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I would like to know how to interpret the following force decomposition (Diagram (2)).

Inclined plane with object

Diagram (1) is the typical force decomposition. Decomposing the force vector this way tells us about how much force is acting on the object ($F_1 = mgsin(\theta)$) to make it slide down the inclined plane and how much force is acting perpendicular to the surface ($F_2 = mgcos(\theta)$).

Now looking at Diagram (2) (assuming it has same inclined angle $\theta$), Force acting on ($-$x) axis is $F_1 = mg tan(\theta)$ and force acting perpendicular to the surface is now $F_2 = mg/cos(\theta)$.

Note: Both diagram's force $mg$ is decomposed such that when added it satisfies a parallelogram shape.

My question is how do we interpret such force decomposition in a physical way?

Diagram (1) is a situation where if you put an object of mass 'm' on a frictionless inclined plane it slides down. In Diagram (2), normal force increased and also force along x-direction increased. Meaning it would slide down faster than diagram (1) despite having same net force of $F_{net} = mg$.

Clearly, I am misunderstanding something. Can someone correct me where I am getting confused?

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    $\begingroup$ You are always allowed to choose weird ways to decompose forces, like in Diagram (2). However, then it will become very irritating and complicated to solve when you choose badly, like in Diagram (2). It is very typical that we want to choose axes that are perpendicular, so that forces in one direction will not come and irritate the equations in another direction, but in Diagram (2) they will all intrude on each other. It can still be solved to give the same physical answers, but it will be annoying. $\endgroup$ Oct 15, 2023 at 8:47
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    $\begingroup$ I'm following the math of why F$_2$ is different in the two diagrams. In diagram (1) it is acting on the object with mass m. What is it acting on in diagram (2)? $\endgroup$ Oct 15, 2023 at 12:23
  • $\begingroup$ @Not_Einstein That is also my question. $\endgroup$
    – Mardia
    Oct 15, 2023 at 16:28

3 Answers 3

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The net force is equal to mass times acceleration. When motion is independent in directions, we can write "Newton's Second Law equations" for each direction. Thus, in the first diagram you have drawn, we describe motion in the parallel and perpendicular axes to the plane. It is not that $F_{\text{net}} = mg$, it is that $F_{\text{net} \parallel} = ma$ and $F_{\text{net} \perp} = 0$.

In the second choice, you have chosen very unnatural axes. The normal force and gravitational force will be have components in the direction of both of the axes you have chose. Namely, if we call your axes 1 and 2, our equations will look like

$$F_{\text{net} 1} = N_1 - F_{g 1} = ma_1$$ and $$F_{\text{net} 2} = N_2 - F_{g 2} = ma_2$$

The magnitude of the forces is not changing, but how you express them in component form is changing.

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In diagram (2) you have shown that the weight of the block $mg$ acting vertically downwards is equivalent to the sum of a force $A = mg \tan \theta$ acting horizontally and a force $B = mg \frac 1 {\cos \theta}$ acting perpendicular to the slope.

This is a correct decomposition, but not very useful since component $A$ acts partly to accelerate the block down the slope and partly to oppose component $B$. To find the net forces acting along and perpendicular to the slope we have to decompose component $A$ again, and we find the force acting down the slope is

$C = A \cos \theta = mg \sin \theta$

and the net force acting perpendicular to the slope is

$\displaystyle D = B - A \sin \theta = mg \left( \frac 1 {\cos \theta} - \frac {\sin^2 \theta} {\cos \theta} \right) = mg \cos \theta$

so we find that $C$ and $D$ are equal to $F_1$ and $F_2$ from diagram (1). So using diagram (2) we reach the same result in two steps that we can reach in one step using diagram (1).

So although it is technically correct to decompose a force (or any other vector) into non-orthogonal components, it is not usually useful to do so.

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  • $\begingroup$ Right. But what would be the physical situation of such case? Since diagram (2) is also a correct force decomposition shouldn't it have a corresponding physical situation? $\endgroup$
    – Mardia
    Oct 15, 2023 at 16:30
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    $\begingroup$ @mardia There is only one physical force here, which is the weight of the block. It is just being decomposed in two different ways. Decomposition of a vector is a purely conceptual process. Neither decomposition is wrong, but one is more useful than the other. $\endgroup$
    – gandalf61
    Oct 15, 2023 at 16:40
  • $\begingroup$ Right. But doesn't decomposing a force vector like diagram 2 give us force acting on x-axis (Thus we can find acceleration in -x direction) and normal force acting on surface (Thus we can find the normal force)? I am confused because like you said there is only one physical force here but it seems like it gives us different force when we find the force in -x direction from diagram 1 which I think it's $F_x = -2mgsin(\theta)cos(\theta)$ but this is different from force in -x direction from diagram 2 which was $F_x=mgtan(\theta)$. $\endgroup$
    – Mardia
    Oct 15, 2023 at 17:11
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in both diagrams for the equations of motion you have to decompose the force components in $~x~,y~$ system

Diagram 1

the force components $$ F_{x1}=m\,g\,\sin(\theta)\quad, F_{y1}=-m\,g\,\cos(\theta)$$

Diagram 2

the force components

$$F_{x2}=u\,\cos(\theta)\quad, F_{y2}=u\,\sin(\theta)+v$$

and with $~F_{x2}=F_{x1}~,F_{y2}=F_{y1}~$ you obtain

$$u=m\,g\,\tan{\theta}\quad,v=-\frac{m\,g}{\cos(\theta)}$$

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