$N _1+ N _2= Mg  N _1=\frac{4 Mg }{9}$

$\left(\tau_{ N }=0\right) N _1(50)= N _2(40)  N _2=\frac{5 Mg }{9}$

$5 N _1=4 N _2$

$f_{l_k}=\mu_k N_1 \quad f_{l_L}=\mu_S N_1$

$f_{i_k}=0.32 N _1 \quad f_{l_1}=0.4 N _1$

$\overline{f_{2 k}=0.32 N _2} \quad \overline{f_{2 L }=0.4 N _2}$

Suppose $x _{ L }=$ distance of left finger from centre when right finger starts moring

$\left(\tau_{ n }=0\right)_{\text {about cratre }} \Rightarrow N _1 x _{ L }= N _2(40)$

$f_{ K _1}= f _{ L _2} \quad \Rightarrow 0.32 N _1=0.40 N _2$

$4 N _1=5 N _2$

$N _{ I } x _{ L }=\frac{4 N _1}{5}(40)$

$x _{ L }=32$

Now $x _{ R }=$ distance when right finger stops and left finger starts moring

$\left(\tau_{ n }=0\right)_{\text {atout centre }} \Rightarrow N _1 x _{ L }= N _2\left( x _{ R }\right)$

$f _{ L _1}= f _{ K _2} \Rightarrow 0.4 N _1=0.32 N _2$

$5 N _1=4 N _2$

$\frac{4 N _2}{5}(32)= N _2 x _{ R }$

$x _{ R }=\frac{128}{5}=25.6 cm$