# Thread: Vectors in the Plane

1. ## Vectors in the Plane

I'm having trouble with the algebra of this word problem.

Resultant Force: Find the angle between the forces given the magnitude of their resultant. Force 1 is a vector in the direction of the positive x-axis and force 2 is a vector at an angle theta with the positive x-axis.

Force 1: 45 pounds
Force 2: 60 pounds
Resultant Force: 90 pounds

I'm stuck on solving for the resultant vector component using Force 2's vector component of,

Force 2 = w = 60<cos theta, sin theta>

Would you please write out every step along the way? I really need to learn this so I can pass my test on monday.

2. Are you given a direction for the Resultant Force?

3. Ok it's quite simple, it goes like:
$F_1+F_2=(45i+0j) + (60\cos\theta\,i+60\sin\theta\,j)=(45+60\cos\theta )\,i + 60\sin\theta\,j$ which is equal to the resultant vector.
Then it follows that $(45+60\cos\theta)^2 \ + \ (60\sin\theta)^2=90^2$ and thus by expanding the squares we get $45^2+60^2\cos^2\theta+(45)(60)(2)\cos\theta+60^2\, \sin^2\theta=90^2$ and by manipulation we get that $\cos\theta=\frac{-11}{24} \Leftrightarrow \theta=117.27 \mbox{degrees}$.

4. Originally Posted by rebghb
Ok it's quite simple, it goes like:
$F_1+F_2=(45i+0j) + (60\cos\theta\,i+60\sin\theta\,j)=(45+60\cos\theta )\,i + 60\sin\theta\,j$ which is equal to the resultant vector.
Then it follows that $(45+60\cos\theta)^2 \ + \ (60\sin\theta)^2=90^2$ and thus by expanding the squares we get $45^2+60^2\cos^2\theta+(45)(60)(2)\cos\theta+60^2\, \sin^2\theta=90^2$ and by manipulation we get that $\cos\theta=\frac{-11}{24} \Leftrightarrow \theta=117.27 \mbox{degrees}$.

magnitude of (F1+F2)= √(45+60cos theta)^2 + (60 sin theta)^2) = 90 is absolutely correct. I don't get the in between process of how the book got the answer of theta equals 62.7 degrees. Can you expand your explanation
of the manipulation process more please?

√( square root )

5. the answer of theta equals 62.7 degrees.
Yeah yeah it's right... You see I forgot to add a second minus sign, u see $180 - 117.27 = 62.7$ Here's a little more into it:

$45^2+60^2\cos^2\theta+(45)(60)(2)\cos\theta+60^2\, \sin^2\theta=90^2$ $\Leftrightarrow 45^2+60^2\underbrace{(\cos^2\theta+\sin^2\theta)}+ 5400\cos\theta=90^2$ where the underbraced expression gives 1. Move everything to the right you get $\cos\theta=\frac{90^2-45^2-60^2}{5400}=\frac{11}{24}$ implies that $\theta=62.72038$...

P.S. Don't do my mistake in your test