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Another Physics Problem

  1. Mar 26, 2003 #1
    Here's the problem:

    A balloon is descending with constant acceleration, a, (which of course is less than g). The weight of the balloon, with its basket and contents, is w1. In order to cause the balloon to accelerate upward with the same accelerations, a, what weight (call it w2 in your solution) of ballast needs to be released? Neglect air resistance. (You need to put in an unknown buoyant force on the balloon.)

    The way I figure it there are two forces acting on the balloon there's the F up, and w1 or (w1 - w2) acting downward. When I use Newton's second law with just these two forces I get w2 = 2ma. The answer that I'm supposed to get is w2 = 2w1a/(g+a).

    F - w1 = -ma, F - (w1 - w2) = ma

    F = w1 - ma ===> w1 - ma -w1 + w2 = ma ==> w2 = 2ma.

    I'm missing something. Any suggestions would be greatly appreciated.

    Thanks
     
  2. jcsd
  3. Mar 26, 2003 #2
    balloon going down with acceleration a:
    F1 = -w1*(a+g)

    you want to go up with accelation a so you need 2 times force F1 in opposite direction:
    F2 = 2F1=2w1(a+g)

    got stuck here :(

    the fault in your solution is this i think: F - w1 = -ma, that isn't right. and using w1 and m is also confusing i think
     
  4. Mar 27, 2003 #3
    Actually the bouyant force remains unchanged. The downward force changes from w1 to w1 - w2, so the mass, in the case of the ascending balloon, changes to (m1 - m2)/g.

    f - w1 = m(-a)
    f - (w1-w2) = (m1-m2)a/g

    the 'a' is the same, but in the opposite direction, and everything looks right, but it doesn't turn into the equations it's supposed to for w2.
     
  5. Mar 28, 2003 #4
    very confusing all those w's and m's

    but i think F-w1 != m(-a)
     
  6. Mar 29, 2003 #5
    OK w1 = m1g.

    Newton's Second Law: F = ma

    The only forces on the balloon in the case where it's descending are the buoyant force F and the weight of the balloon so:

    F - m1g = m1(-a) or F - w1 = m1(-a)
     
  7. Apr 3, 2003 #6
    I finally figured this out. Shame on me. It took way too long!!!

    Using Newton's Second Law.

    Balloon descending: F - m1g = m1(-a)
    Balloon rising: F - (m1 - m2)g = (m1 - m2)a

    m1g - m1a = (m1 - m2)a + (ma - M2)g
    m1g - m1g = m1a - m2a + m1g - m2g
    -2m1a = -m2(g + a) Multiply both sides by g
    2m1ga = m2g(g + a)
    2w1a/(g + a) = w2
     
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