A block of mass `m` rests on a horizontal floor for which coefficient of friction is `mu` . Find the magnitude and the direction of force that should be applied to just move the block so that the applied force is minimum.

Options

A. `mumg`
B. `(sqrt(1+mu^(2)))/(mu)mg`
C. `musqrt(1+mu^(2))mg`
D. `(mu mg)/(sqrt(1+mu^(2)))`

Answer

D

Explanation

Suppose the force F is applied at an angle `theta` with the horizontal as shown in adjoining figure For vertical equilibrium,

`R+F sin theta=mg`
`"or "R=mg-F sin theta" ….(i)"`
While for horizontal motion
`F cos theta ge f_(1)" or "F cos theta ge muR" ......(ii)"`
From eqns. (i) and (ii), we get,
`F cos theta ge mu (mg-F sin theta)`
`"or "F ge(mum g)/((cos theta+mu sin theta))`
For the force F to be the minimum `(cos theta+mu sin theta)`
must be maximum.
`"i.e, "(d)/(d theta)(cos theta+mu sin theta)=0`
`"or "-sin theta+mu cos theta=0`
`"i.e, "tan theta =mu or theta=tan^(-1)(mu)`

`therefore" "sin theta =(mu)/(sqrt(1+mu^(2)))and cos theta=(1)/(sqrt(1+mu^(2)))`
`therefore" "F ge (mumg)/((1)/(sqrt(1+mu^(2)))+(mu)/(sqrt(1+mu^(2))))`
`therefore" "F_("min")=(mumg)/(sqrt(1+mu^(2)))`