# Find coefficient of friction

1. Jun 1, 2007

### rajames429

I am not a student, but once took physics in college more that 20 years ago.

I have the following problem that I am having trouble with:

A book has a mass of 400 grams. When you slide the book against the floor with a force of 5 N, it accelerated at a rate of -1.5 m/s^2.

What would the coefficient of friction be between the book and the floor?

So, my solution seems to be:

Where F is force
N is normal force
FF is force of friction
u is coefficient of friction

I know F = ma, N = -mg, and FF = uN, so that FF = -umg.

Is this just a matter of setting ma = -umg, then a = -ug, and then u = -a/g ?

So given that g = 9.8 m/s^2:

I get an answer of 0.15, is this correct?

Is this the correct way to approach this problem?

Is the 5 N force just unneeded information here?

2. Jun 1, 2007

### ice109

yea you've got the right idea except -a is not 1.5 it's 11, it's acceleration caused by the force of friction

u= -(11)/9.8 = 1.22

3. Jun 1, 2007

### fatal1

Yea, I think the 5N part is suppose to be a trap to get people to think that a force of 5N is applied continuously so they use Fa + Ff = Fnet

4. Jun 1, 2007

### CaptainZappo

Assuming you're designating the direction of the 5 N force to be positive, how could the acceleration possibly be negative?

Forgive me if I'm missing something obvious; it is getting late.

5. Jun 1, 2007

### trajan22

Well, not quite, the force applied is not the same as the force of friction because there is an acceleration. The only time the force of friction is the same as the force applied is when acceleration is zero.

The reason they give you an initial force is so you can find the -acceleration due to the force of friction. Basically this is what you do:
F=ma( your variable is a) as you will see the acceleration is different than that of the actual acceleration.
Now subtract this value from the actual acceleration to find the -acceleration due to friction. You can now find the force due to friction again using F=ma you have now found the force of friction and can solve for the coefficient of friction.
(for convenience I changed the axis and made the initial acceleration +.)

Last edited: Jun 2, 2007
6. Jun 1, 2007

### ice109

theres no such thing as deceleration

7. Jun 1, 2007

### trajan22

What do you mean there is no deceleration? If the force applied produces a larger acceleration than is present. Then that means in effect the force of friction must be slowing the object down or decelerating it.

Is there a better way you could put it? Sometimes I'm a little confusing.

Last edited: Jun 2, 2007
8. Jun 2, 2007

### ice109

acceleration can be positive or negative, deceleration is a misnomer

9. Jun 2, 2007

### trajan22

Ok Ill concede. But I was just trying to avoid directional confusion. Since the problem starts with an - direction for acceleration, and to simplify it I did the problem assuming a positive acceleration.

Edit* there it should be fixed.