# Magnitude of acceleration of an object on inclined plane

## I want to know the acceleration of an object down an inclined plane without breaking the gravitational force vector into its components that are parallel and perpendicular to the plane.

Consider the following situation where the mass of the object is $$\text{m}$$, the angle of the inclined plane from the horizontal is $$\theta$$ and the normal force is represented as $$N$$ .

In this case, $$\vec{N}$$ can be broken down into its components : $$N_x\hat{i}$$ and $$N_y\hat{j}$$.

Here, $$N_y=N\cos\theta=\text{m}g$$.
From this, we find that, $$N_x=N\sin\theta=\text{m}g\tan\theta$$ and $$N=\frac{\text{m}g}{\cos\theta}$$.

Thus, the net force acting on the object at any instant is $$\text{m}g\tan\theta$$ in the direction of positive $$x$$ axis, but the object actually happens to slide because the gravitational force pulls the object in the direction of negative $$y$$ axis with the magnitude $$\text{m}g$$ as it happens to move in the direction of positive $$x$$ axis.

So how would we calculate the actual acceleration of the object down the plane in such an instance without breaking the gravitational force vector into its components that are parallel and perpendicular to the plane?

### Additional question : Why does the magnitude of the normal force differ in the case discussed above and when breaking the gravitational force vector into its components that are parallel and perpendicular to the plane?

Talking about our case, we get the magnitude of the normal force ($$N$$) to be $$\frac{\text{m}g}{\cos\theta}$$ while in the latter case, the magnitude ($$N$$) turns out to be $$\text{m}g\cos\theta$$.

I've learnt that the magnitude of a vector doesn't change even if we change our inertial frame of reference, so why does it differ here? Am I thinking something wrong?

Am I thinking something wrong?

The mistake in your thinking is this $$N_y=N\cos\theta=\text{m}g$$.

It assumes no vertical acceleration, but when the object slides down the plane, it is accelerating downwards.

An alternative way is to think of the mass sliding down the plane a distance $$d$$, the accelerating force $$F$$ down the plane does work $$Fd$$

The gravitational potential energy lost is $$mgd\sin\theta$$ (as the drop in height is $$d\sin\theta$$), and the normal force does no work on the mass, as it doesn't move in the direction of that force...

So

$$mgd\sin\theta = Fd$$

the accelerating force is $$F =mg\sin\theta$$ and the acceleration is $$g\sin\theta$$.

• I haven't yet learnt about potential/kinetic energy and work done, will be learning it soon, till that I need more help on this question from what I know. I don't understand why N_y can't be mg. From what I think, the downwards acceleration is due to gravity. The y component of the normal force causes horizontal acceleration on the right side and at the same instant, as the object moves horizontally without any vertical acceleration, it is pulled down by gravity causing it to slide down the plane. What am I thinking wrong here? Commented Oct 7, 2021 at 14:26
• @ silica 19 probably best to think of gravity as causing the object to press onto the plane as well as pull it down the plane. The pressing onto the plane causes the normal reaction force and that has a component to the right (and a component up). The net result is that the object accelerates down and also to the right, i.e. down the plane. The normal force is calculated from the fact that the component of the weight pulling it onto the plane is balanced by $N$ i.e. $N=mg\cos\theta$ All the best with it. Commented Oct 7, 2021 at 18:45
• I understand how $N=\text{m}g\cos\theta$ when $\text{m}g$ is broken down into components that are parallel and perpendicular to the plane, but why can't we explain the magnitude of acceleration without breaking it as such? Please reconsider my question. What I understand is that the downwards pull of $\text{m}g$ is balanced by the y component ($N_y$) of the normal force ($N$ ), thus $N_y=\text{m}g$ so $N=\frac{\text{m}g}{\cos\theta}$. My understanding of the block sliding is that as it moves right due to $N_x$, it is pulled downwards by $\text{m}{g}$ . What is wrong in thinking it this way? Commented Oct 8, 2021 at 6:08
• @ silica 19 . The downwards pull of $mg$ is not balanced by $N_y$, there is an acceleration downwards. (1 second later the object would be lower than before). What is balanced are the forces in the direction perpendicular to the plane, as there is no acceleration that way. Hope it makes sense after you've had time to think about it, but not sure what else to say.... All the best Commented Oct 8, 2021 at 7:10
• @silica 19 The final thing to say is that when applying $F=ma$ you have to pick a definite direction. Perpendicular to the plane there is no acceleration (the only acceleration is down the plane), so $N-mg\cos\theta = 0$. When applying $F=ma$ vertically it's $mg - N\cos\theta = ma_{down}$ and as mentioned before, there is a component of the acceleration downwards (the block becomes becomes lower with time). Commented Oct 8, 2021 at 11:21