mr.p: Good morning.
Today we're going to do a typical introductory vector
addition problem where we need to use component vectors.
Here we go.
Bo: Hey guys.
Billy: Hey Bo. Bobby: Hi Bo.
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mr.p: Bo, could you please read the problem?
Bo: Slow Velocity Racer races 50.0 cm East,
then turns 35 degrees North of East and scoots for 40.0 cm.
She then turns and moseys another 30.0 cm North.
What was her total displacement?
mr.p: There are clearly vectors in this problem however
the probem description does not give these vectors names.
And this is something you will periodically run
into when you're solving problems.
I do suggest that you always give your vectors names
because this will make it easier to identify
which vectors you are working with,
while you are solving the problem.
I have named the vectors A, B and C.
Vector A is 50.0 cm East.
Vector B is 40.0 cm and an angle of 35 degrees North of East
and Vector C is 30.0 cm North.
We are trying to find the total displacement
for Slow Velocity Racer, which works out to be the
sum of the 3 vectors, A, B and C,
which is the resultant vector R.
Now, whenever you are doing a vector addition problem
you need to draw a vector diagram.
I start by drawing vector A, which is 50 cm East
and notice that this vector is an arrow,
to indicate its direction.
And it has a label A with a vector symbol over it
to identify that this arrow is representing vector A.
I place the tail of vector B on the tip of vector A
because this is tip-to-tail vector addition.
Notice that B is about 4/5ths the length of A
because it represents 40 cm in length
instead of 50 cm in length.
And we have our angle here, which is 35 degrees North
from East or North of East.
And vector C, the tail of vector C is on the tip of vector B.
C is Northward and is only 3/5ths the length of A
because it represents 30 cm in length.
We are looking for the total displacement,
which is this resultant vector R.
Which is the displacement from the initial point
to the final point.
And the arrow head goes right there on the
final point again because this arrow indicates
the direction of the total displacement,
which is from the initial point to the final point.
Now, some students want to put this arrow head
on the resultant vector down here instead.
Now, I've drawn it this way, which is wrong,
please make sure you realize that.
Because a lot ot students want to draw it this way
and I think it's because vector A goes this way,
B this way, C this way
and you want to go back to the beginning.
It makes you feel all warm and fuzzy inside.
Then you have this loop and you can just
keep going around the loop, but that's wrong.
Remember, this resultant vector is the sum of A, B and C
so it's going to start at the beginning of A
and end at the end of C.
So the resultant vector goes from
the initial to the final point.
And we indicate that by having the arrow head go
from the initial, to the final point.
So this resultant vector arrow head
should go up here on the final point.
Another thing I want to point out is that a lot of
students want to draw their vector diagrams like this.
You guys would be like, "Ooo Mr. P., Mr. P.
"I drew my vector diagram."
And I'm like, "No, no.
"No you didn't, that ain't no vector diagram."
Class, what is this?
Bo: That is a quadrilateral.
Billy: You could also call it a quadrangle
or a tetragon but quadrilat--
Quadrilateral is probably what it's most often called.
mr.p: Yes, you can call this a quadrilateral, a quadrangle,
a tetragon, or even just a polygon.
But you certainly cannot call it a vector diagram.
This is a vector diagram.
Notice arrows indicating the direction of the vectors
and labels to identify the vector names.
Please draw vector diagrams, not polygons.
Billy: I'm in the math club.
mr.p: At this point, we can't use the Pythagorean theorem,
or SOH CAH TOA because we don't even have a triangle.
Bo: We could draw a line from the start of vector A
to the end of vector B
and then we could use the law of cosines to determine the
length of that line and then we could--
mr.p: Bo, please stop.
Absolutely Bo, I'm sure we could find
a way to draw some lines up here
and use the law of sines and the law of cosines
and all those things to solve this particular problem.
But rather than give you an individualized approach to
an individual problem, I always try to give you guys
a general approach that will work in all situations.
In the long run, you will all understand it better.
The general approach to solving vector addition problems
is to use what we learned in our last lesson about
breaking or resolving vectors into their component vectors.
