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

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Homework Help Overview

The discussion revolves around understanding the forces exerted by winds of different speeds, specifically comparing the effects of 60 mph and 120 mph winds. Participants are exploring the relationship between mass, air density, and wind speed in the context of physics principles related to momentum and kinetic energy.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to derive expressions for mass in terms of air density and wind speed. There is confusion regarding the use of variables for momentum and volume, leading to discussions about the correct formulas and their implications.

Discussion Status

The discussion is active, with participants providing guidance on the relationships between variables and clarifying the use of notation. Some have expressed confusion over the terminology and notation used in the equations, while others are working through the implications of these relationships.

Contextual Notes

There is a noted confusion regarding the use of symbols for different physical quantities, which may affect participants' understanding of the problem. The discussion reflects a collaborative effort to clarify these concepts without reaching a definitive conclusion.

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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