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°=10√2N.
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) ,
10√2×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!
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 →F⋅ˆx, and hopefully you know that the value of a dot product like →F⋅ˆx 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=→F⋅ˆxFy=→F⋅ˆy
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=1√2(ˆx+ˆy)
You can double-check this by verifying that the x component of →Fa+→Fb is what you expect (→Fa+→Fb)⋅ˆx=([→F⋅ˆa]ˆa+[→F⋅ˆb]ˆb)⋅ˆx=([→F⋅1√2(ˆx+ˆy)]1√2(ˆx+ˆy)+[→F⋅1√2(ˆy−ˆx)]1√2(ˆy−ˆx))⋅ˆx=12([→F⋅ˆx+→F⋅ˆy](ˆx⋅ˆx+ˆy⋅ˆx)+[→F⋅ˆy−→F⋅ˆx](ˆy⋅ˆx−ˆx⋅ˆx))=12([Fx+Fy](1)+[Fy−Fx](−1))=12(2Fx)=Fx
No comments:
Post a Comment