A train consists of an engine and three cars tied closely together. The mass of

Question

A train consists of an engine and three cars tied closely together. The mass of the whole train is 1.2 times 10^5 kg; the engine alone has a mass of 80,000 kg and each car has the same mass. Through its wheels, the engine can exert a horizontal force of 2.7 times 10^4 N on Earth (a) When the engineer wishes to accelerate forward, in which direction should the engine exert its force on Earth? forward backward What is the magnitude of the force exerted by Earth on the engine then? N (b) hen the engineer acts to accelerate the train, what is the force on the set of three cars and what is the subsequent acceleration of these cars? (Treat the three cars as a unified whole.) N m/s^2 (c) When the engineer acts to accelerate the train, what are all the horizontal forces on, and subsequent acceleration of the second car of the train? force from car ahead N force from car behind N resulting acceleration m/s^2

Explanation / Answer

(a) When the engine rotates its wheels, the wheels apply force in the backward direction on the Earth. The earthpushes back on the wheel with an equal and opposite (forwarddirected) reaction force by Newton's 3rd law. Thus, the earthpushes the engine forward with a force

F = 2.7*10^4 N

This reaction force is possible onlydue to the presence of STATIC FRICTION between the wheel and the earth. Thus, the train can move forward only because of STATIC FRICTION which means the wheel is NOT SLIDING and so no energy islost due to friction and no work is done by friction. So, friction just transfers the force applied by the engine on the earth, back to the engine and hence makes the engine move forward. Since the engine is connected to the remaining three cars, they have no choice but to move along at the same acceleration as the engine,'a'.

We have,

1.2*10^5 = 3m + 80000

mass of each car m = 13333.33 kg

(b)

Since the engine and the 3 cars are connected, they'll all havethe same acceleration 'a', and we can consider the train as one single mass M = 1.2*105 kg.

The net force acting on this single mass M is the force F exertedby the earth on the engine.

So, by Newton's 2nd Law, acceleration of train (and the engine andeach of the cars) is given by

a = F / M = (2.7*104)/(1.2*105) = 0.225 m/s^2

from the free body diagram 1

the net force equations will be

T1 = 3m*a = 3* 13333.33 *0.225 = 9000 N -------(1)

Thus, the force on the set of three cars is

T1 = 9000 N

(c)

from the free body diagram of car2 and car1

the net force equations will be:

For car 1:

T1 - T2 = m*a
So, T2 = T1 - ma = 9000 - 13333.33 *0.225  = 6000 N

Thus, force from car ahead = T2 = 6000 N.

For car 2:

T2 - T3 = ma
So, T3 = T2 - ma = 6000 - 13333.33 *0.225 = 3000 N

Thus, force from car behind = T3 = 3000 N.

Resulting acceleration of car 2 will be the same as the other carsand the engine = a = 0.225 m/s2.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.