Question

Wedge is fixed on a horizontal surface. Triangular block A of mass $M$ is pulled upward by applying a constant force $F$ parallel to the incline of the wedge as shown in the figure and there is no friction between the wedge and the block A, while coefficient of friction between A and block B of mass $m$ is $\mu .$ If there is no relative motion between A and B, then frictional force developed between A and B is

• A. $\left[\frac{F+(m-M)gsin\theta }{m+M}\right]mcos\theta$
• B. $\mu mg$
• C. $\left[\frac{F-(m+M)gsin\theta }{m+M}\right]mcos\theta$
• D. $\frac{\mu mg}{2}$

Answer 1

The correct option is C $\left[\frac{F-(m+M)gsin\theta }{m+M}\right]mcos\theta$



Taking all three block as system, consider the force balance along the direction of the wedge surface

$F-Mgsin\theta -mgsin\theta =(M+m)a$

where $a$ is the net acceleration of the two blocks along the wedge surface.

$a=\left[\frac{F-(m+M)gsin\theta }{m+M}\right]$

Now consider the forces on block B in reference frame of B


Considering the frame of reference of block B, $ma$ is the pseudoforce (net force acting on block B as found in the previous step). Since friction is static (there is no relative motion between A and B), it must balance the horizontal component of force acting on B.

So, $f=macos\theta$
$=\left[\frac{F-(m+M)gsin\theta }{m+M}\right]mcos\theta$