48. In the solution to exercise 4, we found that the force provided by the wind needed to equal F=µkmg.
In this situation, we have a much smaller value of µk(0.10) and a much larger mass (one hundred stones
and the layer of ice). The layer of ice has a mass of
mice =917 kg/m3(400 m ×500 m ×0.0040 m)
which yields mice =7.34 ×105kg. This added to the mass of the hundred stones (at 20 kg each) comes
to m=7.36 ×105kg.
(a) Setting F=D(for Drag force) we use Eq. 6-14 to find the wind speed valong the ground (which
actually is relative to the moving stone, but we assume the stone is moving slowly enough that this
does not invalidate the result):
v=µkmg
4CiceρAice
=(0.10) (7.36 ×105) (9.8)
4(0.002)(1.21)(400 ×500)
which yields v= 19 m/s which converts to v=69 km/h.
(b) and (c) Doubling our previous result, we find the reported speed to be 139 km/h, which is a
reasonable for a storm winds. (A category 5 hurricane has speeds on the order of 2.6×102m/s.)