# Need basic understanding of force and friction

1. Oct 10, 2005

### Balthamos

Hello,

I am having some trouble understanding this whole concept about friction as well as force (Newton's 2nd law). What I was told is that Fnet=Ma, which I'm guessing stands for sum force = mass * acceleration.

What I wasn't told is how the units are dealt with. I am very used to dimensional analysis, and that isn't helping me one bit. What exactly do these units need to be? I was thinking:
F=N
M=kg
a=m/s^2

As for friction, I'm not even sure where to begin with that. I don't think that I was given any specific formula.

Thanks,
~Balthamos

2. Oct 10, 2005

### Staff: Mentor

That's Newton's 2nd law of motion. It applies for all forces, not just friction.

Those are the standard units. Note that N = kg-m/s^2.

As far as dimensions go, Force has dimension of $M L T^{-2}$.

Friction is a contact force between two surfaces that resists slipping between those surfaces. It has the same units as any force. Generally, there are two kinds of frictions you'll have to worry about: static friction, in which the surfaces do not slip, and kinetic friction, in which the two surfaces slide over each other. Look them up!

3. Oct 10, 2005

### arildno

Yes, so you have: N=kg*m/s^2

4. Oct 10, 2005

### Balthamos

I know what friction is, but how am i actually supposed to implement it into the equation? Like in a problem that asks: Find the the acceleration of the 10 kg sled w/ force of 180 N (also pulling at an angle, but I can figure that out), if the friction to be overcome is 15N (if F > static friction?)
So I would say:
F=Ma
180=10a
a=18 m/s^2

Then what? Where do I stick in the friction? Would I subtract 15N from 180N because of the resistance or what?

~Balthamos

5. Oct 10, 2005

### Staff: Mentor

The friction acts parallel to the surfaces, opposing the motion of the sled. You will include it as one of the forces acting on the sled.

6. Oct 10, 2005

### cscott

I think you would substract the 15N from the x component of the 180N applied force.

Last edited: Oct 10, 2005
7. Oct 10, 2005

### Balthamos

All they give me is the following:
sled=10kg (pulled across level ground)
sled's rope makes 30 degree angle with ground
rope pulls on sled with 180N

What is this $$F_f = \mu F_N$$ you speak of? Would that be the total friction = something * net force?

~Balthamos

8. Oct 10, 2005

### cscott

Disregard that equation if you haven't learnt it yet.

I would find the x component of the applied force, and substract it the 15N force of friction from it to get the net force in the x direction.

9. Oct 10, 2005

### Balthamos

No, please, explain. There are problems on the worksheet that ask for coefficients of friction, and my teacher doesn't explain everything well enough. Sooner or later, she will tell us about this equation (hopefully within the 2 days left before the test)

~Balthamos

10. Oct 10, 2005

### cscott

OK, well if you have a force acting at an angle, then the y-component will affect affect the force of friction because it changes the normal force.

So,

\begin{align*} F_y &= F \sin \theta \\ F_x &= F \cos \theta \\ \end{align*}

\begin{align*} F_N &= mg - F_y \\ F_f &= \mu F_N \\ \Sigma F_x &= F_x - F_f = ma \\ \end{align*}

If say I was pulling a sled with the rope at an angle I would be slightly lifting the sled which means it's pushing less against the ground and the ground pushes less back at it (normal force). A change in normal force affects friction "through" $F_f = \mu F_N$ where $\mu$ is calculated experimentaly. You see?

Last edited: Oct 10, 2005