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Building an ice rink - question about water pressure exerted on a wall

  1. Nov 29, 2009 #1
    Hi physics folks-

    Every year for the past 3 years, I have built an ice rink in my back yard. Unfortunately, I do not have the flattest back yard, in fact it has quite a downward slope and then a steep uphill slope. But, I'm determined the make the biggest rink I can reasonably make, given what I have to work with.

    So, the challenge I end up with is that the side that is furthest downhill ends up being about 30" high, with the pressure of several thousand gallons of water pushing on that wall. It is a struggle, all winter long (and one I sometimes lose) to keep that wall up. I have tried building various bracing structures, and this year I've pounded 2x4 stakes into the ground at 6 different points along that side, and I'm going to build 45 degree supports in addition. I have one more idea, and this idea is what brings me here.

    My thought is to add a "ramp" of filler up against that wall in such a way that I create something of a bowl instead of a vertical dam. This would have the benefit of a) using less water, which therefore means less water will be exerting pressure on the wall, and b) exerting pressure downward on the filler material instead of on the wall. Here is an attempt to diagram it.

    This diagram represents previous years:

    G = ground
    W = wall
    * = water
    X = filler


    This diagram represents my idea to add filler:


    So will adding filler up against the wall help to reduce the force exerted on the wall by the water?

    Thanks in advance for your help. I am an engineer myself, but in Computer Science, and it's been a long time since I've taken those physics classes where I would be able to calculate something like this myself. Let me know if anything in my explanation is unclear.

  2. jcsd
  3. Nov 30, 2009 #2
    Just a humble physics student here, but perhaps I can help get the discussion started. In the simple case of a completely level rink, the pressure per unit area on your wall is simply [tex] P = \rho g y [/tex], where rho is the density of water, g is the gravitational acceleration, and y is the depth of the rink. In that case, your solution would be very effective, because you're adding strength to the wall proportionally to the increase in pressure.

    The problem I see is that when the rink is titled at an angle, you're now dealing with a two-dimensional problem. There is going to be the pressure component caused by the water in the y direction (vertically), but there is also going to be an x component (horizontally) proportional to the severity of the incline and the length of the rink in that direction. Then, referenced to the lower end of the rink, the pressure equation would become something like [tex]\rho g(ycos(\theta) + x sin(\theta))[/tex] where theta is the angle of the incline with respect to the horizontal. In that case the horizontal component is exerting all its pressure evenly across the wall, so I don't believe the ramp of filler will help for that component. Basically the wall will have to be thick enough from top to bottom to support that component of the force.

    If the slope is totally unavoidable, my intuition would be to use the ramp of filler on the inside to handle the first part of the force, and then brace the upper part of the wall from the outside, where it is weakest with respect to the second component of the force.
  4. Nov 30, 2009 #3
    I maybe I’m missing something but it sounds to me digging down 30 inches on the high side of the hill might be the answer. It would use a lot less water and you would not have the support problem. I don’t know you’re exact situation and this may not be possible.
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