Hope you can help me. I found a question exactly like mine, but I didn’t agree with the answer. If you could explain the concept, I’d appreciate it. Thanks.
A sports car of mass 950 kg (including the driver) crosses the rounded top of a hill (radius = 95 m) at 22 m/s. Determine (a) the normal force exerted by the road on the car, (b) the normal force exerted by the car on the 72-kg driver, and (c) the car speed at which the normal force on the driver equals zero.
The answer I found said that you have to set up an equation as:
mg – Fn = m(v^2 / r), where mg = weight, Fn = normal force, and m(v^2 / r) = the centripetal acceleration.
However, when you’re on a flat road, the normal force always equals the weight of the car, right? Otherwise the car would be lifting off the ground or crashing into the earth. So, if this is true, then the answer to (a) must be 9310 N (the weight of the car), (b) must be 705.6 N (the weight of the person), and (c) must be an impossible situation. Hope you can correct me if I’m wrong, thanks in advance for the help.
K, thanks for the explanation =)
Tags: Circular, Force, Help, laws, Motion, Need, Newton's, Normal, Physics, Question, Uniform
“However, when you’re on a flat road, the normal force always equals the weight of the car, right?”
your not on a flat road your on a rounded hill with a radius r so the problem will be treated using circular motion. there isn’t really any place where the road is flat so Fn and mg are always acting at some angle depending on where you look at the car. As a whole Fn acts up and mg acts down and a=v^2/r.
For part c think of it like this. The car is moving so fast that gravity doesn’t have enough time to push the car into the road and for the road to exert force on the car.
Think of it like your running sooo fast that you could run across water.
> “…when you’re on a flat road, the normal force always equals the weight of the car, right?”
Not necessarily. It’s more correct to say that, if the car is travelling horizontally, AND the car is not accelerating vertically, AND the only vertical forces acting on the car are gravity and the normal force, THEN the normal force equals the weight of the car.
But in this case, there IS a vertical acceleration of the car, because the car is going over the (persumably circular) top of the hill. ANY time something goes in a circular path, it is accelerating toward the center of the circle, in the amount of v²/r.
Therefore, this car IS accelerating vertically even though it’s (momentarily) travelling horizontally. In this case, the normal force DOES NOT equal the car’s weight.
Perhaps a more intuitive way to envision it is that the car seems to “lift up” a bit as it rounds the top of the hill. The shock absorbers and tires relax a little bit, as if some weight has been taken off the car. This reflects the fact that the normal force (the upward pressure from the road) has been reduced somewhat.
To get the normal force exerted on the car, use F=ma:
Vertical forces on car:
(m_car)g (downward)
F_normal (upward)
Net upward force = F_normal – (m_car)g
Use F=ma
Net upward force = m_car * × (net upward acceleration)
F_normal – (m_car)g = m_car × (-v²/r)
From there, you can solve for F_normal:
F_normal = (m_car)(g – v²/r)
For part (b), the analysis is the same, but use m_driver instead of m_car:
F_normal_on_driver = (m_driver)(g – vsup2;/r)
For part (c), use the above equation, setting F_normal_on_driver = 0, and solve for “v”. This essentially corresponds to the case where the driver is going around the top fast enought to “feel weightless.”