The question pose is as follows : The horse pulls the cart by a force F₁ in the forward direction. From the third law of motion the cart pulls the horse by an equal force F₁=F₂ in the backward direction . The sum of these forces is , therefore , zero . Why should then the cart accelerate forward ?

Try to locate the mistake in the argument. According to our scheme, we could first decide the system. We can take the horse as the system or the cart as the system. Suppose you take the cart as the system .Then the force on the cart is F₁ in the forward direction .How much is this acceleration ? Take the mass of the cart to be Mr.’s the acceleration of the cart a = F₁/M_C in the forward direction ? Think carefully . We shall return to this question.

Let us now try to understand the motion the horse. This time we consider the forces on the horse. The forward force F₁ by the gorse acts on the cart and it should not be taken into account when we discuss the motion of the horse. The force on the horse by the cart is F₂ in the backward direction. Why does the horse go in the forward direction when whipped ? The horse exerts a force on the cart in the forward direction and hence the cart is accelerated forward . But the cart exerts the equal force on the horse in the backward direction ? Where are we wrong ? We have not considered all forces acting on the horse. The road pushes the horse by a force P which has a forward component. This force acts on the horse and we must add this force when we discuss the motion of the horse. The horse accelerates forward if the forward component f of the force P exceeds F₂ acceleration of the horse is (f-F₂)/Howe should make sure that all the forces acting on the system are added .note that the force of gravity acting on the horse has no forward component.

Going back to the previous paragraph the acceleration of the cart may not be F₁/M_C.The road exerts a force Q on the cart which may have a backward component f’ .The total force on the cart is then a = (F₁-f’)/M_C in the forward direction .

The forces f and f’ are self adjustable and they so adjust their values that (F₁-f’)/M_C = (f-F₂)/M_H . The acceleration of the horse and that of the cart are equal in magnitude and direction and hence they move together .

So, once again we remind you that only the forces on the system are to be considered to discuss the motion of the system and all the forces acting on the system are to be considered Only then apply F=ma