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Tricky question?

  1. Sep 29, 2004 #1
    tricky question???

    This question comes from an old test that a University in my area gives out to highschools to see who is out there, and possibly give them a scholarship.

    A person is on a boat in the ocean next to an island. On that island there is another person and a tree. There is a rope tied between the boat and the tree. The person on the boat can pull himself in at 1 m/s. and the person on the island can pull the boat in at 1 m/s. If both people pull at the same time on the rope, how fast will the boat be moving towards the island.

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    I was told that the answer is that the boat will move at 1 m/s. Logically to me the boat would move at 2m/s. I don't see where I am going wrong. :confused:
     
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  3. Sep 29, 2004 #2

    NoTime

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    If you call the limiting factor the amount of force that a person can produce, then 1m/s would be correct.
    If you call the limiting factor the dexterity of the person (how fast the people can manipulate a piece of rope) then 2m/s would be correct.

    In real life either limit could take precedence.
    If its labled a physics test, then it would probably be best to assume a force problem.
     
  4. Oct 12, 2004 #3
    i am not precicly sure what you mean by limiting factor but what I am getting at is they each can pull with a force say 500N that allows the boat to move at 1 m/s. so that would mean that both pulling would still be 1 m/s correct?
     
  5. Oct 12, 2004 #4

    rcgldr

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    Problem is poorly worded. One, it doesn't state that there is only one rope, only that one rope is attached to a tree. It doesn't explain how the person on the island is pulling the boat in (another rope?). Lastly, as already posted it doesn't explain the reason for the limit of 1m/s for the pull.

    In a real world situation, it's probably a combination of force and power that are the limiting factors. If each person only has to pull 1/2 as fast or slower, then they could use shorter pull strokes, combining the strength of both arms more often and with better pull leverage than what could be acheived at 1 m / s, so the boat would end up faster the 1 m/s but less than 2 m /s .

    I've never liked "problems" like these, the given answers usually don't match reality.
     
    Last edited: Oct 12, 2004
  6. Oct 14, 2004 #5
    alright, fine! lets make it 2 winches both pulling in the same rope. each can pull in the boat on their own 1 m/s with say, a force of 500N. So when both are pulling in with their 500N each will the boat be moving 1m/s or 2m/s
     
  7. Oct 14, 2004 #6
    To me it seems like 1 m/s is right. It says the person on the boat can pull HIMSELF at 1 m/s and the person on the island can pull the BOAT 1 m/s. So they each pull a component of the man and boat together, at 1 m/s.

    Edit: I may be completely stupid and wrong, it just seems worded funny.
     
    Last edited: Oct 14, 2004
  8. Oct 16, 2004 #7
    anyone with a more confident answer?
     
  9. Oct 16, 2004 #8
    So far, the rope is anchored to a tree.
    So, now if only one pulls the boat moves at 1 m/s. Each person can exert a force of x, with corresponding tension of -T. (Not sure about my terminology there)
    Going by the bolded, I would assume both literally pull at the same time. Since each person exerts the same force of x, in order to pull the boat they must not only pull the mass of the boat but pull back against the tension in the rope created by the other person. So, I think you end up with a force equation (once again, not sure about my terminology) of: -T + -T (for total tension in the rope) = x + x (for the total force on the rope) or -2T = 2x, or a tension/force of -T = x, which is the same as if only one of the men is pulling.

    As far as a confident answer goes, I would wait to see if anyone else comments as to the correctness of the above...something about it doesn't quite look right.
     
  10. Oct 17, 2004 #9

    ehild

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    I think it is more realistic that the force of the person is limited.

    The trick is in the formulation that the man can pull the boat with 1 m/s.
    No force is needed to move something with a constant velocity, in principle. The boat moves with constant velocity. The forces acting on each must cancel.
    Consider the forces acting on the boat. One is the tension in the rope. The other is the drag from the water. The drag depends on the speed of the boat, and increases with increasing speed. Assume that the drag is simply D=Bv, with B a constant. The tension should be equal to the drag, T=Bv. The man on the island pulling the rope experiences the same force T if the rope has no mass, and he exerts an equal force T on the rope. Tmax=B*1(m/s). If the speed of the boat would increase to 2 m/s the man should exert twice as much force. T=B*2(m/s). But his maximum force was Tmax so the speed can not be bigger, either the other man in the boat pulls or not. It would be more useful when he row and just fixed the rope to the boat. :smile:



    ehild
     
  11. Oct 17, 2004 #10
    The person in the boat exerts, via the rope, a force on the tree that makes the boat move. The tree will feel the same force as the man in the boat, it may bent al little until the elastic force of the tree equals the force that the man is pulling with. It does of course not matter whether the rope is tied to a tree or held by a man on the island, and in order not to be dragged into the sea the man on the island will have to exert the same force on the rope as the man in the boat.

    so if they both pull with a force that makes the boat move at 1m/s then the boat will move 1 m/s.
     
