# Coefficient of Kinetic Friction from M, Angle, Accel.

1. Feb 23, 2012

### Flash70

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

m = 52kg
theta = 10.3deg
t = 5.8s
d = 3.63 m

2. Relevant equations

3. The attempt at a solution

Ok so here are my attempts. If they are way off, take it easy on me :). I am not very good at this. Am I missing negatives or am I way off on this? Are my trigs set up correctly?
Thanks for you help.

Try 1) ma = mg (sin10.3) - Umg(cos10.3)
Try 2)
ma = 52kg * (3.63/5.8^2)
mg = 52kg*9.8
mg(sin10.3)+ma = Force of friction
mg(cos10.3) = Force Normal
muk = force of friction / force normal

2. Feb 23, 2012

### PeterO

Try to break it down into simpler pieces.

If the box covers 3.63 m in 5.8 seconds, what acceleration has it had?

For that acceleration, what has been the net force on the 52 kg box.

Since the box is on 10.3 degree slope, what is the component of the weight force parallel to the slope? perpendicular to the slope?

What must be the size of the friction force in order to get the net force you got in step 2 ?

Using F = μR what must the coefficient of friction be. [remember R is the reaction force from the slope, which balances the weight component perpendicular to the slope.]

3. Feb 23, 2012

### Flash70

I hate to ask this but would it be possible for you to explain it in a different way?
For some reason I am not following what you are saying. I believe I have the answers to the first three questions you asked in my Try 2, but am kind of confused.

4. Feb 23, 2012

### PeterO

I described it the way I would do it.

Your answer was a mixture of formulas, with no indication of what you were planning to do with each step - that is why I couldn't comment directly on your work.

5. Feb 23, 2012

### Flash70

Ok I apologize for that. I solved for ma and mg in the beginning then just plugged those numbers into the formulas I wrote. It will appear like this:

509.6 * sin(10.3) + 5.61117 = 96.7288 force of friction
509.6 * cos(10.3) = 501.388 normal force
96.7288/501.388 = .1929

I feel as if the equation I was using in Try 1 may have been more on the money, but it is wrong. I used : ma = mg (sin10.3) - Umg(cos10.3)
5.61117 = (52kg*9.8)(sin10.3) - U(52kg*9.8*cos10.3)

6. Feb 23, 2012

### PeterO

I think you acceleration figure is wrong.

You said ma = 52kg * (3.63/5.8^2)

You effectively have claimed a = s/t2 which means s = at2

That does not gel with one of the formulas s = ut + 0.5at2

7. Feb 24, 2012

### Flash70

You are absolutely right. I was stuck using the wrong thing from last chapter but I am so used to using it I couldn't recognize it!

Thank you Peter!