0

I'm new here and this is my first post so please forgive me if I've not posted it in the right place!

It's an Engineering related question but as maths is involved I thought it may be appropriate to ask here...

I'm given the co-ordinates of a massless quadrilateral (z) with 4 mass particles attached to the vertexes; I need to find the co-ordinates of the centre of gravity (or mass?).

Now, I know that if (z) were a uniform mass it would be a case of finding the sum of the x and y co-ordinates and dividing by the number of vertexes. The question also defines (z) as being a triangular system, so I'm wondering if that is a hint in how to find the COG.

Any suggestions or advice on how to go about solving it would be really appreciated.

Thanks,

Jack

Martin Sleziak
  • 53,687
Jack
  • 1
  • 2
    I think this might be better asked on http://physics.stackexchange.com/ – Newb Nov 02 '13 at 22:00
  • Ah, yes that would probably be a better (more suitable) place. I'll delete this and ask it there... thank you – Jack Nov 02 '13 at 22:04

3 Answers3

1

You can solve it from first principles. Suppose that masses $m_1,m_2,m_3$, and $m_4$ are located at $\langle 0,0\rangle,\langle a,0\rangle,\langle 0,b\rangle$, and $\langle a,b\rangle$, respectively. (You can always impose such a coordinate system.) Suppose that the centre of mass is at $\langle\bar x,\bar y\rangle$; then the moments of the system about the lines $x=\bar x$ and $y=\bar y$ must be $0$. The moment about $x=\bar x$ is

$$m_1(0-\bar x)+m_2(a-\bar x)+m_3(0-\bar x)+m_4(a-\bar x)=a(m_2+m_4)-m\bar x\;,$$

where $m=m_1+m_2+m_3+m_4$, so $$\bar x=\frac{a(m_2+m_4)}m\;.$$

I’ll leave $\bar y$ to you; the computation is very similar.

Brian M. Scott
  • 616,228
0

In physics the center of mass of a mass, or a collection of discrete masses, is the singular point where the weighed relative position of the spread out mass resides. The formula used for finding the center of mass is

R = (1/M) X (∑mi ri ) as i changes from 1 to n

Where the system includes “n” number of masses, each with some mass m that are located in space with coordinates ri, i = 1, …, n. The coordinates R then represents the position of the center of mass

This is a very scientific, mathematical definition.

To simplify it, if you were to take a rectangular card board, you would intuitively know that the center of mass is the point where the diagonals intersect. Or, if you take a tennis ball, you know it is right at the center of the sphere. If you were to take 2 similar tennis balls and put them 1 m apart, you would know that the center of mass of this system (i.e. 2 balls) would be their center. If one of these balls was heavier, the COM will shift towards the heavier ball. But what if you were to take an irregular mass like a cricket bat, the center of mass would probably reside below the center of the bat, in the lower half (since the mass of the handle is relatively less). You would need to divide the entire bat into infinitesimally small pieces and use integral calculus to find the center of mass.

The concept of center of mass can be very powerful if you have to deal with an object for purpose of solving a problem in Physics (often pertaining to the Newton’s laws of motion, linear momentum or angular momentum), you can assume that the entire mass of the object resides at this point. This therefore can then be treated as a point object.

Watch this video made by me to understand this concept better-

Center of Mass and How to Find

Vish
  • 21
0

The centre of mass is the point where the moment of mass is zero.

If $\xi$ is the x-coordinate, and $d m$ is an increment of mass, then one seeks a value where $\int \xi dm = 0$. For a general coordinate, it suffices to simply evaluate $(\int \xi dm) /(\int dm) = \xi_0$, and $\xi_0$ is the required x coordinate.

Experiments with an equalateral triangle would show that while there are planes dividing the mass into two, these do not always cross at a 'centre' of gravity.

wendy.krieger
  • 6,856