1. ## Vector Problem

Im a bit stuck on this question.

In the first part, i worked out that a vector had the component form:

61.3 j + 56.1 i

For the next part, i need to write it in magnitude and direction form.
Im unsure on how to do this.

I know that:

61.3 = m sin d
56.1 = m cos d

m is the magnitude in newtons
d is the direction in degrees

How could i solve this simultaniously?

2. Hello, nugiboy!

You should be familiar with certain formulas . . .

In the first part, i worked out that a vector had the component form: . $61.3 j + 56.1 i$

For the next part, i need to write it in magnitude and direction form.
Im unsure on how to do this.

I know that: . $\begin{array}{ccc}61.3 &= &m\sin d \\ 56.1 & = & m\cos d\end{array}$

$m$ is the magnitude in newtons
$d$ is the direction in degrees
If we don't know the formulas, we can still work it out . .

Divide the two equations: . $\frac{m\sin d}{m\cos d} \:=\:\frac{61.3}{56.1}$

. . This gives us: . $\tan d \:=\:1.092691622$

Hence: . $d \;=\;\tan^{-1}(1.092591622) \;=\;47.53614518^o$

. . Therefore: . $\boxed{d \;\approx\;47.5^o}$

For the magnitude, think of the Pythagorean Theorem.

. . $m^2 \:=\:(61.3)^2 + (56.1)^2 \;=\;6904.9$

Hence: . $m \;=\;\sqrt{6904.9} \;=\;83.09572793$

. . Therefore: . $\boxed{m \;\approx\;83.1}$

3. Originally Posted by Soroban
Hello, nugiboy!

You should be familiar with certain formulas . . .

If we don't know the formulas, we can still work it out . .
I understand how you got the answer, but i was just wondering what you meant by i should be familiar with certain formulas. Is there a simple formula to go from component form to magnitude and angle form, or is the method you showed me the only way?

4. Hello again, nugiboy!

Well, if you learn a few conversion formulas, it'll save a lot of time.
. . But you seem to be familiar with them already.

Going from $(r,\,\theta)$ to $(x,\,y)\!:\;\;\begin{Bmatrix}x & = & r\!\cdot\!\cos\theta \\ y & = & r\!\cdot\!\sin\theta \end{Bmatrix}$

Going from $(x,y)$ to $(r,\,\theta)\!:\;\;\begin{Bmatrix}r & = &\sqrt{x^2+y^2} \\ \theta & =\ & \arctan\left(\frac{y}{x}\right)\end{Bmatrix}$