Fuzzy dice acceleration

[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

 

[Ambitwistor]

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

 

[Kharmon7814]

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

 

[Ambitwistor]

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.

 

[chroot]

Ambi,

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

- Warren

 

[Ambitwistor]

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

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