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.

figure 1: The force vectors on an aeroplane/airplane in steady flight — **figure 1: The force vectors on an aeroplane in steady flight**

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.