Newton's Third Law of Mechanics  

 Newton's third law states that when two bodies interact, they always exert equal and opposite forces on each other. This law is commonly known as the law of action and reaction. For example, a book lying on a table produces a downward force equal to its weight on the table. The third law states that the table exerts equal and opposite force on the book. The opposite force occurs because the weight of the book causes an invisible deformation of the table, which pushes the book upwards like a compressed spring. (Figure 2.)
According to the textbook's definition, every force has an equal and opposite reaction. If body A exerts a force on body B, body B will respond with an equal and opposite force.
When driving a nail with a hammer, for example, into wood, the nail will stop the hammer at the end of its swing. It is completely counterintuitive to believe that the force of the hammer and the counterforce of the nail are equal throughout the contact, not just at the end when the motion stops, which is before the hammer stops. As the hammer advances, the nail "gives way" and drives into the wood, despite the fact that the forces between them remain equal throughout the motion, from initial contact to the end. (Fig. 1.)

It should be noted that Newton's first law also describes the action of equal and opposite forces, but on the same object, and as a result of the balance of forces, the object rests or moves uniformly in a straight line. This is fundamentally different from Newton's third law, which deals with equal and opposite forces, where each force acts upon the other object. As a result, these forces can have varying effects on the objects. Much like the aforementioned nailing. Depending on the vector sum of all forces, the total force on each of these objects could be non-zero.
If the total force on the body is not zero, it changes velocity and moves according to Newton's second law. If the total force on a body is zero, either because there are no forces (which is impossible) or because all forces are perfectly balanced by opposing forces, the body is not accelerating and is said that forces are in equilibrium. Where rest is also a state of unchanged velocity, only that velocity is zero. Conversely, if a body is observed not to change velocity, neither in amount nor direction, we must conclude that the sum of all forces acting on that body is zero.

What force is the counterforce of the normal force of the books on the table?
Electromagnetic contact force of table molecules and atoms. When we put the book on the table, the surface of the book and the surface of the table interact with repulsive contact electromagnetic exchange forces. The reaction of the table cancels the action of gravitational force, so the book does not fall through the table, but rests on it. However, this is not a pair of forces of action and reaction. They are of equal amount according to the 1^st and not according to the 3^rd NL. At the beginning they are not equal and it takes some time for them to equalize and for the book on the table to calm down. Like when we stand on the scale: we wait a bit and read our weight using 1^st NL. The gravitational force with which the Earth attracts the book is equal in amount to the gravitational force with which the book attracts the Earth, but it is in the opposite direction. These are the forces of action and reaction. Action and reaction forces are of the same nature (in this case gravitational) and act on two different bodies. The second pair of forces is of electrical origin. These are the normal force and reaction of the substrate. Normal forcxe is the force with which the book acts on the surface of the table, and the reaction of the surface is the force with which the surface repels the book. So we have two pair of forces.

Couple of forces for deformation of a ball in a collision with a baseball bat
A ball colliding with a baseball bat at a speed v can be replaced by a solid ball of mass m to which a spring of elastic constant k of negligible mass is attached at one end. The spring encounters the bat with its free end. The bat is solid (has a large elastic constant), which means that it can act on the spring if necessary with a large elastic force F₁ with negligible displacement. The bat stops the spring at its left end. Due to inertia, the ball continues to move, which leads to deformation of the spring. This causes an elastic force F = k∙∆x, where ∆x is the shortening of the spring, with which the spring acts on the ball with its other end. This force stops the ball and by 2^nd NL is equal F = m∙a. According to 3^rd NL, the force with which the ball acts on the spring must be equal to the force with which the spring acts on the ball, so F₂ = k∙∆x = m∙a. This is the force that, in conjunction with the force F₁ deforms the spring. In this process, the force F₂ increases, but the same applies to the force F₁, so the equality F₁ = F₂ always holds.

