Solution 15.92 Step 1: FBD (a) - anyplace on the slope:, (b) At point B, (c) At point C
At B
N mg = m
or solving for N
2 vB B
2 vB B
N = mg + m
(1)
From Eq. 1 we see that N>0, thus, the skier does not lose contact with the ground and the limiting condition on acceleration at point B is
aB = 2g
At C:
or
2 vB 2g B
(2)
N + mg = m
or, solving for N
2 vC C
N = mg m
2 vC C
(3)
Thus, at C it is possible for N to become negative and hence, for the skier can lose contact with the slope. The limiting condition on acceleration in this case is that
aC =
2 vC g C
(4)
At this point we need to obtain relationships hA, between hB, hC and the velocities at points B and C. Since N does no work and mg is conservative we use conservation of energy between A and B and then between A and C. Thus,
1
TA + VA = TB + VB 0 1/2 mv2
(5)
Using the ground as the datum for potential energy, then Eq. 5 gives us
mghA =
or
1 2 mv B + mghB 2
(7)
(6)
2 v B = 2 g (hA hB )
Substitutin...