Dude this is physics, not maths, that's why you're not getting any help, try the physics forum. It's all just pythagoras and inverse-square law though.

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Oh stuff it, I'll just tell you:

a) F = (G.m₁.m₂)/x²

F = [(6.67300 × 10-11) x 1 x 2)] / 1

F = 1.3346 x 10^-10 N

b) F = [(6.67300 × 10^-11) x 3] + [(6.67300 × 10^-11)x 4)]/cos45° + [(6.67300 × 10^-11)x 2)]/cos45°

F = 2.0019 x 10^-10 + (2.6692 x 10^-10)/cos45 + (1.3346 x 10^-10)/cos45

F = 2.0019 x 10^-10 + 3.775 x 10^-10 + 1.887 x 10^-10

F = 7.6639 x 10^-10 N (I hope)

c) Probably somewhere just 'above' the centre of the square, as in closer to the top line than the bottom one.

This is probably all completely wrong but it was fun trying. :biggrin:

Reply 4

ApeXaviour

7

For part b you'll need to divide up the forces into their horizontal and vertical componants, add them up. Then us pythagarous' rule to find the resultant "net" force.

Joe's way for that part is close but not completely accurate.

Reply 5

Popcorn

ApeXaviour

For part b you'll need to divide up the forces into their horizontal and vertical componants, add them up. Then us pythagarous' rule to find the resultant "net" force.

Joe's way for that part is close but not completely accurate.

Yes you're correct.
Joe_87 assumed that the resultant force acted in the same direction as the force due to gravitational attractions between 1kg and 3kg masses.

Reply 6

random123

2

Oh well it was worth a try. I'm guessing this is degree level stuff because I certainly haven't come across anything of this complexity on my A level course.