Problems

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Solution:

First observe that many forces are equal by symmetry. Also notice that \(A\) and \(B\) are trying to roll in opposite directions, therefore there is no friction between \(A\) and \(B\). Considering the system as a whole \(R_1 = \frac32 mg\). \begin{align*} \text{N2}(\uparrow,C): && 0 &= -mg +2R_3\cos 30^{\circ} + 2F_{CA} \cos 60^{\circ} \\ \Rightarrow && mg &= \sqrt{3}R_3 + F_{CA} \\ \\ \text{N2}(\uparrow, A): && 0 &= -mg + \frac32mg-R_3\cos 30^{\circ} -F_{AC} \cos 60^\circ \\ \Rightarrow && mg &= \sqrt{3}R_3+F_{AC} \\ \text{N2}(\rightarrow, A): && 0 &= F_{AC}\cos 30^{\circ}+F_1-R_3 \cos 60^\circ -R_2 \\ \Rightarrow && 0 &=F_{AC}\sqrt{3}+2F_1-R_3-2R_2 \\ \overset{\curvearrowleft}{A}: && 0 &= F_1 - F_{AC} \end{align*} Since \(F_1 = F_{AC} = F_{CA}\) we can rewrite everything in terms of \(F= F_1\) so. \begin{align*} && mg &= \sqrt{3}R_3 + F \\ && 0 &= (2+\sqrt{3})F -R_3-2R_2 \\ \Rightarrow && (2+\sqrt3)F &\geq R_3 \\ \Rightarrow && F &\geq (2-\sqrt{3})R_3 \\ \Rightarrow && \mu & \geq 2 - \sqrt{3} \end{align*}