Pulling a yoyo over a surface with tension

[BvU]

I have (from your posts):$$\begin{align*} R\mu(mg-F\sin\theta)-bF &=I\alpha \\ \ &\ \\
I\alpha&={1\over 2} m R^2 \alpha = {R\over 2} ma = {R\over 2} \Bigl ( F\cos\theta - \mu\left ( mg -F\sin\theta\right)\Bigr ) \\ \ &\ \\
2\mu(mg-F\sin\theta)-2b{F/R} &=F\cos\theta - \mu\left ( mg -F\sin\theta\right )\\ \ &\ \\
3\mu mg &= F\Bigl( \cos\theta + 3\mu\sin\theta +2b/R \Bigr ) \\ \ &\ \\
F &= {3\mu mg \over \cos\theta + 3\mu\sin\theta +2b/R }
\end{align*} $$which fits for $\mu=0$ and $\theta = 0$ -- and cw.

Must have been an overlooked minus sign.Not finished for ccw though...

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[Justforthisquestion1]

I have (from your posts):$$\begin{align*} R\mu(mg-F\sin\theta)-bF &=I\alpha \\ \ &\ \\
I\alpha&={1\over 2} m R^2 \alpha = {R\over 2} ma = {R\over 2} \Bigl ( F\cos\theta - \mu\left ( mg -F\sin\theta\right)\Bigr ) \\ \ &\ \\
2\mu(mg-F\sin\theta)-2b{F/R} &=F\cos\theta - \mu\left ( mg -F\sin\theta\right )\\ \ &\ \\
3\mu mg &= F\Bigl( \cos\theta + 3\mu\sin\theta +2b/R \Bigr ) \\ \ &\ \\
F &= {3\mu mg \over \cos\theta + 3\mu\sin\theta +2b/R }
\end{align*} $$which fits for $\mu=0$ and $\theta = 0$ -- and cw.

Must have been an overlooked minus sign.Not finished for ccw though...

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Yes that is correct. Okay i will do it for ccw and then forget about it :D