Newton’s Third Law: Applications, Experiments and Exercises
The Newton’s third law , also called the law of action and reaction indicates that when an object exerts force on the other, the latter also exerts a force on the first magnitude and the same direction and the opposite direction.
A space rocket gets the necessary propulsion thanks to the exhaust gases. Source: Pixabay
Explanation and formulas
The mathematical formulation of Newton’s Third Law is very simple:
F 12 = – F 21
One of the forces is called action and the other is reaction. However, it is necessary to emphasize the importance of this detail: both act on different objects. They also do this simultaneously, although this terminology incorrectly suggests that the action takes place before and the reaction after.
As forces are vectors, they are denoted in bold. This equation indicates that there are two objects: object 1 and object 2. The force F 12 is that exerted by object 1 on object 2. The force F 21 is exerted by object 2 on object 1. The sign (-) indicates that they are opposites.
When looking carefully at Newton’s third law, an important difference is noted in the first two: while they invoke a single object, the third law refers to two different objects.
And if you think about it carefully, interactions require pairs of objects.
This is why the action and reaction forces do not cancel out or balance each other, although they have the same magnitude and direction, but in the opposite direction: they are applied to different bodies.
applications
Interaction Ball – Earth
Here is a very daily application of an interaction related to Newton’s Third Law: a vertically falling ball and the Earth. The ball falls to the ground because the Earth exerts an attractive force, known as gravity. This force causes the ball to fall with a constant acceleration of 9.8 m / s 2 .
However, hardly anyone thinks about the fact that the ball also exerts an attractive force on Earth. Of course, the Earth remains unchanged because its mass is much greater than that of the ball and therefore experiences negligible acceleration.
Another notable point about Newton’s third law is that contact between the two interacting objects is not necessary. It is evident from the example just cited: the ball does not yet make contact with the Earth, but still exerts its attraction. And the ball on Earth too.
A force like gravity, which acts interchangeably, whether there is contact between objects or not, is called “distance action force”. On the other hand, forces such as friction and normal require interacting objects to come into contact, so they are called “contact forces”.
Formulas taken from the example
Returning to the pair of ball-Earth objects, choosing the indices P for the ball and T for the earth and applying Newton’s second law to each participant in this system, you get:
F = m resulting . for
The third law states that:
m P to P = – m T to T
a P = 9.8 m / s 2 directed vertically downwards. As this movement occurs in the vertical direction, vector notation (bold) can be dispensed with; and choosing the up direction as positive and down direction as negative, you have:
at P = 9.8 m / s 2
m T ≈ 6 x 10 24 kg
Regardless of the ball’s mass, the Earth’s acceleration is zero. This is why it is observed that the ball falls towards the Earth and not vice versa.
rocket operation
Rockets are a good example of applying Newton’s third law. The rocket shown in the image at the beginning rises thanks to the propulsion of hot gases at high speed.
Many believe this is because these gases somehow “rest” in the atmosphere or the ground to support and propel the rocket. It doesn’t work like that.
Just as the rocket exerts a force on the gases and expels them backwards, the gases exert a force on the rocket, which has the same modulus but in the opposite direction. This force is what gives the rocket its upward acceleration.
If you don’t have a rocket handy, there are other ways to prove that Newton’s Third Law works to provide propulsion. Water rockets can be constructed, in which the necessary impulse is provided by the water expelled by means of a pressurized gas.
Note that launching a water rocket takes time and requires many precautions.
use of skates
A more accessible and immediate way to check the effect of Newton’s Third Law is to place a pair of skates and push them against a wall.
Most of the time, the ability to exert force is associated with objects that are in motion, but the truth is that immobile objects can exert forces as well. The skater is pushed backwards thanks to the force that the immobile wall exerts on him.
The contact surfaces exert (normal) contact forces against each other. When a book is supported on a horizontal table, it exerts a vertical force called normal on it. The book exerts a vertical force on the table of the same numerical value and opposite direction.
Experience for kids: skaters
Children and adults can easily experiment with Newton’s third law and verify that the forces of action and reaction do not cancel each other out and are capable of providing motion.
Two skaters on ice or on a very smooth surface can push each other and experience movement in the opposite direction, with or without the same mass, thanks to the law of action and reaction.
