In order to find the direction of the net gravitational force, let's draw the forces acting on m1:
The direction F13 is pointing is given by:
[tex]\begin{gathered} \tan (\theta)=\frac{2d}{d} \\ \tan (\theta)=2 \\ \theta=\tan ^{-1}(2) \\ \theta=63.435\degree \end{gathered}[/tex]
Let's find the forces F12 and F13, and decompose the second one in its horizontal and vertical components using this angle:
[tex]\begin{gathered} F12=\frac{G\cdot M\cdot m}{d^2}=\frac{6.674\cdot10^{-11}\cdot0.22\cdot10^{-3}\cdot0.22\cdot10^{-3}}{(3.6\cdot10^{-2})^2}=2.49\cdot10^{-15} \\ F13=\frac{G\cdot M\cdot m}{(\sqrt[]{d^2+(2d)^2_{}})^2_{}}=\frac{6.674\cdot10^{-11}\cdot0.22\cdot10^{-3}\cdot0.22\cdot10^{-3}}{5\cdot(3.6\cdot10^{-2})^2}=4.98\cdot10^{-16} \\ F13_x=F13\cdot\sin (\theta)=4.98\cdot10^{-16}\cdot\sin (63.435\degree)=4.45\cdot10^{-16} \\ F13_y=F13\cdot\cos (\theta)=4.98\cdot10^{-16}\cdot\cos (63.435\degree)=2.23\cdot10^{-16} \end{gathered}[/tex]
Adding the vertical forces, we have:
[tex]F_y=F13_y+F12=2.23\cdot10^{-16}+24.9\cdot10^{-16}=27.13\cdot10^{-16}=2.713\cdot10^{-15}[/tex]
Finally, the direction of the net force is given by:
(Let's use a negative Fy, since it's pointing down)
[tex]\begin{gathered} \theta=\tan ^{-1}(\frac{F_y}{F_x})=\tan ^{-1}(\frac{F_y}{F13_x}) \\ =\tan ^{-1}(\frac{-2.713\cdot10^{-15}}{0.445\cdot10^{-15}}) \\ =\tan ^{-1}(-6.0966) \\ =-80.68\degree \end{gathered}[/tex]
