# The bridge, sagging under the weight of a train weighing 400 tons, forms an arc with a radius of 2000 m

**The bridge, sagging under the weight of a train weighing 400 tons, forms an arc with a radius of 2000 m. Determine the force of pressure of the train in the middle of the bridge. The train speed is considered constant and equal to 20m / s**

m = 400 t = 400,000 kg.

R = 2000 m.

g = 9.8 m / s2.

V = 20 m / s.

V -?

Consider all the forces that act on the train when passing the bridge. It is acted upon by 2 forces: gravity Ft directed vertically downward, force N of bridge pressure directed vertically upward.

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

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

N = Fт – m * a.

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

N = m * g – m * a = m * (g – a).

Since the bridge has a curved shape, the train moves with centripetal acceleration a: a = V2 / R.

N = m * (g – V2 / R).

According to 3 Newton’s laws, the force N with which the bridge presses on the train is equal to the force with which the train presses on the bridge Р: N = P.

P = m * (g – V2 / R).

P = 400000 kg * (9.8 m / s2 – (20 m / s) 2/2000 m) = 3840000 N.

Answer: when passing the middle of the convex bridge, the train acts on it with a force of P = 3,840,000 N.