Unit 1, Part 3 Vectors---AKA “Paying appropriate honor to Trigonometry”
A vector is any quantity that has both a size (a number value) and a direction. Examples of vectors that we have studied so far are velocity, acceleration, and force. Each of these quantities are represented by a numerical size and a direction. So far, we have just used “positive vs. negative” and “North, South, East, or West” for direction. That’s about to change. In this chapter, you will begin to use “degrees away from zero” to relate the direction a vector quantity is acting. You will need to use trigonometry to do this.
In this class we will represent vector quantities by an arrow. The direction the arrow points is the direction the vector quantity acts and numerical value of the vector quantity is written on top of it. There are two parts to a vector: The “head” (the point/tip of the arrow) and the “tail” (the non-point-end of the arrow). Example: If I wanted to represent two different displacements, one acting 10m East (00 ) and the other 30m East (00), I could represent them by the following vectors (arrows):
10m 30m OR
“tail” “head”
We’ll be using the 3600 scale (shown below) to represent the direction vectors act in here the majority of the time. It’s the same Cartesian-coordinate scale you use in math.
900
1800 00
2700
10m
300
Example: A 10m displacement acting at 300 looks something like this:
In many problems in this unit, you will have at least two vectors acting on the same “thing” (like a box, or a car, a football, etc.). They might be force vectors (coming in the next unit), velocity vectors, displacement vectors, etc. When this happens, you will need to “add” the two vectors together in such a way that will create one, single vector (called the “resultant vector”) that causes the same action by itself that the two were doing before. To do this, you MUST follow a very specific set of rules called “vector addition”. Here they are with an example you can follow: ( FYI: This is a very long example with many different parts to it and goes on for a very long time)
Example: What is the “resultant displacement” of two displacement vectors if one is 10m at 00 and the other is 20m at 900?
The first thing you ALWAYS need to do is draw a picture that pictorially represents the scenario. The following diagram represents the diagram of the word-scenario. The diagram below shows how you would draw the original vectors: one vector is a 10m displacement acting at 00 (shown below pointing to the right, which is toward 00 on the 3600 scale) and the other is a 20m displacement acting at 900 (shown below as the vector pointing up, which is 900 on the 3600 scale).
20m
10m
In a situation like this (depending on the physics problem), the first thing any problem might have you do is find the “resultant”. Again, the resultant vector is a single (one) vector whose action will do the exact same thing as any two or more vectors acting on the object will do. The resultant vector ALWAYS includes a numerical value (in m/s, m, m/s2, etc----whatever the units of the two original vectors were) AND an angle of direction that it acts. The angle of direction is always the angle it acts from the horizontal (which is how many degrees it is away from 0o). To add the two original vectors together so you have only ONE “resultant” vector that will do the same thing as the above two do, re-draw the vectors in such a way that their direction doesn’t change, but the head of one is touching the tail of the other. In order to find the “resultant”, the original two vectors must make a 90-degree angle with each other. As long as the original vectors make a 90-degree angle with each other, you can create a single vector (the resultant vector) that will have the same action the other two had combined. It doesn’t matter the order in which you place the original two vectors as long as their directions don’t change and one vector’s head is touching the other one’s tail. You can place the original two vectors from above (the 10m and the 20m vectors) together in either of the following two ways:
10m
20m OR 20m
10m
In order to find the “Resultant Vector” (the single vector that can replace the two vectors), redraw the two original vectors “head-to-tail” and then draw a dashed arrow beginning from the open tail of one and pointing to the open head of the other. The tail of the resultant attaches to the open tail of one of the vectors and the head of the resultant attaches to the open head of the other vector. Since your resultant MUST have a direction, you must have a head on it. Don’t just draw a dashed line….always include the tail and the head.!!!!! The resultant vector’s symbol is a subscript “R”. The resultant vector for the original two forces is shown below. Since I showed you the two different ways that the original vectors can be added (either is correct and will yield the same resultant), I have drawn the resultant for both. Notice that the resultant is the same for both different scenarios (the angle/direction/size). This should tell you that the ORDER in which you add the original vectors doesn’t matter as long as they are drawn “head-to-tail”. You just need to make sure you add them “head-to-tail”. The above two original examples have the “resultant” shown as the bold-type dashed arrow in the diagrams below. Notice that it didn’t matter which original vector’s head I joined to the other’s tail, the resultant is the same angle/direction and length for both (with the angle being the angle it makes with zero degrees).
OR
Now, to figure out the numerical value of the resultant (how many “meters” the resultant displacement will represent) and the direction it acts, you will need to use the Pythagorean theorem and trigonometry. If you look at either drawing of the two original vectors and the resultant, you’ll see that together they form a right triangle. If you know the length of any two sides in a right triangle, you can figure out the length of the third. In our example, the resultant vector is the “hypotenuse” of the triangle, and the other two sides represent 10m and 20m displacement. The Pythagorean theorem is: A2 + B2 = C2. “C” in this equation is the “hypotenuse”, which also represents our resultant force vector. So, if you wanted to figure out the size (numerical value) of the resultant, just plug in the values. It doesn’t matter which side you’re calling “A” or “B”, just as long as you know that the hypotenuse/resultant is “C” in the equation.
