At the competition of horses of heavy harness breeds, one of them transported a load weighing 23 tons.

At the competition of horses of heavy harness breeds, one of them transported a load weighing 23 tons. Find the coefficient of friction if the horse’s traction force is 2.3 kN.

m = 23 t = 23000 kg.

g = 10 m / s2.

Ft = 2.3 kN = 2300 N.

μ -?

We will assume that a horse of a heavy-harness breed transported the load evenly in a straight line. According to 1 Newton’s law, this means that the action on the load of all forces are compensated.

4 forces act on the load: Ft is the horse’s traction force, m * g is the gravity of the load, N is the reaction force of the road surface, Ftr is the friction force.

Ft + m * g + N + Ftr = 0.

ОХ: Ft – Ftr = 0.

OU: N – m * g = 0.

Ft = Ftr.

N = m * g.

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

Fт = μ * m * g.

μ = Fт / m * g.

μ = 2300 N / 23000 kg * 10 m / s2 = 0.01.

Answer: the coefficient of friction is μ = 0.01.