ADVANCED VECTORS
A lot of the formulas we deal with in Physics involve vectors. Velocity is a vector. Acceleration is a vector. Forces are vectors. Momentum is a vector. And so on and so forth. It is very uncommon for only ONE vector to be active in a system. Rather, there are usually two or three or more vectors all occurring simultaneously. This requires us to combine multiple vectors using a Cartesian coordinate system, so that we can deal with a single resultant vector instead of all the smaller pieces.
In order to do this, you must follow four basic steps:
1. Convert all vectors to rectangular form. This means to use the formulas for x and y in order to get the original vectors in horizontal and vertical components.
2. Use negatives were appropriate. We follow the same rules as your math teachers. In the x-dimension, a vector is negative if it points towards the LEFT. In the y-dimension, a vector is negative if it points DOWN.
3. Combine the x-components. Combine the y-components. Now that we have split everything into horizontal and vertical components, we can combine like terms. We use a special symbol: . It is the Greek letter sigma. It means to take the SUM or TOTAL or to COMBINE everything together. What we are doing is calculating X (the total x component) and Y (the total y component).
4. Convert your final answer back into polar form. This means to use the formulas for r and in order to get your final answer as a magnitude and direction.
Let’s try an example. Bob is paddling his canoe to the Northeast. Meanwhile, several factors are complicating our story. The wind is blowing him to the Southeast. A rope is pulling him to the Southwest. And the water current is moving him to the Northwest. So what exactly is going to happen next? Can we predict the overall motion for the canoe? The answer is yes. It just requires vectors.
N
Current moving boat Bob paddling boat
10 m/s 10o 5 m/s
W 75o E
20o
60o
Rope pulling boat
18 m/s
Wind blowing boat
14 m/s
S
We will deal with each vector separately.
PADDLE: x = 5cos75 = 1.29 (right is positive)
y = 5sin75 = 4.83 (up is positive)
WIND: x = 14cos30 = 12.1 (right is positive)
y = 14sin30 = - 7 (down is negative)
But wait!! Where did the angle 30 o come from? The picture shows 60 o . Remember that we always measure from the nearest X-AXIS. In this case, they gave us the angle to the y-axis and so we have to fix it. Be careful with this! It really does matter.
ROPE: x = 18cos20 = -16.9 (left is negative)
y = 18sin20 = -6.16 (down is negative)
CURRENT: x = 10cos80 = -1.74 (left is negative)
y = 10sin80 = 9.85 (up is positive)
Now let’s start putting the pieces together.
We can combine all the horizontal components: X = 1.29 + 12.1 – 16.9 – 1.74 = -5.25 m/s
We can combine all the vertical components:Y = 4.83 -7 – 6.16 + 9.85 = 1.52 m/s
Notice that this will put our resultant vector in QUADRANT II. I know this because the x value is negative and the y value is positive. For those of you that have forgotten your quadrants, let me remind you:
II I
III IV
In this class, all angles are given as positive values. Then we have to identify the quadrant using words. Sometimes you can do this directly. For example, 25o into quadrant IV. Sometimes you can base it off of the x-axis itself. For example, 67 o above the negative x-axis. Or sometimes we use the compass directions. For example, 32o North of East. Any of these would be acceptable angles. But you should NEVER give an angle all by its lonesome. Because then I have no idea which quadrant it belongs to.
Let’s finish the canoe example.
Our final step is to switch back to polar form.
The resultant vector = r = (x 2 + y 2 ) = ( 5.252 + 1.522 ) = 5.47
Notice that I ignored the negative sign. I did this because when you square a number it goes away. So save yourself some time by ignoring it.
The resultant angle = = tan -1 ( y / x ) = tan -1 (1.52 / 5.25 ) = 16.1 o
Notice that I ignored the negative sign again. I only care about the size of my angle. I will then use common sense to place it in the correct quadrant. In this case, the correct quadrant is North of West.
Final answer: The canoe will move 5.47 m/s at an angle 16.1o North of West.
Whew!! That was a lot of work. But we did it. There are a lot of places to make stupid mistakes, but the overall process is very repetitive and very predictable.
Let’s practice a few.
PRACTICE: The wind is pushing a kite with a force of 20 N at an upward angle of 30 degrees. Meanwhile, the string pulls back with a tension of 25 N at a downward angle of 45 degrees. Gravity pulls down on the kite with 8 N. (See diagram)
Convert the string force to rectangular form. ( -17.6, -17.6 )
Wind = 20 N
Convert the wind force to rectangular form. ( 17.3 , 10 )
30 o
45o Convert the gravity force to rectangular form. ( 0 , -8 )
Calculate the total force in the x dimension. ( x= )
String -0.4
= 25 Ngravity = 8 NCalculate the total force in the y dimension. ( y= )
-15.6
ANSWER IS IN QUANDRANT III. We know Convert these back into polar form to get the total force. (r=)
this because of the component vectors.r = 15.6
NEG X and NEG Y.What is the direction of the net force?
= 88.5o below the neg x axis
This kite is going to come crashing straight down unless the wind starts to blow harder!
PRACTICE: No baby steps this time. A sea turtle has been caught in a net. The net pulls the turtle 50 meters at an angle 22 degrees North of West. Meanwhile, a water current pushes the turtle 25 meters at an angle of 50 degrees South of West. A strong wind blows the turtle 12 meters directly South. How far is the turtle now from where he started? What direction did the turtle move? (In the world of physics, direction always means angle.)
50 mnet: [ 50 , 22 o ] becomes ( -46.6 , 18.7 )
current: [25 , 50 o ] becomes ( -16.1 , -19.2 )
wind: [12, 90 o ] becomes ( 0 , -12 )
22o = ( -62.7 , -12.5)
50oANSWER IS IN QUADRANT III. We
r = 63.9 metersknow this because of the components.
25 m = 11.3o degrees South of WestNEG X and NEG Y.
12
In class we did a lab where students had to purposely get three vectors to balance out perfectly. We say that vectors are “balanced” if the resultant vector equals zero. In our lab we created vectors by hanging masses from a large wooden circle. An unbalanced vector makes the board tip over and go crashing to the floor. In order to balance it out we had to place another vector on the other side. In order to make things more complicated, I gave groups three different objects with different masses. The only way to get the board to balance is to get the horizontal and vertical components to cancel each other out perfectly so that there is no leftover force to tip the board.
BALANCED FORCES LAB:
OBJECTIVE: To show mathematically why a system is perfectly balanced.
DESIGN:
y
x
Side ViewTop View
1. Obtain a large wooden circle and balance it on a bar stand.
2. Obtain three masses from the teacher and hang them on your slotted masses.
3. Rearrange the masses until you get the system to balance.
4. Record the masses AND angles at each location. Each slit in the board is 10o .
DATA COLLECTION: Draw a diagram showing the direction (angle) and magnitude (mass) of each hanging
piece. Make sure you include the mass of the hangar itself! It is 50 grams.
Example: 150 g
80o
250 g 350 g
70o 60o
DATA ANALYSIS: 1. Convert the masses into KILOGRAMS. (Because kg is the standard unit)
2. Resolve each piece of the system. (Convert it to rectangular form.)
3. Combine each dimension separately.
CONCLUSION: 1. What was the net vector in the x dimension?
2. What was the net vector in the y dimension?
3. Looking at your answers to part 1 and part 2, any guesses while the system balanced?
4. Why do other angles NOT work? What is it about the system that makes it tip over?
EVALUATION: Do you trust your results? Were there any problems or limitations I should be aware of?