Bobby, could you please break vector B into its
component vectors in the X and Y directions.
Bobby: Oh right, that's a right triangle so--
mr.p: SOH would be fine.
Bobby: Yeah, we could use sine theta equals opposite
over hypotenuse and opposite the angle is
B in the Y direction and the hypotenuse is the vector B,
multiplied by B to solve for vector B in the Y direction
and B sub Y equals B times the sine of theta
or 40 times the sine of 35,
which is 22.943 cm.
mr.p: Correct and Billy, how do we find the component
of vector B in the X direction?
Billy: Cosine theta equals adjacent over hypotenuse,
or the component of B in the X direction over vector B.
Multiplied by B to get B in the X direction
equals B times the cosine of theta, or 40 times the
cosine of 35, which is equal to--
Bobby: 32.766 cm.
Bo: Yeah.
mr.p: Great, now we have the component vectors of B
in the X direction and B in the Y direction.
Bobby, please explain to me why we don't need to
break vectors C and A into their component vectors.
Bobby: Oh uh, because A and C are already directly in
either the X or Y direction and therefore,
they don't need to be broken into
components in the X or Y direction.
mr.p: Yes. Now notice what happens when we re-draw
our vector diagram and instead of using vector B,
we draw in the components of vector B.
Now, we have a right triangle and we can use
SOH CAH TOA and the Pythagorean theorem
to determine our resultant vector.
Now please notice that I've labeled this angle as phi.
This angle right here as phi, which is different than our
angle theta for our vector B, because this is
the angle for the direction of our resultant vector.
Bo, could you please figure out phi,
the angle for our resultant vector.
Bo: We can use tangent phi equals opposite over adjacent
and opposite the angle is C plus B in the Y direction
and adjacent is A plus B in the X direction.
Taking the inverse tangent of the whole equation
gives us that phi equals the inverse tangent of
the quantity of C plus B in the Y direction,
divided by A plus B in the X direction.
With numbers, that is the inverse tangent of 30+22.943
divided by 50+32.766,
which works out to be--
Bobby: 32.606 degrees
and my calculator is in degree mode.
So I know I'm right.
Billy: I'm in the physics club too.
Bo: Excuse me a sec.
What are you babbling about?
Billy: Well what I said was I'm in the math club,
the Latin club and the physics club.
mr.p: Yep, the angle of 32.606 degrees will give us the
direction and now we need the magnitude.
Bobby, could you please find the magnitude?
Bobby: We can use the Pythagorean theorem
so A squared plus B squared equals C squared.
Or, the resultant vector squared equals the quantity of
A plus B in the X direction squared,
plus the quantity B in the Y direction, plus C squared.
Taking the square root of the whole equation gives us
the resultant vector equals the square root of
again, the quantity of A plus B in the X direction squared
plus the quantity B in the Y direction,
plus C, that quantity squared.
With numbers, that is the square root of
50+32.766 squared +22.943+30, that quantity squared,
which works out to be...
Billy: 98.251 cm.
mr.p: And with 2 sig figs, the resultant vector
or total displacement of our Slow Velocity Racer is
98 cm at and angle of 33 degrees North of East
because this angle phi is 33 degrees Northward from East.
Now, let's just review what this means for a moment.
Slow Velocity Racer can travel the vectors A, B and C
or the vectors A, B in the X direction,
B in the Y direction and C.
Or even just the resultant vector R
and she will still start and end in the same location
and therefore have the exact same total displacement.
That is what we just figured out.
Any questions?
Billy: Nope, thanks.
Bobby: I'm good.
mr.p: In that case, let's review.
We started by giving our vectors names.
We then drew our vector diagram,
which had arrows and labels for all of our vectors.
We broke vector B into its components in the X
and Y direction, then we re-drew our vector diagram.
And because we had a right triangle,
then we were able to use the Pythagorean theorem
and SOH CAH TOA to solve for our resultant vector.
Thank you very much for learning with me today.
I enjoy learning with you.