Q.
What should be the maximum value of $M$ (in $kg$ ) so that the $4\, kg$ block does not slip over the $5\, kg$ block? (Take $g=10\, m / s ^{2}$ )
Laws of Motion
Solution:
Common acceleration of system is
$a=\frac{M g}{M+9}$
For $4\, kg$ block we use $f=4 a=\frac{4 M g}{M+9}$
$=16\, N$
$\Rightarrow \frac{4 M g}{M+9}=16$
$\Rightarrow 40 M=16 M+144$
$\Rightarrow M=6\, kg$
