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.

Answer

A

Explanation

Let F be the force applied to move the body at an angle `theta` to the horizontal.
The body will move when `Fcos theta=mu N` …(i)
Applying equilibrium of forces in the vertical direction we get
`Fsin theta+N=mg`
`implies N=mg-Fsintheta` ...(ii)
`implies From (i) and (ii) `F=(mumg)/(cos theta+mu sintheta)` ....(iii)

Differentiating the above equation w.r.t. `theta`, we get
`(dF)/(d theta)=(mumg)/((costheta+musintheta)^2)[-sin theta+mu cos theta]=0`
`implies theta=tan^-1mu`
This is the angle for minimum force.
To find the minimum force substituting these values in equation (iii)
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`sin theta=(mu)/(sqrt(mu^2+1)`, `cos theta=(1)/(sqrt(mu^2+1))`
`F=(mumg)/((1)/(sqrt(mu^2+1))+(mu)/(sqrt(mu^2+1))xxmu)`
`implies F=(mu mg(sqrt(mu^2+1)))/(mu^2+1)=(mumg)/(sqrt(mu^2+1))`
`implies F=mg sin theta`