What is the minimum acceleration with which bar A (figure) 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.

Explanation

Bodies 1 and 2 will remain at rest with respect to bar A for `w_(min)lewlew_(max)`, where `w_(min)` is the sought minimum acceleration of the bar. Beyond the limits there will be a relative motion between bar and the bodies. For `0lewlew_(min)`, the tendency of body 1 in relation to the bar A is to move towards right and is in the opposite sense for `wgew_(max)`. On the basis of above argument the static friction 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).
`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_2lek(N_1+N_2)`
or `f r_1+f r_2lek(mg+mw)` (5)
From (4) and (5)
`m(g-w)lemk(g+w)`, or `wge(g(1-k))/((1+k))`
Hence `w_(min)=(g(1-k))/((1+k))`