Question

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 (Fig. 1.62). 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 ωmax of the axis of the cylinder rolling down the inclined plane?

• A. $\frac{2kg}{2-3k}$
• B. $\frac{3kg}{k-1}$
• C. $\frac{2kg}{2+3k}$
• D. $\frac{3kg}{2-3k}$

Answer 1

The correct option is A $\frac{2kg}{2-3k}$

Here

$f{r}_1+f{r}_2=m{w}_c\to \left(1\right)N_1+N_2-mg-F=0N_1+N_2=mg+t\to \left(2\right)$

Now

$N_{cz}=I_c{B}_z$

We get

$FR-(f{r}_1+f{r}_2)R=\frac{m{R}^2}{2}R=\frac{m{R}^2}{2}\frac{w_c}{R}\to \left(3\right)$

(as for pure rolling)

Solving equations

$F\ge \frac{3kmg}{(2-3k)}$ or $f_{max}=\frac{3kmg}{2-3k}$

$w_{c\left(max\right)}=\frac{k(N_1+N_2)}{m}=\frac{k}{m}[mg+f_{max}]=\frac{k}{m}[mg+\frac{3kgm}{2-3k}]=\frac{2kg}{2-3k}$