Three masses are arranged in the (xy) plane as shown. What is the magnitude of the resulting force on the 2 kg mass at the origin? Universal gravitational constant is 6.6726x10^-11. Answer in units of N.
Points on graph are:
Newton's Law of universal gravitation says the force of attraction of body 1 to body 2 (and attraction of body 2 to body 1) is given by
F = G*m1*m2/r^2
where r is the distance between the 2 bodies. Let's refer to the 3 masses in the order you listed them m1, m2, and m3. In a case like this with 3 masses, you first ignore m3 and find the attraction between m1 and m2. And then ignore m1 and find the attraction between m2 and m3.
We need to do some trigonometry to find the distance and angles involved. To make it easier to talk about without both of us having the same drawing, let's do this. Convert the xy plane to a map so that the first mass is located at a point 5 units north of the origin and 4 units west of the origin. By the way, what are those units, kilometers, inches? I'll assume meters. I admit, I'm fussy about units. The Universal constant should be 6.6726x10^-11 N.m^2/kg^2.
The force on the 2 kg mass by the 8 kg mass is at an angle arctan(4/5) = 38.6o west of north.
The distance between those 2 masses is given by Pythagoras: sqrt(4^2 + 5^2).
The magnitude of the force on m2 due to m1 is given by
F12 = G*8 kg*2 kg/r^2 where r is the result of the above application of Pythagoras.
The component of F12 in the northern direction is F12n = F12*cos38.6
The component of F12 in the western direction is F12w = F12*sin38.6
You can do the same routine to find F32 and the eastern and southern components of F32.
Then it is easy to find the resultant of the western and eastern components F12w and F32e and the resultant of F12n and F32s. It looks as if the 2 results so far will be eastern and southern. Next you find the magnitude and direction of the resultant of those. And you're done.
I hope this helps,