# Fuzzy dice acceleration

• Kharmon7814
In summary, the question is asking for the angle that the fuzzy dice hanging from a car's rearview mirror will make with the vertical as the car accelerates at 3.54m/s^2. Using a force diagram, the equations Fx=Tcos(theta)=M(3.54m/s^2) and Fy=Tsin(Theta)-Mg=0 are derived. To find theta, the equations are divided and the arctangent is taken to solve for the angle. This method can be discussed further in the Homework Help Zone on physicsforums.com.

#### Kharmon7814

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

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

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

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(&theta;)) = &theta;).

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

Ambi,

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

- Warren

$$\begin{equation*} \begin{split} \sum F_x & = T \sin\theta = ma \\ \sum F_y & = T \cos\theta - mg = 0 \Rightarrow T\cos\theta = mg \end{split} \end{equation*}$$

$$\begin{equation*} \begin{split} \tan\theta \equiv \frac{\sin\theta}{\cos\theta} = \frac{T\sin\theta}{T\cos\theta} = \frac{ma}{mg} = \frac{a}{g} \\ \Rightarrow \theta = \arctan(\tan\theta) = \arctan \frac{a}{g} \end{split} \end{equation*}$$

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