Chapter 13, Problem 13.190RP

34,000-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 Bfrom 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.

To determine

The tension in the cable.

Answer to Problem 13.190RP

The tension in the cable is $Q=111596.9lb=111.6kips$

Explanation of Solution

Given information:

Weight of airplane is equal to $34000lb$

Landing speed is equal to $110mi/h$

Travel distance is equal to $d=90ft$

The kinetic energy of a particle is defined as:

$T=\frac{1}{2}m{v}^2$

Principle of work and energy is defined as:

$U_{1\to 2}=T_2-T_1$

Above equation states “If a particle moves from A₁ to A₂ under an action of force F, the work of the force F is equal to the change in kinetic energy.”

Calculation:

Find the mass of airplane,

$m=\frac{W}{g}=\frac{34000lb}{32.2ft/s^2}=1055.9lb.s^2/ft$

The cable tension acts parallel to the cable at the airplane hook.

We know that, the tension $Q$ is constant.

Therefore,

$U_{1\to 2}=-Q\left[\overline{AC}+\overline{BC}-\overline{AB}\right]$

Find $\overline{AC}.$

$\overline{AC}=\overline{BC}=\sqrt{{35}^2+{90}^2}=96.566ft$

Therefore,

$U_{1\to 2}=-Q\left[96.566ft+96.566ft-70ft\right]=-123.132Q$

Apply principle of work and energy:

$U_{1\to 2}=T_2-T_1$

Substitute,

$\begin{array}{l}{U}_{1\to 2}=T_2-T_1\\ -123.132Q=0-\frac{1}{2}m{v}^2\end{array}$

Convert,

$v=110mi/h=161.33ft/s$

Substitute and solve to find tension.

$\begin{array}{l}-123.132Q=0-\frac{1}{2}\left(1055.9lb.s^2/ft\right){\left(161.33ft/s\right)}^2\\ Q=111596.9lb\end{array}$

Convert,

$Q=111596.9lb=111.6kips$

Conclusion:

The tension in the cable is calculated by putting the values of velocity and height in an equation.