Two block A of `6 kg` and B of `4 kg` are placed in contact with each other as shown in figure

There is no friction A and ground and between both the blocks .The coefficient of friction between B and ground is `0.5` A horizontal force `F` is applied on A find the minimum and maximum value of `F` which can be applied so that both blocks can move combinely without any relative between them.

Options

A. `10N, 50 N`
B. `12N, 50 N`
C. `12N, 75 N`
D. None of these

Answer

c

Explanation

`F - N sin 37^(@) = 6a rArr F - (3N)/(5) = 6a`…(i)

`N_(2) = 4g - N cos 37^(@) = 40 - (4N)/(5)`…(ii)
`N sin 37^(@) - f = 4a`…(iii)
From Eqs (i) and (iii) `F - f= 10a`
`rArr F - m N_(2) = 10a`….(iv)
`rArr F - m[40 - (4N)/(5)] = 10a`....(v)
Put the value of `N` from Eqs (i) in (v) and also put the value of `m` to get `a = (5F - 60)/(42)`
Now to start motion `a gt0 rArr F gt 12 N`
So the minimum force `F` to just start the motion is `12 N`
Now miximum `F` will be when `N_(2)` just becomes zero Then from Eqs(ii) `N = 50 N`
From Eqs (i) and (iv) we get `F = 75 N` (by putting `N_(2) = 0)` If we apply `F gt 75 N ` then B will start sliding up on A but we do want this