Problem source

A rough ring of radius $a$ is fixed so that it lies in a plane inclined at an angle $\alpha$ to the horizontal. A uniform heavy rod of length $b(>a)$ has one end smoothly pivoted at the centre of the ring, so that the rod is free to move in any direction. It rests on the circumference	of the ring, making an angle $\theta$ with the radius to the highest point on the circumference. Find the relation between $\alpha,\theta$ and the coefficient of friction, $\mu,$ which must hold when the rod is in limiting equilibrium. 

Solution source

\begin{center}
    \begin{tikzpicture}[scale=2]
        \def\tilt{-40}
        \def\h{0};
        \def\k{0};
        \def\a{1};
        \def\b{2};


        \def\top{100};
        \def\bottom{280};
        \def\side{190};

        \def\pole{140};

        \coordinate (O) at (\h, \k);
        \coordinate (T) at ({\h + \a*cos(\top)*cos(\tilt) - \b*sin(\top)*sin(\tilt)},{\k + \a*cos(\top)*sin(\tilt)+\b*sin(\top)*cos(\tilt)});
        \coordinate (S) at ({\h + \a*cos(\side)*cos(\tilt) - \b*sin(\side)*sin(\tilt)},{\k + \a*cos(\side)*sin(\tilt)+\b*sin(\side)*cos(\tilt)});
        \coordinate (B) at ({\h + \a*cos(\bottom)*cos(\tilt) - \b*sin(\bottom)*sin(\tilt)},{\k + \a*cos(\bottom)*sin(\tilt)+\b*sin(\bottom)*cos(\tilt)});
        \coordinate (P) at ({\h + \a*cos(\pole)*cos(\tilt) - \b*sin(\pole)*sin(\tilt)},{\k + \a*cos(\pole)*sin(\tilt)+\b*sin(\pole)*cos(\tilt)});
        \coordinate (Pend) at ($(O)!1.5!(P)$);

        \coordinate (A) at (-2, -2);

        \coordinate (X) at (0, -1);

        
        \filldraw (O) circle (1pt);
        \node at (O) [right] {$O$};
        \filldraw (T) circle (1pt);
        % \filldraw (B) circle (1pt);
        \draw (O) -- (Pend);
        \draw[dashed] (O) -- (T);

        \draw[dashed] (O) -- (B) -- (X);

        \pic [draw, angle radius=0.8cm, "$\theta$"] {angle = T--O--Pend};
        \pic [draw, angle radius=0.8cm, "$\alpha$"] {angle = X--B--O};
        
        \draw[domain = 0:360, samples=180, variable = \x]  plot ({ \h + \a*cos(\x)*cos(\tilt) - \b*sin(\x)*sin(\tilt)},{\k + \a*cos(\x)*sin(\tilt)+\b*sin(\x)*cos(\tilt)});

        \draw[-latex, blue, ultra thick] ($(O)!0.75!(P)$) -- ++(0, -0.75) node[right] {$mg$};
        \draw[-latex, blue, ultra thick] (P) -- ++(-0.2, 0.5) node[right] {$R$};
        \draw[-latex, blue, ultra thick] (P) -- ++(0.5, 0.2) node[right] {$\mu R$};
        
        \draw[-latex, red, ultra thick] (A) -- ++($0.25*(T)-0.25*(O)$) node[above] {$x$};
        \draw[-latex, red, ultra thick] (A) -- ++($0.5*(S)-0.5*(O)$) node[left] {$y$};
        \draw[-latex, red, ultra thick] (A) -- ++(-0.2, 0.5) node[left] {$z$};

        
    \end{tikzpicture}
\end{center}

It is important to define clear coordinate axes, so let the $x$-axis point up the line of greatest slope of the ring. The $z$-axis perpendicular to the ring, and the $y$-axis perpendicular to both of these.

Our method is going to be to take moments about $O$ to avoid worrying about the force at the pivot.

There are $3$ forces we need to worry about:

\begin{itemize}
\item The mass of the rod
\item The reaction where it meets the ring
\item The friction at the ring
\end{itemize}

In our coordinate frame, the reaction will act in the $z$-direction, $\displaystyle \begin{pmatrix} 0 \\ 0 \\ R \end{pmatrix}$, the friction force will act in the $x-y$ plane: $\displaystyle \begin{pmatrix} \mu R \sin \theta \\ -\mu R \cos \theta  \\ 0 \end{pmatrix}$. We don't know the mass, but we know it will be acting "vertically", so $\cos \alpha$ of it will act in the $z$-axis and $\sin \alpha$ will act in the $y$-axis, ie it will act parallel to $\displaystyle  \begin{pmatrix} \sin \alpha \\ 0 \\ \cos \alpha \end{pmatrix}$.

When taking moments, we need to consider $\mathbf{r}$ the direction of the rod. This will be $\displaystyle \begin{pmatrix} \cos \theta \\ \sin \theta \\ 0 \end{pmatrix}$.

The moment of the weight will all be parallel to $\mathbf{r} \times \begin{pmatrix} \sin \alpha \\ 0 \\ \cos \alpha \end{pmatrix}$. Similarly the moments of the contact forces will be $\mathbf{r} \times \begin{pmatrix}  \mu R \sin \theta \\ -\mu R \cos \theta  \\ R  \end{pmatrix}$. Since these moments sum to $\mathbf{0}$ as we are in equilibrium, these vectors must be parallel. 

Therefore it is sufficient to check the vector triple product, 
\begin{align*}
&& 0 &= \begin{pmatrix} \cos \theta \\ \sin \theta \\ 0 \end{pmatrix} \cdot \left ( \begin{pmatrix} \sin \alpha \\ 0 \\ \cos \alpha \end{pmatrix} \times \begin{pmatrix}  \mu  \sin \theta \\ -\mu  \cos \theta  \\ 1  \end{pmatrix} \right ) \\
&&&=  \cos \theta (\mu \cos \theta \cos \alpha)-\sin \theta (\sin \alpha - \mu \sin \theta \cos \alpha) \\
&&&= \mu((\sin^2 \theta+\cos^2 \theta) \cos \alpha) -\sin \theta \sin \alpha \\
\Rightarrow && \mu  &= \tan \alpha \sin \theta
\end{align*}