Question

A cube of mass $m$ is placed on top of a wedge of mass $2m$ , as shown in figure. There is no friction between the cube and the wedge. The minimum co-efficient of friction between the wedge and the horizontal surface, so that the wedge does not move is

• A. $0.1$
• B. $0.2$
• C. $0.3$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

Answer: (B)

The normal reaction between the wedge and the block is
$N^{{}^{\prime }}=mgcos\theta$
So the net horizontal force on the wedge is
$F=mgcos\theta sin\theta$
The normal reaction between the wedge and the ground is
$N=2mg+mgco{s}^2\theta$
$N=2mg+mgco{s}^2\theta$
Now, $F=\mu N$ gives
$\mu =\frac{F}{N}=\frac{mgcos\theta sin\theta }{2mg+mgco{s}^2\theta }$
$=\frac{cos\theta sin\theta }{2+co{s}^2\theta }$
For $\theta =45^o,\mu =0.2$