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Man climbs rope ladder attached to balloon with acceleration relative to ladder

  1. Apr 6, 2009 #1

    txy

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    1. The problem statement, all variables and given/known data

    There is a balloon of mass Mb. A rope ladder of negligible mass is hung from it. A man of mass m stands on the rope ladder. A buoyant force F acts on the balloon, causing the man-balloon-ladder system to accelerate upwards. Now, the man climbs up the rope ladder towards the balloon with an acceleration of am relative to the rope ladder. Find the acceleration relative to the ground of
    1. the center of mass of the man-balloon-ladder system;
    2. the balloon.
    The acceleration due to gravity is g.


    2. Relevant equations

    I think it's just intelligent application of Newton's Second Law of motion and concepts of the center of mass.


    3. The attempt at a solution

    My book provided the answers, but did not state clearly which expression belongs to which acceleration.
    There is a [tex]\frac{F - m a_{m}}{M_{b} + m} - g[/tex]
    and a [tex]\frac{F}{M_{b} + m} - g[/tex] .

    I think the second expression is for the acceleration of the center of mass. If I consider the whole system, I get
    [tex]F - (M_{b} + m) g = (M_{b} + m) a_{c}[/tex], where [tex]a_{c}[/tex] is the acceleration of the center of mass.

    I'm not sure how to obtain the second acceleration expression.
     
  2. jcsd
  3. Apr 6, 2009 #2

    Doc Al

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    Staff: Mentor

    Right.

    Hint: If the acceleration of the balloon with respect to the ground is "a", what's the acceleration of the man with respect to the ground? Use those results to express the acceleration of the center of mass in terms of "a" and am.
     
  4. Apr 6, 2009 #3

    txy

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    Oops I didn't phrase my question properly. I should have written "I'm not sure how to obtain the other acceleration expression." to avoid ambiguity. Thankfully you understood what I meant.

    Why haven't I thought of finding the acceleration of man with respect to the ground before?

    If a = acceleration of balloon with respect to ground,
    then
    acceleration of man with respect to ground = a + am .

    So net force on center of mass = net force on whole system = vector sum of net forces on individual objects in the system.
    So
    [tex](M_{b} + m) a_{c} = M_{b} a + m(a + a_{m}) = (M_{b} + m) a + m a_{m}[/tex]
    And so
    [tex]a = a_{c} - (\frac{m}{M_{b} + m}) a_{m}[/tex]
    Then I sub in the expression for ac that I've obtained earlier and I'll get the expression given in the answer.

    At first I wasn't sure about how the acceleration of center of mass is like the "weighted average" of the accelerations of the individual objects in the system. Now I understand. I hope my steps above are correct. Thanks a lot for pointing me in the right direction, with regards to the relative acceleration thing.
     
    Last edited: Apr 6, 2009
  5. Apr 6, 2009 #4

    Doc Al

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    Staff: Mentor

    Perfectly correct.
     
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