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Take a look at the following picture

enter image description here

Obviously, constructing $u-v$ axes allow us to resolve the force into two components. My question is what are the rules for constructing such axes? It seems to me it is a matter of trial and error approach as long as the parallelogram is successfully drawn.

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    $\begingroup$ As long as you obey normal laws of vector maths, you can construct any coordinates you want. Whether it helps you to do it one way is a different question. $\endgroup$ – Kyle Kanos Aug 5 '19 at 12:34
  • $\begingroup$ Problem solving strategies should never be called "rules". It is greatly misleading, and I am sure it is what makes students overthink simple problems. You should tackle each physics problem with a fresh mind ready to use insight and experience from similar problems in the past. You should not think "if I have this set of rules then I can handle any physics problem". Do what is best for the problem. Don't worry about following rules. $\endgroup$ – Aaron Stevens Aug 5 '19 at 13:08
  • $\begingroup$ @AaronStevens, the reason I'm asking this because the author of a very well known textbook draws these axes without further explanation. It is not clear why the axes are drawn in a specific structure. I assumed the author wrongly assumes students are aware of certain mathematical rules. $\endgroup$ – CroCo Aug 5 '19 at 13:18
  • $\begingroup$ I am sure there is a reason. Perhaps the question/example is interested in components of the force along those axes? There are many possibilities based on the picture. Without further information I cannot say if they expect you to actually break the force into components along those directions, or if it is there to illustrate some given information in the problem, etc. A full statement of the problem might be helpful here. $\endgroup$ – Aaron Stevens Aug 5 '19 at 13:21
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    $\begingroup$ It would help enormously if the source of the diagram and the text is revealed. The question could then be answered with reference to the example which has been given. $\endgroup$ – Farcher Aug 5 '19 at 13:43
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The choice of axes is usually dependent on the problem in hand.

In two dimensions all you need to define is two non-parallel axes which are in the plane containing the forces.
It is often the case that the two axes are at right angles to one another but they do not need to be so.

One way to understand what is need is to think how you might represent the coordinates of a point on a plane. All you need is two axes but they do not necessary have to be at right angles to one another but they must not be parallel to one another.

In the example that you have given the directions of the two axes might have been chosen to be parallel to the girders shown in the diagram.
The components of force $F$ in the $\hat u$ and $\hat v$ directions would therefore be the longitudinal forces acting on the girders.

So the relationship between the force $\vec F$ and the components of the force in the two directions defined by the axes $F_{\rm u}$ and $F_{\rm v}$ would be $\vec F = F_{\rm u} \hat u + F_{\rm v} \hat v$ which as a diagram might be drawn as a triangle of parallelogram of forces.

enter image description here

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The main rule is that you (usually) wish to resolve force components along Cartesian coordinate axes. This means that the axes should be perpendicular to one another.

Many - especially the elementary - functions in mathematics are rather "simple" in Cartesian coordinate systems. You are welcome to create other non-Cartesian axes, but then you can't directly use many of your usual functions, geometric and trigonometric formulas and the like such as the sine, cosine and tangent relations, dot- and cross-product formulas etc.

You can still use those formulas, of course, by first converting non-Cartesian coordinates to Cartesian ones, and then plugging them in. This might not be the easiest way to go about solving a task, so it is usually recommended to keep component axes Cartesian.

How you place and angle your Cartesian coordinate axes is then entirely up to you. Place them in whichever manner that seems to simplify your work the most. In your example I would just keep the axes horizontal and vertical as always, since that causes one of the vectors to be parallel to an axis and thus we have reduced the number of sine and cosine terms that will appear in the resolved components.

(The second-most relevant coordinate system that you might want to use in certain specific "spherical" cases is the polar coordinate system. In such fairly rare situations (depending on your field or work/study) it might be the case that the resolved components become so much simpler in polar coordinates that it is worth the extra coordinate conversion step.)

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  • $\begingroup$ Dot & cross products only work in Cartesian!? $\endgroup$ – Kyle Kanos Aug 5 '19 at 12:35
  • $\begingroup$ @KyleKanos Well, no, they always work, but the usual plug-and-play Cartesian expressions/formulas will be different and not nearly as "simple" in other coordinate systems. Other versions of the formulas or an extra step with a coordinate conversion would then be necessary. $\endgroup$ – Steeven Aug 5 '19 at 12:47
  • $\begingroup$ Okay, so saying that you can't use them (directly or not) is wrong (or at the very least, greatly misleading). $\endgroup$ – Kyle Kanos Aug 5 '19 at 12:55
  • $\begingroup$ @KyleKanos I completely agree, and I do not hope my answer gives that impression. I have clarified a few sentences to be sure. $\endgroup$ – Steeven Aug 5 '19 at 12:58
  • $\begingroup$ Note that the OP has revealed in the comments that this question specifically asks about the force components along these axes. The axes are not used to solve any problems beyond this. $\endgroup$ – Aaron Stevens Aug 5 '19 at 13:44

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