Hi there!
A)
At this point, the pilot is experiencing a DOWNWARD NET centripetal force (ALWAYS points towards the center of the circle) as well as a downward force due to gravity. Additionally, we can say that the normal force exerted by the seat on the pilot is its apparent weight.
[tex]\Sigma F = F_g + F_N[/tex]
**The net force is the centripetal force.
We can assign both to be positive since they work in the same direction. (Assigning down as positive in THIS situation will make things easier).
We know that:
[tex]F_c = \frac{mv^2}{r}[/tex]
Fc = Centripetal force (N)
m = mass (kg)
v = velocity (m/s)
r = radius (m)
[tex]F_g = mg[/tex]
Fg = Force due to gravity (N)
m = mass (kg)
g = acceleration due to gravity (9.81 m/s²)
Solve for mass. (Given weight)
[tex]m = \frac{F_g}{g}\\\\m = \frac{719}{9.8} = 73.367 kg[/tex]
Now, we have everything we need to solve.
[tex]\frac{mv^2}{r} = F_g + F_N \\\\\frac{mv^2}{r} - F_g = F_N\\\\F_N = \frac{(73.367)(225^2)}{3110} - 719 = \boxed{475.278 N}[/tex]
B)
Weightlessness is achieved when the pilot is in free fall. In this instance, FN (Normal force by the seat) = 0 N.
We can use the same setup as above.
[tex]\frac{mv^2}{r} = F_g + 0 (F_N)[/tex]
[tex]\frac{mv^2}{r} = mg[/tex]
Solve for 'v'.
[tex]\frac{v^2}{r} = g\\\\v = \sqrt{gr} = \sqrt{(9.81)(3110)} = \boxed{174.669 \frac{m}{s}}[/tex]
C)
At the BOTTOM, we have the force of gravity working DOWNWARD (Negative in this instance), while the centripetal force (NET FORCE) and normal force are upward (Positive). Do another summation of forces:
[tex]\Sigma F = F_N - F_g\\\\\\frac{mv^2}{r} + F_g = F_N \\\\F_N = \frac{(73.367)(225^2)}{3110} + 719 = \boxed{1913.278 N}[/tex]
