A sports car of mass 1400 kg (including the driver) crosses the rounded top of a hill (radius = 93 m) at 27 m/s.
1) Determine the normal force exerted by the road on the car.
2) Determine the normal force exerted by the car on the 75 kg driver.
3) Determine the car speed at which the normal force on the driver equals zero.
I will be happy to critique your thinking.
I can't seem to get past the first question. I tried squaring the velocity, dividing that by the radius, and then multiplying all of that by the weight of the car. That answer wasn't right.
Of course it isn't right. Why did you multiply centripetal acceleration by a force?
It idea is to look at the net force downward by the car.
Interesting. Use the centrifugal equation and multiply it with the third derivative of the acceleration formula you have memorized.
To determine the normal force exerted by the road on the car, we can use the concept of centripetal force. When a car travels in a circular path on top of a hill, the net force acting on it must provide the required centripetal force to keep it in a circular motion. In this case, the centripetal force is provided by the normal force from the road. So, to find the normal force:
1) Determine the centripetal force:
The centripetal force can be calculated using the formula:
F_c = m * v^2 / r
where F_c is the centripetal force, m is the mass of the car, v is the velocity of the car, and r is the radius of the curve.
Substituting the given values:
F_c = (1400 kg) * (27 m/s)^2 / 93 m
Simplifying, we find:
F_c ≈ 10957 N
Since the normal force on the car is equal to the centripetal force, the normal force exerted by the road on the car is approximately 10957 N.
2) Determine the normal force exerted by the car on the driver:
The normal force between two objects in contact is equal in magnitude and opposite in direction. Since the normal force exerted by the road on the car is directed upward, the normal force exerted by the car on the driver is directed downward and has the same magnitude. Therefore, the normal force exerted by the car on the driver is also approximately 10957 N.
3) Determine the car speed at which the normal force on the driver equals zero:
The normal force on the driver will become zero when the car and the driver are in free fall, experiencing weightlessness. This happens at the top of the hill when the forces acting on the driver are reduced to only the gravitational force. Therefore, at this point, the gravitational force on the driver will be equal to his weight:
mg = F_g
where m is the mass of the driver (75 kg), g is the acceleration due to gravity (9.8 m/s^2), and F_g is the gravitational force.
Solving for speed(v):
v = √(rg)
Substituting the given values:
v = √(93 m * 9.8 m/s^2)
Simplifying, we find:
v ≈ 30.4 m/s
Therefore, the car's speed at which the normal force on the driver equals zero is approximately 30.4 m/s.