Posted by **Jack** on Saturday, October 13, 2007 at 4:49pm.

Driving in your car with a constant speed of v=13.0 m/s, you encounter a bump in the road that has a circular cross section. If the radius of curvature of the bump is r=33.2 m, calculate the force the car sear exerts on a 72.7 kg person as the car passes over the top of the bump.

No clue...

- Physics -
**Count Iblis**, Saturday, October 13, 2007 at 5:56pm
When the car is over the top of the bump it is accelerating downward. The acceleration is

a = v^2/r = (13 m/s)^2/(33.2 m)

= 5.09 m/s^2

72.7 kg person is thus accelerated downward at this acceleration.

Newton's second law says that:

F = m a

Let'S apply this to the person. There are two forces that act on the person. There is the gravitational force exerted by the entire Earth on the person. This force is directed downward and has a magnitude of:

F_g = 72.7 kg * g = 713 N

And there is the foce exerted by the chair F_ch

If the take the downward direction to be positive, we have:

(F_g + F_ch) = 72.7 kg * 5.09 m/s^2 --->

F_ch = -343 N

The minus sign means that it is directed upward.

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