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Help with Newton's third law

  1. Feb 15, 2010 #1
    1. 2. On a horizontal road, a small sedan is pulling a large station wagon that has run out of gas. While the sedan is speeding up to high way speeds.

    A. The sedan pulls harder on the station wagon than the station wagon pulls on the sedan; otherwise there would be no acceleration.

    B. The station wagon pulls just as hard on the sedan as the sedan pulls on the station wagon, no matter what the acceleration is.

    C. It is not possible to know which car pulls harder on the other without knowing their masses.



    2. Relevant equations Newton's third law.



    3. The attempt at a solution I chose B, due to Newton's third law, and action-reaction pairs. The sedan pulls harder on the rope (if there is one) than the state wagon pulls on the rope. That's how you get forward acceleration, right?
     
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  3. Feb 15, 2010 #2

    kuruman

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    If the sedan is pulling the station wagon there must be something in between, say a rope. Let's assume for the moment that your analysis is correct and the sedan pulls harder on the rope than the station wagon. Say the sedan pulls on the rope with a force of 1200 N and the station wagon pulls on the rope with 1000 N. Now answer me this: what do you think is the tension in the rope?
     
  4. Feb 15, 2010 #3
    Wouldn't that depend on where on the rope you were talking about? On the right side, it would be 1200 N, and on the left it would be 1000 N. In the middle, it would be 1100 N, right? The tension would only be the same throughout the rope if it was in equilibrium.
     
  5. Feb 16, 2010 #4

    kuruman

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    Nope. If the force on one end of the rope were different from the other end, the two ends would have different accelerations and their relative velocity would be changing continuously, i.e. the rope would be stretching continuously.

    Speaking of "equilibrium" what do you think the tension in the rope would have to be if the sedan and station wagon were moving at constant velocity, say 40 miles per hour? Would that change if the two were moving at constant 50 miles per hour?
     
  6. Feb 16, 2010 #5
    The question is very general so the rope between the 2 cars can be assumed to be massless so the tension at the 2 ends of the rope will be same thus the 2 cars are applying equal and opposite forces on each other .

    If you assume the rope to have mass then there is a completely different story. you got to think of the rope as a object on which the working car is appling force and the rope is applying force on the wrecked car.
     
  7. Feb 16, 2010 #6
    OK, well, t=F. In the case of constant velocity, acceleration is zero. T=F=m(0). Thus, Tension would be zero, in both cases of constant velocity.

    Would the tension in the rope be 200 N? 1200 being pulled on the right side, 1000 on the left side. 1200-1000?
     
  8. Feb 16, 2010 #7

    kuruman

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    No. The point I wanted to make is that Newton's 3rd Law says that if the sedan is pulling on the station wagon with a forward force of 1200 N, then the station wagon is pulling on the sedan with a backward force of 1200 N. That's Newton's 3rd Law.

    Well, you might ask, how could then the station wagon accelerate? The two vehicle system experiences a net external force of say 2000 N provided by the sedan's engine. The acceleration of the two vehicle system is then

    a = 2000/(msedan+mwagon)

    and the net force on the sedan is 2000 - 1200 = 800 N whereas the net force on the wagon is just the tension, 1200 N.
     
  9. Feb 16, 2010 #8
    OK, I'm trying to understand this. You said that the sedan pulls on the rope with a force of 1200 N, and the station wagon pulls on the rope with a force of 1000 N. Then, the rope pulls on the sedan with a backwards force of 1200 N, and the rope pulls on the station wagon with a force of 1000 N.

    The tension in the rope is 1200 N? What happened to the 1000 N?

    Oh, and was my statement about constant velocity wrong too?
     
  10. Feb 16, 2010 #9

    kuruman

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    I said "Let's assume for the moment that your analysis is correct and the sedan pulls harder on the rope than the station wagon. Say the sedan pulls on the rope with a force of 1200 N and the station wagon pulls on the rope with 1000 N." What "we assumed for the moment" is not true because it leads to an illogical conclusion. Therefore your analysis is incorrect.

    I am trying to make you understand that the force that the sedan exerts on the wagon is the same as the force that the wagon exerts on the sedan is the same as the tension in the rope.
     
  11. Feb 16, 2010 #10
    Oh, OK! So, it's not possible for a rope that is in equilibrium to have different tension in the rope. So, you can't have 1200 N on one end and 1000 N on the other end, they have to be the same? That was your point? That having 1000 N on one end wouldn't make sense in this case?

    So, B is correct, right?
     
  12. Feb 16, 2010 #11

    kuruman

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    Yes, B is correct but you need to understand why.
     
  13. Feb 16, 2010 #12
    Well, I know Newton's third law. Any action, has an equal and opposite reaction. Whatever force the sedan places on the station wagon, the station wagon places and equal and opposite force back at it. In the case of a rope, if it is massless, it must have the same tension throughout it, thus, the 1200 N produced by the sedan will run throughout the massless rope. Is that good?
     
  14. Feb 16, 2010 #13

    kuruman

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    That is good.
     
  15. Feb 16, 2010 #14
    Thanks!
     
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