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:

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

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.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

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