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Newton's Third Law and the Horse Cart Problem

  1. Dec 3, 2014 #1
    Hi,
    I am facing some issues while dwelling upon the application of Newton's third law of motion to the famous "Horse-Cart" Problem.Anybody who is clueless about what I am talking about please follow the link:

    https://www.lhup.edu/~dsimanek/physics/horsecart.htm

    As one can see the whole system has been sub divided into two sub systems.For the sub system comprising the horse and the ground we have two forces acting on the horse:

    1. The horse pushes the ground with it hooves and so does the ground equally to the horse in an opposite direction.

    2.The cart is pulled by the horse (say with the help of a cable connecting the horse and the cart) and so the cart pulls back the horse with an equal and opposite reaction force.This would the tension in the cable.

    Referring to the link above from the free body diagram of the horse the first force is D and the second force is C.

    Again for the sub system comprising of the cart and the ground we have two forces acting on the cart:

    1.The friction between the wheels and the ground in a direction opposite to impending motion or motion.
    2.The force on the cart by the horse i.e, the tension in the cable connecting the horse and the cart.

    From the free body diagram of the cart the first force is A and the second force is B.

    Now, the force applied by the ground on the horse should be equal to the force applied by the horse on the cart as both horse and cart are constrained to each other.(I suppose I am correct here?)

    So, D=B.

    Again according to the Newton's third law of motion, B=-C.

    Thus we should have D=-C.

    Apparently the analysis discards any possibility of motion for the horse cart system!

    Could anyone outline where did I went wrong?




     
    Last edited: Dec 3, 2014
  2. jcsd
  3. Dec 3, 2014 #2
    The force exerted by the ground on the horse is greater than the force exerted by the cart on the horse. So the horse is able to accelerate. I'll leave it up to you to figure out the corresponding situation for the cart.

    Chet
     
  4. Dec 3, 2014 #3
    Theoretically yes!The force on the horse by the ground should be greater than the force on the horse by the cart for the horse to accelerate.But this is the very point I am having confusion with as the tension in the cable should be equal to the force on the horse by the ground.Consequently the tension(which is the reaction force by the cart on the horse this time) should be equally applicable to the horse.Thus the force on the horse by the ground seems to be equal in magnitude to the reactive force by the cart on the horse at all times!

    Where could I be possibly wrong?
     
  5. Dec 3, 2014 #4
    Why do you feel that the force on the horse by the ground has to equal the force on the horse by the cart?

    Chet
     
  6. Dec 3, 2014 #5
    Force by the ground on the horse=Force by the horse on the cart (as both horse and cart are constrained to each other)

    Am I wrong while affirming the first statement?

    If not, force on cart by the horse= - (force on horse by the cart).

    Thus, force by the ground on the horse = - (force on horse by the cart).
     
  7. Dec 3, 2014 #6

    Nugatory

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    Staff: Mentor

    Yes, you are wrong.
    The only time the magnitude of the two forces will be equal is when the horse and the cart are both moving at a constant speed (any constant speed, including zero) instead of accelerating. Even then, it will not be because the horse and the cart are constrained to move together; it will be because the horse isn't accelerating so the force of the rope on the horse must cancel out the the force of the ground on the horse.

    The constraint imposed by tying the cart and the horse together is just that they must both have the same acceleration.
     
  8. Dec 3, 2014 #7
    If they both have the same acceleration that implies they are acted upon by different different forces as they have different masses.So for the cart which is generally heavier would require more force for the same acceleration as the horse.

    It should be only possible if the force by the ground on the horse is enough to accelerate the combined mass of the horse and the cart to the required acceleration.Conversely,the force contributes more towards accelerating the cart than it does to the horse.So for constrained systems as the horse and the cart the net forward force seems to be in-equally shared.This is where I precisely went wrong!

    Thank You
    Nugatory
     
    Last edited: Dec 3, 2014
  9. Dec 3, 2014 #8
    By the way I couldn't understand your argument!You say at first "The only time the magnitude of the two forces will be equal is when the horse and the cart are both moving at a constant speed (any constant speed, including zero) instead of accelerating.".And then you contradict yourself again!

    "Even then, it will not be because the horse and the cart are constrained to move together; it will be because the horse isn't accelerating so the force of the rope on the horse must cancel out the the force of the ground on the horse."
     
  10. Dec 3, 2014 #9
    I don't see any contradiction in what Nugatory said. But you were implying earlier that, in order for the horse to remiain intact, the forces on it must sum to zero. There is no basis for this.
     
  11. Dec 3, 2014 #10
    Yes I wrongly assumed the condition of the force applied by the ground to the horse being equal to the force on the cart by the horse.Apparently it seemed so as the horse and the cart were tied to each other.But Nugatory pointed out correctly the constraint necessarily ensures a common acceleration for both the horse and the cart and the intuition of the equal transmission of force was stupid indeed!

    In other words the tension in the cable connecting the horse and the cart(which I previously assumed to be equal to the force on the horse by the ground) is actually less than that and hence the horse accelerates.

    Sometimes I feel like banging my head against a wall:s:s
     
  12. Dec 3, 2014 #11

    Danger

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    Gold Member

    Then you need this smilie instead: :headbang:

    (Click the "Old School" tab in the smilies menu to access it. :D)
     
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