A thin uniform equilateral plates 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 between plate and floor is $μ,$ then the least angle $θ$ its base AB can make the horizontal surface is $\cot^{- 1} [x μ + \frac{1}{\sqrt{y}}] .$ Then $x + y$ is equal to______

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answer is 5.

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Detailed Solution

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

$I = \frac{a}{\sqrt{3}}, N_{1} = M g$ and $f_{1} = μ N_{1} = N \Rightarrow N_{2} = μ M g$

Balancing the torque about point A,

$M g \times l \cos (30^{0} + θ) - N_{2} \times a \sin θ = 0 \Rightarrow M g \times \frac{a}{\sqrt{3}} \cos (30^{0} + θ) - μ M g \times a \sin θ = 0$

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

On solving further, we get

$θ = \cot^{- 1} [2 μ + \frac{1}{\sqrt{3}}] \Rightarrow x = 2$ and $y = 3 \Rightarrow x + y = 5$

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