A car has a mass of 1500 kg and travels with a speed of 5 m/s. The car passes over the top of a convex bridge that has a radius of curvature equal to 10 m. What is the force of the bridge on the car as it passes over the top?
According to Newton’s Second Law for uniform circular motion, the net force acting on the car equals mar.
Fnet = mar
The expression Fnet = mar is a vector equation.
In the y-direction, there are only FN and mg.
The positive y-axis is directed toward the center of curvature.
mg – FN = mar (1)
We know that ar = v2/r, then we can rewrite the equation (1) as
mg – FN = mv2/r
FN = mg – mv2/r
FN = m(g – v2/r)
We substitute the values and we get
FN = 1500(9.8 – 52/10) = 10950 N