Wednesday, March 21, 2018

forces - Why taking components of a component of a vector is invalid?


Suppose there's a force F of magnitude 10 newtons in the direction of positive y-axis acting on a particle A. I know that the particle would not experience any force in the positive x-direction because,


component of force F in the positive x-direction =|F|cos90°=0.


But if we first take the component of F at an angle of 45° to the positive y-axis, it is,


|F|cos45°=102N.


If we now consider this a vector (can we? that is the main question), we can find its component in the positive x-direction (it makes an angle of 45° with the positive x-axis) ,


102×cos45°=5N. According to this result, the particle should experience a force of 5 newtons in the positive x-direction, which is impossible. I want to know what all is wrong in the above reasoning. Thanks!Graphic representation



Answer



The most straightforward answer is no, you cannot consider a component to be a vector. A vector is something which has an associated direction, but a component has no direction. It's just a number. For example, if F is a vector, the component of F in the ˆx direction is Fˆx, and hopefully you know that the value of a dot product like Fˆx is just a number with no associated direction.


However... Martijn Pot's answer talks about how you've neglected to consider the other "component", the one that points in the 135 direction (with the angle measured from the x axis as is conventional). I wouldn't use the word "component" in that context, but the answer raises a good point.



Let me explain, starting with some background information: when you have a 2D vector F, you can determine its x and y components using the definitions Fx=FˆxFy=Fˆy

and then you can construct a vector from each of those components as follows: Fx=(Fˆx)magnitudeˆxˆydirectionFy=(Fˆy)magnitudeˆydirection
The vector Fx is called the "projection of F on to ˆx" (or the "projection of F on to the x axis"), and similarly for Fy. When you add all the projections together, you get the original vector: Fx+Fy=F
You presumably know about the expression of a vector in terms of its components, F=Fxˆx+Fyˆy; well, this last equation is saying the same thing. Check that they are equivalent, if it's not obvious to you.


Now for the key fact: when you write a vector in this way, there will be one projection (or component) for each dimension of the space. You have to include all of them! You can't write F=Fxˆx, for example; you need to include both the x and y components. Two components for two dimensions.


Moving on: you can also find the components (or projections) of a vector in a different basis, or in other words, using a different set of axes instead of x and y. In your question, you used a basis vector which is angled at 45 to the x axis and 45 to the y axis. Let's call this basis vector ˆa. Hopefully it makes sense that you can write it as ˆa=12(ˆx+ˆy)

When you take the projection of F on ˆa (or in other words, take the a-component of F and "treat it as a vector"), you are calculating Fa=(Fˆa)ˆa
But you can't then expect Fa to be equal to F. That would be like expecting Fx alone to equal F. This is a two-dimensional space, so you need two basis vectors. The other one has to be perpendicular to ˆa, for example pointing at 135 from the x axis. Let's call that ˆb. You can write it as ˆb=12(ˆyˆx)
If you take the projection of F on ˆb, it will be Fb=(Fˆb)ˆb
and you can reconstruct the original vector by adding both components F=Fa+Fb


You can double-check this by verifying that the x component of Fa+Fb is what you expect (Fa+Fb)ˆx=([Fˆa]ˆa+[Fˆb]ˆb)ˆx=([F12(ˆx+ˆy)]12(ˆx+ˆy)+[F12(ˆyˆx)]12(ˆyˆx))ˆx=12([Fˆx+Fˆy](ˆxˆx+ˆyˆx)+[FˆyFˆx](ˆyˆxˆxˆx))=12([Fx+Fy](1)+[FyFx](1))=12(2Fx)=Fx

The procedure you asked about in your question, where you calculate the x component of the a projection, is equivalent to ignoring the b projection, as Martijn pointed out. If you do that, here's how the math works out: Faˆx=([Fˆa]ˆa]ˆx=([F12(ˆx+ˆy)]12(ˆx+ˆy))ˆx=12([Fˆx+Fˆy](ˆxˆx+ˆyˆx))=12([Fx+Fy](1))=12(Fx+Fy)
So this happens to give you the average of the x and y components. In your example, F has a y component of 10 N, so you get 12(0 N+10 N)=5 N out of the calculation. But it doesn't mean anything physically.


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