Question

A wedge of mass $M=2kg$ is placed over a rough horizontal surface having coefficient of friction $\mu ,$ a block of mass $m=1kg$ is placed over this wedge as shown in figure. If coefficient of friction between wedge and block is ${\mu }_1=0.8$ , then the minimum value of $\mu$ so that wedge does not slip over a horizontal surface, is

• A. $0.8$
• B. $0.75$
• C. $0.5$
• D. zero

Answer 1

The correct option is D zero

$\begin{array}{c}a=\frac{mg\sin {37}^0-\mu mg\cos {37}^0}{m}\\ =g\left[\frac{3}{5}-0.8\times \frac{4}{5}\right]\\ =negative\\ \therefore theblockwillnotmove\\ \therefore evenif\mu =0thewedgewillnotslip\\ Hence,\\ optionDiscorrectanswer.\end{array}$