Question

Should I use the coefficient of friction?:

@mathmate is 90kg and skiing down a 35-degree hill. Coefficient of friction between mathmate and snow is 0.05. Wut's mathmate's friction if he's going at terminal velocity?

Diagram:

Answer 1

2 Methods I know:

Method 1: We know what 'g' is, (900 N), so we can easily find 'g' in the x-direction. And if it's at terminal velocity, 'g' in the x-direction is the same as the force of friction.

Method 2: We can also find 'g' in the y-direction, which is the same as the Normal Force, and calcualate the force of friction using the coefficient of friction (0.05).
Friction = Normal * 0.05

Both methods seem to be different though?

Answer 2

I don't see why there would be a terminal velocity (i.e. velocity that does not vary with time). As long as it is on the incline, with the coef. of friction being 0.05, I see it accelerating continuously... unless I misunderstand the question?

Answer 3

Assume it's at terminal velocity so the net force = 0.

Answer 4

OH I AM SORRY

Answer 5

the downplane force is mgsin(35) > friction = \(\mu\)mg cos(35), so net force cannot be zero.

Answer 6

Yeah, I messed up the problem, I edited it, I meant to find the force of friction :b

Answer 7

but yeah, that's what I meant when I used the friction formula, it doens't equal to 0, so should I ignore that method then?

Answer 8

Ah, then your idea works, by equation downplane force with the friction force. Equation given above! lol

Answer 9

is your x-axis along the plane?

Answer 10

Alright well good enough. I can find 'your' terminal velocity with the drag force equation myself :b
\[D = \frac{1}{4}Area~v^{2}\]

Answer 11

Am I on a parachute?

Answer 12

Nah, you're imaginary :)

Answer 13

If it's sliding down the inclined plane, then area does not matter, at least theoretically.

Answer 14

Area of a human body. Oh that's what you meant.

Answer 15

We modeled it as a box in class.

Answer 16

So you're onto air resistance, then there is more than friction, and the equation would be non-linear because one portion of resisting force varies with v^2.

Answer 17

wuuaaa....there's only 3 variables in that drag-force equation though.
Drag = force of friction.
Area = whatever your body area is.
And that solves for velocity right

Answer 18

I should rename body area to surface area interacting with the air. **

Answer 19

not if you include friction, which is independent of velocity, that makes it a non-linear equation. But it won't be difficult to solve because the air-resistance contributes only at high-velocities.

Answer 20

So would you be ready to post the equilibrium equation, with air-resistance included?

Answer 21

Is that not \[D = \frac{1}{4}Area~v^{2}\] If it's another equation, my class probably hasn't gotten there yet.

Answer 22

Especially not the 'density of air' formula. We haven't gotten that practical.

Answer 23

Any model is good enough to illustrate the idea. But are you sure no constant is missing?
Does D represent a force?
So this is a term that adds to the friction and acceleration equation then.
mg sin(35)=(1/4)A v^2+mgcos(35)

Answer 24

Ohh...you saying the air force resistance ADDS ON to the friction of the skiier and the snow.

Answer 25

Rip my head.....this is when I my head starts to explode. Just leave this question.........leave it.................I'm never coming back to this again.

Answer 26

Yep, they both resist motion!

But hang on, the equation is not more difficult to solve!

Answer 27

Sigh! :(