Find the minimum force required to keep a block of mass `m` in equilibrium on a rough inclined plane with inclination `alpha` and coefficient of friction `mu( lt tan alpha)`.

Explanation

In order to achieve a minimum of `F`, it should be directed as shown in the figure. The value of `theta` can be found using the `FBD` of the block.
Normal to the incline,
`N=mg cos alpha+F sin theta`…..(`i`)
Along the incline
`muN+F cos theta= mg sin alpha`....(`ii`)
From (`i`) and (`ii`) , we get
`F=(mg(sin alpha- mu cos alpha))/(cos theta+mu sin theta)`
For `F` to be minimum
`(d)/(d theta)(cos theta+mu sin theta)=0`
`rArr mu=tan theta`
`F_(min)=(mg(sin alpha-mu cos alpha)sqrt(1+mu^(2)))/(1+mu)`