A wedge of mass 2m and a cube of mass m are shown in figure. Between cube and wedge there is no friction. The minimum coefficient of friction between wedge and ground, so that wedge does not move is

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answer is 0.2.

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Detailed Solution

$\begin{array}{cl} \sum F_{y} = 0 \\ \Rightarrow & N_{1} - mgcos θ = 0 \\ \Rightarrow & N_{1} = mgcos θ \end{array}$
For wedge, $\sum F_{y} = 0$
$\begin{array}{l} + N_{2} - 2 mg - N_{1} \cos θ = 0 \\ \Rightarrow N_{2} = 2 mg + {mgcos}^{2} θ \end{array}$
$\begin{array}{l} \sum F_{x} = 0 \\ \Rightarrow N_{1} \sin θ - {μN}_{2} = 0 \\ \Rightarrow mgcos θsin θ = μ (2 mg + {mgcos}^{2} θ) \\ \Rightarrow μ = \frac{\sin θcos θ}{2 + \cos^{2} θ} = \frac{\sin 45^{\circ} \cos 45^{\circ}}{2 + \cos^{2} 45^{0}} = 0 .2 \end{array}$

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