\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\begin{document}
\parindent=0pt


{\bf Question}

Three particles of equal mass $m$ are placed at the corners of an
equilateral triangle of side length $d.$  Find the total
gravitational force on each mass.

\vspace{.25in}

{\bf Answer}

\setlength{\unitlength}{.5in}
\begin{picture}(4,5)
\put(0,1){\line(1,0){3}}

\put(-.1,.9){$\bullet$}

\put(2.9,.9){$\bullet$}

\put(0,1){\line(1,2){1.5}}

\put(3,1){\line(-1,2){1.5}}

\put(1.4,3.9){$\bullet$}

\put(1.5,4){\line(0,-1){.25}}

\put(1.5,3.5){\line(0,-1){.25}}

\put(1.5,3){\line(0,-1){.25}}

\put(1.5,2.5){\line(0,-1){.25}}

\put(1.5,2){\line(0,-1){.25}}

\put(1.5,1.5){\line(0,-1){.25}}

\put(.65,2.4){\line(2,-1){.9}}

\put(.8,2.7){\line(2,-1){.3}}

\put(1.1,2.55){\line(-1,-2){.15}}

\put(0,0.8){\makebox(0,0){$B$}}

\put(3,0.8){\makebox(0,0){$C$}}

\put(1.5,4.2){\makebox(0,0){$A$}}

\put(.6,2.6){\makebox(0,0){$D$}}

\put(1.65,2){\makebox(0,0){$O$}}

\put(1.4,3.5){\makebox(0,0){$\frac{\pi}{6}$}}

\end{picture}

\setlength{\unitlength}{.5in}
\begin{picture}(8,4)

\put(0,3){\makebox(0,0){At A:}}

\put(0,1){\vector(0,1){0.25}}

\put(0,1){\vector(1,0){0.25}}

\put(0.4,1){\makebox(0,0){\bf i}}

\put(0,1.5){\makebox(0,0){\bf j}}

\put(1.9,1.9){$\bullet$}

\put(2,2){\line(0,-1){1}}

\put(2,2){\vector(0,-1){0.5}}

\put(2,2){\line(-2,-1){1}}

\put(2,2){\vector(-2,-1){0.5}}

\put(2,2){\line(1,0){0.3}}

\put(2.6,2){\line(1,0){0.3}}

\put(2,2){\line(0,1){0.3}}

\put(2,2.6){\line(0,1){0.3}}

\put(2,2){\line(2,1){1}}

\put(3,2.2){\makebox(0,0){$\frac{\pi}{6}$}}

\put(2.15,2.3){\makebox(0,0){$\frac{\pi}{3}$}}

\put(2.3,1.5){\makebox(0,0){{\bf F}$_B$}}

\put(1.5,2){\makebox(0,0){{\bf F}$_C$}}

\put(5,3){\makebox(0,0)[l]{{\bf F}$_C$ = -{\bf F}$_j$}}

\put(5,2){\makebox(0,0)[l]{{\bf F}$_B$ = -F({\bf i} $\cos
\frac{\pi}{6} + {\bf j} \sin \frac{\pi}{6}$)}}

\end{picture}


Total force on $A$: $\ds {\bf F} = {\bf F}_B + {\bf F}_C = -F{\bf
j} - F\left({\bf i} \frac{\sqrt 3}{2} + {\bf j} \frac{1}{2}\right)
= -\frac{F}{2} \sqrt 3 ( {\bf i} + \sqrt 3 {\bf j})$

Magnitude of the force at $A$: $\ds = \frac{F}{2} \sqrt 3 \times 2
= \sqrt 3 \frac{Gm^2}{d^2}$

By symmetry the forces on each of the masses has the same
magnitude.




\end{document}