Two touching blocks 1 and 2 are placed on an | vedclass

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(D) Let $a$ be the common acceleration of the two blocks.For the first block $(m_1)$: The forces acting along the incline are the component of gravity

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$\frac{1}{4} \frac{m_1 m_2}{m_1+m_2} \mu g$

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(D) Let $a$ be the common acceleration of the two blocks.For the first block $(m_1)$: The forces acting along the incline are the component of gravity $m_1 g \sin 60^{\circ}$ downwards, friction force $f_1 = \mu_1 N_1 = 1.5 \mu m_1 g \cos 60^{\circ}$ upwards, and the reaction force $R$ from the second block acting upwards.Equation of motion: $m_1 g \sin 60^{\circ} - 1.5 \mu m_1 g \cos 60^{\circ} + R = m_1 a$  --- $(i)$For the second block $(m_2)$: The forces acting along the incline are the component of gravity $m_2 g \sin 60^{\circ}$ downwards, friction force $f_2 = \mu_2 N_2 = \mu m_2 g \cos 60^{\circ}$ upwards, and the reaction force $R$ from the first block acting downwards.Equation of motion: $m_2 g \sin 60^{\circ} - \mu m_2 g \cos 60^{\circ} - R = m_2 a$  --- $(ii)$From $(i)$, $a = g \sin 60^{\circ} - 1.5 \mu g \cos 60^{\circ} + \frac{R}{m_1}$.From $(ii)$, $a = g \sin 60^{\circ} - \mu g \cos 60^{\circ} - \frac{R}{m_2}$.Equating the two expressions for $a$:$g \sin 60^{\circ} - 1.5 \mu g \cos 60^{\circ} + \frac{R}{m_1} = g \sin 60^{\circ} - \mu g \cos 60^{\circ} - \frac{R}{m_2}$$R \left( \frac{1}{m_1} + \frac{1}{m_2} \right) = 1.5 \mu g \cos 60^{\circ} - \mu g \cos 60^{\circ} = 0.5 \mu g \cos 60^{\circ}$$R \left( \frac{m_1 + m_2}{m_1 m_2} \right) = 0.5 \mu g \left( \frac{1}{2} \right) = 0.25 \mu g = \frac{1}{4} \mu g$$R = \frac{1}{4} \frac{m_1 m_2}{m_1 + m_2} \mu g$.

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