A thin uniform equilateral plate rests in a vertical plane with one of its vertex $A$ on a rough horizontal floor and another vertex $B$ on a smooth vertical wall. If the coefficient of friction $μ = \frac{1}{\sqrt{3}}$ , then the least angle $θ$ its base $AB$ can make with the horizontal surface is

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TestHub · Accepted Answer

Let us assume that the side-length of the triangle is $a$ , then

$l = \frac{a}{\sqrt{3}}$

$N_{1} = M g$

$f_{1} = μ N_{1} = N_{2}$

$N_{2} = μ M g$

Balancing the torque about point $A$

$M g \times l \cos (30 ° + θ) - N_{2} \times a \sin θ = 0$

$\Rightarrow M g \times \frac{a}{\sqrt{3}} \cos (30 ° + θ) - μ M g \times a \sin θ = 0$

$\frac{1}{\sqrt{3}} (\frac{\sqrt{3}}{2} \cos θ - \frac{1}{2} \sin θ) = μ \sin θ$

On solving further we get

$θ = \cot^{- 1} [2 μ + \frac{1}{\sqrt{3}}]$

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