A block of mass `m` rests on a horizontal floor for which coefficient of friction is `mu` . Find the magnitude and the direction of force that should be applied to just move the block so that the applied force is minimum.

Explanation

(a) If force is applied horizontally
`(f_(1))_(max) = mu N = mu mg` (i)
(b) If the block is pushed by applying a force at an angle with horizontal
`N = mg + F sin theta`
`(f_(2))_(max) = mu N = mu (mg + F sin theta)` (ii)
(c ) If the block is pulled by applying a force at an angle with horizontal
`N + F sin theta = mg`
`rArr N = mg - F sin theta`
`(f_(3))_(max) = mu N = mu N = mu(mg - F sin theta)` (iii)
`(f_(3))_(max) lt (f_(1))_(max) lt (f_(2))_(max)`
To move a block , the friction is minimum , when the block is pulled by applying a force at an angle with horizontal
`N + F sin theta = mg rArr N = mg - F sin theta`
The block is about to move if
`F cos theta = f_(max) = mu N = mu(mg - F sin theta)`
`F = (mu mg)/(cos theta + mu sin theta)`
Let the denominator `k = cos theta + mu sin theta`
For `F` to be minimum , `k` must be maximum.
`rArr (dk)/(d theta) = - sin theta + mu cos theta = 0`
`tan theta = mu rArr theta = tan^(-1) (mu)`
`sin theta = (mu)/(sqrt(1 + mu^(2))`
`cos theta = (mu)/(sqrt(1 + mu^(2))`
`F_(min) = (mu mg)/((1)/(sqrt(1 + mu^(2))) + (mu xx mu)/(sqrt(1 + mu^(2)))) = (mu mg)/(sqrt( 1+ mu^(2)))`