Schaum's 3000 Solved Problems in Physics · Problem 4.6 · Newton's laws

Locomotive with rolling friction

Most of the engine's push goes into speeding up. A slice goes into fighting the rails.

How much force does it take to give a 20 000-kg locomotive an acceleration of 1.5 m/s² on a level track with a coefficient of rolling friction of 0.03?as printed, page 51, Fig. 4-1

Tap any symbol, here or in a formula, to see what it is and its value.

The picture
20 000 kg N = 196 kN mg = 196 kN F = 35.9 kN μN = 5.9 kN a = 1.5 m/s²
engine force35.9 kN
of which fights friction5.9 kN
of which speeds it up30 kN
Feel the numbers
196 kNweight of the locomotive

Twenty tonnes. The track pushes up with exactly this much, which is what makes the friction 3% of it.

5.9 kNrolling friction

Only a sixth of the engine's effort. Steel wheels on steel rails roll very easily; that is the whole point of railways.

35.9 kNengine force

About the weight of three and a half cars, pulling forward. At 1.5 m/s² the train reaches 15 m/s, about 33 mph, in ten seconds.

The walkthrough

Press play and it reads straight through, scrolling as it goes. Space pauses. ← → move a step. ▶ on a step plays from there.

Every symbol
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Narrated by the Listen voice. Timing comes from the narration itself, so the light lands on the part of the formula that is being said.
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