Two wedges each of mass 600 g are placed next to each other on a rough horizontal surface. The coefficient of static friction between the wedges and the surface is 0.4. A cube of mass ‘M’ is balanced on the wedges as shown in the figure. If there is no friction between the cube and wedges, the largest mass ‘M’ of the cube that can be balanced without motion of the wedges is $\frac{n}{10} k g,$ find $n$

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Detailed Solution

Given, $μ = 0.4, m = 600 g m = 0.6 k g$

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$2 N \cos 45 ° = M g \Rightarrow \sqrt{2} N = M g \Rightarrow N = \frac{M g}{\sqrt{2}} f_{L} = μ (N \cos θ + m g) f_{L} = μ (\frac{M g}{\sqrt{2}} \times \frac{1}{\sqrt{2}} + m g) = μ g (\frac{M}{2} + m) ............... (i)$

If wedge is balanced,
$F = N \sin θ = \frac{M g}{\sqrt{2}} \sin 45 ° = \frac{M g}{2} ............... (i i)$
From equation (i) & (ii),
$\frac{μ M g}{2} + μ m g = \frac{M g}{2} μ m g = \frac{M g}{2} - \frac{μ M g}{2} \Rightarrow M = \frac{2 μ m}{(1 - μ)} = \frac{2 \times 0.4 \times 0.6}{1 - 0.4} M = 0.8 k g$