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I belong to the aerospace community, and a basic concept among the people of this community is that an aircraft is called to be in 'steady flight' when the lift force balances out the weight of the aircraft (ofcourse, the thrust must also balance out the drag, but that is not the subject of discussion here), as shown in the image below.

enter image description here

Here, the weight of the aircraft is assumed to be m*g, where m is the mass of the aircraft while g is the gravitational acceleration (almost 9.81 m/s^2). The net force on the aircraft (in the vertical axis) becomes equal to zero, and hence there is no net acceleration.

Now what is the definition of weight? Does it always need to be equal to m*g, or it can be m*a as well, where a is not equal to g (when the aircraft is accelerating upwards)? If the weight is equal to m*g always, then where does the inertial forces come from when there is an acceleration upwards? If the weight of the aircraft can be m*a (where a is greater than g), then why does the aircraft even accelerate upwards since Lift will always be equal to weight?

Rameez Ul Haq
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  • The image portrays straight and level flight. Lift is not always equal in magnitude to weight and is not always anti-parallel to weight. Thrust is not always equal in magnitude to drag and is not always anti-parallel to drag. Lift/weight are not always orthogonal to thrust/drag. – David Hammen Jan 19 '22 at 09:44
  • Ofcourse it is not. I posted this condition for simplification because its a physcis community and not an aviation community. – Rameez Ul Haq Jan 19 '22 at 12:37
  • You specifically asked then why does the aircraft even accelerate upwards since Lift will always be equal to weight? The answer is simple: Lift is not always equal to weight. – David Hammen Jan 19 '22 at 13:02
  • @DavidHammen, this is not an aviation question. I know the complete physics of why and how it would accelerate upwards, I think you didn't understand what I was looking for. I wanted to understand if the weight of aircraft will change when it accelerates upwards? Re-read the last paragraph i have written in my posted question, kindly. – Rameez Ul Haq Jan 19 '22 at 16:12
  • @RameezUlHaq The weight of the aircraft will not change, it will always be $mg$. In order for the plane to accelerate upwards there must be a net upward force, i.e. there must be an upward force greater in magnitude than the weight. For the plane to accelerate upwards at acceleration $a$, the upward force must be $F_\text{up} = ma + mg$. – Marko Gulin Jan 19 '22 at 16:18

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I am going to explain this on a much simpler example - an elevator accelerating upwards. Imagine a person standing on a scale in an elevator accelerating upwards. What would the scale show?

To answer this question, we must observe the person from an inertial reference frame such as Earth. For an observer on ground, the resultant (net) force on the person in the elevator is:

$$\vec{F}_\text{net} = m a \hat{\jmath}$$

where positive direction of the vertical $\hat{\jmath}$ axis is defined to be upwards (away from Earth's center). In terms of the second Newton's law

$$\sum_{i} \vec{F}_i = m \vec{a}$$

the resultant force is on the right-hand side of the equation since that is a consequence, and all forces that act on the person are on the left-hand side of the equation. When thinking about motion, you should never mix causation (left-hand side) and consequence (right-hand side)! Now what forces act on the person in the elevator? That would be the gravitational force by Earth and normal force by the floor (scale), hence

$$\vec{n} + \vec{w} = \vec{F}_\text{net}$$

where the scale measures magnitude of the normal force. Since the gravitational force is defined as:

$$\vec{w} = -mg \hat{\jmath}$$

the normal force in this example is

$$\vec{n} = m(a+g) \hat{\jmath}$$

In other words, the scale would measure

$$N = m(a+g)$$

which is also known as the apparent weight. Person in the elevator would feel heavier when accelerating upwards and lighter when accelerating downwards. In an extreme case in which the elevator accelerates downwards at $\vec{a} = -g \hat{\jmath}$, which corresponds to free fall, the person would not feel any weight.

Now what is the definition of weight?

There is no strict definition for weight or apparent weight. See related discussion: Is there a formal definition for apparent weight? In the context of this example, I called the gravitational force weight which you will find in most textbooks, but as far as I know there is no strict definition for the weight. You should not be worried about the terminology, but rather understand what forces act on an object and then draw your conclusions from there.

Marko Gulin
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