# Aircraft force in g's experienced by pilot

Given:

The acceleration of high-speed aircraft is sometimes expressed in g’s (in multiples of the standard acceleration of gravity). Determine the upward force, in $N$, that a $70\cdot kg$ man would experience in an aircraft whose acceleration is $6\cdot g’s$.

My Solution:

I account for the normal acceleration due to gravity by subtracting $1\cdot g$ from the upward acceleration of the aircraft. This is what we would see if we had drawn a free body diagram (not shown)...

$$F=ma=m\times(6g-1g)=70\cdot kg\times5\times9.81\frac{m}{s^2}=3433.5\cdot N$$

Solution in Text:

However the solution in the text does not do this...

$$F=ma=m\times6g=70\cdot kg\times6\times9.81\frac{m}{s^2}=4120.2\cdot N$$

Question

Why is the net upward acceleration not $5\cdot g's$?.

The acceleration is given. However, the upward force is

$6mg = F_{up} - mg$

$F_{up} = 7\cdot mg$

In case of no acceleration:

$0 = F_{up,normally} - mg ==> F_{up,normally} = mg$

If we neglect the gravitational force experienced every day by this man, then we get

$6mg = F_{up}$

The question can't be answered as stated. Acceleration is a vector quantity, but only its magnitude is given. In other words, in which direction is the aircraft accelerating by 6 g?

Since the question only asked about the upward force on the pilot, only the vertical component of the acceleration is relevant. It therefore makes a huge difference whether the acceleration is vertical or horizontal. In the pure horizontal case, for example, the upward force on the pilot is only due to gravity, (70 kg)(9.8 m/s²) = 686 N.

Olin answered it correctly. The pilot, with no acceleration, experiences 1 g of force. Add 6 g's upward acceleration and he will experience 7 g's of force. The DIRECTION of the acceleration must be specified in order to give an answer to the question. I would bet that the question assumes an upward acceleration, but it doesn't say.

Well, the concept of g force measurement goes by measuring the net acceleration considering all effects. That is assume an object in free fall it will have a 0g acceleration, meaning Force= M(g-a), where a is pseudo acceleration=g. Hence F = M*A = M(g-g), A=The acceleration in g. Therefore, A=0g, which is freefall acceleration.

From Wikipedia:

Gravitation acting alone does not produce a g-force, even though g-forces are expressed in multiples of the acceleration of a standard gravity. Thus, the standard gravitational acceleration at the Earth's surface produces g-force only indirectly, as a result of resistance to it by mechanical forces. These mechanical forces actually produce the g-force acceleration on a mass. For example, the 1 g force on an object sitting on the Earth's surface is caused by mechanical force exerted in the upward direction by the ground, keeping the object from going into free-fall. The upward contact-force from the ground ensures that an object at rest on the Earth's surface is accelerating relative to the free-fall condition. (Free fall is the path that the object would follow when falling freely toward the Earth's center). Stress inside the object is ensured from the fact that the ground contact forces are transmitted only from the point of contact with the ground.

Objects allowed to free-fall in an inertial trajectory under the influence of gravitation only, feel no g-force acceleration, a condition known as zero-g (which means zero g-force). This is demonstrated by the "zero-g" conditions inside a freely falling elevator falling toward the Earth's center (in vacuum), or (to good approximation) conditions inside a spacecraft in Earth orbit. These are examples of coordinate acceleration (a change in velocity) without a sensation of weight. The experience of no g-force (zero-g), however it is produced, is synonymous with weightlessness.