# A train weighing 10 6 kg moves with an acceleration of 0.05 m / s2. Find the traction force if the coefficient

**A train weighing 10 6 kg moves with an acceleration of 0.05 m / s2. Find the traction force if the coefficient of friction is 0.03.**

m = 10 ^ 6 kg.

g = 10 m / s2.

a = 0.05 m / s2.

μ = 0.03.

Fт -?

Let us write Newton’s 2 law in vector form when the train is moving: m * a = Ft + m * g + N + Ftr, where Ft is the traction force of the train, m * g is the force of gravity, N is the reaction force of the rails, Ftr is the friction force.

ОХ: m * a = Fт – Fтр.

OU: 0 = – m * g + N.

Ft = m * a + Ftr.

N = m * g.

The friction force Ffr is determined by the formula: Ffr = μ * N = μ * m * g.

The value of the traction force of the train will be determined by the formula: Ft = m * a + μ * m * g = m * (a + μ * g).

Ft = 10 ^ 6 kg * (0.05 m / s2 + 0.03 * 10 m / s2) = 350,000 N.

Answer: the train has a traction force Ft = 350,000 N.