# How do gravity and normal force balance on an inclined plane?

Consider a block placed on an inclined plane. Drawing a free body diagram and applying Newtons laws, I get $N\cos A=mg$ as well as $N=mg\cos A$. Here one equation contradicts the other, could anyone explain this?

• NcosA ≠ mg, else there would be no component of force in vertical direction, resulting in the body remaining stationary always. That's not the case here as you don't have friction Oct 25 '15 at 15:24

Your mistake is that your coordinate systems are a bit messy. The two components are not drawn along coordinate axes. And if they are, then that coordinate system has axes that are not perpendicular.

• gravity $mg$ and the component $N\cos A$ will cancel out, and then the parallel component of the normal force is left and will cause acceleration down the incline.
• the normal force $N$ and the component $mg \cos A$ will cancel each other out, and then the rest of the gravity is a component causing an acceleration down the incline.

You can consider each case separately, since you are allowed to set the coordinate system as you want. But you cannot combine them because then you have a coordinate system in which the axis directions are not perpendicular. Then each axis interferes with the other and does not uniquely describe the components along it.

• I see, you have "the component" written in front of $N\cos A$. My mistake. You should write "perpendicular component" as I scanned over that and didn't parse it. I'll delete my previous comment Oct 25 '15 at 10:35

Well, the problems that you are having here is that you are assuming that the forces will balance. Its true that $N = mg\cos (A)$, since the block cannot move perpendicular (normal) to the plane. However, the block can move parallel to the plane, meaning that there can be a net force in the horizontal and vertical directions (if there is no friction), and the block will slide down the plane. Therefore, the two forces, $N\cos(A)$ and $mg$ are in general not equal.

We know that $N = mg\cos(A)$ since the block cannot move perpendicular to the wedge. Therefore, we can write $N\cos(A) = mg\cos^2(A)$. We can see that these two expressions aren't equal unless the wedge is flat $A=0$. In this case, we know that the block will not have a net acceleration, since the wedge has no angle. For $A>0$, this is not true, and we do expect the block to accelerate down the wedge due to the force of gravity.
But here : There is a component of acceleration in direction chosen (vertical) therefore the equation becomes $$mg-N\cos A=mg-mg\cos A. \cos A=mg\sin^2 A$$