Determine the maximum speed with which the jeep can travel o

Determine the maximum speed with which the jeep can travel over the crest of the hill without losing contact with the road. A pilot weighs 150 lb and is traveling at a constant speed of 120 ft/s. Determine the magnitude of the upward normal force he exerts on his seat when he is upside down at point A.

Solution

1. There are two forces on the car. They are Gravitational force and centrifugal force.

Gravitational force acts on the car in the downward direction and it gives an acceleration of acceleration due to gravity which is g = 32.2 ft/s^2.

Centrifugal force acts on the car in the upward direction out of the circle shown and it gives an acceleration given by

a = V^2/R

For the car to have maximum velocity beore losing contact with the ground, both the foces must balance each other.

Thus both the accelerations must be equal.

===> a = g ===> V^2/R = 32.2

===> V^2/250 = 32.2

===> V = 89.722 ft/s

Thus the car can travel at a maximum velocity of V = 89.722 ft/s before losing contact with the ground.

2. At the point A, the pilot is upside down and there are two forces on him. They are Gravitational force and Centrifugal force.

Centrifugal force acts upwards out of the circle and is given by

Centrifugal force, Fc = mV^2/R

Gravitational force acts downwards on the pilot and is given by

Gravitational force, Fg = mg

Net force between these two forces is the normal force exerted by the pilot on his seat

Thus Normal force on the pilot\'s seat = Centrifugal force - Gravitational force

Fn = Fc - Fg

===> Fn = mV^2/R - mg

===> Fn = m*(V^2/R - g)

===> Fn = 150 lb * {(120^2/400)-32.2}ft/s^2

===> Fn = 570 lb-ft/s^2 or 17.702 lbf

Thus the normal force exerted by the pilot on his seat is Fn = 570 lb-ft/s^2 or 17.702 lbf