# Coefficient of Kinetic Friction

MG5
Heres the problem,

A 1,440-N crate is being pushed across a level floor at a constant speed by a force of 270 N at an angle of 20.0° below the horizontal

(a) What is the coefficient of kinetic friction between the crate and the floor?

I know how to do this. The answer is .165

Heres how I solved it..

270sin20 = 92 N
270cos20 = 254 N

1440N + 92N = 1532N

Ff=uk(n)

254N=uk(1532)

uk= .165

The only thing I don't quite get is why do I use 254N, the product of 270cos20, as the Ff and why is 1532N the normal force?

Thanks.

## Answers and Replies

azizlwl
The only thing I don't quite get is why do I use 254N, the product of 270cos20, as the Ff and why is 1532N the normal force?
...................
You have chose the values. Surely you have reasons for choosing them.
I think the questions should be are your reasonings correct?

MG5
The only thing I don't quite get is why do I use 254N, the product of 270cos20, as the Ff and why is 1532N the normal force?
...................
You have chose the values. Surely you have reasons for choosing them.
I think the questions should be are your reasonings correct?

I knew someone was going to say this. The reason I don't know why I'm using those numbers that I solved for is because I saw the solution to the problem worked out. I understand everything but why those numbers are used for Ff and n. Other than that everything else seems easy and makes sense.