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December 18, 2014

December 18, 2014

Posted by **Ryan** on Monday, December 12, 2011 at 4:26pm.

around a vertical circular loop which has a

radius of 1.2 km.

What is the magnitude of the normal force

on the 100 kg pilot of this plane at the bottom

of this loop? The acceleration of gravity is

9.8 m/s

2

.

Answer in units of k

- physics -
**drwls**, Monday, December 12, 2011 at 6:49pmM g + M V^2/R

R = 1200 m

"Units of k" makes no sense. The answer will not be a temperature; it will be a force, in Newtons.

- physics -
**Edwin**, Wednesday, October 24, 2012 at 6:35pmIt will not be answered in units of k. It is asking for a force, so the units will be in Newtons. (I assume you meant "kN", which would be kiloNewtons.)

The forces acting on the pilot at the bottom of the loop are gravity and the normal force. However, they are pulling in opposite directions (gravity pulls away from the center of the loop, while the normal force pulls toward the center), so the magnitudes of both forces need to be subtracted to get the total net force (F = N - mg).

m = 100 kg

v = 243 m/s

r = 1.2 km (1200 m)

g = 9.8 m/s

With the information given, the total net force can be calculated using the centripetal force formula: F = m * (v^2 / r)

F = 100 * (243^2 / 1200) = 4920.75 N

The problem is only asking for the magnitude of the normal force. As I stated earlier, two forces are acting on the pilot at the bottom of the loop (gravity and the normal force). The force of gravity is calculated with this formula: F = mg

F = 100 * 9.8 = 980 N

Now, take another look at the net force equation.

F = Normal force - mg

Since gravity is pulling away from the center of the loop, it is the negative force, which is why it is subtracted from the normal force to get the net force. So to find the normal force, the force of gravity needs to be added to the net force.

F + mg = Normal force

4920.75 + 980 = 5900.75 N

And since the problem asks for the force in kiloNewtons...

5900.75 N / 1000 = 5.90075 kN

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