Post
by **diliu** » Thu Nov 03, 2011 1:06 pm

Considering the axial symmetry of this system, we firstly vertically divide it into two equal parts, and discuss the left one.

Clearly, there is a left-horizontal fore $$$f$$$ exerts on the half-cube, to keep the half-system balanced, $$f$$ is equal to the friction comes from the floor, which is right toward.

Let the force between the half-cube and the left-wedge be $$$N$$$, which is perpendicular to the interface, to be balanced we have:

$$$N\cos(45^\circ)=\frac{1}{2}Mg$$$.

$$$N\sin(45^\circ)=f$$$.

For static fraction we need:

$$$f\leq\mu(m+\frac{1}{2}M)g$$$

Note that $$$\cos(45^\circ)=\sin(45^\circ)$$$, combining the equations above, we have:

$$$\frac{1}{2}Mg\leq\mu(mg+\frac{1}{2}Mg)$$$.

Thus we get, if $$$\mu\geq1$$$, this inequality is naturally assured, there is no constraint on the value of $$$M$$$; when $$\mu<1$$, we get the answer "D"