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4-2.Friction
medium
The coefficient of static friction, $\mu _s$ between block $A$ of mass $2\,kg$ and the table as shown in the figure is $0.2$. What would be the maximum mass value of block $B$ so that the two blocks $B$ so that the two blocks do not move? The string and the pulley are assumed to be smooth and masseless ....... $kg$ $(g = 10\,m/s^2)$
A
$2.0$
B
$4.0$
C
$0.2$
D
$0.4$
Solution
Let the mass of the block $B$ is $M$.
In equilibrium of block $B$ implies $T=M g \ldots(i)$
If block $A$ does not move, then $T=f_{s}$
where $f_{s}=$ frictional force $=\mu_{S} N=\mu_{s} m g$
implies $T=\mu_{s} m g \ldots(i i)$
Thus, from Eqs. $( i )$ and $(ii),$ we have
$M=\mu_{s} m g$ or $M=\mu_{s} m$
Given: $\mu_{s}=0.2, m=2 k g$
$\therefore M=0.2 \times 2=0.4 k g$
Standard 11
Physics
