Calculating Force of Hail Stones on a Roof

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To calculate the force of hailstones on a roof during a storm, the mass of each hailstone must first be determined, which is approximately 0.833 g based on its volume. The total number of hailstones impacting the roof can be derived from the density of 110 hailstones per cubic meter and the roof's area of 180 m². The force exerted by the hailstones can be calculated using the formula F = ma, where 'a' is the acceleration of the hailstones upon impact, which is influenced by their velocity of 14 m/s. Understanding the relationship between momentum and impulse may also aid in solving the problem. The discussion emphasizes the importance of correctly calculating mass and considering the instantaneous number of hailstones impacting the roof.
binicle
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I cannot know how to calculate the force of the hail :bugeye:
During a violent thunderstorm, hail the size of marbles (diameter = 1.20 cm) falls at a speed of 14.0 m/s. There are estimated to be 110.0 hailstones per cubic meter of air. Assume that, as for ice, 1.00 cm3 of hail has a mass of 0.920 g. What is the mass of each hailstone?
What is the magnitude of the force exerted by the hail on a 9.0 m x 20.0 m flat roof during the storm ? Ignore the bounce of the hail on impact.

please help me,
Thanks in advance.
 
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Hi, binicle, welcome to the site!
What work have you done on this problem so far?
What values do you know?
What equations might be useful?
 
hi, i only calculated the mass of the hail,
1cm^3 has a mass of 0.920g
--> the hail has volume pi.d^3/6 = 0.905
--> the mass of hail is 1.017
Actually, i have no idea about how to calculate the force!
i think F = mg , but it is wrong.
 
Last edited:
binicle said:
hi, i only calculated the mass of the hail,
1cm^3 has a mass of 0.920g
--> the hail has volume pi.d^3/6 = 0.905
--> the mass of hail is 1.017
Actually, i have no idea about how to calculate the force!
i think F = mg , but it is wrong.

You might want to think more carefully about your mass calculation. It looks to me like that has units of g/cm^4. Hint: Mass would have units of grams.

F = ma, always. So once you have the mass right, see if you can figure out what the acceleration of the hail would be when it hits the roof.

Dorothy
 
i have just calculated again,
V = 0.905 cm^3
and mass = 0.905 x 0.920 = 0.833 g.
F = ma , but i don't know how to find acceleration of the hail? is it a = g =9.81 ?
Also, where will we use v=14 m/s and 9.0 m x 20.0 m flat roof and 110 hails in 1 m^3? I can not figure out anything related to those number.
Because i have one try left to submit answer, please help me.
Thank you.
 
NOTE: I do I know how to reach the final answer. But I think I can go one step further than what you have so far.
If there are 110 hailstones/m^3 of air, then every m^3 of air has a mass of (.833g/hailstone)(110hailstones/m^3 air). What to do next, I have no idea. You'd think a=-g, but the velocity obviously is important as well... What section of physics is this question testing you on?
 
binicle said:
i have just calculated again,
V = 0.905 cm^3
and mass = 0.905 x 0.920 = 0.833 g.
F = ma , but i don't know how to find acceleration of the hail? is it a = g =9.81 ?
Also, where will we use v=14 m/s and 9.0 m x 20.0 m flat roof and 110 hails in 1 m^3? I can not figure out anything related to those number.
Because i have one try left to submit answer, please help me.
Thank you.

The total force on the roof will be the sum of all the forces of all the hails that are hitting it at a given moment.

So you still need to calculate the force of one hail to do that.

If a car is traveling at 30 m/s, and two seconds later it is stopped, it had to accelerate to do that. Could you calculate the acceleration of the car? If it hit something that made it stop in two seconds, what force would it have exerted on the thing it hit?

Unfortunately, I have to log off now, but I hope this gives you enough to solve your problem. If not, maybe someone else will step into help.

Dorothy
 
actually, i don't know how many hail will exert on the roof, because 110 hails in 1m^3 doesn't say anything about the number of the hails.
 
It does, actually. Can you use work/kinetic energy to solve this? You never said where you were in your study of physics.
 
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hi, I'm studing momentum & impulse,
i think the force exerted by the hail at time t is the force of the number of hails instantaneously exerting the roof at time t. Not all the hails in a volume V, just the hails in the surface.
 
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