A76 kg pilot at an airshow performs a loop de loop with his plane. at the bottom of the 52-m radius loop, the plane is moving 48 m/s. determine the normal force acting upon the pilot.

The solution for this problem is:

At bottom, both centrifugal force and weight are performing in the same direction. 
So we can say that, force would be computed by f = mg + mv^2 /r

= (76 x 9.81) + (76 x 48 x 48 / 52) 
= 745.6 + 3367.4 
Normal force = 4113 N is the answer

see earlier

answersysvol bloat;