- #1
Argonaut
- 45
- 24
- Homework Statement
- [ Young & Freedman - University Physics 13E, Ex 6.104]
An airplane in flight is subject to an air resistance force proportional to the square of its speed v. But there is an additional resistive force because the airplane has wings. Air flowing over the wings is pushed down and slightly forward, so from Newton's third law the air exerts a force on the wings and airplane that is up and slightly backward (Fig. P6.104). The upward force is the lift force that keeps the airplane aloft, and the backward force is called induced drag. At flying speeds, induced drag is inversely proportional to $v^2$, so the total air resistance force can be expressed by ##F_{air} = \alpha v^2 + \beta /v{^2}##, where ##\alpha## and ##\beta## are positive constants that depend on the shape and size of the airplane and the density of the air. For a Cessna 150, a small single-engine airplane, ##\alpha = 0.30 ~\rm{N} \cdot ~\rm{s^{2}/m^{2}}## and ##\beta = 3.5 \times 10^5 ~\rm{N} \cdot ~\rm{m^2/s^2}##. In steady flight, the engine must provide a forward force that exactly balances the air resistance force. (a) Calculate the speed (in km/h) at which this airplane will have the maximum range (that is, travel the greatest distance) for a given quantity of fuel. (b) Calculate the speed (in km/h) for which the airplane will have the maximum endurance (that is, remain in the air the longest time).
- Relevant Equations
- Work, kinetic force, power
Is my solution correct? (I only have answers to odd-numbered exercises.)
Is it a good solution or have I overcomplicated things?
(a)
The forward force provided by the engine balances the air resistance force, so ##F_{engine}=F_{air} = \alpha v^2 + \beta /v{^2}##.
Let ##W_{engine}## be the energy content of the given quantity of fuel. Then ##W_{engine} = F_{engine}d ## where ##d## is range. So
$$ d = \frac{W_{engine}}{F_{engine}} = \frac{W_{engine}}{\alpha v^2 + \beta /v{^2}} $$
We want to maximise ##d##. We can achieve that if we minimise ##F_{engine}##, since ##W_{engine}## is a constant.
We use calculus to minimise it. Let ##f(v) = \alpha v^2 + \beta /v{^2}##. Then ##f'(v) = 2\alpha - \frac{2\beta}{v^3}##. We find the minimum value by setting ##f'(v) = 0## and rearranging it to express ##v##, we obtain
$$
v = \left(\frac{\beta}{\alpha}\right)^{1/4} = \left(\frac{3.5 \times 10^5 ~\rm{N} ~\rm{m^2/s^2}}{0.30 ~\rm{N} ~\rm{s^{2}/m^{2}}}\right)^{1/4} = 33 ~\rm{m/s} = 120 ~\rm{km/h}
$$
Thus the airplane will achieve the maximum range travelling at a speed of ##120 ~\rm{km/h}##.
(b)
##P_{av} = \frac{\Delta W}{\Delta t}## and we want to maximise ##\Delta t##. We can achieve this by minimising ##P##, since ##\Delta W = W_{engine}## is a constant. We use calculus to minimise ##P##. Let ##P = g(v)##. Then ##g'(v) = 3\alpha v^2-\frac{\beta}{v^2}##. Setting ##g'(v)=0## and rearranging it to express ##v##, we obtain
$$
v= \left(\frac{\beta}{3\alpha}\right)^{1/4} = \left(\frac{3.5 \times 10^5 ~\rm{N} ~\rm{m^2/s^2}}{3(0.30 ~\rm{N} ~\rm{s^{2}/m^{2}})}\right)^{1/4} = 25 ~\rm{m/s} = 90 ~\rm{km/h}
$$
Therefore, the airplane will achieve maximum time in air at a speed of ##90 ~\rm{km/h}##.