Four 7.5 kg spheres are located at the corners of a square of side 0.72m?

Question

Four 7.5 kg spheres are located at the corners of a square of side 0.72m. Calculate the magnitude and direction of the total gravitational force exerted on one sphere by the other three.

What is the magnitude

I have worked on this forever now it seems and I have tried so many different formulas and just cant seem to figure out the answer

I do know the direction is toward the center of the square just need to know the magnitude

Answer 1

Gravitational force = k*mass1*mass/(radius^2)

k= 6.67 × 10−11 N m2 kg-2.

r1=r2=0.72 meters

r3=(0.72^2+0.72^2)^0.5 = 1.02

force 1 and two are at 45 degree angles. They are found by the equation;

f= 6.67 × 10−11  (4.5)(4.5)/(0.72)^2= 2.61X10-9N

These are at 90 degrees from each other. Right between them is the following force;

f= 6.67 × 10−11  (4.5)(4.5)/(1.02)^2 = 1.30 X 10^-9 N

All these forces are pointing toward from the other spheres. They all average out to going toward the center of the square.

The magnitude is first found by applying the Pythagoras theorem. The resultant is towards the center. The magnitude is

2,61e-9^2+2.61e-9^2 = 3.68e-9 N

Add that to the other force going toward the center;

3.68e-9+1.30e-9=4.98e-9N