Hurricane forces — Comparing the force from a 60 mph wind to a 120 mph wind

AI Thread Summary
The discussion revolves around understanding the relationship between wind speed and the force exerted by wind, specifically comparing 60 mph and 120 mph winds. Participants clarify that mass (m) can be expressed in terms of air density (ρ) and volume (V), leading to the equation m = ρV. There is confusion regarding the use of symbols for momentum and volume, with participants emphasizing the importance of using distinct variables to avoid misunderstandings. The kinetic energy equation is also discussed, with corrections made to ensure clarity in the representation of variables. Ultimately, the conversation highlights the need for precision in mathematical notation when discussing physical concepts.
Shaye
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Homework Statement
The arrival of a hurricane is predicted with wind speeds of 120 mph (miles per hour). The force exerted by a 60 mph gale on the sides of a train locomotive has been measured and found to be F. Will the hurricane forces on the same locomotive be:

1. The same as F
2. Twice as large as F
3. Three times as large as F
4. Four times as large as F
Relevant Equations
F=Δp/Δt and Δp = Δ(mv)
The marker wrote that the answer is 4 and it's because m and v double. I don't understand how m doubles??
 
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Can you find an expression for m in terms of the air density and the wind speed?
 
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Orodruin said:
Can you find an expression for m in terms of the air density and the wind speed?
The only thing I can think of is p=m/v and m=pv and ke = 1/2(pv)*v^2...but I'm still at a loss
 
Shaye said:
The only thing I can think of is p=m/v and m=pv and ke = 1/2(pv)*v^2...but I'm still at a loss
So, you now have impulse = mv and ##m = \rho v A## ... (you need the area to make a mass/time out of the density.
 
Shaye said:
The only thing I can think of is p=m/v and m=pv and ke = 1/2(pv)*v^2...but I'm still at a loss
That confused me. p is usually used for momentum and lowercase v for velocity, leading to equations like p=mv. So I think you mean ##m=\rho V##, V being volume. But then you have ##ke = 1/2(pv)*v^2##, which uses v for velocity and volume, so write ##KE = 1/2(\rho V)*v^2##.
Next is to consider what volume of air impinges upon the train in time t.
 
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haruspex said:
That confused me. p is usually used for momentum and lowercase v for velocity, leading to equations like p=mv. So I think you mean ##m=\rho V##, V being volume. But then you have ##ke = 1/2(pv)*v^2##, which uses v for velocity and volume, so write ##KE = 1/2(\rho V)*v^2##.
Next is to consider what volume of air impinges upon the train in time t.
Oh yes, using v instead of V for volume clearly confused me too to see what I expected to see. @Shaye Never use the same letter for different quantities in the same problem.
 
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