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Making a box to always fall on the same side

  1. Jun 11, 2012 #1
    Say I have a cuboid box (eg 120 * 60 * 10cm (lbh)). Is there a way I can add non-uniform mass to it to force the center of gravity on a preferred point (eg making one half heavier than other) - and then when I drop it horizontally or vertically, it always touches the ground at same side (but may be different edges)?

    Eg: A shuttle cock always falls down on its heavier side. I understand aero-dynamics are involved. I want to know if the same can be done for a box. The box may also have non-uniform thickness or smoothened edges if that would force a solution. Thanks.
  2. jcsd
  3. Jun 11, 2012 #2


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    Assuming the height is sufficient to reach the equilibrium velocity*, adding enough mass to the preferred side (and making the other parts very light) should work. It might be quite unstable if 10cm are the "height".

    *the shuttle cock has a high air resistance and reaches this quite fast. If you drop it from 1-2cm above the ground, it will not always land on the right side. With 1m, it does.
  4. Jun 11, 2012 #3
    Thank you for your reply. Wanted to confirm this:
    The height from which the object will be dropped is 1m.

    1) Will this happen even if the mass is distributed evenly? Eg: I may choose to distribute it such that the mass rises upto 2 cm or I may also distribute it unevenly such that it rises to 5cm on one side and 1 cm on the other.

    My understanding - If the mass is distributed evenly, the center of gravity will still be in the center (but may be located lower or higher depending on which side is facing the ground). In that case, the object may not be unstable when dropped from a horizontal position (0 degrees)

    2) Is there a way to design such that equilibrium velocity will be obtained quickly?

    Basically my intention is to design a very small box container (around 120 * 60 * 15mm) that can hold upto 200g mass, and will not get damaged when dropped. I plan to keep a soft material on just one side with the objective that, the object when dropped will always hit at that side.
  5. Jun 12, 2012 #4
    Doesn't the shuttlecock orient itself because the air resistance on the feathers is different than on the tip? If you create a weight distribution *inside* of an otherwise symmetrical box, you have not created any difference in air resistance between light side and heavy side. With equal air resistance (dictated by shape of box), light side and heavy side fall at same rate right? (equivalence principal).
  6. Jun 12, 2012 #5
    I agree on the equivalence principal. But wanted to check what would be the scenario if an object is unstable when held horizontally. Will it try to get into a stable position even when falling down in motion?
  7. Jun 12, 2012 #6
    Just realized that I got this *dead wrong*.

    The acceleration of a box subject to some wind resistance force (f) will be:
    a = g - f/m

    The bigger its mass, the closer its free fall is to g.

    A bowling ball size rubber ball could be held aloft using a leaf blower (large f). An actual bowling ball would come crashing right down.
  8. Jun 12, 2012 #7


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    In this case, it will usually tend to get a small front area (~60mm x 15mm with the dimensions of post 3, not very stable). Opposite sides are perfectly symmetric.

    Make the object light and with a lot of air resistance.
    However, one heavy side (the one which should be the bottom) will help to reach the correct orientation quickly.
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