Please help with physics force problem!?

Question

A jet traveling at a speed of 2.30 *10^2 m/s executes a vertical loop with a radius of 7.00 *10^2 m. Find the magnitude of the force of the seat on a 70.0 kg pilot at the following positions.
(a) the top of the loop
____ N

(b) the bottom of the loop
____ N

Thank you so much in advance! For some reason I cannot figure out this problem and I really need to know how to do it because I have to do other problems just like it. Thanks again!

Answer 1

Keep it simple. There's only two things going on in this problem: uniform circular motion and gravity.

First you have to figure out what force is necessary to keep the pilot in circular motion in this loop. The net force on him has to keep him in the exact same path as the plane, and there are only two ways where forces can act on him. One is gravity pulling him down at all times. The other one is from the seat keeping him in circular motion. The sum of the forces have to equal the centripetal force necessary to stay in the loop.

So first figure out the centripetal force necessary to keep a body in this loop:

F_cent=m*V^2/r
F_cent=(70)*(230)^2/(700)
F_cent=5290N

The net force on the body has to be 5290N pointing to the center of the loop in order to keep the body in circular motion.

The other force in this problem is gravity. Gravity is always acting in the downward direction, regardless where the pilot is.

F_grav=m*g
F_grav=70*9.8
F_grav=686N

(a) At the top of the loop, gravity is pointing down at 686N. Total centripetal force necessary is 5290N pointing down. The only way to make this happen is the seat pushing the pilot down (the seat is pushing the pilot down toward the earth because he's upside down) 4604N. Let's take forces as positive when it points up into the sky and forces as negative when they point down to the ground:

F_cent=F_grav+F_seat
-5290N=-686N+F_seat
F_seat=-4604N

The seat is pushing the pilot down toward the ground at 4604N when he's at the top of the loop.

(b) Assuming the pilot is at the bottom of the loop just starting his ascend into the loop, the same forces apply except with centripetal force in different direction. F_cent now points up into the sky while gravity is still pointing down. Use the same sign convention for forces as the first part.

F_cent=F_grav+F_seat
5290N=-686N+F_seat
F_seat=5976N

The seat is pushing the pilot up into the sky at 5976N.

If the pilot is at the bottom of the loop after having executed his loop and is coming out of it into a level flight path at the same speed, then at the bottom of the loop, centripetal force will disappear. The only thing that needs to happen is the seat pushing him up 686N so he doesn't fall down due to gravity.

The equations are the same. You simply have centripetal force changing directions while gravity stays the same.