A uniform solid cylinder of mass `m` rests on two horizontal planks. A thread is wound on the cylinder. The hanging end of the thread is pulled vertically down with a constant force `F`.

Find the maximum magnitude of the force `F` which may be applied without bringing about any sliding of the cylinder, if the coefficient of friction between the plank and the cylinder is equal to `mu`. What is the maximum acceleration of the centre of mass over the planks?

Explanation

The force acting on the cylinder are a `mg`, downwards b. `F`, applied force downwards c `N`, reaction of each plank upwards d.`f_(r)` frictional force (not limiting) between each plank and the cylinder to the right.
Considered vertical motion
`F+mg=2N`.............i
Considering horizontal motion
`2f_(r)=ma`.........ii
where `a` is the horizontal acceleration of the centre of mass of the cylinder to the right. considering rotational dynamics of the cylinder about its axis.
`tau=FR-2f_(r)/R=(1/2mR^(2))xxalpha`
`(F-2f_(r))R=1/2mR^(2)xxa/R` (`:'=alpha=alphaR`)
`F-2f_(r)=1/2ma`
`F=1/2ma+2f_(r)=3/2ma`..........iii
[`:.2f_(r)=ma` by Eq. ii ]
`F=3f_(r)`
Now `f_(r)lef_(lim)` or `f_(r)lemu`(`:'f_("lin")=muN`)
`1/3Flemu(F+mg)/2`
`2Fle3muF+mumg`
`Fle(2mu mg)/(2-3mu)`
`F_(max)=(2mumg)/(2-3mu)`
When the force is maximum, acceleration is automatically maximum.
`a_(max)=(2F_(max))/(3m)=(2mug)/(2-3mu)`