# Homework Help: Help with Vectors: Force of Friction problem

1. Oct 28, 2012

### LeonM

"Mr Mcgoo pulls a sled with several children on it at a constant speed. If he pulls with a force of 329.461 N at an angle of 17° above the horizontal, what is the force of friction on the sled?"

Not sure if i even drew the picture right(is it like a triangle, but it doesn't say hill, don't understand the whole "horizontal" thing), but i cant figure how to solve this.

I drew a picture(which may have been incorrect) and identified the Fρ(parallel force) as being 329.461 N and i need to find Ff. I also have the knowledge that Ff is the equal and opposite of Fρ.

2. Oct 28, 2012

### PhanthomJay

Hi Leon welcome to PF!

Looks like perhaps the sled is being pulled along a level surface, and the pulling force of 329 N is applied at an angle 17 degrees above the horizontal. Wording is not fully clear, but that is how I would read it. What is the component of that pulling force in the horizontal direction?

3. Oct 28, 2012

### LeonM

Not sure what you mean by component? Do you mean direction, if so, im not sure, just assuming east.

4. Oct 29, 2012

### PhanthomJay

Yes, any vector directed in two dimensions can be thought of as having an influence in two different directions. That is, it can be thought of as having two parts. Each part of a two-dimensional vector is known as a component. If the sled is moving left to right (west to east), the force F, applied 17 degrees above the horizontal, can be broken up into its easterly and northerly components, using basic trig and the properties of a right triangle. What is the easterly component of F? What is the direction of the friction force? Is the friction force equal to the easterly component of F? Why?

5. Oct 29, 2012

### LeonM

Right, Right. So the east component would be 329.461 N, which would give you the adjacent side of 17°. After finding all the sides I will have my Fp which would be the hypotenuse, and Ff is equal or opposite Fp(my answer). If im not mistaken?

6. Oct 29, 2012

### PhanthomJay

First, I am not sure if you are describing the problem correctly. How exactly is it worded? I don't know why the problem would give an applied force accurate to the nearest thousandth of a Newton. The applied force is the hypotenuse. The horizontal component must be calculated.