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Fuzzy dice acceleration

  1. Nov 15, 2003 #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
     
  2. jcsd
  3. Nov 15, 2003 #2
    Divide the two equations and take the arctangent. (And post in the homework help forums.)
     
  4. Nov 15, 2003 #3
    Thanks but why does that work. (Where is the Homework forum?)[?] [?]
     
  5. Nov 15, 2003 #4
    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.
     
  6. Nov 15, 2003 #5

    chroot

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    Staff Emeritus
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    Gold Member

    Ambi,

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

    - Warren
     
  7. Nov 15, 2003 #6
    [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]
     
    Last edited: Nov 15, 2003
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