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Fuzzy dice acceleration

  • #1
I'm not sure I have enough info but the question goes like this...Fuzzy dice are hanging from a rearview mirror. The car accelerates at 3.54m/s^2 What angle will the dice make with the vertical? I have a force diagram showing of the dice Mg points down. T points up to the right in the 1st quadrant. For the summation of forces I have Fx=Tcos(theta)=M(3.54m/s^2) and Fy=Tsin(Theta)-Mg=0. What am I missing? Not sure how to find theta..Thanks
 

Answers and Replies

  • #2
841
1
Divide the two equations and take the arctangent. (And post in the homework help forums.)
 
  • #3
Thanks but why does that work. (Where is the Homework forum?)[?] [?]
 
  • #4
841
1
If you divide the two equations, the T's cancel out and you get a sine divided by a cosine, which is a tangent. The arctangent inverts the tangent in order to solve for the angle (arctan(tan(θ)) = θ).

The Homework Help Zone is near the bottom of the main physicsforums.com page.
 
  • #5
chroot
Staff Emeritus
Science Advisor
Gold Member
10,226
34
Ambi,

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]
 
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