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Another basic question

  1. Jul 28, 2011 #1
    Assume there is a boat of length 4 m weighing 40 kg. One man of weight 50 kg is at one end (A) and the other of weight 65 kg is at the other end (B).
    Both men now move towards the centre of the boat.
    Will the boat move in the water and if so by what distance ? (assume no friction between water & boat)

    Taking origin at A, COM originally is 22/15 from A.

    Since no external force acts on the system the COM remain stationary. Using this principle, assuming that boat moves by 'x' towards the left of the origin, it is obtained from the eqn for COM that
    0 = 50(2-x) + 60(-2-x) + 40(-x) / (150)
    Solving this eqn gives x = 2/15 i.e. boat moves 2/15 m towards the left.

    THE DOUBT IS WHY DID THE BOAT MOVE WHEN NO EXTERNAL FORCE ACTED ON IT ?
     
  2. jcsd
  3. Jul 28, 2011 #2

    A.T.

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    What is "the boat"?

    - Only the boat itself? Then there was an external force from the passengers.

    - Boat & passengers? Then it didn't move, because it's COM stayed stationary.
     
  4. Jul 28, 2011 #3
    Boat will move even if these two guys don't move (they are positioned at each end) because their weights are different thus center of gravity of whole system is not at exactly in the middle of the boat. Boat will move with a constant acceleration in this case (assuming no sinking). When these guys move toward each other, acceleration will decrease because they will be moving to a single point where in the end center of gravity of the system will be exactly in the middle of the boat and acceleration will decrease to zero so equilibrium.
     
  5. Jul 28, 2011 #4

    A.T.

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    So you have invented a boat propulsion that doesn't require any fuel, just an off center COM?
     
  6. Jul 28, 2011 #5
    I didn't invent it and it was there for a long time. Make the math, acceleration is very very small in this system, considering friction etc it might not even be observable from macroscopic point of view. But it is there.
     
  7. Jul 28, 2011 #6
    Also I would like to add that here in Turkey we're implementing a fully electric vehicle which has far more better efficiency characteristics than any other vehicle on the planet and the system is based on changing the center of gravity. Segway's also base on same principle, this is a known fact.
     
  8. Jul 28, 2011 #7
    Sorry to derail, but when you say a physical quantity is small or large, you must compare it to some other physical quantity of the same kind. What exactly is this acceleration small compared to?
     
  9. Jul 28, 2011 #8

    A.T.

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    How about you doing the math, and showing us that a boat will accelerate, just because the guy sitting at one end is heavier than the guy on the other end
     
    Last edited: Jul 28, 2011
  10. Jul 28, 2011 #9

    Doc Al

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    While there's no external force on the system of boat+men, there is certainly an external force on the boat itself. The men push against it when they move.

    (Oops... A.T. already made that very point.)
     
  11. Jul 28, 2011 #10
    My expression was wrong, the system will have a constant velocity rather than having an acceleration. I will proof and post it here.

    That's also another aspect but what I am talking about is the initial imbalance due to weight difference resulting in a motion.

    What I meant by small is we might not perceive the movement of the boat via our sensory organs. Observation of water waves might yield another vectors to work on. On the other hand I realized that my expression was wrong, it doesn't accelerate but it will move with a constant velocity.
     
  12. Jul 28, 2011 #11

    Doc Al

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    So, if the men and boat were initially at rest they'd remain at rest (until they start walking around the boat). I don't think we need a proof of that. :wink:
     
  13. Jul 28, 2011 #12
    :smile:

    Imagine boat is empty no one abroad. We put heavy man to one end and light man to other end simultaneously. Initially what I mentioned will occur.

    Men walking towards each other will also effect the velocity since weights are not equal, change in center of gravity will be towards the heavy guy till they meet each other assuming they are walking at the same speed.
     
  14. Jul 28, 2011 #13
    The only relevant quantities that I can see are the masses of the two persons [itex]m_{1}, m_{2}[/itex] and the boat [itex]M[/itex], the acceleration due to gravity [itex]g[/itex] and the length of the boat [itex]L[/itex]. The only thing I can say by dimensional analysis is that the velocity of the boat must be expressible as:

    [tex]
    v = f(\frac{m_{1}}{M}, \frac{m_{2}}{M}) \, \sqrt{g \, L}
    [/tex]

    Can you tell us the explicit form of the numerical function [itex]f(x, y)[/itex]?
     
  15. Jul 28, 2011 #14

    Doc Al

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    Why? What physics principle are you invoking?
     
  16. Jul 28, 2011 #15

    xts

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    First of all:
    Forum Rules: Do not post your homework/school-type questions here!

    Thus I may assume you do not ask how to solve primary school problem of centre of mass, but you expect some deeper view of the problem.

    1. Practical answer: the boat will move upside down, and both men fall into water.

    2. You've all forgotten about mass of the water displaced by the boat (equal to sum of masses of both men and boat itself), which must be moved as the boat moves, but in opposite direction. But that is just a first order approximation - to make this harder to calculate, the boat is asymetrically submerged at the beginning (men at its ends have different weights), and more symmetrically at the end. In order to compute that you have to know boat profile geometry.
     
  17. Jul 28, 2011 #16

    Doc Al

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    The poster's question was a general one and was answered with the very first response.
     
  18. Jul 28, 2011 #17
    Since there is no external force acting on the system of boat and men, the entire system should remain stationary with respect to an observer on the shore since the boat was stationary in the water to begin with. The total momentum of the boat and men system remains perfectly the same before and after the men move in the boat. All the while assuming ideal conditions of no water and wind resistances, and no stiring of water that carries away some momentum of the boat and men system.

    When both men start moving in the boat, the momentum change of the men is transferred to the boat, causing the boat to move with respect to the observer on the shore. Lets analyse the system when the men of mass m1 and m2 are moving at constant velocities v1 and v2, and the boat of mass M moves at velocities vb, all velocities are measured with respect to the shore observer.

    Due to the conservation of linear momentum without external force,

    m1v1 + m2v2 + Mvb = 0

    Integrate both sides over the same period of time t, and results in the following relationship among the displacements l1, l2 and lb of the men and boat with respect to the shore observer.

    m1l1 + m2l2 + Mlb = 0

    Before the men move, the position of the center of mass of the men and boat system is
    X0 = [m1l10 + m2l20 + Mlb0] / (m1 + m2 + M)

    After the men move, the position of the center of mass of the men and boat system is
    X = [m1(l10 + l1) + m2(l20 + l2) + M(lb0 + lb)] / (m1 + m2 + M)

    But it has been shown that m1l1 + m2l2 + Mlb = 0, thus, X = X0. Hence the center of mass remain at the same position and stationary with respect to the shore observer.

    When both men stopped moving, the momentum change of both men will be transferred to the boat and caused it to stopped as well. The final position of the boat could be different, but the center of mass remain at the same position.
     
  19. Jul 28, 2011 #18
    You forgot to take into account buoyancy which will effect the behavior of the system depending on the surface characteristics. Trick of the problem is that it's on water and buoyancy will dynamically respond to stimuli by men aboard. Symbolic solution to problem will be a bit more complicated than that.

    Newton's 2nd law, which is apparently not enough to generate a solution in this case.
     
  20. Jul 28, 2011 #19
    Do you know the direction of the buoyancy force?
     
  21. Jul 28, 2011 #20
    It is opposite to gravity in sign for newtonian fluids, but will it have a homogeneous distribution along the boats bottom surface ?
     
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