Question

If the coefficient of friction between $A$ and $B$ is $\mu$, the maximum horizontal acceleration of the wedge $A$ for which $B$ will remain at rest w.r.t the wedge is :

• A. $\mu\, g$
• B. $g\left(\frac{1+\mu}{1-\mu}\right)$
• C. $\frac{ g }{\mu}$
• D. $g \left(\frac{1-\mu}{1+\mu}\right)$

Answer 1

Answer: (B)

$N=m g \cos 45^{\circ}+m a \sin 45^{\circ}$
$N=\frac{m g+m a}{\sqrt{2}}$
$ma \cos 45^{\circ}=m g \sin 45^{\circ}+\mu N\,\,\,...(2)$
Put the value of $N$
$\frac{m a}{\sqrt{2}}=\frac{m g}{\sqrt{2}}+\frac{\mu(m g+m a)}{\sqrt{2}} $
$a=g\left(\frac{1+\mu}{1-\mu}\right)$