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 ?

  • $\begingroup$ Limit cases are often useful for checking. -> Imagine the inclined plane is indeed vertical... $\endgroup$ Commented Oct 8, 2015 at 16:59
  • $\begingroup$ 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. $\endgroup$ Commented Oct 8, 2015 at 23:33

1 Answer 1


Check out the solution in the image. It will be dependent on the inclination of the plane. Here, a is the horizontal acceleration of the inclined plane. solution

  • 1
    $\begingroup$ But the center of mass of your system is not at rest so it's wrong to say it's not moving down ! $\endgroup$ Commented Oct 8, 2015 at 16:54
  • $\begingroup$ 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. $\endgroup$ Commented Oct 8, 2015 at 23:29
  • $\begingroup$ I'm not concerned with the direction parallel to the plane, bcz I have balanced the forces perpendicular to the plane, in the changed reference. $\endgroup$
    – Arkya
    Commented Oct 9, 2015 at 5:05
  • $\begingroup$ I guess you have found the normal force between the inclined plane and the block. I had asked for N between wedge and ground. Given that you have found N(s), so is the N between wedge and ground N(s)+Mg ? Please explain. Thanks @ArkyaChatterjee $\endgroup$
    – user94848
    Commented Oct 11, 2015 at 16:09
  • $\begingroup$ You'll find the answer to your question in the solution I posted. I have clearly mentioned the relation between N and N_s $\endgroup$
    – Arkya
    Commented Oct 11, 2015 at 16:13

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