# Sled on an inclined plane

1. Dec 12, 2008

### cartoonorange

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

A sled weighing 250 N rests on a 15° incline, held in place by static friction (Figure 5-61). The coefficient of static friction is 0.5.

a)What is the magnitude of the static friction on the sled?

The sled is now pulled up the incline at constant speed by a child. The child weighs 510 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.

b)What is the magnitude of the force exerted on the child by the incline?

2. Relevant equations
F=ma
F=us*N

3. The attempt at a solution
For a, I found Normal force: 250cos(15)=241.5
Then to find magnitude: 241.5*.5=120.7, But it is not right.

For b, i foind Fgy: 510cos(15)=492.6
492.6*.5=246.3, but it is also wrong

2. Dec 12, 2008

### LowlyPion

In a) you must remember that Friction is only a maximum force. If the down hill force of gravity doesn't exceed the maximum then the sled remains static.

What you calculated is the maximum amount of frictional force. Not what is only required.

3. Dec 12, 2008

### cartoonorange

I figured out a, but i am still lost on b

4. Dec 12, 2008

### Elbobo

Are you sure you've included every force acting upon the child in your free-body diagram?

5. Dec 12, 2008

### LowlyPion

You need to consider all the forces acting on the child.

This includes gravity of course - a can't miss 510N force. And that resolves into the normal and parallel with the incline. The normal force is what they are interested in. Nevermind that the parallel force resolves itself into the downward incline pull of gravity from the child and sled and the frictional resistance generated by the snow boots.

If the rope of the sled is being pulled at 30° to the incline upward with 100N, then what does that mean in terms of any force acting on the child?

6. Dec 13, 2008

### cartoonorange

Thanks, I got it, I was not include the tension from the rope