Consider the situation as shown in the figure. The coefficient of the friction is `0.5` for all surfaces.
(a) In figure (i) , if `F = 15 N` , find the friction force acting on the upper block.
(b) In figure (ii) , find the minimum value of `F` to move the lower block. The upper block is connected to a rod and held at rest. Find the force applied by rod on the upper block.

Explanation

(a) `uarr : N_(0) - 50 = 0`
`rArr N_(0) = 50 N`
`f_(max) = mu N_(0)`
`= 0.5 xx 50 = 25 N`
Since the applied force `15 N lt 25 N`
The system will not move . The upper block is at rest hence the friction on it is zero.
(b) `2 kg : uarr : N_(0) = 20 N`
`(f_(1))_(max) = mu N_(0) = 0.5 xx 20`
`= 10 N`
`T = (f_(1))_(max) = 10 N`
`3 kg : uarr : N' - 30 - N_(0) = 0`
`rArr n' = 50 N`
`(f_(2))_(max) = mu N' = 0.5 xx 50`
`= 25 N`