1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Walking on a board on frictionless surface

  1. Aug 14, 2008 #1
    1. The problem statement, all variables and given/known data
    A 100 kg man stands on the left end of a 100 kg board 2 meters long which sits on a frictionless surface as shown. Where will the left end of the board come to rest if the man walks to the right end of the board and stops?
    [​IMG]

    2. Relevant equations
    p=mv


    3. The attempt at a solution
    I can only think of using momentum for this problem but I don't know how to apply it here.
    The answer is that the left end of the board will be at the -1 m mark
     
    Last edited: Aug 14, 2008
  2. jcsd
  3. Aug 14, 2008 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi black_squirrel! :smile:

    Use good ol' Newton's first law:

    The man and the board together are a body on which no horizontal external forces are acting.

    So what happens to their centre of mass? :smile:
     
  4. Aug 14, 2008 #3
    okay so i guess the momentum of the center of mass would be conserved. But in the beginning, the center of mass is .5 meters from the left end of the board. When the guy walks over, the center of mass changes to 1.5 meters from the left end of the board. The left end of the board is initially at the "0" mark and there is a 1 m shift in the position of the center of mass. I see the answer is right there staring me in the face but I'm having perspective issues. The center of mass changed so what do we compare it to and what is the momentum relative to? A little ocnfused here. I'd appreciate if someone could explain this pleaase
     
  5. Aug 14, 2008 #4
    You were on the right track with conservation of momentum. Forget about what the CG is doing.

    Think about their relationship; mb x vb = mm x vm

    Their masses are equal right? So what does that say about their velocities?
     
  6. Aug 14, 2008 #5

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    You do not need to use velocities of the man or the board to solve the problem. Tiny Tim was spot-on. Before the man starts walking the momentum of the combined man-board system is zero. There are no external forces acting on the combined man-board system, so the momentum of combined man-board system is always zero. Momentum is mass times the time derivative of the center of mass. Since the mass is constant, the center of mass is not moving.
     
  7. Aug 14, 2008 #6

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    :biggrin: Woohoo! :biggrin:

    Hi black_squirrel! :smile:

    I wonder whether you've been confused about where the man is walking to?

    Remember, he's not going to the end of where the board is now, but to the end of where it will be! :wink:
     
  8. Aug 14, 2008 #7
    It is? I thought it was mass times the derivative of its position and orientation? The mass is constant, but the center of mass IS moving. Whether you consider the sum of the board and the man or consider them separately the center of mass is still moving. You dont need to know the values of the velocities but you do need to know their relationship to each other.
     
  9. Aug 14, 2008 #8
    As stated above, the momentum of the center of mass is zero, since
    * it is initially zero
    * there is not any force applied on the complete system (man + board).

    So the center of mass is fixed. The man can move, the board can move, but the center of mass of the complete system will stay where it was originally.

    From that, you can derive a relationship between the two velocities and complete the problem.
     
  10. Aug 14, 2008 #9

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    No, it is not. Assume we have a system of N constant mass particles that interact with one another but not with anything else and do so in concordance with Newton's laws of motion. In other words, the only forces acting on the particles are internal; there are no external forces. The linear momentum of the system as a whole is the sum of the linear momentum of each particle in the system:

    [tex]\mathbf p_{\text{sys}} = \sum_{i=1}^N m_i \frac {d \mathbf r_i}{dt}[/tex]

    Differentiating with respect to time,

    [tex]\frac{d\mathbf p_{\text{sys}}}{dt} =
    \sum_{i=1}^N m_i \frac {d^2 \mathbf r_i}{dt^2}[/tex]

    By Newton's second law, the force acting on the ith particle is

    [tex]m_i\frac {d^2 \mathbf r_i}{dt^2} = \sum_{j=1,j\ne i}^N\mathbf F_{ij}[/tex]

    where [itex]\mathbf F_{ij}[/itex] is the force on the ith particle due to the jth particle. A particle applies zero net force on itself, or [itex]\mathbf F_{ii} = 0[/itex]. With this, the above becomes

    [tex]m_i\frac {d^2 \mathbf r_i}{dt^2} = \sum_{j=1}^N\mathbf F_{ij}[/tex]

    Applying this to the time derivative of the system momentum vector,

    [tex]\frac{d\mathbf p_{\text{sys}}}{dt} =
    \sum_{i=1}^N \sum_{j=1}^N \mathbf F_{ij}[/tex]

    Rewriting the sum by pairing the [itex]\mathbf F_{ij}[/itex] and [itex]\mathbf F_{ji}[/itex] terms,

    [tex]\frac{d\mathbf p_{\text{sys}}}{dt} =
    \sum_{i=1}^N \sum_{j=1}^i \mathbf F_{ij} + \mathbf F_{ji}[/tex]

    By Newton's third law, [itex]\mathbf F_{ji} =- \mathbf F_{ij}[/itex], and thus

    [tex]\frac{d\mathbf p_{\text{sys}}}{dt} = 0[/tex]

    In other words, the momentum of a system of particles that is not subject to any external forces is conserved regardless of any internal interactions amongst the particles so long as those internal interactions obey Newton's third law.

    In the problem at hand, the initial velocities of the man and board are both zero and the initial total momentum is thus zero. Since no external forces act on the man-board system, the total momentum is constant: zero. The combined center of mass does not move.
     
    Last edited: Aug 14, 2008
  11. Aug 14, 2008 #10
    Actually the center of mass didn't change. What is confusing you I think is that the configuration of the masses changed, but the center of mass did not. Also, you need to keep focused on what the scale reads, not what the man sees relative to the board. Yes, in the old configuration, the center of mass is 0.5 from the left of the board. But on the scale, this is at 0.5 meters. Yes, in the new configuration the center of mass is 1.5 meters from the left of the board, but that isn't what counts. If the left of the board is now at -1 on the scale, then with respect to the scale, the center of mass of the system is still 0.5 meters.
     
  12. Aug 14, 2008 #11
    Woops, I was wrong. For some reason I thought the mass was moving in the Y direction as the man walked, but it isn't. For some reason I thought as the man walked he would be bringing the CGs closer together but that is not the case. Not that it has anything to do with this, I just like being specific. Carry on.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Walking on a board on frictionless surface
  1. Frictionless surface? (Replies: 1)

  2. Frictionless Surface (Replies: 7)

Loading...