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das1130 das1130
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12 years ago
A surfer "hangs ten," and accelerates down the sloping face of a wave. If the surfer's acceleration is 3.25 m/s2 and friction can be ignored, what is the angle at which the face of the wave is inclined above the horizontal?
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wrote...
#1
12 years ago
I'm not sure what "hangs ten" means, but the force accelerating him down the slope is the component of his weight "mg" that acts down the plane. If wave makes angle "A" with horizontal, then his weight will make angle "A" with a perpendicular line to the wave surface.
So the component of weight down the plane is, mgSin(A) and from 2nd law;

mgSin(A) = ma

Sin(A) = a/g

          = 3.25/9.8

A = 19.4 deg
wrote...
#2
12 years ago
Basically you treat the wave like an incline plane.  So the portion of gravity acting is F = m*g*sin(theta), or m*a = m*g*sin(theta)

masses cancel and you are left with a = g*sin(theta)

sin(theta) = a/g
theta = arcsin(a/g) = arcsin(3.25/9.8)
theta = 19.4 degrees

hope that helps!
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