Question

If the coefficient of friction between $A$ and $B$ is $\mu$, the minimum 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

The correct option is D $g\left(\frac{1-\mu }{1+\mu }\right)$

Solving in wedge's frame,
To keep block $B$ at rest, from FBD of block $B$ we get

$ma\sin {45}^{\circ }+f_s=mg\sin {45}^{\circ }$ ... (1)
$R=mg\cos {45}^{\circ }+ma\cos {45}^{\circ }$ ... (2)
From (1) & (2) and using $f_s={\mu }_{s}R$
$ma\sin {45}^{\circ }+\mu (mg\cos {45}^{\circ }+ma\cos {45}^{\circ })=mg\sin {45}^{\circ }$
$ma+\mu (mg+ma)=mg$
$a(1+\mu )=g(1-\mu )$
$a=g\left(\frac{1-\mu }{1+\mu }\right)$