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A uniform lamina is in the shape of an equilateral triangle $ABC$, and rests in equilibrium inside a vertical circular hoop with $B$ and $C$ in contact with the hoop, and $A$ at the centre of the circle. Contact is smooth at $C$ and rough at $B$, where the coefficient of friction is $μ$. The lamina is in limiting equilibrium with $B$ lower than $C$, and the axis of symmetry through $A$ makes an angle with the vertical. By first taking moments about $A$, show that

$$tanθ=\frac{μ√3}{√3-2μ}$$

The lamina $ABC$ is placed so that $θ = 0$, and the hoop is slowly rotated about its centre in a vertical plane so that $θ$ increases. Find the value, $θ_o$, of $θ$ beyond which the lamina will topple about $B$, if it has not slid already. Also find the least value of $μ$ for which $θ$ will reach the value of $θ_o$ without the lamina sliding.

The axis of symmetry, which is from $A$ to the base of the triangle (which forms a chord of the circle (hoop)) has length $\frac{√3r}{2}$. The normal reaction is in line with $A$ and the frictional force is at right angles to this at $B$. The lamina will have weight $W$.

J132
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2 Answers2

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Let the distance from any vertex of $ABC$ to the centre of mass of$ABC$ (the centre of the triangle) be $d$. There is a force $F$ acting along $AB$ which gives rise to the frictional force $\mu F$. Taking moments about $A$, since there is limiting equilibrium we get \begin{equation} dW\sin\theta=r\mu F. \tag{1} \label{eq:one} \end{equation} Resolving around $C$ we get a moment due to the weight of $$dW\cos\left(\frac{\pi}{6}-\theta\right)$$ which is balanced by the force $F$ at $B$ resolved orthogonal to $CB$, so \begin{equation} dW\cos\left(\frac{\pi}{6}-\theta\right) = rF\frac{\sqrt{3}}{2}. \tag{2} \label{eq:two} \end{equation} Solving \eqref{eq:one}, \eqref{eq:two} we get the answer.

If $ABC$ hasn't already slid, then it will topple when the centre of mass is over $B$, i.e. for $\theta > \theta_0 = \frac{\pi}{3}$. To find the least value of $\mu$, simply solve for it given $\theta=\frac{\pi}{3}$.

Chris Kerridge
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  • Thanks for your response, a good answer and a good way to model this problem. I have an alternative answer above. – J132 Sep 19 '15 at 19:36
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I have modeled it slightly differently.

I have a normal reaction $R$ acting along $AB$ towards $A$. The frictional force $μR$ orthogonal to this pointing to the left and $W$ vertically downwards from the centre of $BC$.

So I have:

Moments about $A$: $$\frac{r√3}{2}W=rμRsinθ$$ So $$W=\frac{2rμRsinθ}{r√3}$$

Resolving vertically (with R pointing towards the right and μR to the left (at an angle $θ$)): $$W=Rsinθ - μRcosθ$$

So $$\frac{2rμRsinθ}{r√3}= Rsinθ - μRcosθ$$

Which gives $$2rμsinθ - r√3sinθ = -r√3μcosθ$$

Which rearranges to give $$tanθ = \frac{μ√3}{√3-2μ}$$

For the value of $μ$, I have:

$$tan(60)=\frac{μ√3}{√3-2μ}$$

$$√3(√3-2μ)=μ√3$$ $$μ=\frac{3}{√3+2√3}=\frac{√3}{3}$$

J132
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