Momentum/Impulse- Calculating Average Force of Raindrops

[MadTay]

Homework Statement

The problem says:
"A record rainstorm produced 304.8 mm (approximately 1 ft)
of rain in 42 min. Estimate the average force that the rain exerted
on the roof of a house that measures 10 m * 16 m. Indicate
any assumptions you made. (Note: density of water is 1000 kg/m³"

Homework Equations

Volume= lwh
Favg=ΔP/ΔT
ΔP= mΔV

The Attempt at a Solution

I determined the mass of the rain that fell in the 42 minute/2520 second interval to be 48,768kg and the mass of rain to fall in one second to be 19.35kg.
From there I am absolutely at a loss of what to do to figure out the average force. Assuming that the final velocity of the rain is 0, I'm thinking that I need the initial velocity of the rain in order to figure out the change in momentum.
The answer itself isn't a problem, as I was given a key, but I'm not sure if I should use the mass of all of the rain and 2520 seconds as Δt or use 19.35kg and 1 second as Δt. I'm just really not sure where to go after calculating the masses of the amounts of rain.
The key says the answer is 2x10^-4 N, but I can't figure out how to get to that answer.

 

[Bystander]

thinking that I need the initial velocity of the rain

Which is absolutely correct. Large raindrops in a torrential downpour might be moving 10 m/s. That and the mass rate you've calculated eventually give you a number that's for force ~ 2, but the order of magnitude is off by a factor of 6. Whoever solved it for the answer key probably used 10 m/s for the fall rate, but mixed grams and tons.

 

[MadTay]

Which is absolutely correct. Large raindrops in a torrential downpour might be moving 10 m/s. That and the mass rate you've calculated eventually give you a number that's for force ~ 2, but the order of magnitude is off by a factor of 6. Whoever solved it for the answer key probably used 10 m/s for the fall rate, but mixed grams and tons.

Whew! Glad to know it wasn't a lack of knowledge on my part that was stopping me from getting to that answer. Thank you very much for your help!

 

[DEvens]

The 10 m/s seems reasonable.

http://wxguys.ssec.wisc.edu/2013/09/10/how-fast-do-raindrops-fall/

The smaller drops presumably make up a small fraction of the total mass of water.

 

[HalfLostAndSearching]

I would first clarify the problem statement and assumptions with the person who gave it to me. The problem does not specify whether the roof is flat or sloped, and it does not specify whether the rain is falling directly onto the roof or at an angle. These factors could affect the average force calculation.

Assuming a flat roof and direct rainfall, I would approach the problem as follows:

1. Calculate the total mass of the rain that fell on the roof in 42 minutes:
304.8 mm = 0.3048 m
Volume of rain = (10 m)(16 m)(0.3048 m) = 48.768 m^3
Mass of rain = (48.768 m^3)(1000 kg/m^3) = 48,768 kg

2. Calculate the average force exerted by the rain on the roof:
Favg = ΔP/Δt
ΔP = mΔv
Δv = 0 - initial velocity (assuming initial velocity is 0)
Δt = 42 minutes = 2520 seconds
Favg = (48,768 kg)(0 - 0)/2520 s
Favg = 0 N

This result seems strange, as it suggests that the rain did not exert any force on the roof. However, this is because we are considering the entire mass of rain that fell in 42 minutes. In reality, the rain is falling continuously, so we need to consider the average force exerted in one second.

3. Calculate the average force exerted in one second:
Favg = ΔP/Δt
ΔP = mΔv
Δv = 0 - initial velocity (assuming initial velocity is 0)
Δt = 1 second
Favg = (19.35 kg)(0 - 0)/1 s
Favg = 0 N

Again, this result seems strange, but it is because we are assuming the initial velocity is 0. In reality, the rain is falling with some initial velocity.

4. Use the given answer of 2x10^-4 N as a guide to find the initial velocity:
Favg = ΔP/Δt
ΔP = mΔv
Δv = final velocity - initial velocity
Favg = (19.35 kg)(0 - initial