We know from experience that a donkey can pull a cart. But Newton's 3rd law says that the cart acts on the donkey with the same force in the opposite direction.
Let's see what forces act on the donkey-cart system. The weight of the donkey and the reaction of the ground are two forces in opposite directions and cancel each other out. The same conclusion applies to the forces between the cart and the donkey. The weight of the cart and the reaction of the ground on the car are also canceled. To pull the cart there must be an unbalanced force.
The donkey shoves its hooves at the ground, pushing the Earth back. As a reaction, a forward-directed frictional force appears between the hoof and the ground. The donkey pulls the cart by overcoming the rolling frictional force of the wheels, which acts in the opposite direction to the cart's movement. When the friction under the hooves is greater than the rolling friction of the wheels, the cart moves. Therefore, the resultant force acting on the system (donkey and cart) is equal to the sum of all forces acting on that system. The only two forces that do not cancel each other out are the forces of the frictional reaction of ground on the donkey's hooves and the forces of rolling friction of the wheels (red and dark blue arrows in Figure 3.).

If there was ice under the donkey's hooves (low friction) and the wheels were in mud (high friction), the donkey would not be able to pull the cart.

Of course, a horse or a donkey pulling a cart does not have "starting blocks" like athletes, which would enable them to "push off" the ground without slipping. To a certain extent, horseshoes help horses on dirt, but on asphalt they even reduce the friction between iron and smooth asphalt.

Rolling friction (of the idle wheels) opposes the car's motion, while the friction between the drive wheels and the road is the traction force that allows the car to move without wheel slippage, provided that the engine force F_Eg is less than the wheel friction:
               F_Eg < µ_s·m·g,
where µ_s is the coefficient of static friction between the wheels and the road (Figure 7). Here we need to distinguish whether we are talking about the coefficient of static (µ_s) or dynamic friction (µ_d) or the coefficient of rolling friction (µ).
Let's assume that the drive wheel turns clockwise (as in Figure 7) and at the same time pushes the road away with the tangential force of the engine F_Eg. Under normal circumstances, the wheel does not slip, which means that at the point of contact the road acts on the wheel with an equal force in the opposite direction. That friction force between the wheels and the road F_fr is the tangential component of the ground reaction. It is the traction force acting on the car in the direction of motion.

The wheel pushes the road, and since the road is attached to the Earth, it pushes the entire Earth. According to Newton's third law, the Earth pushes the wheel away with an equal force in the opposite direction.
               F_fr = F_Eg.
Now we will apply Newton's second law and compare the accelerations, m·a = M_Earth·a_Earth, where m and a are the mass and acceleration of the car, and M_Earth and a_Earth are the mass and acceleration of the Earth. We see that the Earth accelerates with acceleration a_Earth = m·a/M_Earth. Since the mass of the Earth is M_Earth = 6·10²⁴ kg and the mass of the car is m =1500 kg, the acceleration of the Earth is 2.5·10²¹ times less than the acceleration of the car and is completely imperceptible and unmeasurable. We could also say that it is irrelevant, but we would be terribly wrong. No matter how small, that acceleration moves the car. Without it, the car would stay in place. It is not the frictional force between the drive wheels that opposes the car's motion.
The forces opposing the car's motion are the rolling friction force F_wf and air resistance F_Air (Figure 7), where the rolling friction factor of the non-driven idle wheels is about 0.02. The dominant force opposing the car's motion is air resistance, which increases approximately with the square of the speed up to a speed of approximately 100 km/h, and with the third power of the speed for higher speeds. It is also the dominant consumer of fuel, so at a speed of 160 km/h, about 1.5 times more fuel is consumed than at 100 km/h. On the other hand, the actual friction factor between the drive wheel and the dry asphalt road is about 0.9, and on ice it is 0.1.

Convincing students of the ideas contained in Newton's third law — that there is an equal and opposite reaction to every force — can be difficult at times. If object A exerts a force on object B, object B responds with an equal and opposite force. As an auxiliary explanation, it is commonly stated that: if you push the wall with some force, the wall will push you back in the opposite direction with equal force. To overcome this difficulty, the following simple demonstration can help. All of the students in the class should follow the instructions demonstrated with hands, and they repeat them for themselves. All you need for accessories is a wall and, of course, a hand to push it!