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The physics of a seesaw.

  1. Aug 12, 2011 #1
    This is not homework, it is purely for my own enjoyment.

    Say I have a lead sphere with a one meter radius, a mass of 465800N and a terminal velocity of 1290m/s. Say I also have a different spherical body of 685N. I now place the 685N body on one side of a seesaw of 20m in length and a height of 5m at the at the 10m mark.

    How, from this information, can I work out how high I must drop the lead ball from to accelerate the 685N ball to >= 11200m/s? Also what is the relationship between mass and acceleration and how does it transfer to the other side of this idealized seesaw.

    I'd be grateful if you didn't work it out but told me how to, as the working is what I find fun.

    P.S I am ignoring things such as friction and assuming the 685N ball is launching vertically.


  2. jcsd
  3. Aug 12, 2011 #2
    Before you can start calculating a solution, you must set a location for the fulcrum. For example, if the see-saw's fulcrum is put at the halfway (10 m) point, then I believe it is impossible for the heavy ball to launch the light ball at a speed any greater than approx. 2 × 1,290 m/s = 2,580 m/s. (I could be wrong—this is just my gut feeling.) By this reckoning, and making some simplifying assumptions, I would estimate that when the heavy ball lands at its terminal velocity of 1,290 m/s, the fulcrum should be placed approx. 1.87 m from the heavy ball's end of the see-saw in order to launch the light ball at the required 11,200 m/s. (Note: My answer is very approximate and may be refuted by other members.)
  4. Aug 13, 2011 #3
    The fulcrum is the centre - 5m high (I was trying to describe it but did not know the correct word). I would like to change the mass of the object instead of the fulcrum position. Also what is the equation for how fast an object approaches its terminal velocity?
  5. Aug 13, 2011 #4
    Last edited by a moderator: Apr 26, 2017
  6. Aug 13, 2011 #5


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    the affect of air resistance on your lead sphere is negligible even if you were to somehow be able to drop it from the height of Mount Everest to sea level , at which point it might be falling at say about 400m/s. Check it out:

  7. Aug 13, 2011 #6
    Thank you, both of you, that link was very appropriate! I got this far:
    Let it fall from mount Everest and the velocity will be 384.7m/s when it hits the ground. F=ma and a=finalvelocity-startingvelocity/time which would be 0-384.7/the time it takes for the ball to travel 10m which would be 0.025994281s so that = −14801.847 = a. Then * mass(in grams) = -703100751673.72
    The Moment/Torque = Fd (where d = perpendicular distance from force application to fulcrum) =
    -703100751673.72*10 = M = −7.031007517×10¹²
    This is a stupidly large number! What have I done wrong? Also how do I get the other object's velocity from torque?
  8. Aug 13, 2011 #7


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    You might want to think in terms of the velocity ratio. If the end of the seesaw where the lead sphere hits travels 10 meters over a certain time period, what distance does the other end travel over that period?
  9. Aug 13, 2011 #8
    I agree with PhanthomJay—it is the velocity ratio of the two ends of the see-saw that is important. In post #2 I hypothesised that the stationary ball will be launched at no greater than twice the speed of the end of the see-saw on which it sits. For a central fulcrum setup, the heavy falling ball must therefore strike the see-saw at (11,200 / 2) m/s = 5,600 m/s.
  10. Aug 13, 2011 #9


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    This is a very hypothetical problem, of course. But assuming a rigid seesaw, do you think it is able to bring the massive lead sphere to a stop? What do you think might be the approximate linear average speed of the see saw at the lead sphere end in its 10 meter journey? What then would be the speed at the other end with the lighter mass on it? How high will the lighter mass go, neglecting air resistance? What is the acceleration of the lighter mass once it leaves the see saw?
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