Question

The two blocks shown in the figure are not attached. The minimum magnitude of the horizontal force $F$ required to keep the smaller block from slipping down the larger block is -
[Take $g=10{\text{m/s}}^2$]

• A. $472\text{N}$
• B. $672\text{N}$
• C. $372\text{N}$
• D. $572\text{N}$

Answer 1

The correct option is D $572\text{N}$

Let the combined acceleration of the two blocks system be $a$.

FBD of blocks,


From second law of motion,$(\sum F=ma)$, for block $M$

$N=88a........\left(1\right)$
[Along horizontal]

Similarly, for block $m$,

$F-N=16a$ [Along horizontal]

$\Rightarrow F-88a=16a$ [from (1)]

$\Rightarrow F=104a.........\left(2\right)$

Also, for just slipping condition
$f=160$ [Along vertical]

$\Rightarrow \mu N=160$

$\Rightarrow 0.33\times 88a=160$ [from (1)]

$\Rightarrow a=5.5{\text{m/s}}^2$

Putting the value of $a$ in eq.(2)

$F=104\times 5.5$

$\Rightarrow F=572\text{N}$ is the minimum force required.

Why this question ? Concept involved - (1) Newton's law of motions. Tip - If there are two blocks or more which are static with respect to each other, then consider them as a system. Such questions are generally asked in JEE main.