Calculate the Force (F) required to cause the block of mass `m_(1) = 20 kg ` just to slide under the block of mass `m_(2) = 10 kg` [ coefficient of friction `mu = 0*25` for all surfaces ]

Explanation

The lower block will experience following forces :

1. Weight `m_(1)g`
2. `N_(1)` normal contact from floor and `"fr"_(1)` friction force by floor .
3. `N_(2)` normal contact and `"fr"_(2)` friction by upper block. Reaction of these forces will be on upper block.
4. Force F applied to move block horizontally .
The upper block will experience following forces
(i) Tension force due to string
(ii) `N_(2) and " fr" _(2)` contact force by lower block as mentioned above
(iii) Weight
For equilibrium of two blocks
`N_(1) = N_(2) +m_(1)g " " `
` F = "fr"_(1) + "fr"_(2) " "` ...(ii)
` T cos 53^(@) = "fr"_(2) " " ` ...(iii)
` N_(2) + T sin 53^(@) = m_(2)g " "` ....(iv)
For limiting case ` "fr"_(1) = muN_(1) and "fr"_(2) = muN_(2)`
Solving above equations
`N_(2) = 75`
` N_(1)= 275`
` F = mu (N_(1) +N_(2))= 87*5` N.