A particle of mass m rests on a horizontal floor with which it has a coefficient of static friction `mu`. It is desired to make the body move by applying the minimum possible force F. Find the magnitude of F and the direction in which it has to be applied.

Options

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

Answer

A

Explanation

`R=Fsin theta=mg`
`R=mg-F sin theta`
Force of friction, `f=muR=mu (mg-F sin theta)`
The block will move, when
`F cos ge f`
`F cos theta ge mu (mg-sin theta)`
`Fge(mu mg)/(cos theta+mu sin theta)` ....(i)
F will be minimum, when
`cos theta+ mu sin theta=` maximum , for which
`(d)/(d theta)(cos theta + mu sin theta)=0`
or `-sin theta+ mu cos theta=0` or `mu cos theta=sin theta`
or `tan theta=mu` or `theta=tan^(-1)(mu)`
`therefore sin theta=(mu)/sqrt(1+mu^(2))` and `cos theta=(1)/sqrt(1+mu^(2))`
From (ii),
`Fge(mumg)/((1)/sqrt(1+mu^(2))+(mu^(2))/sqrt(1+_mu^(2)))`
`Fge(mumg)/sqrt(1+mu^(2)) thereforeF_("min")=(mumg)/sqrt(1+mu^(2))`