Objects on inclined planes and the normal force [closed]

Question

Say we had an object lying on a inclined plane, at an angle of $\theta$ to the horizontal, and the object has a mass of $m$. If we take the obejct to have an acceleration of 0 perpendicular to the plane, (i.e it accelerates down the slope), we can conclude that $mg\cos\theta$ = $N$ where $N$ is the normal reaction, however I have seen online the expression $N\cos\theta = mg$ and I would like a detailed explanation on why that is, where does the second formula come from and why does it not "agree" with the first one (I dont think they can both be true at the same time thats why I am confused), many thanks in advance!!!

Answer 1

If a car is travelling round a smooth banked curve at an angle $\theta$ to the horizontal and is at a constant height on the bank then its vertical acceleration is zero and by resolving forces vertically we can conclude that

$N \cos \theta = mg$

This may be where you have seen the other expression.

Answer 2

Only gravity $w$ and the normal force $N$ are involved. Set up Newton's 1st law in the perpendicular direction (let's call it the $y$ direction):

$$\sum F_y=0\quad\Leftrightarrow\quad -w_y+N=0 \quad\Leftrightarrow\quad N=w_y.$$

Now, what is $w_y$? Write up the right-angled triangle and you will see that $w_y$ constitutes the adjacent side from the angle $\theta$ in that triangle. Thus, you use the cosine function to retrieve it:

$$w_y=w\cos(\theta)=mg\cos(\theta).$$

So, your first-mentioned expression is correct. Not the second-mentioned one.

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We can do a quick check to see why the second-mentioned expression isn't true.

Let's try to imagine the $y$ axis not perpendicular to the surface but vertical. If we used Newton's 1st law again just like before, then we might expect to get something like this:

$$\sum F_y=0\quad\Leftrightarrow\quad -w+N_y=0 \quad\Leftrightarrow\quad w=N_y.$$

Draw up the right-angled triangle in this scenario, and you will see that $N_y$ is the adjacent side to the angle, so $N_y=N\cos(\theta)$, and we would expect the following result which is your second-mentioned expression: $$w=mg=N\cos(\theta).$$

But note: This second result is incorrect. Because in fact, Newton's 1st law does not apply here. By choosing a vertical direction to resolve the forces along for Newton's law, then there will be a small acceleration component. Only in the direction perpendicular to the acceleration - which was the first scenario - will there be no acceleration component at all, so only then will Newton's first law apply. In this second scenario we actually should have used Newton's 2nd law:

$$\sum F_y=ma_y\quad\Leftrightarrow\quad -w+N_y=ma_y \quad\Leftrightarrow\quad w=mg=N\cos(\theta)+ma\cos(\theta).$$

So, we see that a term was missing from the second expression before it would be correct.

In some other scenarios, where the angle was measured differently or where the acceleration is angled differently, it is possible to achieve your second expression. The other answers provide such other examples.

Answer 3

The second thing you mentioned here that the $cosθ$ component of the normal force is equal to $Mg$ is not actually true because there will be still some net acceleration downwards arised from the two force vectors i.e Gravitational force and cos component of normal force which will depend on the angle of inclination and cause the object to accelerate downwards (along the inclination).

Else it can be stated as $N=Mg$ when $θ$ is absolute 0. The case which you might have seen in which $Ncosθ=Mg$ is already mentioned in other answers which deals with banking of road.