Chapter 13, Problem 13.190RP

A 32,000-Ib 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. Knowing that the piston-cylinder system maintains a constant tension of 85 kips in the cable during the entire landing, determine the landing speed of the airplane if it travels a distance $d=95$ ft after being caught by the cable.

    

To determine

The landing speed of the airplane after it being caught by the cable.

Answer to Problem 13.190RP

$v=102.6mi/h$

Explanation of Solution

Given:

Mass of the airplane $m=\frac{W}{g}=\frac{32,000lb}{32.2ft/s^2}=993.79lb\cdot s^2/ft$

Distance travelled $d=95ft$

Tension in the cable = $85kips$

Work of arresting cable force, $Q=85kips=85000lb$

Calculation:

As the cable is pulled out, the cable acts parallel with the cable at the airplane hook. For a small displacement,

$\Delta U=-Q(\Delta I_{AC})-Q(\Delta I_{BC})$

Here, Q is the work done by the cable force and ?l is the change in length of the cable.

Since Q is constant in both the given conditions

$\begin{array}{l}{U}_{1\to 2}=-Q[\overline{AC}+\overline{BC}-\overline{AB}]\\ \therefore \overline{AC}=\overline{BC}=\sqrt{{(35)}^2+{(95)}^2}=101.24ft\\ U_{1\to 2}=-(85000)[101.24+101.24-70]\\ U_{1\to 2}=-11.261ft\text{.lb}\end{array}$

Now by applying Principle of work and energy:

$\begin{array}{l}{T}_1+U_{1\to 2}=T_2\\ \frac{1}{2}m{v}_1^2+U_{1\to 2}=\frac{1}{2}m{v}_2^2\end{array}$

Since, $v_2=0$ we get

$\begin{array}{l}{v}_1^2=-\frac{2U_{1\to 2}}{m}\\ v_1^2=-\frac{(2)(-11.261)}{993.79}\\ v_1^2=22.663\times {10}^3{f}^{2}t/s^2\\ v=150.54ft/s=102.6mi/h\end{array}$

Conclusion:

The landing speed of the airplane is $v=102.6mi/h.$