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This question already has an answer here:

Please answer the question in detail. I have tried other websites for answers but was not able to understand properly.

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marked as duplicate by David Z Apr 22 '16 at 13:39

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This is a question that many have wondered about at some time, I think...

The action goes from the horse to the cart.
The reaction goes from the cart to the horse.
They are applied to different bodies!
And they also go in different directions. So if you would add them (which makes no sense, since they are applied to different bodies), you'd get zero.

So this is a different concept from: all forces on one body should be balanced, if it moves with constant velocity.
The first statement (action=reaction) is always true, the second (forces are zero) is only true with no acceleration. Those two sound similar, but the only similarity is indeed that they are statements about forces :)

The second statement gets true in the form: if the forces on a body are balanced, it doesn't accelerate. This is a special case of $F=ma$.


Now consider the horse+cart moving with constant velocity.
The forces on the cart are: force from the horse and force of friction. They are equal and opposite -> balanced.
The forces on the horse are: force from the cart and force of friction. They too are balanced.
The forces on the earth are the two forces of friction, which also are balanced! (examine the directions!...).

Now consider the horse+cart starting to move.
The forces on the cart are not balanced, the force from the horse is greater than friction, so it accelerates.
The forces on the horse are not balanced either, the friction is still greater than the force from the cart. The horse, too, has to have a net force forward, so it can accelerate.
So the friction under the horse is much (at least two steps) greater than the friction under the cart.
So there is also a net force on the earth, accelerating it (very little, because of the slightly greater mass :)) backwards.

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  • $\begingroup$ could you please explain who downvoted ... you probably din not read? :) I think the answer is very enlightening to a beginner who was taught badly about those concepts in school. I'm sure it is, indeed :) $\endgroup$ – Ilja Apr 22 '16 at 11:27
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Action has the same intensity as reaction, but the two forces act on two different bodies.

The horse pulls the cart, and the horse force acts on the cart. $$ m_1 \mathbf{a}_{\textrm{cart}} = \mathbf{F}_{\textrm{horse on cart}} + \mathbf{F}_{\textrm{friction on the cart}}+ \textrm{other contributions} $$ Likewise $$ m_2 \mathbf{a}_{\textrm{horse}} = \mathbf{F}_{\textrm{cart on horse}} + \mathbf{F}_{\textrm{friction on the horse}}+ \textrm{other contributions}. $$ It is true that $\mathbf{F}_{\textrm{horse on cart}} = - \mathbf{F}_{\textrm{cart on horse}}$ but nevertheless the other pieces in the equations are different, therefore the solutions are different. If no other contribution is present, then the centre of mass of the entire system moves at constant velocity, which is the requirement for a system of particles in mutual interaction, should no external force act.

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because of frictional force of the ground. the horse is applying force onto the ground and ground's frictional force is applied to the horse.so horse along with cart moves

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