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Normal force on curves
Intermediate
Normal force combines weight component and centripetal requirement.
On a curved path the normal force N adjusts to both gravity and the centripetal need. At the top of a hill N = mg − mv²/r (lighter); at the bottom of a dip N = mg + mv²/r (heavier).
At 15 m/s over a 50 m radius crest, a 1000 kg car has N = 9800 − 4500 = 5300 N.
The car loses contact if v² ≥ gr at a crest.
In more detail
On a curved surface, the normal force is directed perpendicular to the surface, so it contributes to the centripetal force. Its magnitude depends on the curvature and the speed.
Real-world uses
- Motorcycle riders lean into turns to rely on normal force and friction.
- Designing hills and bridges to avoid lift-off at speed.
- A car driving over the top of a hill or through a dip feels a normal force that differs from its weight, which matters for the ride and the grip.
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Frequently asked questions
Why does the normal force change on a hilltop?
At the top of a hill, gravity and the normal force both have components towards the centre, so the normal force drops.
Can a vehicle lose contact with a hilltop?
Yes, if its speed is high enough that the normal force reaches zero.
Why do I feel lighter at the top of a hill?
The normal force is less than your weight because part of gravity provides the centripetal force needed to curve over the hill.
Related topics in Newton's Laws & Forces
Torque and equilibrium · Center of mass · Motion of center of mass · Constraint forces · Newton's laws and inertial frames · Dimensions and limits of Newtonian mechanics