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The horse and cart problem

  1. Sep 10, 2012 #1
    I am extremely sorry for making this thread, as it has been made here numerous times and there are a number of other websites that discuss this apparent problem, but there is something about this that I just can't understand no matter how I think about it.

    Here's how I understand it: A horse is hitched to a cart. The cart will move if the horse is able to move, and for the horse to move, the force that it exerts on the cart should be greater than the force the cart exerts on it.

    But the way I look at it, it is not possible for the horse to pull the cart with a greater force than the cart pulls it in the other direction. The horse gets all his force from the ground, right? It pushes the ground backward and the ground pushes it forward. Now, here's where I think you'll find the mistake that I'm making, but I can't convince myself that it is a mistake (I'm not being willful!); for the force with which the horse is being pushed forward should also be the force with which it pushes the cart forward and consequently, the force with which the cart pulls it backward. I know that this is not what Newton's third law says and this is not the correct action/reaction pair, but I can't think of it any other way. To me, the horse's role is merely to transmit the force it is being pushed forward with by the ground over to the cart. So, let's say, the ground pushes horse with 20 N, the horse pulls cart with 20 N, the cart pulls horse with 20 N. If the force that the ground pushes the horse with is doubled, the force on the horse by the cart and on the cart by the horse should also be doubled and we are still left with 0 net force. I understand that this is not a requirement of Newton's Third law, but in this particular case, I can't make myself believe that it is not so.

    I hope you understand what I'm having trouble understanding, and once again, I'm sincerely very sorry.
  2. jcsd
  3. Sep 10, 2012 #2
    The horse pulls on the cart, but the strap on the cart pulls on the horse, the two forces are acting on different bodies.
    That is why when you pull something, you feel it pulling back at you.
  4. Sep 10, 2012 #3
    Calculate the net forces acting on both objects.
    or even simpler you can treat the horse-cart as one object, because internal forces cancel
  5. Sep 10, 2012 #4
    The statement in red is not true. Starting with an erroneous statement will not help the understanding.
    The motion of a body depends on the forces acting on the body and not on the forces exerted by that body.

    The horse will start to move if the net force on the horse is non-zero.
    In this case, the horse interacts basically with two other systems: the ground and the cart.
    So there are two forces acting on the horse: from the ground and from the cart. If the force (in forward direction) is larger than the force from the cart, the horse will move.
  6. Sep 10, 2012 #5
    Did you forget gravity?

  7. Sep 10, 2012 #6
    Yes. Just thinking about the horizontal forces...
    Without gravity there will be not much force from ground either.
  8. Sep 10, 2012 #7


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    Yes, the force the horse exerts on the ground (though its hooves) is the same as the ground exerts on it. The ground isn't "going anywhere" because the mass of the earth is enormous compared to the mass of the horse! But the horse will move forward (a= f/m) as long as the force the ground exerts on it is greater than the force the cart exerts on the horse. The first depends upon the strength of the horse. The second depends upon the mass of cart and the friction in its wheels and axle.

    No. The horse, if it is, in fact, strong enough to pull the cart, transmits part of the force the ground applies to the horse to the cart, not all of it. The force the horse needs to transmit to the cart, to get it started (where it needs to get an acceleration a= f/m) depends entirely upon the mass of the cart and the friction in the wheels. Of course, it also has to accelerate its own mass so it must apply a force to the ground (so that the ground can apply the force back to the horse) equal to both of those. If the horse is not strong enough to to that then it simply cannot do that. If it is strong enough to accelerate itself and the cart to the desired speed then it only has to apply enough force to overcome friction and maintain that speed.

    Once again- the force the horse needs to apply to the cart depends upon the mass and frictional resistance of the cart. The force the horse can apply (through its legs and hooves to the ground) depends upon its strength and the traction it can get on the ground. Those two are NOT in general the same.
  9. Sep 12, 2012 #8
    So, the horse has to apply a force to the ground and be pushed by it that is equal to the force needed to accelerate itself and then it transmits only part of that force to the cart, because the cart requires a lesser force to accelerate than does the horse. Thus, the pull of the cart experienced by the horse is less than the push of the ground on the horse and so the system moves.

    Am I correct?

    Thanks for the answers everyone, especially HallsofIvy. Your comprehensive response made the most sense to me. I really appreciate it.
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