Thread: i need help with vectors!.......

1. i need help with vectors!.......

Physics:
For the vectors A and B in the figure use the method of components to find the magnitude and direction of a.) THe vector sum A + B, b.) the vector difference A - B, c.) The vector difference B - A.

in the figure
A is 12 m West
B is 18 m 37$\displaystyle \circ$ N of E

im not sure how to do vector sum but i think it will be from the head to tail...
but i dunno how to form it... same as vector difference... can u help me with this?

Physics:
For the vectors A and B in the figure use the method of components to find the magnitude and direction of a.) THe vector sum A + B, b.) the vector difference A - B, c.) The vector difference B - A.

in the figure
A is 12 m West
B is 18 m 37$\displaystyle \circ$ N of E

im not sure how to do vector sum but i think it will be from the head to tail...
but i dunno how to form it... same as vector difference... can u help me with this?

Component method.

A+B
First choose a coordinate system:
I'll take N as +y and E as +x.

Now resolve each vector into components in this coordinate system. ($\displaystyle \theta$ is the angle between the vector and the +x axis.)

$\displaystyle A_x = Acos(\theta) = (12 \, m)cos(180) = -12 \, m$
$\displaystyle A_y = Asin(\theta) = (12 \, m)sin(180) = 0 \, m$
$\displaystyle B_x = Bcos(\theta) = (18 \, m)cos(37) = 14.3754 \, m$
$\displaystyle B_y = Bsin(\theta) = (18 \, m)sin(37) = 10.8327 \, m$

Now, add the x components together to form the x component of the resultant $\displaystyle R_x$ and add the y components together to form the y component of the resultant $\displaystyle R_y$.
$\displaystyle R_x = A_x + B_x = -12 \, m + 14.3754 \, m = 2.3754 \, m$
$\displaystyle R_y = A_y + B_y = 0 \, m + 10.8327 \, m = 10.8327 \, m$

I recommend drawing a sketch at this point, but note that both components of R are positive. Thus R is a vector in Quadrant I. So to find the length and the angle:
$\displaystyle R = \sqrt{R_x^2 + R_y^2} = 11.0901 \, m$
$\displaystyle \theta_R = atn \left | \frac{R_y}{R_x} \right | = 77.6317 ^o$
(Note: Finding $\displaystyle \theta_R$ using the absolute value bars gives the reference angle, from which you can extract the angle if you know what quadrant it is in.)

So A + B is 11 m at 78 degrees above the + x axis, otherwise known as 78 degrees N of E.

-Dan

3. To do A - B consider the following hint:
A - B = A + (-B)

The components of the vector -B are simply the negatives of the components of the vector B.

-Dan

For the vectors A and B in the figure, use the method of components
to find the magnitude and direction of:
. . a) the vector sum $\displaystyle \vec A + \vec B$
. . b) the vector difference $\displaystyle \vec A - \vec B$
. . c) the vector difference $\displaystyle \vec B - \vec A$

In the figure: A is 12 m West, B is 18 m 37° N of E
You're expected to be able write the vectors in component form.

. . . . . $\displaystyle \vec A \:=\:\langle \text{-}12,\:0\rangle\qquad\qquad\vec B \;=\;\langle 18\cos37^o,\;18\sin37^o\rangle$

$\displaystyle (a)\;\;\vec A + \vec B \;=\;\langle\text{-}12- 18\cos37^o,\;18\sin37^o\rangle$

Then: .$\displaystyle |\vec A + \vec B| \;=\;\sqrt{(\text{-}12-18\cos37^o)^2 + (18\sin37^o)^2} \;\approx\;\boxed{28.5}$

And: .$\displaystyle \tan\theta \;=\;\frac{18\sin37^o}{\text{-}12-18\cos37^o} \;=\;-0.410710523\quad\Rightarrow\quad\theta \:\approx\:-22.3^o,\:157.7^o$

Since $\displaystyle \vec A+ \vec B$ is in Quadrant 2: .$\displaystyle \theta \:=\:\boxed{157.7^o}$

Get the idea?

5. thankssss!!!