Daniel B. answered • 12/17/21
A retired computer professional to teach math, physics
As Timothy said, the question is ambiguous and with missing information.
Let me just show you how to calculate the forces acting on a pilot in general.
Hopefully that will help you answer the question once you find out what it is.
Let
m be the mass of the pilot (unknown),
r = 1 km = 1000 m be the radius of the loop,
v = 540 km/h = 540000m/3600s = 150 m/s be the speed of the plane,
g = 9.81 m/s² be gravitational acceleration.
The pilot is subjected to two forces:
1) gravitational force of magnitude mg always pointing down
2) centripetal force of magnitude mv²/r always pointing towards the center of the loop.
At the bottom the two forces are additive, and the pilot is subject to the total force
mg + mv²/r = m(g + v²/r) in downward direction.
At the top the two forces are opposite, and the driver is subject to the force
mg - mv²/r = m(g - v²/r) in downward direction.
(Recall a negative downward force would be a positive upward force.)
This means the pilot is subject to accelerations of (g + v²/r) and (g - v²/r), respectively.
Stresses on pilots are commonly measured in multiples of g.
To get that, you divide the acceleration by the quantity g.
At the bottom:
(g + v²/r)/g = 1 + v²/rg = 1 + 150²/1000×9.81 = 3.3
At the top:
(g - v²/r)/g = 1 - v²/rg = 1 - 150²/1000×9.81 = -1.3
If the pilot were stationary he would be subjected to the normal 1g.
At the bottom the pilot is subjected to 3.3g;
that means he is being pushed into his seat with force more than three times greater than
when stationary.
At the top he is subjected to -1.3g;
the negative sign means that he is being pushed up.
But since he is upside down, this upward force also pushes him into his seat.
And it is doing it with magnitude only slightly greater than when stationary.
