Answer :
Answer:
3472.62 N
Explanation:
First, we need to make the free body diagram of the pilot at the bottom of the loop
Therefore, the net force is equal to
[tex]F_{net}=F_n-mg=ma_c[/tex]In a circular motion, the centripetal acceleration is equal to v²/r, so we will use the following equation
[tex]\begin{gathered} F_n-mg=m\frac{v^2}{r} \\ \\ F_n=m\frac{v^2}{r}+mg \end{gathered}[/tex]Where Fn is the normal force, m is the mass, v is the speed, r is the radius, and g is the gravity.
Replacing m = 62.5 kg, v = 62 m/s, r = 84 m, and g = 9.8 m/s², we get:
[tex]\begin{gathered} F_n=\frac{(62.5\text{ kg\rparen\lparen62 m/s\rparen}^2}{84\text{ m}}+(62.5\text{ kg\rparen\lparen9.8 m/s}^2) \\ \\ F_n=2860.12\text{ N + 612.5 N} \\ F_n=3472.62\text{ N} \end{gathered}[/tex]Therefore, the normal force is 3472.62 N
