Question

An airplane weighing \(40\ \text{kN}\) is landing on a horizontal runway and is retarded by an arresting cable. The tension in the arresting cable at a given instant is \(100\ \text{kN}\), and the cable makes \(10^\circ\) with the runway as shown. Assume engine thrust continues to balance airplane drag. The magnitude of the horizontal load factor is \underline{\hspace{1cm}}. \;(round off to one decimal place) \begin{center} \includegraphics[width=0.5\textwidth]{09.jpeg} \end{center}

Hint:

When thrust and drag balance, the arresting cable's \(\cos\)-component sets the decelerating force. Load factor in any direction is simply \(\text{(net force in that direction)}/W\).

Answer 1

Step 1: Define horizontal load factor.
The horizontal (axial) load factor is the ratio of the net horizontal force to the weight: \[ n_x=\frac{|F_x|}{W}. \]

Step 2: Identify horizontal forces.
Thrust balances drag \(\Rightarrow\) their horizontal effects cancel. Thus the only net horizontal force is the horizontal component of cable tension: \[ F_x=T\cos 10^\circ = 100\cos 10^\circ\ \text{kN}. \]

Step 3: Compute \(n_x\).
\[ n_x=\frac{100\cos 10^\circ}{40} = \frac{100\times 0.9848}{40} \approx 2.462 \;\Rightarrow\; \boxed{2.5}\ \text{(to one decimal place)}. \] (Note) The vertical component of cable tension is balanced by a change in lift; it does not affect the horizontal load factor definition.