Am trying to find how far the centre of mass is skewed from the central distance point. The part I'm unable to grasp here is that I'm working with forces which also push the see-saw upwards (anti-gravity style). Take the following as a basic model ...

basic see-saw model

The basic formula for center of mass: $$x=\frac{\sum{m_ia_i}}{\sum{m_i}}$$

This works fine until I have a situation where some of the $m_i$ are negative. How might I account for this?

example with 'negative' weights

  • $\begingroup$ What is your definition of "centre of mass" in this situation, to start with? If you have 4 antigravity masses and you put your bar on top of the triangle like in your diagrams, it will just float away... Do you mean "at which point on the bar should you attach a hinge to make the bar not want to rotate due to the forces"? $\endgroup$ – BjornW Dec 3 '18 at 10:02
  • $\begingroup$ Good question - your definition does make more sense. I would like to know at which point along the line should the fulcrum/hinge be placed so the bar does not rotate. I acknowledge that there are situations where it's not possible $\endgroup$ – Carl Dec 3 '18 at 10:22

The calculation to find a centre of mass has nothing to do with forces.

Suppose you are asked to find the centre of mass $\bar x$ of the system shown below.

enter image description here

The position of the centre of mass is given by the application of your equation
$$(m_1+m_2+m_3+m_4)\, \bar x = m_1\, (-\frac {3a}{2})+m_2\, (-\frac {a}{2})+m_3\, (\frac {a}{2})+m_4\, (\frac {3a}{2}) $$

One could decide on a different origin

enter image description here

$$(2+1+1+2)\, \bar x = 2\, (0)+1\, (+a)+1\, (+2a)+2\, (+3a) $$

and this gives $\bar x = \frac {3a}{2}$

Secondly, we get a divide by zero situation when $m_i$ are 2, 1, -1, -2 respectively along the line, although I would expect intuitively the answer to be $\frac{3a}{2}$.

I do not know how your intuition gives you an answer of $\frac {3a}{2}$ and I also do not understand the use you have made of forces which are "anti-gravity" style to try and find the position of the centre of mass.

Suppose the forces to the left of the pivot were downwards and those to the right of the pivot were upwards then that would represent a non static equilibrium situation but given the masses as shown this would not alter the position of the centre of mass of the system.

  • $\begingroup$ Thanks. This helped me clarify a couple things. Updated my question. The divide by 0 issue implies there is no centre of gravity in that situation. $\endgroup$ – Carl Dec 3 '18 at 9:42

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