Two blocks A and B of masses m1 and m2 are placed in contact with each other on an inclined plane as shown . The co – efficient of friction of block A and B with inclined surface is $μ_{1}$ and $μ_{2}$ respectively with $μ_{1} > μ_{2} .$ The angle of inclination is $θ .$ Mark the correct option:

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If both blocks slip on incline, then their acceleration is $g \sin θ - (\frac{μ_{1} m_{1} + μ_{2} m_{2}}{m_{1} + m_{2}}) g \cos θ$

Under a situation that $μ_{1} < μ_{2}$ and both blocks slipping on incline, the normal force between the blocks will become zero.

If both the blocks do not slip on the incline, then $\tan θ \leq \frac{μ_{1} m_{1} + μ_{2} m_{2}}{(m_{1} + m_{2})}$

If both blocks slip on incline, then the normal force between the blocks is zero.

answer is A, B, D.

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Detailed Solution

For the blocks to remain stationary on the inclined surface:
$(m_{1} + m_{2}) g s i n θ \leq μ_{1} m_{1} g \cos θ + μ_{2} m_{2} g \cos θ$
$\Rightarrow \tan θ \leq \frac{u_{1} m_{2} + μ_{2} m_{2}}{(m_{1} + m_{2})}$
If N is the force applied by upper block on the lower block, then assuming both are slipping on the incline:
$(m_{1} + m_{2}) g \sin θ - (μ_{1} m_{1} g c o s θ + μ_{2} m_{2} g \cos θ) = (m_{1} + m_{2}) a$
$g \sin θ - \frac{(μ_{1} m_{1} + μ_{2} m_{2})}{(m_{1} + m_{2})} g \cos θ = a$
In this situation, using Newton’s laws for only lower block:
$(m_{1}) g \sin θ + N - (μ_{1} m_{1} s \cos θ) = (m_{1}) a$
Which shows, $N \neq 0$
Under a situation that $μ_{1} < μ_{2}$ with both blocks slipping on incline, the lower block will have more acceleration that upper block causing the blocks to get separated. Thus, the normal force will become zero