# Rain is falling at the rate of 5.0 cm/h and accumulates in a pan. If the raindrops hit at 8.0 m/s, what is the force on the bottom of a 1.0 m^2 pan due to the impacting rain which does not rebound? Water has a mass of 1.00 X 10^3 kg per m^3?

Nov 23, 2017

Given:
$\setminus \rho = 1000$ (kg)/m^3; \qquad v = 8 m/s; \qquad (\Delta h)/(\Delta t) = 5 $\frac{c m}{h r} = 13.9 \setminus \times {10}^{- 6}$ $\frac{m}{s}$
Pressure: $\setminus q \quad P = \frac{F}{A} = \setminus \rho . \frac{\setminus \Delta h}{\setminus \Delta t} . v = 0.111$ $\frac{N}{m} ^ 2$

#### Explanation:

Consider rain drops impacting a surface of area $A$. The rain drops have a speed $v$ and carry a momentum. Upon impact it looses all its momentum. So the change in its momentum is just the momentum with which it impacted.

We do not know the size of each drop. But the rain fall rate allows us to calculate the rate of change of momentum.

Total Force: The total force experience by the surface due to the momentum transfer from the impacting rain drops is:
$\setminus q \quad F = \frac{\setminus \Delta p}{\setminus \Delta t} = \frac{\setminus \Delta \left(m v\right)}{\setminus \Delta t} = \setminus \rho . A . \frac{\setminus \Delta h}{\setminus \Delta t} . v$

Where $\frac{\setminus \Delta h}{\setminus \Delta t}$ is the rainfall rate

Pressure: Pressure is the force per unit area.
$P = \frac{F}{A} = \setminus \rho . \frac{\setminus \Delta h}{\setminus \Delta t} . v = 0.111$ $\frac{N}{m} ^ 2$

Given:
$\setminus \rho = 1000$ (kg)/m^3; \qquad v = 8 m/s; \qquad (\Delta h)/(\Delta t) = 5 $\frac{c m}{h r} = 13.9 \setminus \times {10}^{- 6}$ $\frac{m}{s}$