What Determines the Terminal Speed of a Skier?

[White_Noise]

Homework Statement

What is the terminal speed for an 85.0 kg skier going down a 44 degree snow-covered slope on wooden skis μ= 0.060?

Assume the skier is 1.7m tall and .50m wide

Homework Equations

v = sqrt(4μmg/A)

The Attempt at a Solution

The area of the skier is (1.7m*.5m)=0.85m^2

sqrt (4(.06)(85)(9.8)/0.85) = 15.34 m/s

I tried to find velocity along the slope by calculating 15.34/sin(44) and got 22.08 m/s. This is wrong. I think it's because either I messed up my original vectors or I assumed normal force is equal to gravity which wouldn't be true on the slope (or probably both). I'm not sure how to find normal force or factor it into the problem.

 

[White_Noise]

I think this is probably the only question in Mastering Physics that's not covered somewhere on the interwebs. Bump for great justice. I have faith in you Physics Forums. Let us contribute to the knowledge of humanity, and my grade point average.

 

[mgb_phys]

Terminal velocity is where the downward force from gravity equals the aerodynamic and surface drag

Aero drag = 1/2 rho C A v^2
Friction along the slope you can work out from the coeff and normal force

 

[White_Noise]

rho?

 

[White_Noise]

I don't understand. We are not supposed to use that equation. It has not been presented to us and is not in the chapter. We are supposed to adapt the equation I have posted.

 

[Mark44]

Homework Statement

What is the terminal speed for an 85.0 kg skier going down a 44 degree snow-covered slope on wooden skis μ= 0.060?

Assume the skier is 1.7m tall and .50m wide

Homework Equations

v = sqrt(4μmg/A)

The Attempt at a Solution

The area of the skier is (1.7m*.5m)=0.85m^2

sqrt (4(.06)(85)(9.8)/0.85) = 15.34 m/s

If the formula for v above is the one you're supposed to work with, it looks like all you need to do is just plug in the numbers and evaluate. Your value looks fine to me.

I tried to find velocity along the slope by calculating 15.34/sin(44) and got 22.08 m/s. This is wrong. I think it's because either I messed up my original vectors or I assumed normal force is equal to gravity which wouldn't be true on the slope (or probably both). I'm not sure how to find normal force or factor it into the problem.

 

[White_Noise]

Okay, the problem is simpler than I thought. My problem was that I was trying to rush through the problem without thinking it out thoroughly enough. I tend to get lazy because classical mechanics are not where my scientific interests lie :/

To anyone having trouble with the problem:

Find the force of kinetic friction along the slope:

Ff= μN
normal force along slope = cos(θ)mg

Ff = 599.21*.06 = 35.952N

Find force of gravity along slope:

Fg = sin(θ)mg = 578.65N

Net force = Fg - Ff = 578.65N - 35.952N = ~542.70N

v = sqrt(4mg/A)
v = sqrt((4*542.7)/.85) = ~50.54 m/s

 

[mgb_phys]

Sorry I meant to post a link to the drag equation, I assumed that since you are given the cross section area of the skier you were expected to take rag into account - especialy since it's the main limit on terminal velocity for a skier