# A car with a mass of 1 t moves at a speed of 20 m per second along a convex bridge arc of a circle R

**A car with a mass of 1 t moves at a speed of 20 m per second along a convex bridge arc of a circle R = 100 m with what force the car presses on the bridge at the top point of the bridge**

m = 1 t = 1000 kg.

V = 20 m / s.

g = 10 m / s2.

R = 100 m.

R – ?

When a car moves along a convex bridge, when passing the top point, 2 forces act on it: gravity Ft directed vertically downward, force N of bridge pressure on the car directed vertically upward.

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

Since the centripetal acceleration a is also directed downward, then 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 – V2 / R).

According to Newton’s 3 Laws, the force of action is equal to the force of reaction. The force N with which the bridge presses on the car is equal to the force with which the car presses on the bridge P: N = P.

P = m * (g – V2 / R).

P = 1000 kg * (10 m / s2 – (20 m / s) 2/100 m) = 6000 N.

Answer: at the top of the convex bridge, the car presses on it with a force of P = 6000 N.