# Homework Help: Static and kinetic friction on an incline

1. Oct 7, 2008

### a18c18

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

A sled weighing 210 N rests on a 15° incline, held in place by static friction. The coefficient of static friction is 0.50.

(c) The sled is now pulled up the incline at constant speed by a child. The child weighs 546 N and pulls on the rope with a constant force of 100 N. The rope makes an angle of 30° with the incline and has negligible weight. What is the magnitude of the kinetic friction force on the sled? (Enter 0 if sled does not move.)
N

(d) What is the coefficient of kinetic friction between the sled and the incline? (Enter 0 if sled does not move.)

(e) What is the magnitude of the force exerted on the child by the incline?
N

2. Relevant equations

F(normal)=mg*cos(theta)
f(kinetic)=u(kinetic)*F(normal)

3. The attempt at a solution

x: -f(kinetic)-mg*sin(theta)=-ma
y: F(normal)=mg*cos(theta)

I'm not sure where to go from here.

2. Oct 7, 2008

### nasu

1. For the x direction, you forgot the force applied by the child (100 N).
2. What does it means "at constant speed"? What is the acceleration?
3. Pay attention that you have two angles, the angle of the incline and the angle of the rope. Which one do you call theta?

3. Oct 7, 2008

### a18c18

Okay i retried this and got:

x: f(kinetic)+100*cos30=mg*cos15
y:F(normal)=mg*cos(15)

I got that F(normal)=202.84
then I plugged this into f(kinetic)+100*cos30=mg*cos15 which would be f(kinetic)=202.84-100cos30 which would be 116.24 but this answer was incorrect
Does anyone know why?

4. Oct 8, 2008

### nasu

1. The normal force is not just m*g*cos(15). There is normal component of the tension in the rope.

2. The equation for the x direction looks OK.