Question

Prism 1 with bar 2 of mass $m$ placed on it gets a horizontal acceleration $w$ directed to the left (shown in figure above). At what maximum value of this acceleration will the bar be still stationary relative to the prism, if the coefficient of friction between them $k<\cot \alpha$?

Answer 1

Let us fix (x-y) coordinate system in the frame of ground as shown in figure below. From second law of motion in projection form along y and x axes,
$mg\cos \alpha -N=mw\sin \alpha$
$N=m(g\cos \alpha -w\sin \alpha )$ ......(1)
$mg\sin \alpha +fr=mw\cos \alpha$

$fr=m(w\cos \alpha -g\sin \alpha )$ ......(2)

But $fr\le kN$, so from (1) and (2)
$(w\cos \alpha -g\sin \alpha )\le k(g\cos \alpha -w\sin \alpha )$
$w(\cos \alpha +k\sin \alpha )\le g(k\cos \alpha +\sin \alpha )$
$w\le g\frac{(k\cos \alpha +\sin \alpha )}{\cos \alpha +k\sin \alpha }$,
So, the sought maximum acceleration of the wedge,
$w_{max}=\frac{(k\cos \alpha +\sin \alpha )g}{\cos \alpha +k\sin \alpha }=\frac{(k\cot \alpha +1)g}{\cot \alpha +k}$ where $\cot \alpha >k$