# Confusion on Normal force and its resolution on inclined plane

I had the following doubt regarding normal forces and its resolution on inclined plane. (Also I am quite new to SE, so I have no idea how do we upload diagrams to better represent my arguement).

Consider two cases, in both cases there is a inclined plane with some angle of inclination ($$\theta$$), on which there is a block of mass m that is kept. In Case I, we fix our coordinate axes such that the x-axis lies on the base of the inclined plane, and the y-axis perpendicular to x-axis, and in the next case the x-axis is placed along the inclined plane and the y-axis is perpendicular to the x-axis.

The problem I was facing was that if we consider the normal force in case I and say that the normal force is $$mg cos(\theta)$$ then this is usually accompanied by the reasoning that there is no acceleration in the direction of normal force, therefore net force in that direction is 0, and hence acceleration is 0 as well.

But in this coordinate plane,saying that acceleration normal to the inclined plane is 0, is just like saying that its rectangular components is also 0, because if we have a vector whose magnitude is 0, then its rectangular components must also have a 0 magnitude. So in one sense we will be saying that the horizontal component of the block's acceleration is also 0, which is not true.

But in the second case, where we have a inclined coordinate system, we can easily make such a assumption that normal force is mgcos($$\theta$$) because in the inclined coordinate system, due to the normal force being perpendicular to the x-axis (which we placed on the inclined plane), it doesn't have any rectangular x-component and thus we can easily conclude the normal force to be mgcos($$\theta$$).

I don't understand how different coordinate system can predict different normal forces? Where exactly am I going wrong?

## 3 Answers

So in one sense we will be saying that the horizontal component of the block's acceleration is also 0

This would be true if you were to replace 'horizontal component' by 'contribution to the horizontal component'.

However there is an acceleration down the plane, and that has a horizontal component.

The net force normal to the plane is 0 not $$mg cos(\theta)$$. The block moving on a one-dimensional plane has only one degree of freedom - along the plane. And the force along the plane is $$mg sin(\theta)$$

I don't understand how different coordinate system can predict different normal forces?

They don't, so your confusion is warranted.

Let's break it down in each case. First, the typical scenario where the axes are aligned with the inclined plane at angle $$\theta$$ where $$\hat x$$ is parallel to the incline and $$\hat y$$ is perpendicular to the incline. Then for each direction we have Newton's second law telling us that

$$mg\sin\theta=ma$$ $$N-mg\cos\theta=0$$

Notice that $$N$$, the magnitude of the normal force, is not accompanied with a trigonometric function. This is because the normal force points entirely along the y-axis, and so it has a component of $$N$$ in the y-direction and a component of $$0$$ in the x-direction. On the other hand, the weight force with magnitude $$mg$$ is not aligned with the axes, and in this case it has a component of $$mg\sin\theta$$ in the x-direction and a component of $$-mg\cos\theta$$ along the y-direction. And finally, since the acceleration is along the incline, it has a component of $$a$$ in the x-direction and $$0$$ in the y-direction.

Now, let's look at the same scenario where the axes are aligned such that $$\hat x$$ is horizontal and $$\hat y$$ is vertical. Then for each direction we have Newton's second law telling us that

$$N\sin\theta=ma\cos\theta$$ $$N\cos\theta-mg=-ma\sin\theta$$

Notice that $$N$$, the magnitude of the normal force, is now accompanied with trigonometric functions. This is because the normal force points perpendicular to the incline, and so it has components of $$N\sin\theta$$ in the x-direction and a component of $$N\cos\theta$$ in the y-direction. On the other hand, the weight force with magnitude $$mg$$ is aligned with the axes, and in this case it has a component of $$0$$ in the x-direction and a component of $$mg$$ along the y-direction. And finally, since the acceleration is along the incline, it has a component of $$a\cos\theta$$ in the x-direction and $$-a\sin\theta$$ in the y-direction.

Both of these sets of equations are saying the same thing. We can show this by taking the first set of equations and multiplying the first by $$\cos\theta$$ and the second by $$\sin\theta$$

$$mg\sin\theta\cos\theta=ma\cos\theta$$ $$N\sin\theta-mg\sin\theta\cos\theta=0$$

Now add them together: $$N\sin\theta=ma\cos\theta$$

This is the first equation from the second case.

And then, go back to the first set of equations and multiply the first by $$-\sin\theta$$ and the second by $$\cos\theta$$

$$-mg\sin^2\theta=-ma\sin\theta$$ $$N\cos\theta-mg\cos^2\theta=0$$

Now add them together: $$N\cos\theta-mg=-ma\sin\theta$$

This is the second equation from the second case.

(Note that there is a faster way to do this using rotation matrices all at once)

So, in conclusion, $$N=mg\cos\theta$$ is always true for this system. It does not depend on the axes you use to analyze it.