Misunderstanding on Newton's Third Law

Question

I am confused on how objects are able to move, even though Newton's third law states that every action has an equal and opposite reaction.

I feel it would be easier to explain my misunderstanding in an example: If a horse is harnessed to a carriage, and applies a force on the carriage, the carriage applies a force back onto the horse. I understand that the "action" and "reaction" forces act on different objects, but wouldn't the "reaction" force applied back on the horse prevent the horse from moving? The horse applies a force onto the ground to start moving, but immediately after, this force transfers to the harness, and the harness pushes back onto the horse--no movement happens.

Answer 1

If the force applied to the horse by the carriage was equal to the force applied onto the ground by the horse, the horse would not move. But they are different if the carriage is accelerating.

To clear out the confusion, I recommend individually considering the free body diagrams for each object you're studying. You will immediately notice why things move the way they do.

The answer to the question why things accelerate is invariably that they have a net force acting upon them. (Philosophically, the relation is actually backwards, i.e. force is defined to be the rate of change in momentum.)

Answer 2

As you noted, the key issue is that Newton's Third Law is a statement about forces on two different objects: $$ \vec{F}_\text{A on B} = - \vec{F}_\text{B on A} $$ Therefore you will never see these two forces within the same free-body diagram; they cannot "cancel out".

Draw the free-body diagram for the horse. There are: a tension force of the carriage backwards on the horse; a force of gravity downward on the horse; a normal force of the ground upward on the horse; and a static frictional force of the ground on the horse, pointing forward. If we state these forces with explicit reference to "object acting on object", they are: $$ \vec{F}_{T, \text{ carriage on horse}}\,, \, \, \vec{F}_{g, \text{ earth on horse}}\,, \, \, \vec{F}_{N, \text{ ground on horse}}\,, \, \, \vec{F}_{fs, \text{ ground on horse}} $$ Note that all of these forces belong on the free-body diagram of the horse, and are all "on horse". The acceleration of the horse will be determined, through Newton's Second Law, by the vector sum of these forces. In particular, for the horse to accelerate forward, the fourth of these forces must be larger in magnitude than the first. Those two forces are in no way related by Newton's Third Law.

The Newton's Third Law pairs of each of those forces are easy to write down (just reverse the "object on object", but maintain the type of force): $$ \vec{F}_{T, \text{ horse on carriage}}\,, \, \, \vec{F}_{g, \text{ horse on earth}}\,, \, \, \vec{F}_{N, \text{ horse on ground}}\,, \, \, \vec{F}_{fs, \text{ horse on ground}} $$ The first two forces belong in the free-body diagrams of the carriage and the (whole) earth, respectively, while the last two are on the free-body diagram of the ground. The acceleration of each of those objects will be due to the sum of forces in their own free-body diagrams. Again, Newton's Third Law plays no role in their dynamics.

The necessity of Newton's Third Law is to extend the dynamics of a single particle (which is prescribed by Newton's Second Law) to the dynamics of systems of particles. From this simple connection you can build up a theory the predicts the dynamics of rigid objects, clouds of particles interacting, fluids, and galaxies of stars.

Answer 3

but wouldn't the "reaction" force applied back on the horse prevent the horse from moving?

It depends on what other forces act on the horse.

If the static friction force between the horses feet and the ground equals the reaction force of the carriage on the horse, then the net force on the horse will be zero and the horse will not move per Newton's second law.

On the other hand, if the static friction force between the horse and the ground exceeds the maximum possible static friction force of $\mu_{s}Mg$, then the horse will accelerate in the direction of the reaction force of the carriage. An possible example is if the horse is standing on ice while the carriage is on dry ground.

Hope this helps.