Question

If the coefficient of friction between the wedge and the block is $\mu$, then the maximum acceleration of the wedge for which the block will remain at rest with respect to it is

• A. $2g\left(\frac{1+\mu }{1-\mu }\right)$
• B. $g\left(\frac{1+\mu }{1-\mu }\right)$
• C. $g\left(\frac{1-\mu }{1+\mu }\right)$
• D. $2g\left(\frac{1-\mu }{1+\mu }\right)$

Answer 1

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

FBD of block with respect to accelerating frame i.e. wedge is: -


At equilibrium,
$N=mg\cos {45}^{\circ }+ma\sin {45}^{\circ }$ .......(1)
& $f+mg\sin {45}^{\circ }=ma\cos {45}^{\circ }$ .......(2)
$\Rightarrow \mu N=ma\cos {45}^{\circ }-mg\sin {45}^{\circ }$
[for maximum acceleration, we have to take limiting case]
$\Rightarrow \mu (mg\cos {45}^{\circ }+ma\sin {45}^{\circ })=ma\cos {45}^{\circ }-mg\sin {45}^{\circ }$
[from (1)]
$\Rightarrow a=\frac{\mu g\cos {45}^{\circ }+g\sin {45}^{\circ }}{\cos {45}^{\circ }-\mu \sin {45}^{\circ }}$
$\Rightarrow a=g\left(\frac{1+\mu }{1-\mu }\right)$