Question

Two wooden blocks are moving on smooth horizontal surface such that the mass $m$ remains stationary with respect to the block of mass $M$ as shown. Find the force exerted by the $M$ on $m$.

• A. $(M+m)g\tan \beta$
• B. $Mg\tan \beta$
• C. $mg\sec \beta$
• D. $(M+m)g\sec \beta$

Answer 1

The correct option is C $mg\sec \beta$

As per the question we have to find the normal force between the blocks
FBD of $m$

Let the system accelerate with $a$
Since, $m$ remain at rest w.r.t. $M$
$\Rightarrow mg\sin \beta =ma\cos \beta \Rightarrow a=g\tan \beta$

Resolving the forces perpendicular to the plane and solving the equation for that, we get
$N=mg\cos \beta +ma\sin \beta$
$=mg\cos \beta +m\sin \beta \left(g\tan \beta \right)$
$=mg\cos \beta +\frac{mg{\sin }^2\beta }{\cos \beta }$
$=\frac{mg}{\cos \beta }=mg\sec \beta$
Thus, the force exerted by the $M$ on $m$ is $mg\sec \beta$