Download
1 / 52
520 likes | 612 Views
Begin the slide show. When right triangles go wrong. Subtitle: Non-right Triangle Vector Addition. Question:. Why in the name of all that is good would someone want to do something like THAT?. Answer: Because there is no law that states vectors must add up to make right triangles.
E N D
When right triangles go wrong. Subtitle: Non-right Triangle Vector Addition Question: Why in the name of all that is good would someone want to do something like THAT?
Answer: Because there is no law that states vectors must add up to make right triangles. (Oh, but if only there were.)
CONSIDER THE FOLLOWING... An ant walks 2.00 m 25° N of E , then turns and walks 4.00 m 20° E of N. The total displacement of the ant… dt 4.00 m 2.00 m …can not be found using right-triangle math because WE DON’T HAVE A RIGHT TRIANGLE!
An ant walks 2.00 m 25° N of E , then turns and walks 4.00 m 20° E of N. The total displacement of the ant… We can add the two individual displacement vectors together by first separating them into pieces, called x- & y-components This can’t be solved using our right-triangle math because it isn’t a RIGHT TRIANGLE!
Into WHAT????????? COMPONENTS Every vector can be thought of as pointing somewhat horizontally…. [This is the black vector’s shadow on the y-axis] [This is the black vector’s shadow on the x-axis] …and somewhat vertically. They’re kind of like the vector’s shadows.
If we add the x- and y-components together… they create the original vector… …and it makes a right triangle!
Just a few things to keep in mind... EAST is considered positive. WEST is considered negative. Since X-component vectors can point either EAST or WEST…
Just a few things to keep in mind... SOUTH is negative. NORTH is positive. Since Y-component vectors can point either NORTH or SOUTH…
This vector has a POSITIVE x-component... …and a NEGATIVE y-component.
This vector has a NEGATIVE x-component... …and a POSITIVE y-component.
So, let’s practice what you just learned… COME ON…IT’LL BE FUN!!!! Complete # 1 – 6 on the worksheet your teacher has just given you. When you are finished, check your answers by continuing through this presentation.
The fact that the x- & y-components of a vector add up to create a right triangle is REALLY cool because we can calculate the magnitudes (lengths) of these component vectors using trig functions. (YEA!!!!!!)
R y θ x First, we label the triangle as so… R = the magnitude of the vector θ = the angle the vector makes relative to a horizontal (east-west) line x = the vector’s x-component y = the vector’s y-component
R y θ x …then, using our beloved trig functions: and we can calculate the magnitudes of the components: and
That’s how we can resolve (break) any vector into its x- & y-components. A word of caution, however…
R y θ x In order for the component equations, y = R sinθandx = R cosθ to give correct values for the x- & y-components, θ must be a “horizontal” angle. (an angle measured relative to a horizontal line)
In the compass direction, 55° N of E, the 55° angle is referenced to the EAST; therefore, it is a “horizontal” angle. 55˚ If the direction of a vector is 38° W of S, the 38° angle is referenced to the SOUTH -- it is an angle measured relative to a vertical line. 52˚ 38˚ In order to obtain correct values from the component equations, you must use its complementary angle, 52°.
Now complete # 7 – 9 on the worksheet. Again, when you are finished, check your answers by continuing through this presentation.
7) For the vector 1350 ft, 30° N of E… R = 1350 ft θ = 30°
8) For the vector 14.5 km, 20° W of S… R = 14.5 km θ = 70°
9) For the vector 2400 m, S… R = 2400 m θ = 90°
Now, let’s try resolving some vectors into their x-y components. Remember, the components are labeled ‘+’ or ‘-’, showing their directions. Also, θ must be an angle to the horizontal. Complete # 10 – 15 on the worksheet. The answers can be found at the end of the worksheet. When you are finished and have checked your work, continue on with this presentation.
R y θ x You use the following equations: y = R sinθandx = R cosθ to calculate the x- & y-components of a vector θ must be a “horizontal” angle. (an angle measured relative to a horizontal line)
Complete # 10 – 15 on the worksheet. The answers can be found at the end of the worksheet. When you are finished and have checked your work, continue on with this presentation. Remember, the components are labeled ‘+’ or ‘-’, showing their directions. ‘+’ is for East (x) and North (y) ‘-’ is for West (x) and South (y)
So, what can we do with all this stuff? Well, let’s go back to the ant and resolve (break) its two displacement vectors into components.
This was the situation... An ant walks 2.00 m 25° N of E , then turns and walks 4.00 m 20° E of N. The total displacement of the ant… dt 4.00 m 2.00 m R1 = 2.00 m, 25° N of E R2 = 4.00 m, 20° E of N
R1 = 2.00 m, 25° N of E 0.84524 m 25° 1.81262 m x = R cosθ = (2.00 m) cos 25° = +1.81262 m y = R sinθ = (2.00 m) sin 25° = +0.84524 m
R2 = 4.00 m, 20° E of N 3.75877 m θ = 70˚ 1.36808 m x = R cosθ = (4.00 m) cos 70° = +1.36808 m y = R sinθ = (4.00 m) sin 70° = +3.75877 m
So, you have broken the two individual displacement vectors into components. Now we can add the x-components together to get a TOTAL X-COMPONENT; adding the y-components together will likewise give a TOTAL Y-COMPONENT. Let’s review first…
R1 = 2.00 m, 25° N of E 0.84524 m 25° 1.81262 m x = R cosθ = (2.00 m) cos 25° = +1.81262 m y = R sinθ = (2.00 m) sin 25° = +0.84524 m
R2 = 4.00 m, 20° E of N 3.75877 m 1.36808 m x = R cosθ = (4.00 m) cos 70° = +1.36808 m y = R sinθ = (4.00 m) sin 70° = +3.75877 m
x y R1 R2 +1.81262 m +1.36808 m +0.84524 m +3.75877 m We have the following information:
Now we have the following information: x y Adding the x-components together and the y-components together will produce a TOTAL x- and y-component; these are the components of the resultant. R1 R2 +1.81262 m +1.36808 m +0.84524 m +3.75877 m
+3.18070 m +4.60401 m x-component of resultant y-component of resultant x y R1 R2 +1.81262 m +1.36808 m +0.84524 m +3.75877 m
θ Now that we know the x- and y-components of the resultant (the total displacement of the ant) we can put those components together to create the actual displacement vector. dT 4.60401 m 3.18070 m
The Pythagorean theorem will produce the magnitude of dT: c2 = a2 + b2 (dT)2 = (3.18070 m)2 + (4.60401 m)2 dT = 5.59587 m 5.60 m A trig function will produce the angle, θ: tanθ = (y/x) θ = tan-1 (4.60401 m / 3.18070 m) = 55º
55° 55º 55º 55º Of course, ‘55º’ is an ambiguous direction. Since there are 4 axes on the Cartesian coordinate system, there are 8 possible 55º angles. …and there are 4 others (which I won’t bother to show you). To identify which angle we want, we can use compass directions (N,S,E,W)
dT 4.60401 m θ 3.18070 m From the diagram we can see that the angle is referenced to the +x axis, which we refer to as EAST. The vector dT is 55° north of the east line; therefore, the direction of the dT vector would be 55° North of East
We started with the following vector addition situation… dt 4.00 m 2.00 m …which did NOT make a right triangle.
Then we broke each of the individual vectors ( the black ones) into x- and y-components… And now we have a right triangle we can analyze! dt …and added them together to get x- and y-components for the total displacement vector.
Yeah, baby! Let’s give it a try! Complete #16 on your worksheet. (Check back here for the solution to the problem when you are finished.)
