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Finding MY center of mass!

  1. Oct 28, 2005 #1
    Ok so we were told to find our center of mass using the reaction board method, but modified. We just do a pushup position with our hands on a scale, I have a electronic one and toes on the floor.

    My values obitained are:
    the scale reading in a pushup position=48.9kg
    horizontal distance from hands to my toes=112cm


    Locate my center of mass:
    Ok so, the sum of torques=(F1 x d1) + (F2 x d2)
    F1=reaction force from scale
    F2=weight force
    d1=112 cm
    d2=my unknown value!

    0=(48.9 kg x 9.81)(1.12m) + [-(68kg x 9.81)(d2)] (negative because it's going clockwise)
    667.08 (d2)=537.27
    d2=0.805m or 80.5 cm.

    Alright, that means my center of mass is 80.5 cm from my toes, making it around the belly area?:uhh:

    Second part of the question says what happens if a 20kg backpack is placed halfeway between you hips and shoulders. I can't "do" this, I must only calculate, so can't find reaction force with scale.
    So anyone have ideas for this one?

    I think I can do the same thing? So :
    Sum of torque=(F1 x d1) +(F2 x d2) + (F3 x d3)? F3 being the reaction force of my back against the back pack on me. But that doesn't work, because that just means the torque force for F1 and F2 will cancel and my d3 can't be 0. So how does this work?

    Thanks again!
    Last edited by a moderator: Apr 21, 2017
  2. jcsd
  3. Oct 29, 2005 #2
    Anyone with ideas? bump!
  4. Oct 29, 2005 #3


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    F1 will increase with the backpack on. I think you are suppose to choose a d3 according to your body and work out what the new reaction force of the scale , F1, will be. Once you've done that you can go back and recalculate your new center of mass with the backpack on.
    Last edited: Oct 29, 2005
  5. Oct 29, 2005 #4

    I don't get it, are you suggesting I should just choose an arbitrary value for d3? A value that lies somewhere in between my shoulder and hips?
  6. Oct 29, 2005 #5


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    The location of a composite object's center-of-mass is given by
    M_total x L_com = sum of (m_i x l_i) for all items.
    Once you know your c.o.m. location and your mass,
    just add the backpack mass multiplied by its location
    (you have to decide on a coordinate axis origin).
    Don't forget that M_total also increases!
  7. Oct 30, 2005 #6
    Ok...so from your formula you're saying that:

    Mass total (ie me and the backpack) x the length of C.o.m. which I already found=sum of (M_i? x l_i?> I don't get this part. what is i?

    Also, I don't know the value for the distance from the center of mass of backpack to the fulcrum or my feet. Can you clarify a bit?
  8. Oct 30, 2005 #7


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    BOTH quantities on the left-hand side refer to the entire (new) object:

    (M_total = mass_you + m_backpack) x (L_new_total_com) =

    (m_you x location_com_you) + (m_backpack x location_com_bkpk)

    each term on the right-hand side is about one item in the total,
    the same kind of operation (weighted average of location) as left-h-side.
    You MUST use same coordinate system for all locations (say, from feet)

    How far from your feet are your hips? how far from feet are shoulders?
    where is the location that's "halfway between" them?

    By the way, One way to tell whether a "formula" is worth remembering
    (or whether it is a special-case derived for one instance)
    is that each term should be the same kind of quantity,
    and each variable in that term should refer to the same item.
    Last edited: Oct 30, 2005
  9. Oct 30, 2005 #8
    could I not use this formula:

    X (center of mass)=(m1)(X1)+(m2)(X2)/ (m1 +m2)

    m2 being my backpack,
    m1 being me
  10. Oct 30, 2005 #9
    Ahhh, my sticky point was the distance from my backpack to my toes. Thought I had to calculate that :rofl: :rofl: . Ok, I think I got this. Thanks for the help.
  11. Oct 30, 2005 #10


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    Look at the formula you just wrote.

    It comes from the equation I wrote to you at 10:00 .

    But the original equation is already generally true,
    (it doesn't have to be modified if there's a 3rd or 4th item)
    and all terms look "similar", even the left-hand side!
    and the terms are vectors, like locations should be,
    not distances (which are always positive).

    It comes straight from the *concept* of center-of-mass,
    "what's the location of that entire mass"?
    Last edited: Oct 30, 2005
  12. Oct 30, 2005 #11
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