Calculating Normal Force on a Driver in Circular Motion: Hill and Valley Scenario

In this problem, we analyze the forces acting on a driver of a car moving at a constant speed over a circular hill and into a valley. We aim to determine the normal force exerted on the driver by the car seat at the bottom of the valley. The driver has a mass of 70 kg, and it is given that at the top of the hill, the normal force is zero. Additionally, the radius of curvature is the same for both the hill and the valley.

Problem Approach

1. Understanding the Forces:

• • At the top of the hill, the normal force (Fn) from the seat on the driver is zero. This implies that the only force acting on the driver is their weight (mg), which provides the centripetal force needed for circular motion.
• At the bottom of the valley, two forces act on the driver:
  • • The gravitational force (mg) acting downwards.
  • The normal force (Fn) from the seat acting upwards.

1. Application of Newton's Second Law:

• At the Top of the Hill:
  • • The forces in play are the gravitational force (mg) and the centripetal force (Fc).
• Using Newton's second law, the equation is:
  • • Fn - mg = -Fc
• Given that Fn = 0 at the top of the hill, the equation simplifies to:
  • • Fc = mg
  • This indicates that the centripetal force is provided entirely by the weight of the driver, confirming that the speed is such that Fc = 70 kg * 9.8 m/s² = 686 N.
• At the Bottom of the Valley:
  • • The forces acting are still the gravitational force (mg) downwards and the normal force (Fn) upwards. However, at this point, the centripetal force (Fc) is directed upwards.
• Using Newton's second law, the equation becomes:
  • • Fn - mg = Fc
• Rearranging to solve for the normal force (Fn):
  • • Fn = mg + Fc
• Substituting the known values:
  • Fn = 70 kg * 9.8 m/s² + 686 N = 686 N + 686 N = 1372 N

Thus, the normal force on the driver at the bottom of the valley is 1372 N, which means the driver experiences a force twice his weight.