1. Dec 9, 2007

### anonymous820

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

a skier stands at the top of a slope. the skier is able to remain stationary on a horizontal surface without using her poles. she can remain stationary until there is an incline of 6 degrees.

WHAT IS THE COEFFICIENT OF STATIC FRICITON FOR THE SKIER?

2. Relevant equations

no relevant equations in notes.

3. The attempt at a solution

the solution is clearly impossible because the only information that they give u is 6 degrees.

2. Dec 9, 2007

### cristo

Staff Emeritus
Really? I don't think it's clearly impossible! You are told that the skier can stay stationary until the slope reaches 6 degrees; that is, when the slope is 6 degrees, the friction between the skier's skis and the slope is at its maximum when the slope has gradient of 6 degrees. Can you carry on? Try drawing a diagram.

3. Dec 9, 2007

### anonymous820

i did draw a diagram. i put the incline as 6 degrees. Fn going up. Fy going down. Ff going opposite the skier. Fg perpendicular. Fx from Fy to Fg. and 6 degrees between Fy and Fg. if any of that makes any sense. i just dont know where to go from there.

4. Dec 9, 2007

### cristo

Staff Emeritus
Well, you know that, when the slope is 6 degrees, the component of the weight of the skier acting down the slope is balanced by the frictional force up the slope between the skiers skis and the snow.

5. Dec 9, 2007

### anonymous820

so Ff=Fx? then sin 6 degrees=Fx/Fg? but there is no Fg its F(9.81) because they dont give u a mass?

6. Dec 9, 2007

### cristo

Staff Emeritus
Right, you need to explain your notation. Here's how I see what you mean. Ff is the frictional force? If so, then do you have an equation that gives you the frictional force in terms of the skier? What is Fx and Fg? You need to calculate the component of the skier's weight (mg) that is acting parallel to the slope. This would be equal to mg.sin6, so I guess that's what you mean by Fx (although your notation is not the best).

Don't worry about the fact that you aren't given the mass; just leave it as m in the equations, and it should all work out.

7. Dec 9, 2007

### anonymous820

yeah. Ff is the frictional force. Ff=coefficient of friction*normal force. Fx is the x component and Fg is the force due to gravity.

so if Fg=mg, i put it as 9.81m (mass), then i plugged that into the equation sin6=Fx/Fg, and got Fx to equal 1.03m?

this stuff makes no sense to me.

8. Dec 9, 2007

### cristo

Staff Emeritus
So Fx is the x component taking the coordinate system parallel and perpendicular to the slope; i.e. Fx is the component of the weight acting down the slope?

Ok, so now you've got a value for it. Now use the fact that Fx is balanced by Ff. What is the normal force? Can you put this into the equation Fx=Ff and obtain a result for the coefficient of friction?

9. Dec 9, 2007

### anonymous820

yes. Fx is the component of the weight acting down the slope.

which is "1.03m". so Fx=Ff from equilibrium? then Ff equals 1.03m too?

Ff=coefficient of friction*normal force?

wait, so then how do i find Fn? the same way?

heres a visual i did. hope it helps.

http://i188.photobucket.com/albums/z175/anonymous820/idk/physics.jpg [Broken]

Last edited by a moderator: May 3, 2017
10. Dec 9, 2007

### cristo

Staff Emeritus
Fn is always equal to Fy in these questions. Can you find Fy? Note that the green triangle is a right angled triangle with the top angle equal to 6 degrees.

11. Dec 9, 2007

### anonymous820

yeah. sorry i forgot to include that. so it would be tan6=Fx/Fy --> tan6=1.03m/Fy --> Fy and Fn=9.8m, m being mass, not meters.

so then u put it in the friciton formula --> 1.03m=coefficient of friciton*9.8m --> and the coefficient of friction comes out to be 0.12? which seems right because coefficients of friciton are usually decimals. and the m's cancelled out. right?

12. Dec 9, 2007

### cristo

Staff Emeritus
That looks ok. You should really carry through the question using more significant figures than you state your answer in, though. Try doing that, and see if it changes the result.

Either way, you have completed the hard part of the problem (putting the numbers back in shouldn't take more than a minute or so).

13. Dec 9, 2007

### anonymous820

thats how the teacher wants us to round it though. so it wont matter to him. wow. its actually making more sense now. thanks for all ur help.

14. Dec 9, 2007

### cristo

Staff Emeritus
I realise he probably asked you for your final answer correct to 3sf, but the point is that by rounding earlier, it can affect the final answer.

Anyway, that was just a "heads up." Oh, and you're welcome!