# The force vectors on an aeroplane in steady flight & numerical

In this post, we will study a probable set of force vectors on an aeroplane in steady flight. The forces acting on an aeroplane (airplane) are shown in Figure 1.

## force vectors on an aeroplane in steady flight

Let’s list down the force components acting on the airplane when it’s having a steady flight.

The magnitude (strength) of the forces are indicated by

- T: the thrust provided by the engines,
- W: the weight,
- D: the drag (acting against the direction of flight) and
- L: the lift (taken perpendicular to the path.)

In a more realistic situation force vectors in three dimensions would need to be considered. But here, for simplicity, we are considering only 2 dimensions.

### resolving forces

**As the plane is in steady flight the sum of the forces in any direction is zero.**

(If this were not the case, then, by Newton’s second law, the non-zero resultant force would cause the aeroplane to accelerate.)

So, resolving forces in the direction of the path:**T cos α − D −W sin β = 0** …………………….. (1)

Then, resolving forces perpendicular to the path:**T sin α + L −W cos β = 0** ……………………… (2)

Now, let’s take a * use case* based numerical problem where we will be using the above equations to find the unknown parameter value.

## Numerical problem **based force vectors on an aeroplane in steady flight**

**If the plane has mass 72 000 tonnes, the drag is 130 kN, the lift is 690 kN and β = 6° find the magnitude of the thrust and the value of α to maintain steady flight.**

Solution:

**T cos α = D + W sin β **= 130000 + (72000)(9.81) sin 6* °* = 203830.54 …. (1)

**T sin α = W cos β – L** = (72000)(9.81) cos 6

*− 690000 = 12450.71…..(2)*

**°**Hence, **tan** **α** = **T sin α** / **T cos α** = 12450.71 / 203830.54 =0.061084

**α** = 3.50 **°**

and consequently, for the thrust: (using equation 1 or 2 and the value of **α** )

T = 204210 N.