# What objects' masses do we need to take into account when calculating force?

In the following problem:

A steam engine of mass $3\times 10^4\ \mathrm{kg}$ pulls two wagons each of mass $2\times 10^4\ \mathrm{kg}$ with an acceleration of $0.2\ \mathrm{m/s^2}$. Neglecting frictional force, calculate the force exerted by the engine and the force experienced by the wagons.

To solve this problem we have to use the $F=ma$ formula, right? So to calculate the force exerted by the engine we have to only take the mass and acceleration of the engine in the formula. We don't have to take mass of the wagons, right?

My physics teacher says that to calculate the force exerted by the engine we have to take the mass of both the engine and the wagon, but I don't understand why we have to take the mass of the engine into account. When I asked him this he just said that we have to, and obviously I am not satisfied with his answer. So who is correct, me or my physics teacher, and if the physics teacher is, then why?

Well, you can think of this as a car pulling a wagon with $a = 0.2 \text{ m/s}^2$. Naturally, the car's motor needs to generate enough force to accelerate both the car itself and the wagon it is pulling. This is true of the situation in this question. The steam engine needs to generate enough force to accelerate both itself and the two wagons at $a = 0.2 \text{ m/s}^2$. So you need to take into account both the engine's mass and the mass of the two wagons. Consider what would happen if there weren't any wagons pulled by the engine. Would the engine then require the same amount of force to accelerate?

Another way to think of it is considering the engine and two wagons as one big block. Then the force needed to accelerate the block would be its total mass (mass of both the wagons and the engine) times its acceleration.

Of course, when you are calculating the force experienced by the wagons, things are a little different.

Choosing your system carefully and drawing free-body diagrams is crucial!

It is often easiest to start with defining your system as one big block. You can do this because the engine and wagons are all connected by rigid steel locks that cannot compress or stretch, meaning that if one car moves, the others have to move along at the exact same rate.

If we define our system as one big block, the total mass is $m = 70000 \text{ kg}$. Since we know $a$, we can easily find that $F = 14000 \text{ N}$ in the horizontal direction. Since the engine is the only object in our system that can generate force, we know that $F$ is the force exerted by the engine.

Now look only at car 2, the car at the end of the train. The force acting on it in the horizontal direction is not $F$. It is actually a different force, $T_2$, exerted by the steel lock between car 1 and car 2. We still know, however, that all cars accelerate at $0.2 \text{ m/s}^2$. Therefore: \begin{align} \sum F &= ma \\ T_2 &= \left(2 \times 10^4 \text{ kg}\right)\left(0.2 \text{ m/s}^2\right) \\ &= 4000 \text{ N} \\ \end{align}

. . . meaning the force experienced by the last car is quite different from the force exerted by the engine!

Now look only at car 1, the car in the middle. By Newton's third law, it is being pulled backwards with force $T_2$ from car 2. However, it is also being pulled forward by a force $T_1$ exerted by the steel lock between car 1 and the engine. Thus:

\begin{align} \sum F &= ma \\ T_1 - T_2 &= \left(2 \times 10^4 \text{ kg}\right)\left(0.2 \text{ m/s}^2\right) \\ T_1 &= 8000 \text{ N} \\ \end{align}

Finally, let's check our work by looking at a system that only includes the engine. It is experiencing $F$ from itself in the positive direction. By Newton's third law, it is also being pulled backwards by $T_1$ from car 1. \begin{align} \sum F &= ma \\ F - T_1 &= \left(3 \times 10^4 \text{ kg}\right)a \\ a &= 0.2 \text{ m/s}^2\\ \end{align}

Sure enough, we have found the correct acceleration.