2007-05-23 04:17:14 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Lift is an aerodynamic force that would depend on the geometry of the body against the wind.
The push force however if exceeded could make a person airborne. A person of your description has a frontal area of about 0.7 m^2
A force of friction is say f=0.5 m g
200lb(mass)=90.72kg
now f = 0.5 x 90.72 x 9.81 m/s^2 = 445N
A 35m/s wind or 126km/hour ( about 80mph)
will produce a force on the body = 490N
Than means it will slide a person across ground with little effort.
A wind over 100 – 120 mph would probably make one 'fly'.
2007-05-23 04:29:10 · answer #1 · answered by Edward 7 · 1⤊ 0⤋
The person will lift off when F > W; where F is the vertical upward force of the wind on the person and W is the vertical downward weight of the person. In which case F - W = f = ma > 0; so there is a net force f acting upward on the mass of the person to give it an upward acceleration.
In other words, to lift a person off the ground, the upward velocity of the wind must be sufficient to yield an upward force of a bit over 150 to 200 lbs...the range of weights you gave. That upward force, coming from wind, is called drag. Its equation is F = 1/2 rho Cd A v^2; where rho is the density of the air, Cd is a coefficient of drag, A is the cross-sectional area of the person, and v is the vertically upward wind velocity that will give us F > W. For a free falling person, v ~ 120 mph when F = W so the velocity stays constant and we call it "terminal velocity."
So, we know the vertical velocity to begin lifting a person will be at least 120 mph. But, and this is a BIG BUT, that is not necessarily the "wind speed" you are looking for. For one thing, in practical terms, how would that vertical velocity be generated? Clearly, it would be just the vertical component of the real wind velocity (your wind speed), which will have a horizontal component as well as vertical. Thus, V^2 = v^2 + u^2; where V is the wind velocity, v is the vertical component that lifts the person, and u is the horizontal component that pushes the person along the ground. That push would be P = 1/2 rho Cp A' u^2; where the Cp and A' are different from Cd and A because the person is not symetrical and the coefficients and cross-sectional areas vary according to the relative directions of the wind. That is P <> F generally speaking; so u <> v, again, generally speaking.
Thus, the best we can say in response to your question is that V > v because it also includes u in the vector sum. And u > 0 generally speaking. So, the wind speed needed to lift someone is somewhat greater than the terminal velocity of around 120 mph. And, you must admit, that would be some blow.
2007-05-23 05:27:22 · answer #2 · answered by oldprof 7 · 0⤊ 0⤋
(I assumed that the wind is blowing straight up,)
That would be equal to or greater than the terminal velocity of the person. On average it is around 200 km/h(124mph)
You can find the equation at the source:
2007-05-23 04:26:27 · answer #3 · answered by Blank 2 · 1⤊ 0⤋