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What can be said about two 10kg masses that are 50 cm apart?

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kkittkatt095 | eNotes Newbie

Posted May 9, 2013 at 8:13 PM via web

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What can be said about two 10kg masses that are 50 cm apart?

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oldnick | Valedictorian

Posted May 9, 2013 at 10:45 PM (Answer #1)

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The question isn' clear at all. However you can consider the two masses a part of a inertial system . Then , Physics, with some rule, tell us we can see them as a lonely mass (mass sum) that in a case of dynamic action, have a behaviour run according its centrer of mass.

Determinated a system of riferiment, let be `M_1` and `M_2`  the two mass. As far as I could understand the mass laying on the same plane, then called  `G_1`  and `G_2`  their relative centers of gravity, and let it be `O` the original point. According rules:

`(M_1 xx OG_1 +M_2 xx OG_2)/(M_1+M_2)= OG`  where `G`  stands for the gravity center of the system.

Then you can think system as a lonely gravitiy and study its motion as all the mass is tookl all around `G` .

Of course the sysyem has to be hard, (its clear that if masses moved according force effetcs, changing their relative positions  `G ` changes too and no more system has to be commpared to an lonely mass)

  

 

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oldnick | Valedictorian

Posted May 9, 2013 at 11:00 PM (Answer #2)

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If  you consider  a point of riferiment `O`  between `M_1`  and `M_2`  at 5mts from the two mass, for the rule above we have:

`G=(5xx25-5xx(-25))/(2(10+10))=0`

So `G `  is exactly in the midle.

Then  you can  consider  the two masses of 10 Kgs as one of 20 Kg in the midest of,  far `+- 25` meters away .

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oldnick | Valedictorian

Posted May 9, 2013 at 11:35 PM (Answer #3)

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This is the graph of the value of `OG`  left the mass `M_1=`  `10 Kgs`  e  changing `M_2` . Note , for a mass  `M_2=10 Kgs`  the center `G ` is in the middle ( `y=0` ), as we have calculated. 

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mathsworkmusic | (Level 3) Associate Educator

Posted May 13, 2013 at 7:33 PM (Answer #5)

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Newton's law of universal gravitation states that

Every point mass attracts every single other point mass by a force pointing along the line intersecting both points. The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them [1].

We can write this as

`F = G ((m_1m_2)/r^2)`

where  `G`  is the universal gravitational constant, equal to

6.674 x 10^(-11) Nm^2 kg^(-2)  (Cavendish, 1798).

Therefore, if we have two particles mass 10kg, 50cm apart, the gravitational force between them is

`F = G(((10)(10))/50^2) = G(100/2500) = G/25` Newtons

`= ` 2.670 x 10^(-12) N

Since  `F = ma`   (Newton's 2nd law of motion) each mass is accelerating

toward the other at the same rate equal to

` ``a = F/m_1 = ` 2.670 x 10^(-13) ms^(-2)

This is of course very small relative to the distance between them :-

5 x 10^(-1) m

The gravitational force between them is F = 2.670 x 10^(-12) N.

Their centre of mass, considering them as a unit, is at the midpoint

between them.

--------------------------------------------

References

[1] Page 297 in H W Turnbull (ed.), Correspondence of Isaac Newton, Vol 2 (1676-1687), (Cambridge University Press, 1960), document #235, 24 November 1679.

 

 

 

 

 

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