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Bowl of water A bowl full of water is sitting out in a pouring rainstorm. Its surface area is 500 cm2. The rain is coming straight down at 5 m/s at a rate of 10−3g/cm2 s. If the excess water drips out of the bowl with negligible velocity, find the force on the bowl due to the falling rain. What is the force if the bowl is moving uniformly upward at 2 m/s?

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umohduke14

Answer: 4.9 x 10-3 N

Explanation:

A = 500cm^2 = 5 x 10^-2 m^2

V = 5 m/s

R = 10^-3 g/cm^2.sec = 10^-2kg/m^2 . sec

Prain water = R / V = 10^-2 / 5 = 2 x 10-3 kg/m^3

For the stationary bowl,

dm/dt =pAv= RA

F= dp/dt = (dm/dt) v = RAv = 2.5 x 10^-3 N

Bowl moving upwards to speed u = 2 m/s

dm/dt = pA ( v + u) / v

F = dp/dt = (dm/dt)(v+u) = RA (v+u)^2 / v = 4.9 x 10^-3 N

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