Find Pressure Exerted on Roof by Raindrops in Tropical Storm

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SUMMARY

The discussion focuses on calculating the pressure exerted on a roof by raindrops during a tropical storm, where 40mm of rain falls in one hour at an average speed of 10m/s. The relevant equation for pressure is derived from the relationship between force and area, specifically P = F/A. To find the mass of the rainfall impacting the roof, participants clarify that 40mm translates to 40 liters per square meter per hour, which can be converted to mass per second per square meter. This mass, combined with the velocity of the raindrops, allows for the calculation of pressure using the equation P = (dm/dt) * v / A.

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Homework Statement


In a tropical rainstorm, 40mm of rain fell in one hour. Assuming that the raindrops struck an adequately drained roof normally with an average speed of 10m/s, find the pressure exerted on the roof by the rain.


Homework Equations


(delta)p = F(delta)t


The Attempt at a Solution



(delta)p = 10 x mass of rainfall striking the roof = F(roof on raindrop) x (delta)t
F(roof on raindrop) = F(raindrop on roof)
Pressure = F(raindrop on roof) / Area of the roof

I have difficulty deriving the expression for mass of raindrop striking the roof. I also do not understand how is "44mm of rain" relevant to the question.
 
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What you need the "40 mm of rain" for is to figure out how much rain falls in a given timespan.
From the 40mm (meaning 40 liters per square meter per hour) of rain, obtain the mass of the falling water (per second per square meter). Now, you know how much water falls (the mass), and you know how fast it is falling, so you can calculate the pressure from the equation
P = F/A = \frac{dp}{dt}/A = \frac{dm/dt}{A} v
as you can see, you will have all the necessairy quantities.
 

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