written 6.7 years ago by | modified 2.6 years ago by |
written 2.6 years ago by | • modified 2.6 years ago |
Solution :
Given :
Weight of kite, $W=0.8\mathrm{~kgf} = 0.8 \times 9.81 = 7.848 \mathrm{~N}$
Angle made by kite with horizontal $=10^{\circ}$
Angle made by string with horizontal $ = 45 ^ { \circ }$
$ C_{D}=0.6 $
$C_{L}=0.8$
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
$$T \cos 45^{\circ} =0.3 \times 78.48=23.544 $$ $\begin{aligned} \qquad\qquad T &=\frac{23.544}{\cos 45^{\circ}}=33.3 \mathrm{~N} \end{aligned}$