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Problem Text
A small light ring is attached to the end $A$ of a uniform rod $AB$ of weight $W$ and length $2a$. The ring can slide on a rough horizontal rail. One end of a light inextensible string of length $2a$ is attached to the rod at $B$ and the other end is attached to a point $C$ on the rail so that the rod makes an angle of $\theta$ with the rail, where $0 < \theta < 90^{\circ}$. The rod hangs in the same vertical plane as the rail. A force of $kW$ acts vertically downwards on the rod at $B$ and the rod is in equilibrium. \begin{questionparts} \item You are given that the string will break if the tension $T$ is greater than $W$. Show that (assuming that the ring does not slip) the string will break if $$2k + 1 > 4 \sin \theta.$$ \item Show that (assuming that the string does not break) the ring will slip if $$2k + 1 > (2k + 3)\mu \tan \theta,$$ where $\mu$ is the coefficient of friction between the rail and the ring. \item You are now given that $\mu \tan \theta < 1$. Show that, when $k$ is increased gradually from zero, the ring will slip before the string breaks if $$\mu < \frac{2 \cos \theta}{1 + 2 \sin \theta}.$$ \end{questionparts}
Solution (Optional)
\begin{center} \begin{tikzpicture} \def\a{-2}; \def\b{-1}; \coordinate (A) at (\a, 0); \coordinate (B) at (0, \b); \coordinate (C) at (-\a, 0); \draw[thin] ($(A)!-0.5!(C)$) -- ($(A)!1.5!(C)$); \filldraw (A) circle (1.5pt) node[above left] {$A$}; \filldraw (B) circle (1.5pt) node[above] {$B$}; \filldraw (C) circle (1.5pt) node[above] {$C$}; \draw[ultra thick] (A) -- (B); \draw[thick] (B) -- (C); \draw[-latex, ultra thick, blue] (A) -- ++(0, 1) node[above] {$R$}; \draw[-latex, ultra thick, blue] (A) -- ++(-1, 0) node[above] {$F_A$}; \draw[-latex, ultra thick, blue] (B) -- ++(0, -1) node[below] {$kW$}; \draw[-latex, ultra thick, blue] (B) -- ($(B)!0.5!(C)$) node[right] {$T$}; \draw[-latex, ultra thick, blue] ($(A)!0.5!(B)$) -- ++(0, -1.5) node[below] {$W$}; \pic [draw, angle radius=1.2cm, angle eccentricity=.6, "$\theta$"] {angle = B--A--C}; \end{tikzpicture} \end{center} \begin{questionparts} \item $\,$ \begin{align*} \overset{\curvearrowright}{A}:&& W \cos \theta \cdot a + kW \cos \theta \cdot 2a - T \cos \theta \sin \theta \cdot 2a - T \sin \theta \cos \theta \cdot 2a &= 0 \\ && (2k+1) \cos \theta W &= T \cos \theta \cdot 4 \sin \theta \\ \Rightarrow && T &= \frac{2k+1}{4 \sin \theta} W \\ \Rightarrow && \text{breaks if }\quad \quad 2k+1 &> 4 \sin \theta \end{align*} \item $\,$ \begin{align*} \text{N2}(\uparrow): && R - W - kW - T \sin \theta &= 0 \\ \Rightarrow && R &= (k+1)W - T \sin \theta \\ &&&= (k+1)W - \frac{2k+1}{4} W \\ &&&= \frac{2k+3}{4}W \\ \text{N2}(\leftarrow): && F_A - T \cos \theta &= 0 \\ \Rightarrow && F_A &= \frac{2k+1}{4 }\cot \theta \\ \Rightarrow && \text{slips if }\quad \quad\quad \quad\quad \quad F_A &> \mu R \\ \Rightarrow && \text{slips if }\quad \quad \frac{2k+1}{4 }\cot \theta &> \mu \frac{2k+3}{4}W \\ \Rightarrow && 2k+1 &> (2k+3) \mu \tan \theta \end{align*} \item The condition for breaking is $k > 2\sin \theta -\frac12$. The condition for slipping is equivalent to: \begin{align*} && 2k+1 &> (2k+3) \mu \tan \theta \\ \Leftrightarrow && 2k(1- \mu \tan \theta) &> 3 \mu \tan \theta-1 \\ \Leftrightarrow && k &> \frac{3 \mu \tan \theta-1}{2(1- \mu \tan \theta)} \end{align*} Therefore we will slip first if: \begin{align*} && \frac{3 \mu \tan \theta-1}{2(1- \mu \tan \theta)} &< 2 \sin \theta - \frac12 \\ \Leftrightarrow && 3 \mu \tan \theta-1 &< 4 \sin \theta (1- \mu \tan \theta) - (1- \mu \tan \theta) \\ &&&=4 \sin \theta - 1 + \mu \tan \theta (1-4 \sin \theta) \\ \Leftrightarrow && 3 \mu \tan \theta &< 4 \sin \theta + \mu \tan \theta (1- 4 \sin \theta) \\ \Leftrightarrow && \mu \tan \theta(3-1+4\sin \theta) &< 4 \sin \theta \\ \Leftrightarrow && \mu &< \frac{2 \cos \theta}{1+2 \sin \theta} \end{align*} \end{questionparts}
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