Calculating a push force using area, speed, and density?

  • #1

Homework Statement


The doorway in the previous question measures 1.06 m x 2.04 m, and the wind blows parallel to the wall surface at 3.89 m.s-1. Calculate the force pushing the curtains out of the doorway. The density of air is 1.29 kg.m-3.

Known data:
A = (1.06m)(2.04m) = 2.162 m^2
v(wind) = 3.89 ms-1
air density = 1.29 kg.m-3

Homework Equations


F=ma
P=F/A
change in P = (density)(g)(h) ???

The Attempt at a Solution


change in P = (density)(g)(h)
= (1.29) (9.8) (2.04)
= 25.79 Pa

P = F/A
F = PA
= (25.79)(2.1624)
= 55.77 N

I know my method is most likely wrong but I'm not too sure what other equations I could use, especially one that involves the velocity of the wind.
 

Answers and Replies

  • #2
34,458
10,572
Hint: Bernoulli discovered something about the pressure of moving fluids.
 
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  • #3
So using Bernoulli's equation:

P1 + (½ * ρ * v1^2) + (½ * ρ * g * h1) = P2 + (½ * ρ * v2^2) + (½ * ρ * g * h2)

Because the wind is flowing parallel, it would be considered horizontal allowing me to cancel out the heights giving me the equation of

P1 + (½ * ρ * v1^2) = P2 + (½ * ρ * v2^2)

Now i'm expecting to be able to calculate P1 and then use F=PA to calculate the final answer (I'm given the area and would calculate P).

However, because this is a before and after type equation, how would I calculate P1 without a value for P2? I am completely stumped. Thanks for your help!
 
  • #4
34,458
10,572
You get a pressure difference, that is sufficient. The absolute pressure does not matter (apart from its influence on the density, but that is taken into account already).
 
  • #5
I'm really sorry but I think I am confusing myself now.

I've rearranged P1 + (½ * ρ * v1^2) = P2 + (½ * ρ * v2^2) to become
P1-P2= ½ * ρ ( v2^2 - V1^2). This would allow me to get a change in pressure.

However, Velocity is not changing so i would be getting a 0 value for v2^2 - V1^2 and wouldn't be able to progress the equation. Am I still on the right track?
 
  • #6
34,458
10,572
There is no changing velocity, but there is a different velocity. One velocity is zero, the other is not.
 
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  • #7
Oh yes, I see. Thankyou very much!

Using that then:

P1-P2= ½ * ρ ( v2^2 - V1^2)
change in pressure = 1/2 (1.29)(3.89^2)
= 9.7602045 Pa.

Am I now right to believe that I can use this value in P=F/A to calculate force?
It is a change in pressure so I'm unsure what 'types' of pressure to use in this calculation and why absolute pressure is not considered.

Edit: My calculations for force using P=F/A gives me an answer of 21.105 N
 
  • #8
34,458
10,572
If you have the same pressure on both sides it doesn't influence the door. Only the pressure difference matters.

Looks fine.
 
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  • #9
Thank you so much! I really appreciate the help :)
 

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