In the situration shown in the figure the friction coefficient between `M` and the horizontal surface is `mu`. The force `F` is applied at an angle `theta` with vertical. The correct statements are

Options

A. If `theta gt tan^(-1)mu` the block cannot be pushed forward for any value of `F`
B. If `theta lt tan^(-1)mu` the block cannot be pushed forwards for any value of `F`
C. As `theta` decreases the magnitude of force needed to just push the block `M` forward increases
D. None of these

Answer

B, C

Explanation

`N = F cos theta+Mg` …(i)
`f_(max)=(mu(M-2m))/(3)`……(ii)
To just push the block `F sin theta =f_(max)`
`rArr F sin theta mu (F cos theta+Mg)rArr F=(muMg)/(sin theta-mu cos theta)`
`rArr sin theta-mu cos theta gt 0 rArr tan theta gt mu`
`rArr tan theta gt tan (tan^(-1)mu)rArr theta gt tan^(-1)mu`
Hence the block can be pushed forward only if `theta gt tan^(-1)mu`.
Again as `theta` decreases `sin theta` decrease while `cos theta` increases, therefore, `sin theta-mu cos theta`
decreases. Hence `(muMg)/(sin theta-mu cos theta)` increases.