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Air streams horizontally past a small airplane`s wing such that the speed is 70.0m/s over the top surface and 60.0m/s past the bottom surface. if the plane has a wing area of 16.2 m^2 on the top and on the bottom , what is the net vertical force that the air exerts on the airplane? The density of the air is 1.20kg/m^3
Air streams horizontally past a small airplane`s wing such that the speedis 70.0m/s over the top surface and 60.0m/s past the bottom surface. ifthe plane has a wing area of 16.2 m^2 on the top and on the bottom , what is the net vertical force that the air exerts on the airplane? The density of the air is 1.20kg/m^3

```
6 years ago

```							    This is Bernoulli's principle problem (you need to assume laminar flow; otherwise the calculation is near impossible). Bernoulli's principle says, (1/2)*density*(flow velocity)^2 + density*g*height + Pressure = constant. Between the top of the wing and the bottom, the height difference should
be small enough that we can ignore it. Then, we have (rho = density, v =
flow velocity, and P = pressure) (1/2)*rho*v^2 + P = constant. When the air gets cut through by the airplane wing, although their flow
velocity changes, we have no reason to think that the constant on the
right hand side (which is related to energy conservation, by the way)
will change. So, that must mean that the pressure should change to
compensate for the change in velocity, so that the constant remains,
well constant. Putting this into equation, we have, (1/2)*rho*v_below^2 + P_below = (1/2)*rho*v_above^2 + P_above. P_above. - P_below = 1/2 *rho*(v_below^2 - v_above^2 )P_above. - P_below = 1/2 *1.2(60^2 - 70^2) = ) = 0.6*(3600-4900) = -780 N/m^2or we can say P_net = 780 N/m^2We know everything necessary to calculate P_above - P_below. Once we
have that, then we use the fact that F_net = area*P_net, = 16.2*780 = 12636 NThanks & Regards,Nirmal SinghAskiitians Faculty

```
6 years ago
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