the horizontal. What is the coefficient of kinetic friction between the box and the ramp?

I have been trying to figure this out for a while. Heh.

So far I drew my picture. One thing I am confused about is what direction friction is in.
I thought that since the box is sliding downward with a force, then the frictional force is up and opposite of that sliding force. I might have this mixed up. I don't know if there is a force pushing it down.

Then I broke it up into X component forces:

Fa - friction (making the downward direction positive) + Wx = ma
Then I became confused about the Fa and wondered if it even belongs.

So then:
-f + wx = ma
-Mu(mg) + mgsin35 = ma (then cancel out the masses)
Mu(g) + gsin35 = a

Mu(g) = a - gsin35
Mu = a - sin35

I have done something wrong. If anyone can help me figure this out, I would really appreciate it. I have tried my best!

 

[OlderDan]

There are three forces in the problem, one of which you are going to break into components.

1) Gravity (weight)- points straight down- break it into a component parallel to the plane and a component perpendicular to the plane

2) Normal force- points opposite and is equal in magnitude to the gravity component perpendicular to the plane.

3) Friction- points up the plane and is equal to the magnitude of the normal force times the coefficient of friction. The sum of the opposing forces (difference of magnitude) of the gravity component parallel to the plane and the friction force is the net force acting on the box, so it is equal to the mass of the box times the acceleration.

If you get the directions of those forces right, and you can do the trig right to break the gravity into two components, you will get the answer.

 

[thepatientmental]

First of all, good job on drawing a picture and setting up the forces in the x-direction. You are correct that friction acts in the opposite direction of the sliding force, which in this case is down the ramp. So, the frictional force would be up the ramp.

Now, let's take a closer look at your equation -Mu(mg) + mgsin35 = ma. This is close, but there are a few things to note:
1. The acceleration in this equation should be the acceleration down the ramp, not just any acceleration. In this case, it is given as 0.65 m/s^2 directed down the ramp. So, the acceleration (a) should be 0.65 m/s^2, not just any variable.
2. The weight (W) should be broken up into its components, Wx and Wy. Wx is the component of weight that acts down the ramp, and Wy is the component of weight that acts perpendicular to the ramp. In this case, Wy is equal to mgcos35. So, your equation should be:

-Ff + Wx = ma

Now, we can plug in the values we know:
-Mu(mgcos35) + mgsin35 = ma

And we know that a = 0.65 m/s^2, so we can solve for Mu:
-Mu(mgcos35) + mgsin35 = 0.65m
-Mu(9.8cos35) + 9.8sin35 = 0.65
-Mu(8.04) + 5.63 = 0.65
-Mu(8.04) = -4.98
Mu = 0.62

So, the coefficient of kinetic friction between the box and the ramp is approximately 0.62. I hope this helps clear things up for you!