The car at the top of the convex bridge has a speed of 72 km / h, which should be the radius of curvature

The car at the top of the convex bridge has a speed of 72 km / h, which should be the radius of curvature of the bridge so that the driver at the top of the bridge is in zero gravity.

V = 72 km / h = 20 m / s.

P = 0 N.

g = 9.8 m / s2.

R -?

Weightlessness is a state of a body in which its weight P is 0: P = 0.

When passing the upper point of the convex bridge, 2 forces act on the car: gravity Ft directed vertically downward, force N of the axle pressure on the car wheels directed vertically upward.

m * a = m * g + N – 2 Newton’s law in vector form.

For projections onto the vertical axis 2, Newton’s law will take the form: – m * a = – Fт + N.

N = Fт – m * a.

The force of gravity Ft is determined by the formula: Ft = m * g.

N = m * g – m * a = m * (g – a).

The centripetal acceleration a is expressed by the formula: a = V2 / R.

N = m * (g – V ^ 2 / R).

According to Newton’s 3 laws: N = R.

m * (g – V ^ 2 / R) = 0.

g = V ^ 2 / R.

R = V ^ 2 / g.

R = (20 m / s) ^ 2 / 9.8 m / s2 = 40.8 m.

Answer: the radius of curvature of the bridge should be R = 40.8 m.