A train has one locomotive and 25 cars. Each of the cars has a mass of 55,000 kg.

Question

The train needs to go up a hill that is angled 1.1 degrees up from the horizontal. What does the mass of the locomotive have to be for it to climb the hill at constant speed? You can assume all the coefficients of friction are steel-to-steel.

Answer 1

Let

m = 25×55,000 = 1,375,000 kg be the total mass of all the cars,

M (to be calculated) be the mass of the locomotive,

α = 1.1° be the angle of the hill,

g be gravitational acceleration,

μ = 0.1 be the coefficient of kinetic friction.

In general, the weight F of an object on a incline can be decomposed into two perpendicular forces:

- Fsin(α) is the component of the weight parallel to the slope, and that is the force pulling the object downhill.

- Fcos(α) is the component of the weight normal to the slope, and that is the force responsible for friction;

the force of friction acting against sliding has magnitude Fcos(α)μ.

The total weight of the train is (m+M)g.

The force pulling it downhill is (m+M)gsin(α).

Only the weight Mg of the locomotive provides friction preventing the train from sliding down.

(The weight of the other cars would contribute to friction only if their brakes were applied.)

So the force of friction is Mgcos(α)μ.

To avoid sliding, the force friction must be at least the force pulling the train down:

Mgcos(α)μ ≥ (m+M)gsin(α)

This can be simplified into

M ≥ msin(α)/(cos(α)μ - sin(α))

Substituting actual numbers

M ≥ 1,375,000×sin(1.1°)/(cos(1.1°)×0.1 - sin(1.1°)) = 326,753 kg