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(figure).
Find the maximum magnitude of the force F which still does not bring about any sliding of the cylinder, if the coefficient of friction between the cylinder and the planks is equal to k. What is the acceleration `w_(max)` of the axis of the cylinder rolling down the inclined plane?

Explanation

Let us sketch the force diagram for solid cylinder and apply Newton's second law in projection form along x and y axes (figure):
`f r_1+f r_2=mw_c` (1)
and `N_1+N_2-mg-F=0`
or `N_1+N_2=mg+F` (2)
Now choosing positive direction of `varphi` as shown in the figure and using `N_(cz)=I_cbeta_z`,
we get
`FR-(f r_1+f r_2)R=(mR^2)/(2)beta=(mR^2)/(2)(w_c)/(R)` (3)
[as for pure rolling `w_c=betaR`]. In addition to,
`f r_1+f r_2lek(N_1+N_2)` (4)
Solving the Eqs., we get
`Fle(3kmg)/((2-3k))`, or `F_(max)=(3kmg)/(2-3k)`
and `w_(c(max))=(k(N_1+N_2))/(m)`
`=k/m[mg+F_(max)]=k/m[mg+(3kmg)/(2-3k)]=(2kg)/(2-3k)`