# Example: Free body diagrams

Published 2008-10-15 | Author: Kjell Magne Fauske

Illustration of the classical “bodies connected by a string” problem. A great advantage of using TikZ for drawing illustrations like this, is that the drawings can be parameterized. In this example the inclination angle has been parameterized. PGF’s mathematical engine is then used to do the necessary calculations.

Note that this example uses the shorthand scope notation in some places.

Do you have a question regarding this example, TikZ or LaTeX in general? Just ask in the LaTeX Forum.
Oder frag auf Deutsch auf TeXwelt.de. En français: TeXnique.fr.

\documentclass{article}

\usepackage{tikz}
\usetikzlibrary{scopes}
\begin{document}

\def\iangle{35} % Angle of the inclined plane

\def\down{-90}
\def\arcr{0.5cm} % Radius of the arc used to indicate angles

\begin{tikzpicture}[
force/.style={>=latex,draw=blue,fill=blue},
axis/.style={densely dashed,gray,font=\small},
M/.style={rectangle,draw,fill=lightgray,minimum size=0.5cm,thin},
m/.style={rectangle,draw=black,fill=lightgray,minimum size=0.3cm,thin},
plane/.style={draw=black,fill=blue!10},
string/.style={draw=red, thick},
pulley/.style={thick},
]

\matrix[column sep=1cm] {
%% Sketch
\draw[plane] (0,-1) coordinate (base)
-- coordinate[pos=0.5] (mid) ++(\iangle:3) coordinate (top)
|- (base) -- cycle;
\path (mid) node[M,rotate=\iangle,yshift=0.25cm] (M) {};
\draw[pulley] (top) -- ++(\iangle:0.25) circle (0.25cm)
++ (90-\iangle:0.5) coordinate (pulley);
\draw[string] (M.east) -- ++(\iangle:1.5cm) arc (90+\iangle:0:0.25)
-- ++(0,-1) node[m] {};

\draw[->] (base)++(\arcr,0) arc (0:\iangle:\arcr);
\path (base)++(\iangle*0.5:\arcr+5pt) node {$\alpha$};
%%

&
%% Free body diagram of M
\begin{scope}[rotate=\iangle]
\node[M,transform shape] (M) {};
% Draw axes and help lines

{[axis,->]
\draw (0,-1) -- (0,2) node[right] {$+y$};
\draw (M) -- ++(2,0) node[right] {$+x$};
% Indicate angle. The code is a bit awkward.

\draw[solid,shorten >=0.5pt] (\down-\iangle:\arcr)
arc(\down-\iangle:\down:\arcr);
\node at (\down-0.5*\iangle:1.3*\arcr) {$\alpha$};
}

% Forces
{[force,->]
% Assuming that Mg = 1. The normal force will therefore be cos(alpha)
\draw (M.center) -- ++(0,{cos(\iangle)}) node[above right] {$N$};
\draw (M.west) -- ++(-1,0) node[left] {$f_R$};
\draw (M.east) -- ++(1,0) node[above] {$T$};
}

\end{scope}
% Draw gravity force. The code is put outside the rotated
% scope for simplicity. No need to do any angle calculations.
\draw[force,->] (M.center) -- ++(0,-1) node[below] {$Mg$};
%%

&
%%%
% Free body diagram of m
\node[m] (m) {};
\draw[axis,->] (m) -- ++(0,-2) node[left] {$+$};
{[force,->]
\draw (m.north) -- ++(0,1) node[above] {$T'$};
\draw (m.south) -- ++(0,-1) node[right] {$mg$};
}

\\
};
\end{tikzpicture}

\end{document}


• #1 Jason Waskiewicz, February 10, 2010 at 4:14 a.m.

This was a great help to me in Physics recently. I was able to modify the bare bones of this to illustrate the forces on a ramp for my high school students.

Thank-you for contributing it. Pictures like this are a huge help to those of us who are learning.

• #2 ozan, June 11, 2010 at 2:33 a.m.

Why do I feel so mortal when I see so few lines of code do such great things?

• #3 Kai, May 9, 2013 at 6:53 a.m.

really nice example! Excellent coding!