What is the magnitude of the normal force on the pilot of the plane at the bottom of the loop in the following case?
An airplane moves at 171 m/s as it travels around a vertical circular loop which has a radius of 3 km. What is the magnitude of the normal force on the 150 kg pilot of this plane at the bottom of the loop? The acceleration of gravity is 9.8 m/s2.
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The pilot is flying the plane in a vertical circular loop that has a radius of 3 km. The speed of the plane is 171 m/s. The normal force on the pilot at the bottom of the loop is required.
At the bottom of the loop there are two forces acting on the pilot. One is the gravitational force of attraction due to the Earth that acts downwards. The other is the centrifugal force with which the plot is pushed outwards that also acts in the downward direction. The total normal force acting on the pilot is the sum of the two forces.
As the pilot has a mass of 150 kg, the gravitational force given by m*g = 150*9.8 = 1470 N = 1.47 kN. The centrifugal force on the pilot due to the circular path being followed by the pilot is given by m*v^2/r = 150*171^2/3000 = 1.46205 kN.
The total normal force on the pilot is equal to 1.47 + 1.46205 = 2.93205 kN.
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