Pulling a yoyo over a surface with tension

[Justforthisquestion1]

How does this $F_{\text {max }}$ follow from #24 ?

#\ ##

First of all i used $a$ = $alpha$ * $R$
So i switched the $a$ in m * $a$ in Newtons second law for $alpha$ * $R$
so now i have
$m * alpha * R$ = F * cos(theta) - F(s)
then i simply adjust that to
$alpha$ = (F * cos(theta) - F(s)) / (m * R)

Next i insert this $alpha$ in line 24 and solve for F

 

[BvU]

I chose my coordinate system with i_hat to the right j_hat down. Doesnt that make cw positive?

I see. Upside down. You in Australia ? -- never mind, not problematic in this exercise, I think...

Easy check: you claim you did $\theta =0$ successfully. So that must have been $F_{\text {max} }=\displaystyle {3\mu mg \over \cos\theta+3\mu \sin \theta-2b/R}\quad$ ?

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[Justforthisquestion1]

I see. Upside down. You in Australia ? -- never mind, not problematic in this exercise, I think...

Easy check: you claim you did $\theta =0$ successfully. So that must have been $F_{\text {max} }=\displaystyle {3\mu mg \over \cos\theta+3\mu \sin \theta-2b/R}\quad$ ?

$\$

Almost!
The -2b are the problem!
it should be +2b!
Thank you! i didnt think of this easy yet effective method to test this solution
Do you have any idea where the mistake could come from? If not then i will write my solution on a paper and scan it!

 

[Justforthisquestion1]

Almost!
The -2b are the problem!
it should be +2b!
Thank you! i didnt think of this easy yet effective method to test this solution
Do you have any idea where the mistake could come from? If not then i will write my solution on a paper and scan it!

In my mind, the solution for $theta = 0$ should be
##Fmax= (mu * m * g * 3*R) / (2 * b +R)

 

[Justforthisquestion1]

I see. Upside down. You in Australia ? -- never mind, not problematic in this exercise, I think...

Easy check: you claim you did $\theta =0$ successfully. So that must have been $F_{\text {max} }=\displaystyle {3\mu mg \over \cos\theta+3\mu \sin \theta-2b/R}\quad$ ?

$\$

Got it thank you so much
Now i can forget about yoyos forever!!!

 

[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...

$\$

 

[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...

$\$

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