A cubical block of side 'a' is held at rest, against a rough vertical wall by applying a force `F` acting along the centre. The mass of the block is `m`. Taking acceleration due to gravity as 'g' , determine
`(i)` The minimum coefficient of friction between the block and wall
`(ii)` The torque by normal reaction about the centre of mass

Explanation

The situation is shown in the figure.

Let `C` be the cenre. For equilirbium of the block
`sumF_(x)=0`, `sumF_(y)=0`, `sumtau=0`
`impliesF-N=0`, `f_(s)=mg`, `f_(s)xx(a)/(2)+tau_(N)=0`
Here, clockwise torque has been taken as negative.
Note: Normal reaction due to the wall `N`, does not pass through the centre `C`.
Now, `N=F`
`f_(s)=mg`
`(i)` As `f_(s) le muN`
`impliesmg le mu F`
or `mu ge (mg)/(F)`
`(ii)` `-f_(s)(a)/(2)+tau_(N)=0`
`tau_(N)=f_(s)xx(a)/(2)=mg.(a)/(2)` (in counter clockwise sense)