Consider two skaters with a very different mass. They are in the middle of an ice rink with negligible friction and are initially at rest. At any time they push each other with constant force with the palms of their hands. How will the two move?
It’s important to note that since it’s a frictionless surface, the only unbalanced forces are the forces that skaters apply to each other. Although weight and normal work on both, these forces balance each other out; otherwise, skaters would accelerate in the vertical direction.
Formulas applied in this example
Newton’s third law states that:
F 12 = – F 21
That is, the force exerted by the 1 in 2 skater is equal in magnitude to that exerted by 2 in 1, with the same direction and the opposite direction. Note that these forces are applied to different objects, just as the forces were on the ball and on the Earth in the conceptual example above.
m 1 to 1 = -m 2 to 2
As the forces are opposite, the accelerations they cause will also be, but their magnitudes will be different, because each skater has a different mass. Let’s see the acceleration acquired by the first skater:
So the next move is the separation of the two skaters in opposite directions. In principle, the skaters were at rest in the middle of the track. Each exerts a force on the other that provides acceleration while the hands are in contact and the momentum lasts.
After that, the skaters move away from each other with a uniform rectilinear movement, no longer acting with unbalanced forces. Each skater’s speed will be different if their masses are too.
Exercise solved
To solve problems in which Newton’s laws must be applied, it is necessary to carefully draw the forces acting on the object. This drawing is called a “free-body diagram” or “isolated body diagram”. This diagram should not show the forces the body exerts on other objects.
If there is more than one object involved in the problem, it is necessary to draw a free-body diagram for each of the objects, remembering that action-reaction couples act on different bodies.
1- The skaters from the previous section have respective masses m 1 = 50 kg and 2 = 80 kg. They push each other with a constant force of 200 N. The push lasts 0.40 seconds. To locate:
a) The acceleration that each skater acquires thanks to the impulse.
b) The speed of each when they separate
Solution
a) Take the positive horizontal direction from left to right. Applying Newton’s second law with the values provided by the statement, we have:
F 21 = m 1 to 1
From where:
b) To calculate the velocity they carry at the time of separation, the kinematic equations of uniformly accelerated rectilinear motion are used:
The initial speed is 0 as they were at rest in the middle of the track:
v f = in
v f1 = a 1 t = -4 m / s 2 . 0.40 s = -1.6 m / s
v f2 = a 2 t = +2.5 m / s 2 . 0.40 s = +1 m / s
Results
As expected, person 1 being lighter acquires greater acceleration and therefore greater speed. Now note the following about the product of the mass for each skater’s speed:
m 1 v 1 = 50 kg. (-1.6 m / s) = – 80 kg.m / s
m 2 v 2 = 80 kg. 1 m / s = +80 kg.m / s
The sum of the two products is 0. The product of mass and velocity is called momentum P. It is a vector with the same direction and direction as velocity. When skaters were at rest and with their hands in contact, it was assumed that they formed the same object whose amount of movement was:
P o = (m 1 + m 2 ) v o = 0
After completion of the push, the movement amount of the skater system remains 0. Therefore, the movement amount is retained.
Examples of Newton’s Third Law in Everyday Life
Walk
Walking is one of the most everyday actions that can be taken. If carefully observed, the action of walking requires pushing the foot against the ground so that it returns equal and opposite force to the walker’s foot.
It is precisely this force that allows people to walk. In flight, birds exert force in the air and the air pushes the wings so that the bird is pushed forward.
movement of a car
In a car, the wheels exert forces on the sidewalk. Thanks to the reaction of the pavement, it exerts on the forces of the tires that propel the car forward.
Sport
In sports, the action and reaction forces are numerous and have a very active participation.
For example, let’s look at the athlete with their foot on a starting block. The block provides a normal force in response to the athlete’s thrust on it. The result of this normality and the runner’s weight result in a horizontal force that allows the athlete to advance.
fire hoses
Another example where Newton’s third law is present is in firefighters holding fire hoses. The end of these large hoses has a handle on the nozzle that the firefighter must hold when the jet of water exits, to prevent the recoil that occurs when the water exits at full speed.
For the same reason, it is convenient to tie boats to the dock before leaving them, because when taken to reach the dock, the boat receives a force that pulls it away from it.