  12. Oct 17, 2004 #11

    rcgldr

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    The issue is a conflict between a constant force situation and a constant power situation. Winches aren't a good choice because electric motors approximate constant power devices, generating more torque at lower speeds (consuming more amps, and generating more heat). Winches at both ends would pull the boat faster than just one winch and one tree, especially if you're allowed to change the gearing. The answer is the same as my human powered winch scenario I describe below.

    If you want a constant force scenario, replace the tree with a rocket that generates 500n of force and basically remains motionless like the tree.

    This is supposed to be a physics problem, so I prefer a more realistic constant power result to this question, and will also assume that drag increases with the square of the speed, a reasonable assumption (at higher speeds, a boat planes, greatly reducing drag, but I don't think this is an issue around 1m/s).

    Lets say that instead of pulling on the rope directly, the rope is spooled onto a hub driven by a pedal and gear system, like a modified 10 speed bicycle. The man on the is pulling the boat by pedaling the human powered winch setup, his 500nm/s of power are distributed as 500n at 1m/s. The tree just provides a resistive force of 500n, but does no work since it doesn't move the rope. Next, replace the tree with another man with the same human powered winch setup, as the guy on the boat.

    Since the force required to move the boat increases by the square of the speed, some algebra is needed, but the result is that each man distibutes his 500nm/s of power by gearing his winch to produce 793.8n at .630m/s, with a resulting boat speed of 1.26 m/s. This corresponds with the fact that the boat needs 793.8n of force to move at 1.26m/s.
     
    Last edited: Oct 17, 2004
  13. Oct 17, 2004 #12

    NoTime

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    Exactly.

    Agreed :approve: (usually? :rofl: Always maybe? :smile: )

    Confidence in the possible answers is 100%.
    Confidence in the question is nearly zero.
     
  14. Oct 17, 2004 #13
    IS THE ANSWER ZERO!!!


    coz 1-1=0

    and one is pulling the boat and the other is pulling the tree!!
     
  15. Oct 17, 2004 #14
    No...the boat moves, the tree doesn't. Obviously, if the guy in the boat moves toward the island when he pulls on the boat, he's not going to cancel out the effect of the other person pulling- it's not like a tug of war, since the boat isn't anchored.
     
  16. Oct 17, 2004 #15

    NoTime

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    That's funny. :rofl:
    And correct. :wink:

    For the constant force solution (1ms) the distinction between the tree or the person on the beach doing the pulling is indeed zero.
    And if the person on the beach pulls the tree, well the tension in that segment of the rope goes up, but no work gets done. Also zero.
    (within the limits of the question, not real life)
     
  17. Oct 17, 2004 #16
    I could see that would be the case if when one person pulled, the other moved, but that doesn't seem to fit with the fact that no matter who pulls, the guy in the boat is the one who moves. It doesn't make logical sense for him to be able to stop himself from being pulled to shore simply by pulling on the rope- it seems as though it's much the same as saying that you could stop yourself from falling if you pushed upwards on something falling with you with a force equal to the force of gravity, and then grabbed on.
     
  18. Oct 17, 2004 #17

    NoTime

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    There are distinctions, but physics is sometime quite counterintuitive.

    If two people are pulling 1m/s the amount of rope going through each individuals hands is 0.5m/s.
    If one person is pulling the amount of rope going through that persons hands is 1m/s.
    Look up the work equation. If the rate of work stays constant which parameters can you change?

    You can actually do this. :smile:
    The constraints are pretty tight, but certainly possible.
    The limits of your body get exceeded very rapidly.
     
    Last edited: Oct 17, 2004
  19. Oct 18, 2004 #18
    Interesting...OK, I think the reasoning behind all this is starting to make sense. I suppose that in theory it shouldn't be different then if it were a push instead of a pull...it's easier to visualize that as resulting in 0 work done then it is with a rope being pulled on.
     
  20. Oct 18, 2004 #19
    I think the question is simply ment to be about: action = reaction

    If the person in the boat pulls with certain force the tree has to pull with equal force, if instead of the tree a person holds the rope that person has to pull with this force (in order not to be dragged into the sea), so both have to pull with a force that makes the boat move at 1 m/s to move the boat at 1 m/s
     
  21. Oct 18, 2004 #20
    Hmm...actually, that makes even more sense than if the answer was 0...Physics is Phun, do you know for certain that the answer is 1m/s, or is that just what someone else thinks the answer is? This does seem to be a tricky problem.
     
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