When the 150-lb skier is at point A he has a speed of 5 ft/s. When the skier rea

Question

When the 150-lb skier is at point A he has a speed of 5 ft/s. When the skier reaches point B on the smooth slope, the slope follows the cosine curve shown. Suppose that y(x)=62cos(100x)ft, where x is in ft. Neglect friction and air resistance. (Figure 1).

VB = 47.0 ft/s

What is the normal force on the skier's skis at B?

When the 150-lb skier is at point A he has a speed of 5 ft/s. When the skier reaches point B on the smooth slope, the slope follows the cosine curve shown. Suppose that 62cosf Figure 1) r ft, where z is in ft. Neglect friction and air resistance

Explanation / Answer

v = 47 ft/s

dy/dx = -62 x 3.14/100 x sin(180X/100)

= -1.735

d2y/dx2 = -62 x (0.0314)^2 x cos (180X/100)

= -0.0277

R = (1+1.735^2)^1.5 / 0.0277 = 290 ft

N = mgcos@ - mv^2/R

= 150x32cos60 - 150 x 47^2 / 290

= 1257 lb-ft/s^2

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