An ideal string is wrapped several times on a solid cylinder of mass 4 kg and radius 1 m. The pulleys are ideal and the surface between block and ground is smooth. If the torque acting on the cylinder is $\frac{80}{n}$ N-m, then find the value of n. 
 

 

FBD of cylinder and block are as shown by Newton’s laws 
$40-2T-f_s=8a$     (1)
 $T-f_s=4a/2$    (2) 
Subtracting equation (2) from equation (1) 
  $40-3T=6a$
 
  $T=(40-6a)/3$
Also, by $\tau =I\alpha$ , we get 
 $T\times R+f_s\times R=I\frac{3a}{2R}$
  $\Rightarrow T+f_s=3Ia/2R^2$
 $\Rightarrow (40-6a)/3+(40-12a)/3=3Ia/2R^2\Rightarrow a=80/(18+9I/2R^2)$
As   $\tau =I\alpha =I\times (3a/2R)$
  $\therefore \tau =I\alpha =I\times (3a/2R)$
 $\therefore \tau =\frac{3I}{2R}\frac{80}{(18+\frac{9I}{2R^2})}\Rightarrow \tau =\frac{3\times 80}{18+9}=\frac{80}{9}N-m$