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Newton's Laws of a cart pull

  1. Oct 8, 2007 #1
    1. The problem statement, all variables and given/known data
    According to legend, a horse learned Newton's laws. When the horse was told to pull a cart, it refused, saying that if it pulled the cart forward, according to Newton's third law there would be an equal force backwards. Thus, there would be balanced forces, and, according to Newton's second law, the cart wouldn't accelerate. How would you reason with this horse.


    2. Relevant equations
    a=f/m



    3. The attempt at a solution
    Well I don't really know it was in interesting question that I found while reading through them and thought someone on hear might know.
     
  2. jcsd
  3. Oct 8, 2007 #2

    D H

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    What other third law pairs (action/reaction) are in play beside the horse and cart pulling against each other?
     
  4. Oct 9, 2007 #3
    F=ma. the force the horse and cart exert on each other is the same. however, due to the difference in mass between the horse and cart, the acceleration of the 2 parties involved will be different. in this case of course, i would assume that the mass of the cart is greater than that of the horse. so the acceleration of the cart has to be smaller then that of the horse for the force to be equal. thus, the overall velocity would be in favour of the horse.
     
  5. Oct 9, 2007 #4

    D H

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    The separation between the horse and cart is constant. How can "the acceleration of the 2 parties involved" differ?

    What else is happening beside the horse pulling on the cart and the cart pulling on the horse?
     
  6. Oct 9, 2007 #5
    The cart has wheels and the horse has heels. Wheels produce less friction than heels. Think of it this way (values are obviously fantastic): Horse=90kg Cart=120kg Horse exerts 2 000 Newtons of force while cart does not exert any force in the x-axis (only mass*gravity (weight) affects the cart when the horse is not put into the equation). Let's say that the Cart's wheel produce a co-efficient of friction of about Uk/s=.5 while the horse's hooves produce a friction coefficient of Uk/s=.9. So acceleration of entire system is the Sum of All Forces divided by all the masses involves. The sum of all forces in this equation would be Force (2 000 Newtons) minus Force of Friction (this depends on the Normal Force).

    The Normal Force is the force opposite the masses downward acceleration due to mass*gravity and is the minimum upward force exerted by the surface in order to keep the masses from sinking through the surface. The Normal Force upon the horse would be 90*9.8 (9.8 being gravitational acceleration in meters per second, and is positive since Normal Force is going upwards).

    Therfore, the horse's Normal Force (N) would be 882 Newtons. Also the Cart's Normal Force is 120*9.8=1176 Newtons. The Force of Friction is given by Uk/sN. So the Horse's Force of Friction is .9*882=793.8 Newtons and the Cart's is .5*1176=588 Newtons. So we have to find the system's Sum of Forces: 2 000 Newtons (Horse's Force) minus the sum of the Horse's and Cart's Force of Friction (793.8 Newtons+588 Newtons) which is 1 381.8. So 2 000 minus 1 381.8 is 618.2 Newtons (this is. So the acceleration of the system is given Sum of Forces divided by Total masses involved. So we have 618.2/(90+12)=618.2/112=5.52 meters per second, in the positive x direction.

    The acceleration of the the two parties do differ, it is just that their combined accelerations counter the other forces within the system so as to cause the system to move with a uniform velocity (remember: for the rope to have tension, there must be some pulling involved, in other words, there must be some uneven acceleration within the sytem). The horse's acceleration is equal to the Sum of Forces: ( -tension, -friction, and +force exerted) acting upon it divided by its own mass. One force is the tension created by the rope connecting it to the Cart. Tension is defined as the masses the string is pulling (in the acceleration's direction) multipied by the acceleration of the system. In this case the rope is pulling the Cart which is 120kg. So the Tension is 120*5.52= 622.4 Newtons of Tension.And, as we learned above, the Horse's Force of Friction is 793.8 Newtons.

