Q.
In the figure shown, the block of mass $m$ is at rest relative to the wedge of mass $M$ and the wedge is at rest with respect to ground. This implies that
Laws of Motion
Solution:
$N=m g \cos \theta, f=m g \sin \theta$
Net force applied by $M$ on $m$ (or $m$ on $M$ ):
$F =\sqrt{N^{2}+f^{2}}$
$=\sqrt{(m g \cos \theta)^{2}+(m g \sin \theta)^{2}}$
$=m g$
