Question

A block of mass M is kept on as rough horizontal surface. The coefficient of static frictionbetween the block and the surface is `mu`. The block is to be pulled by applying a force to it. What minimum force is needed to slide the block? In which direction should this force act?

Answer 1

Let P bet he force applied to it an angle `theta` From the free body diagram,
`R+P isn theta -mg=0`
`rarr R=-Psin theta+mg`……….i
`muR=Pcostheta `……….ii
From equation i
`mu(mg-Psintheta)-Pcostheta`
`mumg=muPsintheta-Pcostheta`
`rarr P=(mu mg)/(musintheta+costheta)`
Applied force Pshould be minimum, when `m sintheta+costheta` is maximum.
Again `(musintheta+costheta)` is maximum when its derivative is zero
`d/(dtheta) (mu sintheta+costheta)=0` rarr mucostheta-sintheta=0`
rarr `theta=tan^-1 mu`
So, `P=(mumg)/(musintheta+costheta)`
`=(mumg/costheta)/((musintheta)/(costheta)+(costheta)/(costheta))`

(Dividing Numerator and DenomiN/Ator by `costheta), we get
`P=(mumgsectheta)/(1+mutantheta)`
`(mu mgsectheta)/(1+tan^2theta)=(mumg)/(sectheta)`
`=(mumg)/(sqrt(1+tan^2theta)(mu mg)/(1+mu^2)`
Hence minimum fore is `mu(mg)/(sqrt(1+mu^2)` at an angle `theta=tan^-1mu`