Centre of mass/gravity between points?

Question

find the disrance from A of the centre of mass of the given set of particles.

. 2a . 3a . a . < lengths
A_________o____________o______B
. 4m . m < mass

the answer to where the centre of mass is at is 13/5a

i cant work out where this came from i have a sigma formula to work with and got to x bar = 2a4m + 5am but who knows.

Help please and method please would be delightful thank you muchly.

Answer 1

The center of mass is a fictitious point A where all mass can be considered concentrated. A is found by weighting all the elemental mass distances (an) from a zero point a0 where n = 1...N. The weights are no more or less than mn/M; where mn are the mass elements at respective distances an from zero. M = sum of all mass elements mn...mN. The weighting follows because MA = m1a1 + m2a2 + ... + mNaN; where A measured from a0 is the center of mass M, and the mn X an's are each mass element times its distance from the same zero point a0.

It's like getting a bunch of kids (m) at respective distances (a) on one side of the teeter totter fulcrum offset by one big adult (M) at A distance on the other side of the fulcrum to balance the teeter totter. A and the a's are all measured from the imaginary fulcrum, which we set as the zero point a0. From MA = m1a1 + m2a2 + ... + mNaN, we can write A= (m1a1 + m2a2 + ... + mNaN)/M = m1a1/M + m2a2/M + ... + mNaN/M; where each mn/M is the weight for each distance an.

If the elemental distances are all multiples of a1, the first distance, then we can write a2 = 2a1, a3 = 3a1.... Therefore, MA = (m1a1 + m2(2a1) + ... + mN(Na1)) and A = (m1a1 + m22a1 + ... + mNNa1)/M = a1(m1 + 2m2 + ... + NmN)/M If m1 = m2 = ... = mN, then A = a1m(1 + 2 + ... + N)/M = a1(m/M)(1 + 2 + ... + N).

We can find the sum S = (1 + 2 + ... + N) = N(N + 1)/2. In which case A = a1(m/M)N(N + 1)/2 From this, once you know a1, once you know each m, and once you set the number of elemental masses (N), you can use the above to solve for A, the center of mass M.

As said earlier, M is the sum of the elemental masses m. Or M = Nm when all the elemental masses are equal. If total mass M = Nm, we can rewrite A = a1(m/M)N(N + 1)/2 = a1(m/Nm)N(N + 1)/2 = a1(N + 1)/2. In this special case, we only need to know the basic elemental distance a1 and the number of elemental masses N to find A the center of mass M. [NB: All these lie in line, like on the X axis. If the masses are scattered about in three dimensions, the concept, but not these equations, can be expanded in 3D to give a CM in 3D.]

[NB: Your answers puzzle me. I have no clue where they came from because I have little clue what you started with. How many elemental "lengths" of what size (a) do you have. What are the mass values (m) and how many (N) do you have? If you want specific answers you need to ask specific questions with all the assumptions and givens.]

Answer 2

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