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Iron ball hanging from a scale and suspended

  1. Aug 18, 2013 #1
    Iron ball hanging from a scale and suspended....

    1. Iron ball hanging from a scale and suspended in a glass of water filled to the rim, what happens to the reading of the scale, volume of water, and weight of the glass of water?



    2. FB = weight of removed water by the ball



    3. there's a FB that's equal to the weight of the water removed pushing on the ball upwards, then weight - FB is equal to the tension in the thread, which means weight is less, then reading is less, volume of water stays the same because the volume of the water removed is equal to the volume of the ball, weight of the glass is........i have no idea!!!
     
  2. jcsd
  3. Aug 18, 2013 #2

    TSny

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    Note that the glass is originally filled to the rim. So, you need to consider what happens to the water when you immerse the ball.

    Unfortunately, the wording of the problem is not real clear. When they ask for what happens to the weight of the glass of water when the ball is immersed, are they asking for what happens to the force of gravity acting on the glass of water or are they asking for what happens to the reading of a scale if the glass is sitting on the scale? I suspect they want to know what happens to the reading of the scale.
     
  4. Aug 18, 2013 #3
    they want both the reading of the scale regarding the iron ball, and the weight of the glass of water!!!! and i did consider what happened, the ball removes a volume of water that is equal to the volume of the ball
     
  5. Aug 18, 2013 #4

    TSny

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    What do you mean by the "weight of the glass of water"?

    Do you mean the force of gravity acting on the glass of water?

    Or, do you mean the reading of a scale that the glass of water is sitting on? (This scale is different from the scale that the iron ball is hanging from.)
     

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    Last edited: Aug 18, 2013
  6. Aug 18, 2013 #5
    exactly u've drawn the problem perfectly, i mean the force of gravity acting on the glass of water after the ball is suspended in water!
     
  7. Aug 18, 2013 #6

    TSny

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    If that's what you mean, then that should be a very easy question to answer!

    But, it's more interesting to think about what happens to the reading of scale #2.
     
  8. Aug 18, 2013 #7
    i already posted my attempt at the solution! could u plz explain?
     
  9. Aug 18, 2013 #8

    TSny

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    I agree with your answer for what happens to the reading of the scale from which the ball is suspended. (Scale #1)

    The question about what happens to the volume of water is not clear to me. Are they asking about the volume of water that is in the glass before and after the ball is immersed? Or, do they want you to include the volume of water that spills out? I don't know.

    I don't see where you gave an answer regarding what happens to the "weight of the glass of water" when the ball is immersed.
     
  10. Aug 18, 2013 #9
    sorry i meant the level of water not the volume! i didn't give an answer coz i have no idea what happens to it!

    p.s. my textbook says that scale 1 doesn't change so apparently both me and you are wrong!
     
  11. Aug 18, 2013 #10

    TSny

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    Scale 1 definitely changes when the ball is immersed.
     
  12. Aug 19, 2013 #11
    sorry it doesn't say what happens to scale 1, it says scale 2 stays the same because according to newton's third law the ball pushes downward with an equal force to FB making the weight of the glass of water the same
     
  13. Aug 19, 2013 #12
    The answer for the effect on scale 2 can be obtained by three methods.

    Method 1:
    What is the hydrostatic pressure on the bottom of the beaker before the ball is immersed? What is the hydrostatic pressure on the bottom of the beaker after the ball is immersed?

    Method 2:
    If the remaining water in the beaker exerts a net upward force on the immersed ball equal to the volume of water displaced, what is the net downward force that the immersed ball exerts on the remaining water in the beaker? If it is the same, then, as far as the remaining water and beaker are concerned, there might as well be water inside the region occupied by the ball.

    Method 3:
    First, treat the ball as a free body. If W is the weight of the ball, and wD is the weight of the displaced water when the ball is immersed, what is upward force exerted by spring 1?
    Now, treat the ball plus the water in the beaker as a combined free body. What is the upward force exerted by spring 1 on this free body? If w0 is the original weight of water in the beaker and wD is the displaced weight of water when the ball is immersed, what is the force of gravity on this free body? The only other vertical force acting on the free body is the upward force exerted by the base of the beaker. Since this free body is in equilibrium, what is the value of this upward force?

    All three methods give the same answer.

    Chet
     
    Last edited: Aug 19, 2013
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