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I'm not going to post the full question, I just want a general idea of how I should go about solving this type of problem.

There is a square with 4 charged masses on each of the corners, in the center of the square, there is a fifth charged mass. I am supposed to find the net force acting on the center mass ($m_{5}$).

Now I could find the gravitational forces between the center mass and each of the corner masses, split that into x and y components, and then repeat the same process for the electrostatic forces. I'm pretty sure there is an easier, less cumbersome way to do this. Any ideas?

m2--------m1
|          |
|    m5    |
|          |
m3--------m4

None of the charges/masses directly cancel each other out, however the electrostatic ones seem to be small enough to be trivial in the final result.

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2 Answers 2

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A square has its diagonals at right angles. So, find the forces along each diag. i.e., m1 and m3 on m5 for one direction and m2 and m4 on m5 for other. You could actually simpy find difference between, "m1 and m3" and "m2 and m4" and the corresponding charges and use them as resultant mass and charge.

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  • $\begingroup$ So, find the $F_{g}$ of m1 + m3 on m5, and then do the same for m2 and m4? $\endgroup$
    – ta3920
    Commented Sep 9, 2014 at 2:34
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    $\begingroup$ no. G-force by m1 on m5 and m3 on m5 would be in opposite directions (attractive in both cases) so find the force on m5 by $|m_1-m_3|$. It's direction is towards the larger mass. Similarly, for all other forces. $\endgroup$
    – tpb261
    Commented Sep 9, 2014 at 2:37
  • $\begingroup$ Glad, it helped. $\endgroup$
    – tpb261
    Commented Sep 9, 2014 at 2:40
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Use the fact the masses m1 and m3 apply forces along the diagonal. Similarly for the other diagonal. Then convert to the cartesian solution using the two resultants.

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    $\begingroup$ Do you mind elaborating? $\endgroup$
    – ta3920
    Commented Sep 9, 2014 at 2:30
  • $\begingroup$ I missed this answer. What he means is that m1 and m3 cause Gravitational forces in opposite directions on m5, similarly m2 and m4 - but this is perpendicular to the force by m1 and m3. Similarly solve for the electrostatic forces. $\endgroup$
    – tpb261
    Commented Sep 9, 2014 at 2:35
  • $\begingroup$ Sorry I could've been clearer. $\endgroup$
    – Michael
    Commented Sep 9, 2014 at 2:44

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