Question

Ithe coefficient of friction between A and B is $\mu$, the maximum acceleration of the wedge A for which B will remain at rest with respect to the wedge is

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

Answer 1

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

From the free body diagram,

Perpendicular to wedge,

$\Sigma f_y=(mgcos\theta +masin\theta -N)=0$

and

$\Sigma f_x=mgsin\theta +\mu N-macos\theta =0$ (for maximum $a$)

$⟹mgsin\theta +\mu (mgcos\theta +masin\theta )-macos\theta =0$

$\implies a=\dfrac{g\,sin\'theta+\mu g\,cos\theta}{cos\theta-\mu\,sin\theta}

For $\theta ={45}^0$

$a=g\left(\frac{tan{45}^0+\mu }{cot{45}^0-\mu }\right)=g\left(\frac{1+\mu }{1-\mu }\right)$