Question

Two wedges, each of mass $m$, are placed next to each other on a flat horizontal floor. A cube of mass $M$ is balanced on the wedges as shown in the figure. Assume no friction between the cube and the wedges, but a coefficient of static friction $\mu <1$ between the wedges and the floor, What is the largest $M$ that can be balanced as shown without motion of the wedges?

• A. $\frac{m}{\sqrt{2}}$
• B. $\frac{\mu m}{\sqrt{2}}$
• C. $\frac{\mu m}{1-\mu }$
• D. $\frac{2\mu m}{1-\mu }$

Answer 1

The correct option is D $\frac{2\mu m}{1-\mu }$

$2N\cos \theta =Mg$ [$\theta ={45}^{\circ }$]

$\sqrt{2}N=Mg$

$N=\frac{Mg}{\sqrt{2}}$

$f=\mu (N\cos \theta +mg)$

$f=\mu (\frac{Mg}{\sqrt{2}}\times \frac{1}{\sqrt{2}}+mg)=\mu g(\frac{M}{2}+m)$

If wedge is balanced then

$F=N\sin \theta =\frac{Mg}{\sqrt{2}}\sin {45}^{\circ }=\frac{Mg}{2}$

$\frac{\mu Mg}{2}+\mu mg=\frac{Mg}{2}$

$\mu mg=\frac{Mg}{2}$$-\frac{\mu Mg}{2}$

$M=\frac{2\mu m}{(1-\mu )}$