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Conceptual mechanics problems

  1. Dec 12, 2015 #1
    1. The problem statement, all variables and given/known data .

    First problem -
    ?temp_hash=de356a036067d2efad7e2d8b722a042f.png



    2. Relevant equations
    F=Ma

    3. The attempt at a solution

    Thinking in terms of newton's second law ,since there is no external force on the system(fan+sail+boat) ,the boat should not move .

    Thinking in terms of newton's third law ,the fan pushes on the air with some force F (say to the right) .The air pushes on the fan(and boat) to the left with same force F (to the left ). The air pushes on the sail(and boat) to the right with force F . So net force on the boat is zero . Hence the boat should not move .

    Is this reasoning correct ?

    Many Thanks
     

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    Last edited: Dec 12, 2015
  2. jcsd
  3. Dec 12, 2015 #2

    Bystander

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  4. Dec 12, 2015 #3
    Thanks :smile:

    Second problem : If a bird sitting in a closed airtight container starts to fly what happens to the weight of the container (the container is placed on a weight balance).What would happen if it was a cage (i.e air could escape out) ?

    In closed container :

    1) If the bird hovers (flies at the same place) ,then the weight registered would be same . The force of gravity on the bird Mg would be balanced by the upward force of air F i.e F=Mg .The air would be pushed down by the same force F . Doing a force balance on the mass of air ,this downward force on air would be balanced by an upward normal reaction force from the container . The weight registered would again be same .

    2) But if the bird accelerates upwards ,then the force exerted by the air would be more than its weight Mg . This results in an increased force on the bottom of the container . Hence the weight increases during the period the bird accelerates upwards .

    3) If the bird accelerates downwards ,then the force exerted by the air would be less than its weight Mg . This results in decreased force on the bottom of the container . Hence the weight decreases during the period the bird accelerates downwards .

    In open cage :

    If the bird hovers at the same place , then some of the air escapes out of the cage and the weight registered would be less as compared to when the bird was sitting at the bottom .

    Is this reasoning correct ?
     
  5. Dec 12, 2015 #4

    nrqed

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    No, it is incorrect. The boat is not an isolated system and the air molecules "rebound" against the sail (if the molecules would "stick" to the sail, then you would be correct). The molecules push on the right with a force larger than F when the hit the sail because they rebound with a nonzero velocity (if they had a zero velocity after the bounce, then you would be right). There are many classroom demonstrations showing that the boat actually moves forward. The simplest way to see that the boat will actually move forward is to use conservation of momentum. Before turning the fan on, the total momentum is zero. At the end, there is a flow of air backward (going away from the sail), after having rebounded against the sail. So there must be a net momentum of the boat in the direction of the sail, i.e. forward. The boat will move forward. You can also do it in two steps (being pushed by the fan and then rebounding against the sail), considering one air molecule at a time, say. You will find that the boat ends up having a net momentum forward.
     
  6. Dec 12, 2015 #5

    Bystander

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    Yes.
    Depends; toroidal circulation comes into play, as it does in the first case.
     
  7. Dec 12, 2015 #6
    Thanks for replying.

    In the first step ,the boat has a backwards(towards left) momentum and the air has a forward momentum (towards right). In the second step , after the air rebounds ,the boat would have a forward momentum . The result is that even though the boat moves a little forward and backward ,the net effect is that it stays at the same place .

    What do you think ?
     
  8. Dec 12, 2015 #7
    Are reasoning for 1) and 2) correct i.e when the bird hovers at the same place and when it accelerates upwards in the closed container ?
     
  9. Dec 12, 2015 #8

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  10. Dec 12, 2015 #9
    Ok . So , we have definite answers when the container is closed . But if the cage is opened , then weight might increase or decrease .

    Is that correct ?
     
    Last edited: Dec 13, 2015
  11. Dec 12, 2015 #10

    nrqed

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    Close. But my point is that the overall effect is that the boat *will* move forward. As I said, it is easier to do in terms of momentum (but we can also do it in terms of forces).
    Let's call p1 the momentum of one air molecule after it is "ejected" by the fan. At that point, th molecule has a momentum p1 to the right and the boat has a momentum p1 to the left.
    Now the molecule rebounds on the sail. Let's consider an elastic collision, first. Then the molecule goes from having a momentum p1 to the right to having a momentum p1 to the left, therefore its momentum has *changed* by 2 p1 to the left during the "collision". Therefore the boat has seen its momentum to the right increase by 2p1 during that collision.

    So overall, the total change of momentum of the boat is -p1+2p1 = +p1 to the right. So the boat moves forward. This is confirmed by doing the actual experiment.
    Of course, we can simply say that at the very beginning (before the fan is on), the air molecule has no momentum and at the end, it has a momentum p1 to the left so the boat has a momentum p1 to the right.

    In real life, the collision with the sail is not elastic but the air molecules still have a nonzero momentum to the left after rebounding.

    Now, in principle someone could argue: what prevents from the molecule to end up having a zero momentum at the very end? This is not enforced by conservation of momentum but I think that conservation of energy is sufficient to make the case that the energy provided by the fan must translate into overall nonzero motion of th boat and the air molecules at the end (assuming a few reasonable things)
     
  12. Dec 12, 2015 #11
    Nice reasoning :smile:
     
    Last edited: Dec 13, 2015
  13. Dec 12, 2015 #12
    I did not understand your point here . Could you explain a little more ?
     
  14. Dec 13, 2015 #13
    How is momentum approach and Newton's II and III laws giving different answers ? Although the momentum approach makes sense , yet what is the flaw in the OP ?
     
  15. Dec 13, 2015 #14

    ehild

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    The system is not closed. The surrounding air, brought into motion, exerts force on the boat system.
     
  16. Dec 13, 2015 #15
    Thanks for replying.
    OK.

    And what is the flaw in applying Newton's 3rd law as done in the OP.
     
  17. Dec 13, 2015 #16

    haruspex

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    I find this bouncing air argument rather dubious. Any bouncing air molecules would find themselves tangled up in the stream of air from the fan. The air would end up exiting sideways.
    It might work if you had the sail in a curve and directed the air only onto one half of it; the air could follow around the curve and return from the other half. Of course, the sailor would do better just to point the fan out the back of the boat in the first place.
     
  18. Dec 13, 2015 #17

    haruspex

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    Only that it assumes the air, having hit the sail, becomes stationary again (and that's relative to the surrounding air, not relative to the boat).
    It might be easier to think in terms of picking up balls floating near the boat and hurling them at the sail. If they continue to move with the boat, the boat slows down (if it was moving before); if they bounce back just enough that they are stationary relative to the water, no change to the boat's speed; if they bounce back any harder (landing in the water) you go forwards.
     
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