Results 1 to 5 of 5

Math Help - Mechanics Problem

  1. #1
    Ant
    Ant is offline
    Member
    Joined
    Apr 2008
    Posts
    137
    Thanks
    4

    Mechanics Problem

    Question:

    A Balloon of mass 490Kg is descending at a speed of 5 m/s with an acceleration of 0.2m/s^2. What mass must be thrown out to reduce the acell to zero.

    I keep on getting an incorrect answer, but I'm not sure where I'm going wrong.

    Here is how I am (trying) to work out the solution.

    T.F = ma
    mg - Upwards force = ma
    - upwards force = ma - mg
    Upwards force = (ma - mg) * -1

    Then:

    T.F = ma
    mg - upwards force = ma = m*0 = 0
    Mg = Upwards force
    m = upwards force/g

    Original M - new value for M = delta m = amount the mass must be reduced.

    Can anyone enlighten me on where I'm going wrong please, perhaps the upwards force doesn't remain constant, or I'm missing a force in my modelling.

    Any help is very much appreciated!!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,964
    Thanks
    349
    Awards
    1
    Quote Originally Posted by Ant View Post
    Question:

    A Balloon of mass 490Kg is descending at a speed of 5 m/s with an acceleration of 0.2m/s^2. What mass must be thrown out to reduce the acell to zero.

    I keep on getting an incorrect answer, but I'm not sure where I'm going wrong.

    Here is how I am (trying) to work out the solution.

    T.F = ma
    mg - Upwards force = ma
    - upwards force = ma - mg
    Upwards force = (ma - mg) * -1

    Then:

    T.F = ma
    mg - upwards force = ma = m*0 = 0
    Mg = Upwards force
    m = upwards force/g

    Original M - new value for M = delta m = amount the mass must be reduced.

    Can anyone enlighten me on where I'm going wrong please, perhaps the upwards force doesn't remain constant, or I'm missing a force in my modelling.

    Any help is very much appreciated!!
    Call the upward force B, for buoyant force. And call +y downward, as you have already done.

    Then as the balloon is falling
    \sum F_y = mg - B = ma

    B = mg - ma

    Now, throw out a mass \Delta m and do it again. (The buoyant force will be constant as the volume of the balloon has not changed.)
    \sum F_y = (m - \Delta m)g - B = 0

    mg - \Delta m g - (mg - ma) = 0
    (Where a is the acceleration from before.)

    \Delta m = \frac{ma}{g}

    -Dan
    Last edited by topsquark; June 1st 2008 at 12:24 PM. Reason: Fixed the algebra mistake
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Ant
    Ant is offline
    Member
    Joined
    Apr 2008
    Posts
    137
    Thanks
    4
    Thank for the reply!

    so, pluggin numbers in gives:

    Change in Mass = (mg) / a
    = (490 * 9.81) / 0.2
    = 24010 Kg

    isn't this impossible - going from 490kg to minus 23530 Kg?! The answer in my book says there must be a reduction in mass of 10kg. Perhaps there's an error in my book?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,964
    Thanks
    349
    Awards
    1
    Quote Originally Posted by Ant View Post
    Thank for the reply!

    so, pluggin numbers in gives:

    Change in Mass = (mg) / a
    = (490 * 9.81) / 0.2
    = 24010 Kg

    isn't this impossible - going from 490kg to minus 23530 Kg?! The answer in my book says there must be a reduction in mass of 10kg. Perhaps there's an error in my book?
    Quote Originally Posted by topsquark View Post
    mg - \Delta m g - (mg - ma) = 0
    (Where a is the acceleration from before.)

    \Delta m = \frac{mg}{a}
    Tsk tsk! We both should have caught this. The solution to the equation above
    mg - \Delta m g - (mg - ma) = 0
    is
    \Delta m = \frac{ma}{g}
    not what I had written before. Sorry about that!

    I have fixed this in my original post.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Ant
    Ant is offline
    Member
    Joined
    Apr 2008
    Posts
    137
    Thanks
    4
    aah yes!

    Thanks again for you help.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Mechanics Problem!
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: March 4th 2010, 10:46 PM
  2. Mechanics problem!
    Posted in the Math Topics Forum
    Replies: 3
    Last Post: January 20th 2010, 03:05 AM
  3. Mechanics Problem
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: April 12th 2009, 07:52 AM
  4. Mechanics problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 12th 2008, 01:58 PM
  5. Mechanics problem
    Posted in the Advanced Applied Math Forum
    Replies: 5
    Last Post: October 30th 2006, 01:49 PM

Search Tags


/mathhelpforum @mathhelpforum