- #1

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- Thread starter Kharmon7814
- Start date

- #1

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- #2

- 841

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Divide the two equations and take the arctangent. (And post in the homework help forums.)

- #3

- 7

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Thanks but why does that work. (Where is the Homework forum?)[?] [?]

- #4

- 841

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The Homework Help Zone is near the bottom of the main physicsforums.com page.

- #5

chroot

Staff Emeritus

Science Advisor

Gold Member

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Ambi,

Do me a favor and try using some of the LaTeX code! :)

- Warren

Do me a favor and try using some of the LaTeX code! :)

- Warren

- #6

- 841

- 1

[tex]

\begin{equation*}

\begin{split}

\sum F_x & = T \sin\theta = ma \\

\sum F_y & = T \cos\theta - mg = 0

\Rightarrow T\cos\theta = mg

\end{split}

\end{equation*}

[/tex]

[tex]

\begin{equation*}

\begin{split}

\tan\theta \equiv \frac{\sin\theta}{\cos\theta} = \frac{T\sin\theta}{T\cos\theta} = \frac{ma}{mg} = \frac{a}{g} \\

\Rightarrow \theta = \arctan(\tan\theta) = \arctan \frac{a}{g}

\end{split}

\end{equation*}

[/tex]

\begin{equation*}

\begin{split}

\sum F_x & = T \sin\theta = ma \\

\sum F_y & = T \cos\theta - mg = 0

\Rightarrow T\cos\theta = mg

\end{split}

\end{equation*}

[/tex]

[tex]

\begin{equation*}

\begin{split}

\tan\theta \equiv \frac{\sin\theta}{\cos\theta} = \frac{T\sin\theta}{T\cos\theta} = \frac{ma}{mg} = \frac{a}{g} \\

\Rightarrow \theta = \arctan(\tan\theta) = \arctan \frac{a}{g}

\end{split}

\end{equation*}

[/tex]

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