A horse-pulled cart is moving with acceleration a.

If the force exerted by the horse on the cart and the force exerted by the cart on the horse are equal and opposite as required by Newton's third law of motion, then why does the combined system move and accelerate?

We may assume for simplicity that the design of the wheels of the cart is such that there is no drag on the cart because of the road.

If the horse–cart system is moving with acceleration a, then this system must obtain a forward force of magnitude            

(m_horse   +     M_cart) a.

This force is obtained by the horse by pushing the road backwards and the road provides the force F required for driving the horse-cart system forward with acceleration a,            

F = (m_horse   +     M_cart) a.

Let the force with which horse is pulling the cart be T

Therefore,

T = M_cart   a.

The magnitude of the force with which cart will pull the horse towards itself by the Newton's third law of motion will also be T. Therefore, the net force on the horse will be         F –   T.

The magnitude of the force with which cart will pull the horse towards itself by the Newton's third law of motion will also be T. Therefore, the net force on the horse will be                 F –       T.

F –   T   = (m_horse + M_cart) a - T        
             = (m_horse +M_cart) a –   M_carta
             = m_horsea.

Teaching Newton's Laws of Motion

Prof. A. N. Maheshwari