forked from eggy/eifueo
phys: remove decomposition and rephrase for clarity
This commit is contained in:
parent
4393424e4c
commit
d89b2cc3c6
@ -179,9 +179,37 @@ When subtracting a vector, **negate** the vector being subtracted by giving it a
|
|||||||
|
|
||||||
<img src="/resources/images/vector-subtract-direction.png" width=700>(Source: Kognity)</img>
|
<img src="/resources/images/vector-subtract-direction.png" width=700>(Source: Kognity)</img>
|
||||||
|
|
||||||
|
### Adding/subtracting vectors algebraically
|
||||||
|
|
||||||
|
Vectors can be broken up into two vectors (**"components"**) laying on the x- and y-axes via trigonometry such that the resultant of the two components is the original vector. This is especially helpful when adding larger (3+) numbers of vectors.
|
||||||
|
$$\vec{F}_x + \vec{F}_y = \vec{F}$$
|
||||||
|
|
||||||
|
!!! info "Reminder"
|
||||||
|
The **component form** of a vector is expressed as $(|\vec{a}_x|, |\vec{a}_y|)$
|
||||||
|
|
||||||
|
<img src="/resources/images/vector-simple-adding.png" width=700>(Source: Kognity)</img>
|
||||||
|
|
||||||
|
By using the primary trignometric identities:
|
||||||
|
$$
|
||||||
|
|\vec{a}_{x}| = |\vec{a}|\cos\theta_{a} \\
|
||||||
|
|\vec{a}_{y}| = |\vec{a}|\sin\theta_{a}
|
||||||
|
$$
|
||||||
|
|
||||||
|
<img src="/resources/images/vector-decomposition.png" width=700>(Source: Kognity)</img>
|
||||||
|
|
||||||
|
Using their component forms, to:
|
||||||
|
|
||||||
|
- add two vectors, add their x- and y-coordinates together.
|
||||||
|
- subtract two vectors, subtract their x- and y-coordinates together.
|
||||||
|
|
||||||
|
$$
|
||||||
|
(a_{x}, a_{y}) + (b_{x}, b_{y}) = (a_{x} + b_{x}, a_{y} + b_{y}) \\
|
||||||
|
(a_{x}, a_{y}) - (b_{x}, b_{y}) = (a_{x} - b_{x}, a_{y} - b_{y})
|
||||||
|
$$
|
||||||
|
|
||||||
### Parallelogram rule
|
### Parallelogram rule
|
||||||
|
|
||||||
The parallelogram rule states that the sum of two vectors that form two sides of a parallelogram is the diagonal of that parallelogram.
|
The parallelogram rule states that the sum of two vectors that form two sides of a parallelogram is the diagonal of that parallelogram. The **sine** and **cosine laws** can be used to solve for the resultant vector.
|
||||||
|
|
||||||
<img src="/resources/images/vector-parallelogram.png" width=700>(Source: Kognity)</img>
|
<img src="/resources/images/vector-parallelogram.png" width=700>(Source: Kognity)</img>
|
||||||
|
|
||||||
@ -189,34 +217,10 @@ The parallelogram rule states that the sum of two vectors that form two sides of
|
|||||||
|
|
||||||
The product of a vector multiplied by a scalar is a vector with a magnitude of the vector multiplied by the scalar with the same direction as the original vector.
|
The product of a vector multiplied by a scalar is a vector with a magnitude of the vector multiplied by the scalar with the same direction as the original vector.
|
||||||
|
|
||||||
$$v[direction] × s = (v×s)[direction]$$
|
$$\vec{v} × s = (|\vec{v}|×s)[\theta_{v}]$$
|
||||||
|
|
||||||
### Vector decomposition
|
!!! example
|
||||||
|
$$3 \text{ m} · 47 \text{ ms}^{-1}[N20°E] = 141 \text{ ms}^{-1}[N20°E]$$
|
||||||
By breaking up a vector into lengths along the x- and y-axes, the sum of two vectors can be calculated algebraically.
|
|
||||||
|
|
||||||
<img src="/resources/images/vector-decomposition.png" width=700>(Source: Kognity)</img>
|
|
||||||
|
|
||||||
For vector $\vec{a}$ and vector $\vec{b}$:
|
|
||||||
|
|
||||||
$$
|
|
||||||
\vec{a}_{x} = a\cos\theta_{a} \\
|
|
||||||
\vec{a}_{y} = a\sin\theta_{a}
|
|
||||||
$$
|
|
||||||
|
|
||||||
Proof:
|
|
||||||
$|\vec{a}| = \sqrt{a^{2}_{x}+a^{2}_{y}} \\$
|
|
||||||
$= \sqrt{(|\vec{a}|\cos\theta_{a})^2 + (|\vec{a}|\sin\theta{a})^2} \\$
|
|
||||||
$=\sqrt{|\vec{a}|^2(\cos\theta_{a}^2 + \sin\theta_{a}^2)} \\$
|
|
||||||
$=\sqrt{|\vec{a}|^2} \\$
|
|
||||||
$=|\vec{a}|$
|
|
||||||
|
|
||||||
From the diagram above, we can figure out that:
|
|
||||||
|
|
||||||
$$
|
|
||||||
|\vec{r}|=\sqrt{(|\vec{a}|\cos\theta_{a} + |\vec{b}|\cos\theta_{b})^2 + (|\vec{a}|\sin\theta_{a} + |\vec{b}|\cos\theta_{b})^2} \\
|
|
||||||
\theta_{r}=\tan^{-1}(\frac{|\vec{a}|\sin\theta_{a} + |\vec{b}|\sin\theta_{b}}{|\vec{a}|\cos\theta_{a} + |\vec{b}|\cos\theta_{b}})
|
|
||||||
$$
|
|
||||||
|
|
||||||
## Resources
|
## Resources
|
||||||
|
|
||||||
|
Loading…
Reference in New Issue
Block a user