Chapter 9, Problem 66GP

To determine

To Calculate: The magnitude of drag force, $F_D$ and its point of action above the centre of mass.

Answer to Problem 66GP

  $5\times {10}^5\text{N}$ & $2.6\text{m}$

Explanation of Solution

Given:

The thrust force of the engine, $F_T=5\times {10}^5\text{N}$

The mass of the aircraft, $m=67000\text{kg}$

Formula used:

Let the drag force is acting x distance above the centre of mass (CM) of aircraft.

  

From the equilibrium condition in x & y -direction:

  $\begin{array}{l}{\displaystyle \sum F_x}=0\\ F_D-F_T=0\mathrm{....}(1)\end{array}$

  $\begin{array}{l}{\displaystyle \sum F_y=0}\\ F_L-mg=0\mathrm{...}(2)\end{array}$

  $\begin{array}{l}{\displaystyle \sum M_{CM}=0}\\ F_D\times x-F_L\times 3.2+F_T\times 1.6=0\mathrm{....}(3)\end{array}$

Calculation:

Substituting the given values:

From equation (1) & (2):

  $\begin{array}{l}{F}_D=F_T=5\times {10}^{\text{5}}\text{N}\\ F_L=mg=67000(9.8)=6.57\times {10}^5\text{N}\end{array}$

Substitute the value of drag force and lift force in equation (3):

  $\begin{array}{l}(5\times {10}^5)x-(6.57\times {10}^5)3.2+5\times {10}^5(1.6)=0\\ 5x=-8.02+21.024\\ 5x=13\\ x=\frac{13}{5}=2.6\text{m}\end{array}$

Conclusion:

The drag force is equal to the thrust force of the engine and is acting 2.6 m above the centre of mass (CM) of aircraft.