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What makes a material able to spread weight

  1. Dec 15, 2007 #1
    when i was in the army,i saw corp of engenires guys able to walk in a mine field on this foam mattress plates attached to their feet.
    now, i understand that this is possible because the foam spread their weight ( pressure?) along a bigger area,thus reducing pressure,how is the foam able to do it? a thin piece of paper attached to your feet will not have the same pressure reducing effect,....why? does it has to do with thickness?.... would a plate of wood achieve the same "weight distribution" result?
     
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  3. Dec 15, 2007 #2
    [tex] P = \frac{F}{A}[/tex], where F is the force you are imparting, and A is the area over which you are imparting the force.

    If you increase the area, as you would by attaching plates of wood to your shoes, then the pressure under the wood will be less than it would be if you were simply standing normally (imagine an elephant wearing stiletto heels).

    The 'foam mattress' plates you describe, however, would probably have the added effect of reducing the total force across the area by flexing and warping - much like a mattress flexes a certain amount when you lie on it - it cushions the force.

    Paper, however, is too thin to have this cushioning effect - and as for distribution of weight, it would simply bend so that the area in contact with the ground was the same as if you hadn't attached paper to your shoes.

    I think...
     
  4. Dec 15, 2007 #3
    As long as you increase the area over which the force acts, you'll reduce the pressure.

    By placing a large sheet of paper on the mine and then placing your foot on top of it, yes you increase the "area" but you don't increase the area over which your weight acts. The parts of the paper which is not in contact with your foot is not helping in distributing your weight over a wider area.

    So instead lets say we were to place a BIG HUGE solid but weightless cone on top of the mine. Then you could climb on top of this cone and now your weight is evenly distributed over the wide base of the cone, thus reducing the net force acting on the mine.

    As for the "cushioning" element - lets say you take a leap. You first foot lands, but you are on your tippie-toes, your whole foot hasn't made contact with the ground yet. your second food is still in the air. What this means is that all your weight is distributed over that small point your foot is in contact with. This would trigger the mine if it were right below you.

    So the cushion prevents all your weight from being placed on one area in a short amount of time. It gives you a buffer to distribute your weight better.
     
  5. Dec 15, 2007 #4

    cepheid

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    Why does the trigger of the mine (or, for that matter, the cracking of a frozen lake), depend on pressure, rather than weight, as these examples imply? Shouldn't it be that if there is sufficient weight directly on the mine, it will go off?
     
  6. Dec 15, 2007 #5
    I think that has to do with the material's property (max shear stress before its bonds give way). I studied it once, kinda fuzzy now cause its been a while.. maybe someone can elaborate? But I like to think of it as the bonds between the material can only take so much force. so by focusing force on a small area, you focus your force on the material's bonds.

    Also I think I understand the need for cushions a bit better for the mines now. Any outdoor terrain will not be perfectly flat. So lets say you were wearing long wooden boards on the soles of your shoes to distribute your weight, but you stepped on a pebble. so all of a sudden all your weight is again concentrated on the pebble. If the mine was right below the pebble, you're screwed.

    So instead if they use a cushioning/gell-like substance below your boards (like the air bubbles in nike shoes) then they will just "hug" or deform to distribute itself over the pebble and make contact with the ground. Thus you would no longer just be standing ON the pebble but the pebble would become "a part of your shoe".
     
  7. Dec 15, 2007 #6

    Danger

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    It's not the total amount of weight that matters; it's how much of that weight is transferred to the trigger. As an example, a petite woman (think 'Evo') in stilleto heels will cause far more damage (ie: higher pressure) stomping on your foot than an M1 Abrams main battle tank running over it.
    This is the basic principle behind snowshoes and skis.
     
  8. Dec 16, 2007 #7

    cepheid

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    Right of course...what was I thinking? If all of the weight from Evo's leg (and I'm not trying to suggest that it it a substantially large weight or anything), but as I was saying, if all of that weight is applied over a small area, it's more likely that a crack will form at that point in the ice than if the weight is distributed over some larger area. Pressure gives us a useful measure of that. I should know this. It is pretty obvious.

    Still...a battle tank? I don't know anything about military machines, but I would have thought that having your foot run over by a tank tread would be quite devastating (now I'm just nitpicking your example...I understand the idea you were conveying).
     
  9. Dec 16, 2007 #8

    Danger

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    I haven't actually done the calculations, but a buddy of mine did a couple of decades ago. Think of it this way: a 120(?) lb. woman stomping down with all of her weight on a .0625 sq. inch (1/4" x 1/4" spike heel) surface vs. 160 tonnes or so spread over (guestimate) 8,500 sq. inches of surface.
    I'm too drunk to do the math.
     
  10. Dec 16, 2007 #9

    cepheid

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    Hmm....those numbers give:

    1920 psi = 13.2 MPa (heel)

    For the tank I assumed a short ton (US) since we're working in imperial (even though you spelled it tonne, which to me suggests metric tonnes). Which meant that 160 tons = 320,000 lbs

    Anyway, the result was

    37.64 psi = 259.6 kPa

    Note: I was lazy and used the unit conversion widget on my dashboard throughout. Let's hope the thing is accurate.

    Edit: I don't know anything about the imperial system, so it's possible I picked the wrong ton. Again, that was one of the choices available in the unit conversion widget.

    So, in conclusion, if your guesstimate of the tank tread area is in the ballpark, then you're right. It makes a huge difference.
     
  11. Dec 16, 2007 #10

    Danger

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    I was indeed using the metric tonne, being a good little Canuk. I think that an Abrams is actually more around 150 tonnes, but it varies with different configurations, fuel loads, etc.. Since I don't know the actual specs, I guestimated that the track surface in contact with the ground would be 15' x 2' on each side. That could very well be way out of sync with reality.
    Still, though, even if those numbers are made up, it does serve to demonstrate the idea.
     
  12. Dec 16, 2007 #11
    thx everybody...a lot of food for thought
     
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