# Top of massive \cfrac cut off

Working on a physics assignment, where I'm showing my work how I derived an equation from other equations. First time LaTex user, so not sure how to fix this.

In the last fraction, the top part of the top fraction is missing when compiled on Overleaf.

If anyone also knows how to make the parentheses a bit longer so that they better fence each section, that would be wonderful!

We begin with the formula \emph{M}\cdot \emph{g}=$$\displaystyle \emph{F_T}$$
which we can rearrange to:
\begin{center}
\emph{g} = $$\displaystyle \frac{\emph{F_T}}{\emph{M}}$$
\end{center}

Using $$\displaystyle \emph{F_T}$$ = $$\displaystyle \frac{\emph{F_C}}{\sin(\theta)}$$, we get:
\begin{center}
\emph{g} = $\cfrac{( \cfrac{\emph{$F_C$}}{\sin(\theta)})}{\emph{M}}$
\end{center}

\emph{$F_C$} can be replaced as \emph{$F_C$} = $$\displaystyle \frac{\emph{mv^2}}{\emph{r}}$$, giving us :
\begin{center}
\emph{g} = $\cfrac{( \cfrac{(\cfrac{\emph{$mv^2$}}{\emph{r}})}{\sin(\theta)})}{\emph{M}}$
\end{center}

Next we replace \emph{v} from \emph{v} =$$\displaystyle \frac{2\pi\emph{r}}{\emph{T}}$$ :
\begin{center}
\emph{g} = $\cfrac{( \cfrac{(\cfrac{\emph{$(m(\cfrac{2\pi\emph{r}}{\emph{T}})^2)$}}{\emph{r}})}{\sin(\theta)})}{\emph{M}}$
\end{center}

We also eliminate the variable \emph{r} as \emph{r} = \emph{L} \cdot \sin($\theta$) :
\begin{center}
\emph{g} = $\cfrac{( \cfrac{(\cfrac{\emph{$(m(\cfrac{2\pi(\emph{L} \cdot \sin(\theta))}{\emph{T}})^2)$}}{(\emph{L} \cdot \sin(\theta))})}{\sin(\theta)})}{\emph{M}}$
\end{center}

The last variable $\theta$ is replaced with $\arccos(\cfrac{m}{M})$ :
\allowdisplaybreaks
\begin{center}
\emph{g} = $\cfrac{( \cfrac{(\cfrac{\emph{(m(\cfrac{2\pi(\emph{L} \cdot \sin(\arccos(\cfrac{m}{M})))}{\emph{T}})^2)}}{(\emph{L} \cdot \sin(\arccos(\cfrac{m}{M})))})}{\sin(\arccos(\cfrac{m}{M}))})}{\emph{M}}$
\end{center}

• Welcome to the TeX/LATEX user site. I would honestly change everything about your code :-). For what reason do you use \cfrac? For the parenthesis you can use \bigl( and \bigr) for example. – Sebastiano Apr 17 at 22:17
• I honestly didn't understand the formula in your output :-( – Sebastiano Apr 17 at 22:27

You should really consult a manual for starters: typing math in LaTeX is not as difficult as you maybe think it to be.

Besides, eight story fractions are very difficult to interpret and your last equation has indeed redundant terms: you need no arccosine.

\documentclass{article}
\usepackage{amsmath}

\begin{document}

We begin with the formula $$M\cdot g=F_T$$ which we can rearrange to:
$g = \frac{F_T}{M}$
Using $$F_T = \frac{F_C}{\sin(\theta)}$$, we get:
$g = \frac{\dfrac{F_C}{\sin(\theta)}}{M}=\frac{F_C}{M\sin(\theta)}$
$$F_C$$ can be replaced as $$F_C = \frac{mv^2}{r}$$, giving us:
$g = \frac{\dfrac{mv^2}{r\mathstrut}}{M\sin(\theta)}=\frac{mv^2}{Mr\sin(\theta)}$
Next we replace $$v$$ from $$v = \frac{2\pi r}{T}$$:
$g = \frac{m}{Mr\sin(\theta)}\frac{4\pi^2r^2}{T^2}=\frac{4\pi^2rm}{MT^2\sin(\theta)}$
We also eliminate the variable $$r$$ as $$r = L\sin(\theta)$$:
$g = \frac{4\pi^2mL\sin(\theta)}{MT^2\sin(\theta)}=\frac{4\pi^2mL}{MT^2}$
\end{document}


• Good point. So it's not just me who has misread the last equation :-) – Sebastiano Apr 17 at 22:49

Something like this?

\documentclass[a4paper,12pt]{article}
\usepackage{amsmath,amssymb}
\begin{document}
The last variable $\theta$ is replaced with $\arccos(\tfrac{m}{M})$:
$g =\frac{\sin\left(\arccos\left(\frac Mm\right)\right)}{M}$
\end{document}


or this?

\documentclass[a4paper,12pt]{article}
\usepackage{amsmath,amssymb}
\begin{document}
The last variable $\theta$ is replaced with $\arccos(\tfrac{m}{M})$:
$g =\frac{\dfrac{L\cdot\sin\Bigl(\arccos\bigl(\frac Mm\bigr)\Bigr)}{\sin \Bigl(\arccos\left(\frac Mm\right)\Bigr)}}{M}$
\end{document}