Sample Problem: Toy Train

Question: Suppose a child pulls on the toy train below with a force of Fpull = 6 newtons and the cars have masses of m1 = 3 kg, m2 = 2 kg, and m3 = 1 kg. If the friction between the floor and the cars is very small, what is the acceleration of the train, and what is the tension (T12 and T23) in each cable between the cars?

Answer: We can think of this problem two ways: We can think of the child's force Fpull acting on the collection of cars as a whole (and thus ignore the internal forces of tension), and we can think of each car individually (and include the forces of tension through which they interact). By utilizing the first technique, we can easily calculate the acceleration.

Thinking of the collection of cars as a whole, Newton's Second Law (net force on the object = mass of the object * acceleration, where, in this case, the "object" is the entire train) is

Fpull = (m1 + m2 + m3) a,

so a = Fpull / (m1 + m2 + m3) = 6 N / (3 kg + 2 kg + 1 kg) = 6 N / 6 kg = 1 m/s^2 is the train's acceleration.

We can now think of each car individually. For example, we know that the smallest car has an acceleration of 1 m/s^2 (just like the other two). The only force acting on this car is the tension T23 pulling to the right. So, Newton's Second Law for this car is T23 = m3 * a = (1 kg)*(1 m/s^2) = 1 N.

Now that we know T23, we can look at Newton's Second Law on the medium-sized cart. The net force on this cart is T12 - T23, since T12 is pulling right and T23 is pulling left. So, Newton's Second Law gives us T12 - T23 = m2*a. We can rearrange to get T12 = T23 + m2*a = 1 N + (2 kg)*(1 m/s^2) = 3 N.

At this point, we're technically done with the problem, but if we take just a few extra seconds to apply Newton's Second Law to the largest car, we can check our work! The net force on this car Fpull - T12 (since Fpull is pulling right and T12 is pulling left), so Newton's Second Law is Fpull - T12 = m1*a. We can rearrange for a to make sure that we get 1 m/s^2: a = (Fpull - T12) / m1 = (6 N - 3 N) / (3 kg) = 1 m/s^2. Check! Now we know we were correct!

How would the tensions and acceleration change if there were friction between the blocks and the floor? Would the acceleration be larger or smaller? Would the tensions be larger or smaller?