A blocky of mass m rests on a horizotal 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 be applies.

Options

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

Answer

D

Explanation

Suppose the force F is applied at an angle θ with the horizontal as shown in adjoining figure. For vertical equilibrium,
`R=Fsintheta=mg`
or `R=mg-F sintheta` . . . (i)
Whiite for hotizontal motion
`F cos theta ge f_(L) or F cos theta ge muR` . . . (ii)
From eqns. (i) and (ii), we get
`F costheta gemu(mg-F sintheta)`
or `Fge(mumg)/((costheta+musintheta))`
For the force F to be minimum `(costheta+musintheta)` must e maximum,
i.e., `(d)/(d theta)(costheta+musintheta)=0`
or `-sintheta=mucostheta=0`
i.e., `tantheta=mu or theta=tan^(-1)(mu)`
`therefore sintheta=(mu)/(sqrt(1+mu^(2))) and costheta=(1)/(sqrt(1+mu^(2)))`
`F ge (mumg)/((1)/(sqrt(1+mu^(2)))+(mu^(2))/(sqrt(1+mu^(2))))`
`F_("min")=(mumg)/(sqrt(1+mu^(2)))` .