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Vertical rise of liquid

  1. Jan 29, 2016 #1
    1. The problem statement, all variables and given/known data
    in the second picturre, why the zs1 and zs2 is calculated by using 2m and 0.6m respectively ? it should be 0.8m , am i right ? what's the purpose of the author to use 2m and 0.6m ?

    2. Relevant equations


    3. The attempt at a solution
     

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  3. Jan 29, 2016 #2

    haruspex

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    0.8m is the height of the tank. The author is comparing two orientations of the tank. In one orientation the front-to-back distance is 2m, in the other it is 0.6m. The zs values are the corresponding extra heights of the water, so one calculation uses 2m, while the other uses 0.6m.
     
  4. Jan 29, 2016 #3
    why not we use 0.8m instaed of 2m and 0.6m to get zs? the zs is in the same direction as 0.8m , right ?
     
  5. Jan 29, 2016 #4
    can you please sketch out the 3d diagram , so that i can understand better
     
  6. Jan 29, 2016 #5

    HallsofIvy

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    Yes, but that is only relevant to the question whether or not the water pours out of the tank: whether zs< 0.8 or zs> 0.8.
    To calculate the actual increase in the height of the water on the back wall you need to use the distance from front wall to back wall. And that depends upon the orientation of the tank.
     
  7. Jan 29, 2016 #6
    i think i get the idea already . why the distance is distance between the front and back wall ? Isn't teh dstance= distance between the right and left wall as shown in the figure 3-52 ?
     
  8. Jan 29, 2016 #7

    HallsofIvy

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    In which direction is this box moving? How do you distinguish between "front and back" and "side to side"?
     
  9. Jan 29, 2016 #8

    haruspex

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    I was using front-to-back in reference to the direction of movement. In one scenario, front-to-back is 2m (and the 0.6m doesn't matter); in the other it's the other way around.
     
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