In `Q` . 120, the minimum force `F` required to pull it. `(mu=(1)/(2))`is:

Options

A. `(mg)/(2)`
B. `(mg)/(2)costheta`
C. `(mg)/sqrt(5)`
D. None of these

Answer

C

Explanation

For minimum force the object should be pulled at angle equal to angle of friction and applied force should be equal to
implies `tantheta=mu=(1)/(2)`
hence `costheta=(2)/(sqrt(5))` and `sintheta=(1)/(sqrt(5))`
Hence applied force `F=(mg)/(sqrt(5))` .