# Vectors in the Plane

• Apr 25th 2010, 12:34 AM
ugkwan
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.
• Apr 25th 2010, 12:42 AM
Are you given a direction for the Resultant Force?
• Apr 25th 2010, 12:54 AM
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 $\cos\theta=\frac{-11}{24} \Leftrightarrow \theta=117.27 \mbox{degrees}$.
• Apr 25th 2010, 01:17 AM
ugkwan
Quote:

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?

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 $\theta=62.72038$...