(10m)2 + (20m)2 = (Resultant)2
resultant = (10m)2 + (20m)2 , resultant = 22.3m
This means that you could replace the two original displacements that the object originally underwent with a single 22.3m dsiplacement acting at “some angle to 00” (to be determined in a minute…) and the object will have the exact same displacement. Now you need to find the angle of the resultant displacement. In order to do this you need a review of your trigonometry.
Review of Trigonometry
If you have a right triangle (this only works with right triangles) and you know the length of at least two of the sides, you can figure out the angles the sides make with each other. In order to do this, you need to know the “names” of the sides.
a Hyp.
1
b
2
In any right triangle, the hypotenuse is ALWAYS the side directly opposite of the right/900 angle. It is ALWAYS the longest side in the right triangle.
If we are referring about angle “b” in the above triangle, then side 1 would be the side opposite angle b. Side 2 is the side adjacent to angle b (the other side adjacent to angle b is the hypotenuse and that is never referred to as an “adjacent” side). If we’re talking about angle “a”, then side 2 is the side opposite of angle a and side 1 is the side adjacent to angle a. The “opposite” and “adjacent” sides change depending on the particular angle of the triangle you’re talking about. Then you just plug the values into one of the following formulas. (You need to KNOW these; I won’t give them to you on a test….)
sinq = the length of the opposite side
the length of the hypotenuse
cosq= the length of the adjacent side
the length of the hypotenuse
tanq= the length of the opposite side
the length of the adjacent side
In order to find the angle (i.e. “the direction”) of the resultant, you need to know what angle we are referring to. THE ANGLE OF THE RESULTANT IS ALWAYS THE ANGLE IT MAKES WITH ZERO DEGREES. To determine the angle, picture the tail of your resultant vector coming out of the center (origin) of an x-y 3600 coordinate axis system. The angle it makes with the closest horizontal (x-axis) is the angle you are looking for. However, sometimes you might need to do some additional thinking to determine the actual angle your resultant makes with zero degrees,
22.3m 20m
q
10m
In the above example (which is the original example from the start of these notes), the angle we’re looking for is the one with the little “q ” symbol. That is the Greek symbol “Theta” and it stands for “angle”. That particular angle in the above diagram is the one the resultant makes with the 00 horizontal. You can use any of the three trig functions to calculate the angle. I’m going to use the “sin” (pronounced sign or sine) function.
sinq = opp sinq = 20m sinq = 0.896
Hyp 22.3m
The number you just found is the “sin of the angle”, not the angle itself. In order to calculate the actual angle, you need to take the “inverse sin” of that number. Look at your calculator and find the “sin” button. Directly above it is a “sin-1”. That’s the “inverse sin” function. To get the angle, hit the “shift” or “second function” button on your calculator, then hit the “sin” button (with the 0.896 number still in the calculator). The number that pops up should be 63.740. ( If you get some random number other than 63.74, your calculator is probably in RADian mode and you need to switch it to DEGree mode. If you don’t know how to do this, ask me or someone else close by.) You can round that to 640. Some calculator’s require you to put the function in first (inverse sin) and then the number (for example: 1)“shift”, 2)“Sin”, 3)“place in your number”). You’ll just have to work with the above example to figure out what kind of calculator you have. And please make sure your calculator is in “degree” mode, not radian.
So, to complete the original problem, what you have come up with is a resultant vector that has a value of 22.3N directed at 640. If you replace the ORIGINAL two vectors with this ONE resultant, the box will move in the same direction as if the two original vectors were still acting on it.
Here are some practice problems. Remember to always draw a diagram!!!
1. After walking 11 Km due North (900) from camp, a hiker then walks 11 Km due East (00). (a) What distance did the hiker walk, and (b) what was his total displacement from the starting point?
11 Km N
11 Km
W E
S
The diagram for problem #1 is above. Notice here that the way that I have drawn my two 11km vectors and the resultant, the angle the resultant makes with zero degrees is outside of the actual 900 triangle. However, since we need the value of that angle to determine the displacement, you can find the value of the angle inside the triangle and then subtract it from 90 degrees to get the true angle of the resultant (the angle it makes with zero degrees). In this problem, both “sides of” the angle are 11km, so the opposite and the adjacent sides would both be “11 km”.
2. Two boys push on a box. One pushes with a force of 125 N to the East (00) and the other pushes with 165 N to the North (900). What is the resultant (size and direction) force on the box? (Yes, I am aware that we haven’t covered forces, but we will in the next unit so just understand for now that the unit of force is the Newton—N)
Draw a diagram first!!!!
125N
165N
· The horizontal (east) vector is 125N and the vertical (north) vector is 165N (look at the N-S-E-W axis system drawn with the previous practice problem if you don’t understand this).
· The “re-drawn” head-to-tail version with the resultant drawn can look like either of the flowing two diagrams:
125N