A kite weighing 0.8kg effective area= $0.8m^2$. It is maintained in the air at an angle of $10^0$ to the horizontal. The string attached to the kite makes an angle of $45^0$ to the horizontal and at this position, the value of Cd = 0.6, CL = 0.8 respectively. Find the speed of the wind and the tension in the string.$ Take \rho = 1.25 kg/m^3$.

Solution :

Given :

1. Weight of kite, $W=0.8\mathrm{~kgf} = 0.8 \times 9.81 = 7.848 \mathrm{~N}$

2. Angle made by kite with horizontal $=10^{\circ}$

3. Angle made by string with horizontal $ = 45 ^ { \circ }$

4. $ C_{D}=0.6 $

5. $C_{L}=0.8$

6. Density of air, $ \rho=1.25 \mathrm{~kg} / \mathrm{m}^{3}$

Let the speed of the wind $=v \mathrm{~m} / \mathrm{s}$ and tension in string $=T$

The free body diagram for the kite

Drag force,

$F_{D}=$ Component of $T$ along $X-X$ $ \quad\space =T \cos 45^{\circ}$ But drag force $F_D$ is also $\begin{aligned} F_{D} =C_{D} \times A \times \frac{\rho v^{2}}{2} \end{aligned}$ $\begin{aligned} \quad \space &=\frac{0.6 \times 0.8 \times 1.25 \times v^{2}}{2} \end{aligned}$ $\begin{aligned} =0.3 v^{2} \end{aligned}$ Equating the two values of $F_{D},$ we get $$T \cos 45 ^{\circ} =0.3 v^{2}$$..............(i)

lift force from Figure,

$F_{L}=$ Component of T vertically downward + Weight of kite

$\quad\space\space =T \sin 45^{\circ}+7.848$

But lift force $F_L$ also is

$\begin{aligned} F_{L} =\frac{C_{L} \times A \times \rho v^{2}}{2} \end{aligned}$ $\begin{aligned}\quad\space=\frac{0.8 \times 0.8 \times 1.25 \times v^{2}}{2.0}\end{aligned}$ $\begin{aligned} =0.4 v^{2} \end{aligned}$

Equating the two values of $F_{L}, $ we get

$T \sin 45 ^{\circ} +7.848=0.4 v^{2}$

or

$ T \sin 45^{\circ}=0.4 v^{2}-7.848$..........(ii)

$\therefore T \cos 45^{\circ} =T \sin 45^{\circ}$ $\qquad\space \space 0.3 v^{2} =0.4 v^{2}-7.848$ $\qquad \space\space 7.848 =0.4 v^{2}-0.3 v^{2}=0.1 v^{2} $ $\begin{aligned} \qquad\quad\space\space \space v^{2} &=\frac{7.848}{0.1}=78.48\end{aligned}$ $\begin{aligned} \qquad \qquad \space \space&=\sqrt{78.48}=8.86 \mathrm{~m} / \mathrm{s}\\ \end{aligned}$

Subsituting the value of $v^{2} =78.48$ in equation (i), we get

$$T \cos 45^{\circ} =0.3 \times 78.48=23.544 $$ $\begin{aligned} \qquad\qquad T &=\frac{23.544}{\cos 45^{\circ}}=33.3 \mathrm{~N} \end{aligned}$