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Pressure in liquid

  1. Aug 16, 2011 #1
    This started to bother me today. In the first picture, there's a cup wheres a certain amount of water in it. A ball made of aluminium hasnt been dropped in the cup yet.
    In the second picture, the ball has been dropped in.

    The ball has not sinked, a string is holding it.

    Now please tell me, how did the pressure on the botton of the cup created BY THE WATER change?
    I mean, the water level just rose. So the pressure should also.
    But there seems to be another solution to the problem, and I just cant figure it out.

    The equation for liquid pressure is p= RO*g*h.

    Thanks in advance,

    [PLAIN]http://www.cornerpocketpool.com/images/catalog/category57.jpg [Broken]
    http://www.billiardlover.com/mcdermott-pool-cue.html [Broken]
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Aug 16, 2011 #2


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    Your initial solution is correct. Since the depth of the water is now larger, the pressure at the bottom is higher.
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