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Center of Gravity - Product Safety Question

  1. Mar 8, 2007 #1
    1. The problem statement, all variables and given/known data

    Hi. I am a product designer and I'm looking for a way to solve the following: I have a toy bin (see jpeg) which currently does not pass a child safety test as it tips over under certain conditions and I need to change the dimensions so that it will pass this test.

    Dimensions of toy bin: Cylincrical plywood bin 18"h x 16.5"D.
    THere is a removable disc-shaped lid that is recessed into the bin by 1"
    Weight: 18.5 lbs.
    Wall thickness: 0.4"

    Test: The bin is placed on a 15 degree incline. There is a stopper at the low edge so the edge of the bin won't slide.
    A 50lb weight with center of gravity at 1.7" horizontally from outer edge of seating area and 8.7"vertically from seating area. This simulates a child sitting, tipping backward on the bin. The 50lb weight is placed on a 15 degree wedge to keep the weight vertical (and the wedge is accounted for in the COG values)

    Under these conditions, the bin tips over. I need to change the width and/or height (only these variables) so that it will not tip with a 50lb weight at the 15 degree incline.

    2. Relevant equations

    I made this flash program to help me find the right values (includes diagram):

    3. The attempt at a solution

    I realized that this will never work - the 50lb weight will always make it tip over unless the toy bin is very very short or very very wide. I treated the bin as a 2d object with uniform weight distribution (?) and just calculated the areas and weights on either side of the fulcrum at equilibrium.
    This must be the wrong way to go about it! Can anyone help?
  2. jcsd
  3. Mar 8, 2007 #2


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    Instead of a cylinder, try a truncated cone with a 75 degree tilt from the bottom. The bottom of the truncated cone will be larger than the opening at the top. No matter how much weight is placed into this container, it will never tip over on a 15 degree incline.
  4. Mar 8, 2007 #3
    i'm a little confused--is the removeable lid the seat? I assume that the shape can't b changes as suggested above--great idea tho.
  5. Mar 8, 2007 #4
    I'm limited to using the cylindrical shape to match other products in our line. But I thought of a cone too - that idea got nixed unfortunately! Similarly I can't just put a lip at the bottom. The removable lid IS the seat. I didn't put it in my flash equation just trying to simplify. But yes, so the "seating surface" that the weight is placed on is actually 17" from the floor, not 18. Thanks for the replies...
  6. Mar 8, 2007 #5
    shouldn't you have to use the moment of inertia for the cylinder?
  7. Mar 8, 2007 #6
    Unfortunately, I'm so long out of my last physics class that I forget how to calculate that. Because I have the center of gravity for my 50lb weight (the kid-butt simulator), will it be possible to calculate the MOI for that too? What is the purpose of the moment of inertia? duhhh...sorry for the simple questions...
  8. Mar 8, 2007 #7
    frankly I'm not sure if moi is even needed, maybe someone else can lend a hand.
  9. Mar 8, 2007 #8
    ok, thanks for the input.
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