Chapter 13, Problem 13.190RP

A 34,000-lb airplane lands on an aircraft carrier and is caught by an arresting cable. The cable is inextensible and is paid out at A and B from mechanisms located below deck and consisting of pistons moving in long oil-filled cylinders. The piston–cylinder system is designed to maintain a constant tension in the cable. Knowing that the landing speed is 110 mi/h and the airplane travels a distance d = 90 ft after being caught, determine the tension in the cable.

Fig. P13.190

To determine

Find the tension in the cable (F).

Answer to Problem 13.190RP

The tension in the cable (F) is $\underset{\_}{111.6\text{kips}}$.

Explanation of Solution

Given information:

The landing speed $\left(v_0\right)$ is $11\text{0mi/hr}$.

The weight of the airplane (W) is $34000\text{lb}$.

The distance traveled by the airplane (d) is $90\text{ft}$.

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

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

The acceleration due to gravity (g) is $32.2{\text{ft/s}}^{\text{2}}$.

Calculation:

Calculate the mass of the airplane (m) using the formula:

$\begin{array}{c}W=mg\\ m=\frac{W}{g}\end{array}$

Substitute $34000\text{lb}$ for W and $32.2{\text{ft/s}}^{\text{2}}$ for g.

$\begin{array}{c}m=\frac{34000}{32.2}\\ =1055.9\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}\end{array}$

Convert the unit of landing speed $\left(v_0\right)$ from $\text{mi/hr}$ to $\text{ft/s}$:

$v_0={\left(v_0\right)}_{\text{mi/hr}}\times \left(5280\frac{\text{ft}}{\text{mi}}\right)\times \left(\frac{1}{3600}\frac{\text{hr}}{\text{sec}}\right)$

Here, ${\left(v_0\right)}_{\text{mi/hr}}$ is the landing speed in $\text{mi/hr}$.

Substitute $11\text{0mi/hr}$ for ${\left(v_0\right)}_{\text{mi/hr}}$.

$\begin{array}{c}{v}_0=110\times \left(5280\frac{\text{ft}}{\text{mi}}\right)\times \left(\frac{1}{3600}\frac{\text{hr}}{\text{sec}}\right)\\ =161.33\text{ft/s}\end{array}$

Calculate the amount of cable $\left(\Delta d\right)$ using the relation:

$\Delta d=\left(\sqrt{d^2+y^2}\right)-x$

Substitute $35\text{ft}$ for x, $90\text{ft}$ for d, and $35\text{ft}$ for y.

$\begin{array}{c}\Delta d=\left(\sqrt{{90}^2+{35}^2}\right)-35\\ =\sqrt{9325}-35\\ =\text{61}\text{.566ft}\end{array}$

Calculate the initial kinetic energy of the airplane $\left(T_1\right)$ using the formula:

$T_1=\frac{1}{2}m{v}_0^2$

Substitute $1055.9\text{lb}\cdot {\text{s}}^{\text{2}}/\text{ft}$ for m and $161.33\text{ft/s}$ for $v_0$.

$\begin{array}{c}{T}_1=\frac{1}{2}\times 1055.9\times {\left(161.33\text{ft/s}\right)}^2\\ =13741149.41\text{lb}\cdot \text{ft}\end{array}$

The final kinetic energy of the airplane $\left(T_2\right)$ is zero.

Calculate the work done by the airplane $\left(U_{1-2}\right)$ using the relation:

$U_{1-2}=-2F\left(\Delta d\right)$

Substitute $\text{61}\text{.566ft}$ for $\Delta d$.

$\begin{array}{c}{U}_{1-2}=-2F\left(61.566\right)\\ =-123.132F\end{array}$

The expression for the work energy in the airplane as follows:

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

Substitute $13741149.41\text{lb}\cdot \text{ft}$ for $T_1$, $-123.132F$ for $U_{1-2}$, and 0 for $T_2$.

$\begin{array}{l}13741149.41\text{lb}\cdot \text{ft}-123.132F=0\\ 123.132F=13741149.41\\ F=\frac{13741149.41}{123.132}\\ F=111.6\text{kips}\end{array}$

Therefore, the tension in the cable (F) is $\underset{\_}{111.6\text{kips}}$.