A car weighing 1000 kg moves at a speed of 72 km / h on a concave bridge with a radius

A car weighing 1000 kg moves at a speed of 72 km / h on a concave bridge with a radius of 40 m. With what force does the car act on the bridge at its lowest point?

m = 1000 kg.

g = 9.8 m / s2.

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

R = 40 m.

P -?

The force with which the body presses on the support or stretches the suspension is called the body weight P.

Two forces act on the car: the reaction force of the bridge N, directed vertically upwards;

the force of gravity m * g directed vertically downward.

Let’s write Newton’s 2 law for projections on the vertical axis directed vertically upward: m * a = – m * g + N.

N = m * a + m * g = m * (a + g).

Centripetal acceleration, and we will express it by the formula: a = V2 / R.

N = m * (V2 / R + g).

According to Newton’s 3 law, the force of action N is equal to the force of reaction P: P = N.

P = m * (V2 / R + g).

P = 1000 kg * ((20 m / s) 2/40 m + 9.8 m / s2) = 19800 N.

Answer: the weight of the car at the lowest points of the trajectory is P = 19800 N.