A aeroplane with a mass of 9940 kg completes a vertical loop of radius 596m with a speed of 155m/s. What normal force does the airplane seat exert on the 92 kg pilot at the top of the loop and at the bottom of the loop?
Centripetal force = M v^2/R = Weight +/- (seat force)
A + sign applies at the top of the loop and a minus sign applies at the bottom.
The weight is M g. M is the pilot's mass. The airplane's mass does not matter. Solve for the seat force
Thank you for giving both speed and radius this time:
Ac = v^2/r
at the top
force on seat = m (v^2/r - g)
= 92 (155^2/596 -9.8)
= 2806 N
at the bottom
m (v^2/r + g)
=92 (155^2/596 +9.8) = 4610 N
To find the normal force exerted on the pilot at the top and bottom of the loop, we need to analyze the forces acting on the pilot.
At the top of the loop:
At the top of the loop, the pilot is experiencing downward acceleration. The forces acting on the pilot are gravity (mg) pointing downwards and the normal force (N) pointing upwards.
Using Newton's second law (F = ma), we have:
Net force = ma
At the top of the loop, the net force is the difference between the normal force and the weight:
N - mg = m * acceleration
The acceleration at the top of the loop can be calculated using the centripetal acceleration formula:
a = v^2 / r
Here, v is the velocity of the plane and r is the radius of the loop.
Plugging in the values, we have:
N - mg = m * (v^2 / r)
Substituting the given values:
N - (92 kg * 9.8 m/s^2) = 92 kg * [(155 m/s)^2 / 596 m]
Simplifying the equation will give us the value of the normal force at the top of the loop.
Similarly, we can follow the same steps to calculate the normal force at the bottom of the loop. The only difference is that the acceleration is directed upwards, so the equation will be:
N + mg = m * (v^2 / r)
Solving this equation will give us the value of the normal force at the bottom of the loop.