6
$\begingroup$

A while ago someone proposed the following thought experiment to me:

A horse attached to a cart is resting on a horizontal road. If the horse attempts to move by pulling the cart, according to the 3rd Newton's Law, the cart will exert a force equal in magnitude and opposite in direction, cancelling each other out and thus the horse and the cart should not move. And yet it moves (pun intended ;) Why?

I never got a satisfactory answer, my guess is that the answer lies in the frames of reference involved: horse-cart and horse-road. Any ideas?

| cite | improve this question | |
$\endgroup$
10
$\begingroup$

The opposing force always work on a different body, thats why they dont cancel. For instance the moon pulls the earth and the earth pulls the moon with same force. Also, the horse dont move by pulling the cart, but by pulling the earth.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ you mean by "pushing the earth"? $\endgroup$ – user346 Dec 20 '10 at 18:55
  • $\begingroup$ I mean by alternately pulling and pushing the earth $\endgroup$ – TROLLHUNTER Dec 20 '10 at 19:27
  • 1
    $\begingroup$ Ambiguous answer , take Newtonian gravity fir example, two bodies exert equal gravitational forces on each other which would mean neither body should move as they both cancel out each other $\endgroup$ – user1062760 Aug 26 '16 at 1:10
5
$\begingroup$

The horse also exerts a force on the road, and the road reacts on the horse. To compute the movement of the horse, you need to add the 2 forces acting on the horse (by the road and by the cart).

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ So then the horse needs to exerts a force against the road greater than the force exerted on him by the cart right? So that there is a net force in his forward direction? $\endgroup$ – teto Dec 20 '10 at 18:59
  • $\begingroup$ @teto Greater or equal to $\endgroup$ – TROLLHUNTER Dec 20 '10 at 19:30
  • $\begingroup$ @TROLLHUNTER why equal to? won't they cancel out and be left with no acceleration? $\endgroup$ – Reinstate Monica Nov 20 '18 at 18:14
4
$\begingroup$

Let us draw free body diagram for the cart. It is important to note that reaction force is NOT acting on the cart. And that is why my friend it moves.

This does not mean that reaction is absent. It is there but its NOT acting on the cart itself. So not all the forces on the cart are balanced. There is net force.

alt text

This is common misunderstanding. See examples here.

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.