Centripetal force on a curved bridge

[aurao2003]

Homework Statement

When a car is speeding over a curved surface e.g a bridge, it sometimes loses contact with the ground. We have seen this many times in high speed chases and rally driving.

In this scenario, the textbook wrote this equation:
S-mg =Mv^2/r
The RHS side is obviously the equation for centripeta force (F=Mv^2/r).
But why is the resultant force S-mg since S=0? I thought the only relevant force should be the weight which acts downwards thereby forcing the car back on to the road. Also in the context of the centripetal force, is it not the weight that acts as the centripetal force in this case. Therefore I was thinking the equation should be
mg =mv^2/r.

Homework Equations

The Attempt at a Solution

 

[Stonebridge]

Is S the upward force of the road on the car?
If so, the equation should be mg-S=mv²/r
S will be equal to zero just at the point where the car leaves the road. Then, as you rightly say, the centripetal force is provided purely by the car's weight mg.
I agree the equation you quoted doesn't make sense.