Question

Three 3.20-kg masses are at the corners of an equilateral triangle and located in space far from any other masses.

(a) If the sides of the triangle are 2.80 m long, find the magnitude of the net force exerted on each of the three masses.
_____ N

(b) What would your answer to part (a) be if instead the sides of the triangle are doubled in length?
_____ N

Answer 1

using F=G*(m1*m2)/r^2

and noting that the angle subtended by two sides of an equilateral triangle is 60 degrees
If we call one side of the equilateral triangle x
fx=G*3.2^2*(1+cos60) N
and fy=G*3.2^2*(sin60) N

The net is
sqrt(fx^2+fy^2)=
G*3.2^2*sqrt((1+cos60)^2+sin60^2)

F=1.04*10-9 N


___________________-

If the length is doubled to 2 m

The force is reduced by 1/2^2=1/4 as strong

1.11 N
___________________-

Answer 2

force on one block due to other block = G*m1*m2/ r^2 = 8.81* 10^-11 N

force due to the other block will be the same but at an angle of 60 degree

so net force =sqrt ( F1+ F2 + 2F1F2 cos 60 )= sqrt3 F = 15.25 * 10^-11N

on doubling the distance the force will become 1/4 due to the inverse square law

so new resultant force is = 3.81* 10^-11N