# If a body slides down a frictionless inclined plane what will be the net normal force?

If a body(m) slides down a frictionless inclined plane (M), will the net normal force between the ground and the inclined plane be (M+m)g ?

I feel it should be less than (M+m)g. This is because one of the component of the weight of the body(m) which is mgsinΘ does not need to be cancelled by the normal force. So, the normal force should be less than (M+m)g by mgsinΘ.

Actually I had a question in my book which said it should be (M+m)g. So, is it correct ? How ?

• Limit cases are often useful for checking. -> Imagine the inclined plane is indeed vertical... – Fabrice NEYRET Oct 8 '15 at 16:59
• Take the limit as $\theta \rightarrow 0$ and the limit as $\theta \rightarrow \pi/2$. In the former case, you should have the combined effect of both the body and the inclined plane. While in the latter case, the body would just free fall and thus could not exert a normal force on the inclined plane. – honeste_vivere Oct 8 '15 at 23:33

• Your net acceleration vector should be along the plane, not parallel to the surface on which the plane rests. The component of $m \ \mathbf{g}$ that does not cancel with $\mathbf{N}$ is parallel to the inclined plane surface. Since this is equivalent to a net force, there must be an acceleration along that direction. – honeste_vivere Oct 8 '15 at 23:29