    So to find the Sum of Forces acting on the Horse we take the Force exerted and minus from it Force of friction and Force of Tension. We have 2000-(622.4+793.8)= 583.8 Newtons. Divide the Sum of Forces by the Horse's mass and we get the Horse's acceleration. So we have 583.8/90=6.487 meters per second squared.

    Similarly, the only real forces acting upon the Cart are the Force of Friction (which involves gravity) and the Tension in the string which is actually pulling the Cart in the positive x direction. So the Sum of Force for the Cart would be Tension minus Friction, which is 622.4-588=34.4 Newtons. Now we divide this by the Cart's mass to get the Cart's acceleration. We have 34.4/120=0.287 meters per second squared.

    From this we get that the Horse and Cart do not accelerate at the same rate, but notice that the cart's acceleration is much lower than the Horse's acceleration. This uneven acceleration keeps the rope tight, and tension is what makes the Cart move. One way some are confused by this Horse-Cart thing is that they assume the the Cart exerts Force when it does not, it actually absorbs force, since it is the one being pulled. The equal and opposite reaction of the Horse exerting force is the GROUND moving in the opposite direction and the tension and friction trying to pull the horse back. In turn, The Cart is acted upon by Tension pulling it and the Ground moving away and friction trying to keep it from moving. You must remember that the Horse and Cart work as a system. Just as the engines of a rocket and the cargo work as a system to propel the system away from the ground vertically, the Horse (rocket engine) and the cart (cargo) work as a system to propel the system horizontally.

    P.S. When you say seperation is constant, what you are really saying is magnitude of Tension (the rope) is constant.
     
    Last edited: Oct 9, 2007
  7. Oct 9, 2007 #6

    PhanthomJay

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    There are multiple errors in this response. As DH noted , the acceleration of the horse, cart, and horse-cart system, is one and the same. Regardless, the question relates to the understanding of the differences between Newton's 2nd and 3rd laws.
     
  8. Oct 9, 2007 #7
    Please forgive my errors, I was only going on what I was recently taught in class. Can you please point out said errors so that I won't have flawed physics?
     
  9. Oct 9, 2007 #8

    PhanthomJay

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    That's Ok, I sure have made my share. The 2000N force of the horse is actually provided by friction between its hooves and the ground. It pushes backwards with that force, and the friction drives the horse (and cart) forward. The cart's friction force (rolling friction or axle friction) is typically rather small, but lets go with your 0.5 value. So acceleration of entire system is the Sum of All Forces divided by all the masses involves. The sum of all forces in this equation would be Force (2 000 Newtons) minus cartForce of Friction (this depends on the Normal Force).

    The Normal Force is the force opposite the masses downward acceleration due to mass*gravity and is the minimum upward force exerted by the surface in order to keep the masses from sinking through the surface. The Normal Force upon the horse would be 90*9.8 (9.8 being gravitational acceleration in meters per second, and is positive since Normal Force is going upwards).

    Therfore, the horse's Normal Force (N) would be 882 Newtons. Also the Cart's Normal Force is 120*9.8=1176 Newtons. The Force of Friction is given by Uk/sN. So the Cart's friction is .5*1176=588 Newtons. So we have to find the system's Sum of Forces: 2 000 Newtons (Horse's Force) minus the sum of the Cart's Force of Friction. So the acceleration of the system is given Sum of Forces divided by Total masses involved.

    The acceleration of the the two parties do not differ, they are connected so they must move together, accelerating at the same rate. So to find the Sum of Forces acting on the Horse we take the Force exerted and minus from it Force of Tension.

    The only acting upon the Cart are the Force of Friction and the Tension in the string which is actually pulling the Cart in the positive x direction. So the Sum of Force for the Cart would be Tension minus Friction.
    Note also that sibce the horse exerts a force on the cart, then the cart must also exert a force on the horse, equal in magnitude but opposite in direction.
     
    Last edited: Oct 9, 2007
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