Consider system shown below. A horizontal force F is applied to a block X of mass 8 ,kg , such that

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4-2.Friction

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Consider the system shown below. A horizontal force $F$ is applied to a block $X$ of mass $8 \,kg$, such that the block $Y$ of mass $2 \,kg$ adjacent to it does not slip downwards under gravity. There is no friction between the horizontal plane and the base of the block $X$. The coefficient of friction between the surfaces of blocks $X$ and $Y$ is $0.5$. The minimum value of $F$ is ............ $N$ (take, acceleration due to gravity to be $10 \,ms ^{-2}$ )

A

$200$

B

$160$

C

$40$

D

$240$

(KVPY-2014)

Solution

(a)

Acceleration of system of blocks $=\frac{F}{8+2}=\frac{F}{10} \,ms ^{-2}$

Now, from free body diagram below, we see that

Reaction of $X$ and $Y$ is

$R=m_Y a=2 \times \frac{F}{10} \Rightarrow R=\frac{F}{5}$

Block $Y$ does not slips, if

Force of friction $=$ Weight of $Y$

$\Rightarrow \quad \mu R=m_Y g \Rightarrow 0.5 \times \frac{F}{5}=2 \times 10$

So, $\quad F=200 \,N$

Standard 11

Physics

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