Chapter 13, Problem 13.190RP

To determine

The landing speed $\left(v_1\right)$ of the airplane.

Answer to Problem 13.190RP

The landing speed $\left(v_1\right)$ of the airplane is $\underset{\_}{102.6\text{mi}/\text{hr}}$.

Explanation of Solution

Given information:

The weight of the airplane (W) is 32,000 lb.

The distance traveled by the airplane (d) is 95 ft.

The vertical height of BC (x) is $35\text{ft}$.

The vertical height of CA (y) is $35\text{ft}$.

Calculation:

Show the diagram of the airplane and aircraft carrier as in Figure (1).

Calculate the mass of the plane $\left(m\right)$ using the formula:

$m=\frac{W}{g}$

Here, g is the acceleration due to gravity.

Substitute 32,000 lb for $W$ and $32.2\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{c}m=\frac{32,000\text{lb}}{32.2\text{ft}/{\text{s}}^2}\\ =993.79\text{lb}\cdot {\text{s}}^2/\text{ft}\end{array}$

Use the Pythagorean theorem:

Calculate the length of the cable portion $l_{AC}$ and $l_{BC}$.

$\begin{array}{c}{l}_{AC}=l_{BC}\\ =\sqrt{d^2+{\left(\frac{l_{AB}}{2}\right)}^2}\end{array}$

Here, $l_{AC}$ is the inclined length of the cable AC and $l_{BC}$ is the inclined length of the cable.

Substitute 95 ft for $d$ and 70 ft for $l_{AB}$.

$\begin{array}{c}{l}_{AC}=l_{BC}\\ =\sqrt{{\left(95\text{ft}\right)}^2+{\left(\frac{70\text{ft}}{2}\right)}^2}\\ =\sqrt{9025+1225}\\ =101.24\text{ft}\end{array}$

Calculate the work done by the cable at the airplane hook in arresting the plane $\left(U_{1-2}\right)$ using the formula:

$\begin{array}{c}{U}_{1-2}=-Q\left(\Delta l_{AB}\right)\\ =-Q\left(l_{AC}+l_{BC}-l_{AB}\right)\end{array}$

Here, Q is the tension acting of the arresting cable, $\Delta l_{AB}$ is the extension in the arresting cable AB, $l_{AB}$ is the initial length of the cable AB, $l_{AC}$ is the length of the cable portion AC after arresting the airplane, and $l_{BC}$ is the length of the cable portion BC after arresting the airplane.

Substitute 85 kips for $Q$, 70 ft for $l_{AB}$, and $101.24\text{ft}$ for $l_{AC}$ and $l_{BC}$.

$\begin{array}{c}{U}_{1-2}=-\left(85\text{kips}\right)\left(101.24\text{ft}+101.24\text{ft}-70\text{ft}\right)\\ =-\left(85\times \frac{1,000\text{lb}}{1\text{kips}}\right)\left(132.48\right)\\ =-11,260,800\text{ft}\text{.lb}\end{array}$

Write the expression for the initial kinetic energy of the airplane on landing $\left(T_1\right)$ as follows:

$T_1=\frac{1}{2}m{v}_1^2$

Here, $T_1$ is the initial kinetic energy of the airplane on landing and $v_1$ is the initial landing velocity of the airplane.

Write the expression for the final kinetic energy of the airplane after landing $\left(T_2\right)$ as follows:

$T_2=\frac{1}{2}m{v}_2^2$

Here, $T_2$ is the final kinetic energy of the airplane and $v_2$ is the final velocity of the airplane.

Write the expression for the principle of work and energy to the airplane as follows:

$T_1+U_{1-2}=T_2$

Substitute $\frac{1}{2}m{v}_1^2$ for $T_1$ and $\frac{1}{2}m{v}_2^2$ for $T_2$.

$\frac{1}{2}m{v}_1^2+U_{1-2}=\frac{1}{2}m{v}_2^2$

Substitute $993.79\text{lb}\cdot {\text{s}}^2/\text{ft}$ for $m$, $-11,260,800\text{ft}\text{.lb}$ for $U_{1-2}$, and 0 for $v_2$.

$\begin{array}{l}\frac{1}{2}\left(993.79\text{lb}\cdot {\text{s}}^2/\text{ft}\right)v_1^2-11,260,800\text{ft}\text{.lb}=\frac{1}{2}m\left(0\right)\\ \frac{1}{2}\left(993.79\right)v_1^2=11,260,800\\ v_1^2=\frac{\left(11,260,800\right)\left(2\right)}{993.79}\\ v_1^2=22,662.33\end{array}$

$\begin{array}{c}{v}_1=\sqrt{22,662.33}\\ v_1=150.54\text{ft}/\text{s}\\ v_1=150.54\times \left(\frac{\text{0}\text{.000189394}\text{mi}}{1\text{ft}}\right)\times \left(\frac{3,600\text{s}}{1\text{hr}}\right)\\ v_1=102.6\text{mi}/\text{hr}\end{array}$

Therefore, the landing speed $\left(v_1\right)$ of the airplane is $\underset{\_}{102.6\text{mi}/\text{hr}}$.