In downhill speed skiing a skier is retarded by both the air

In downhill speed skiing a skier is retarded by both the air drag force on the body and the kinetic frictional force on the skis. Suppose the slope angle is = 44.0°, the snow is dry snow with a coefficient of kinetic friction k = 0.0410, the mass of the skier and equipment is m = 86.0 kg, the cross-sectional area of the (tucked) skier is A = 1.60 m2, the drag coefficient is C = 0.180, and the air density is 1.20 kg/m3. (a) What is the terminal speed? (b) If a skier can vary C by a slight amount dC by adjusting, say, the hand positions, what is the corresponding variation in the terminal speed? Give your answer in terms of the given values and dC (need derivative for given values).

Solution

a.) The drag force acting on a body of area A and drag coefficient C is given as and air density say :

Fd = 0.5 C A V²

Fd = 0.5 * 0.18 * 1.2 * 1.6 x V^2 = 0.1728V^2

Now, for the inclined plane, the frictional force acting on the skier would be:

F = MgCos44 x Uk = 24.882 N

Now, for terminal velocity to be reached, the net force acting on the skier must be zero.

That is, F + Fd = MgSin44

or 0.1728 x V^2 + 24.882 = 586.055

V = 56.987 m/s is the required terminal speed.

b.) From the equation we had above, we have:

Fd + F = MgSin44

or, 0.5 C A V² + 24.882 = 586.055

V^2 = 584.56 / C

Differentiating both the sides, we get:

2V dV = -(584.56 / C²) dC

Hence, dV = -158.3 dC is the required relation