A car weighing 3.6 tons moves at a speed of 72 km per hour along a concave road profile with a radius of 100 m.

A car weighing 3.6 tons moves at a speed of 72 km per hour along a concave road profile with a radius of 100 m. Determine the pressure force of the car at the lowest point of the concavity of the road.

m = 3.6 t = 3600 kg.

g = 10 m / s2.

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

R = 100 m.

P -?

When the car passes the lower point of the concave bridge, 2 forces act on it: the force of gravity Ft directed vertically downward, the force N with which the surface of the bridge presses on the car directed vertically upward.

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

For projections onto the vertical axis, which is directed vertically upward, 2 Newton’s law will take the form: m * a = – Ft + N.

N = m * a + Fт.

Let us express the force of gravity Ft by the formula: Ft = m * g.

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

We find the centripetal acceleration a by the formula: a = V2 / R.

N = m * (V ^ 2 / R + g).

3 Newton’s law: N = R.

P = m * (V ^ 2 / R + g).

P = 3600 kg * ((20 m / s) ^ 2/100 m + 10 m / s2) = 50400 N.

Answer: at the lowest point of the concave bridge, the force of pressure of the car on the bridge is P = 50400 N.