Question

What is the minimum acceleration with which bar $A$ (shown in figure above) should be shifted horizontally to keep bodies 1 and 2 stationary relative to the bar? The masses of the bodies are equal and the coefficient of friction between the bar and the bodies is equal to $k$. The masses of the pulley and the threads are negligible, the friction in the pulley is absent.

Answer 1

Bodies 1 and 2 will remain at rest with respect to bar $A$ for $w_{min}\le w\le w_{max}$, where $w_{min}$ is the sought minimum acceleration of the bar. Beyond these limits there will be a relative motion between bar and the bodies. For $0\le w\le w_{min}$, the tendency of body 1 in relation to the bar $A$ is to move towards right and is in the opposite sense for $w\ge w_{max}$. On the basis of above argument the static frictions on 2 by $A$ is directed upward and on 1 by $A$ is directed towards left for the purpose of calculating $w_{min}$.
Let us write Newton's second law for bodies 1 and 2 in terms of projection along positive x-axis. (figure shown below).
$T-f{r}_1=mw$ or, $f{r}_1=T-mw$ (1)
$N_2=mw$ (2)
As body 2 has no acceleration in vertical direction, so
$f{r}_2=mg-T$ (3)
From (1) and (3)
$(f{r}_1+f{r}_2)=m(g-w)$ (4)
But $f{r}_1+f{r}_2\le k(N_1+N_2)$
or $f{r}_1+f{r}_2\le k(mg+mw)$ (5)
From (4) and (5)
$m(g-w)\le mk(g+w)$, or $w\le \frac{g(1-k)}{(1+k)}$
Hence $w_{min}=\frac{g(1-k)}{(1+k)}$