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I have a vector $F_e$ and I would like to graph it to its corresponding $X$ and $Y$ components. I know that the $i_y$ component is negative, and there is no $i_x$ component.

$\vec{F_e} = F_x\vec{i_x} + F_y\vec{i_y}$

$\vec{F_e} = F_y\vec{(-i_y)}$

And to get there we have:

$F_{ex}=|\vec{F_{ex}}|\cos(270^\circ)$

$F_{ey}=|\vec{F_{ey}}|\sin(270^\circ)$

My question is when do we use the $\sin(\cdot)$ and when the $\cos(\cdot)$ to find its magnitude. Intuitively $\cos (270^\circ)$ equals $0$ and $\sin (270^\circ)$ equals $-1$. But why didn't we set $F_{ex}$ with $\sin (270^\circ)$. I know that the angle between the $X$ axis and the vector is $270^\circ$.

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3 Answers 3

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If a point is at a unit distance from the origin, and it makes an angle $p$ with the x-axis, then $\cos p$ is defined as the x co-ordinate of that point. $\sin p$ is defined as the y co-ordinate of that point.

If a point is a distance $r$ from the origin and makes an angle $p$ wih the x-axis, then its x and y co-ordinates are $r \cos p$ and $r\sin p$ respectively.

Since you have to find the x-component of the force here, you have to use $|\vec{F}| \cos p$

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  • $\begingroup$ This is the thing i was bothered with, it seems simple but why is cos defined with x and sin with y ? $\endgroup$
    – Edmond Dantes
    Commented Apr 21, 2021 at 5:05
  • $\begingroup$ @MarkoMajstorovic Historically, the x co-ordinate has been named cos, and the y has been named sin. Naming is arbitrary. $\endgroup$
    – Ryder Rude
    Commented Apr 21, 2021 at 5:39
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    $\begingroup$ @MarkoMajstorovic You can think of $\cos {p}$ to mean the sentence: "The x co-ordinate of the point which is at a unit distance from the origin, and makes an angle $p$ with the x-axis". The name $\cos{p}$ has been assigned to this sentence. $\endgroup$
    – Ryder Rude
    Commented Apr 21, 2021 at 5:43
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    $\begingroup$ @MarkoMajstorovic the standard notation measures angles anticlockwise from the x axis. That is an arbitrary decision, but the definition of "cos" and "sin" then mean "use cos for the x axis and sin for the y axis". Note, the angle convention also works for graphing complex numbers. The formula $e^{i\theta} = \cos\theta + i\sin\theta$ is true by definition, so if you graph complex numbers with the real axis horizontal and the imaginary axis vertical, that fixes how $\theta$ is defined (i.e. measure it anticlockwise from the x axis). $\endgroup$
    – alephzero
    Commented Apr 21, 2021 at 12:53
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So to clarify, you're just trying to

  1. plot the vector using it’s x-y components, or
  2. or are you asking how to get the $X$ or $Y$ components of a vector?

If it’s # 1 it seems like you already did it since there’s no $X$ component and the $Y$ component is in the negative direction. If it’s # 2 then you're probably looking for $x=R\cos\theta$ and $y = R\sin\theta$ where $R$ is the force vectors magnitude and theta is the angle between the vector and the $X$ axis.

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  • $\begingroup$ Its the first one, but my question is more of a trigonometric one, why didnt we use sin(270) for the x axis of the magnitude $F_{ex}$ $\endgroup$
    – Edmond Dantes
    Commented Apr 21, 2021 at 4:21
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    $\begingroup$ @MarkoMajstorovic imagine a right triangle with its bottom side on the x-axis. That bottom side is the adjacent and the long side is the hypotenuse. The other end is the opposite, now you can see since it’s aligned, the adjacent is the x and the opposite is the y. The right triangle forms an angle theta with the x axis and by definition sin(theta) =opposite/hypotenuse or y/r $\endgroup$
    – Railmints
    Commented Apr 21, 2021 at 4:31
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Recall the triangle law of vectors.

To find the x and y components, construct a triangle with the vector $F_e$ as the hypotenuse.

This is where trigonometry comes in- since $F_{ex}$ is the base of the triangle and $F_e$ is the hypotenuse, you can now figure out the trigonometric relation between $\theta$, $F_e$ and $F_{ex}$.

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  • $\begingroup$ I mean this is not a homework question more of a how did they get that type of question, and how can i make Fe to be the hypotenuse ? $\endgroup$
    – Edmond Dantes
    Commented Apr 21, 2021 at 4:36
  • $\begingroup$ @MarkoMajstorovic in regular triangles the hypotenuse will be made in a regular form, but in this case it will overlap with y-axis(as you can see) and the x-axis will be reduced to zero. $\endgroup$
    – Sid
    Commented Apr 21, 2021 at 4:42
  • $\begingroup$ To visualise: Imagine a regular right-angled triangle with $\theta = 45 \circ$. Now increase the angle between x-axis and hypotenuse, while keeping the hypotenuse constant. You can see that the length of the base would decrease, right? And at 90 deg., it would be just a line along the y-axis. Increasing it further, the triangle would go into the second quadrant.At 180 deg., it would be just a straight line along the negative x- axis. At 270, just a straight line along negative y-axis. $\endgroup$
    – Sid
    Commented Apr 21, 2021 at 4:47
  • $\begingroup$ @MarkoMajstorovic mathsisfun.com/algebra/trig-interactive-unit-circle.html is a good visualisation $\endgroup$
    – Sid
    Commented Apr 21, 2021 at 4:56

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