Answer:
F=8.0*10^{-10}N
Explanation:
See the attached file for the masses distributions
The force between two masses at distance r is expressed as
[tex]F=\frac{Gm_{1}m_{2} }{r^{2} }\\ G=Gravitional constant \\[/tex]
since the masses are of the same value, the above formula can be reduce to
[tex]F=\frac{Gm^{2}}{r^{2} }\\[/tex]
using vector notation,Let use consider the force on the lower left corner of the mass due to the upper left side of the mass is
[tex]F_{12} =\frac{Gm^{2}}{r^{2} }j\\[/tex]
The force on the lower left corner of the mass due to the lower right side of the mass is
[tex]F_{14} =\frac{Gm^{2}}{r^{2} }i\\[/tex]
The force on the lower left corner of the mass due to the upper right side of the mass is
[tex]F_{13} =\frac{Gm^{2}}{d^{2} }cos\alpha i +\frac{Gm^{2}}{d^{2} }sin\alpha j\\[/tex]
The net force can be express as
[tex]F=\frac{Gm^{2}}{r^{2} }j +\frac{Gm^{2}}{r^{2} }i +\frac{Gm^{2}}{d^{2} }cos\alpha i +\frac{Gm^{2}}{d^{2} }sin\alpha j\\\\F=Gm^{2}[\frac{1}{r^{2}}+ \frac{1}{d^{2}cos\alpha }]i + Gm^{2}[\frac{1}{r^{2}}+ \frac{1}{d^{2}sin\alpha }]j\\\alpha=45^{0}, G=6.67*10^{-11}Nmkg^{-2}[/tex]
if we insert values we arrive at
[tex]F=6.67*10^{-11}*2.5^{2}[\frac{1}{1^{2}}+ \frac{1}{\sqrt{2}^{2}cos45 }]i + 6.67*10^{-11}*2.5^{2}[\frac{1}{1^{2}}+ \frac{1}{\sqrt{2}^{2}sin45}]j\\F=5.643*10^{-10}i+5.643*10^{-10}j[/tex]
if we solve for the magnitude, we arrive at
[tex]F=5.643*10^{-10}i+5.643*10^{-10}j \\F=\sqrt{(5.643*10^{-10})^{2} +(5.643*10^{-10})}^{2} \\F=8.0*10^{-10}[/tex]
Hence the net force on one of the masses is
[tex]F=8.0*10^{-10}N[/tex]
