The force of wind blowing on a wall at an angle

[hyperddude]

Homework Statement

Say there is a wall with width [itex]w[/itex] meters and height [itex]h[/itex] meters. There is a wind with velocity v m/s blowing on the wall at a 45 degree angle. The collisions of the air molecules with the wall are perfectly elastic. What is the magnitude of the force on the wall? (The density of air is 1.2 kg/m^3).

Homework Equations

[itex]Δp = FΔt[/itex] (p is momentum)

The Attempt at a Solution

In one second, an air particle moves v meters. The area of the wall is wh, so the volume of the air that hits a surface in one second is vwh. The density is 1.2kg/m^3, so the mass would be 1.2*vwh. My next step would be to find the change in velocity, which I think is [itex]v\sin{45°} + v\sin{45°} = \sqrt{2}v[/itex].

So, the change in momentum would be [itex]mv = 1.2\sqrt{2}v^2wh[/itex]. The time is 1 second, so the force would be [itex]1.2\sqrt{2}v^2wh[/itex] Newtons. Is my method correct?

 

[Dick]

Homework Statement

Say there is a wall with width [itex]w[/itex] meters and height [itex]h[/itex] meters. There is a wind with velocity v m/s blowing on the wall at a 45 degree angle. The collisions of the air molecules with the wall are perfectly elastic. What is the magnitude of the force on the wall? (The density of air is 1.2 kg/m^3).

Homework Equations

[itex]Δp = FΔt[/itex] (p is momentum)

The Attempt at a Solution

In one second, an air particle moves v meters. The area of the wall is wh, so the volume of the air that hits a surface in one second is vwh. The density is 1.2kg/m^3, so the mass would be 1.2*vwh. My next step would be to find the change in velocity, which I think is [itex]v\sin{45°} + v\sin{45°} = \sqrt{2}v[/itex].

So, the change in momentum would be [itex]mv = 1.2\sqrt{2}v^2wh[/itex]. The time is 1 second, so the force would be [itex]1.2\sqrt{2}v^2wh[/itex] Newtons. Is my method correct?

I think you are overestimating the mass of the air hitting the wall. If the wall were directly facing the wind the volume would be vwh. But it's tilted away from the wind, shouldn't it be less? And I think your answer for the change in velocity came out ok, but you should really be doing it by taking the difference of two vectors.

 

[hyperddude]

I think you are overestimating the mass of the air hitting the wall. If the wall were directly facing the wind the volume would be vwh. But it's tilted away from the wind, shouldn't it be less?

Oh, good catch. I believe the mass should be [itex]1.2*vwh\sin{45°}[/itex]. So the force is [itex]1.2v^2wh[/itex]?

 

[Dick]

I think so.

 

[hyperddude]

Thanks!