The mass of a railway carriage is 60 tons, and the coefficient of friction of the carriage wheels on the railway track is 0.03.

The mass of a railway carriage is 60 tons, and the coefficient of friction of the carriage wheels on the railway track is 0.03. Determine the deformation of the buffer spring that connects the cars, if its stiffness is 40 MN / m.

m = 60 t = 60,000 kg.

g = 10 N / m.

μ = 0.03.

k = 40 MN / m = 40 * 106 N / m.

x -?

According to 1 Newton’s law, the car moves uniformly rectilinearly in the case when the action of forces on it is compensated: Ffr + Ffr + m * g + N = 0, where Ffr is the friction force, Ffr is the elastic force, m * g is the force of gravity, N – the reaction force of the railroad bed.

ОХ: Ftr – Fcont = 0.

OU: m * g – N = 0.

Ftr = Fcont.

m * g = N.

The elastic force Fel is determined by Hooke’s law: Fel = k * x.

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

k * x = μ * m * g.

x = μ * m * g / k.

x = 0.03 * 60,000 kg * 10 N / m / 40 * 106 N / m = 0.00045 m.

Answer: the deformation of the buffer spring is x = 0.00045 m.