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Why in the name of all that is good would someone want to do something like THAT? Question: Non-right Triangle Vector Addition Subtitle: Non-right Triangle.

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Presentation on theme: "Why in the name of all that is good would someone want to do something like THAT? Question: Non-right Triangle Vector Addition Subtitle: Non-right Triangle."— Presentation transcript:

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2 Why in the name of all that is good would someone want to do something like THAT? Question: Non-right Triangle Vector Addition Subtitle: Non-right Triangle Vector Addition

3 Answer: Because there is no law that states vectors must add up to make right triangles. (Oh, but if only there were.)

4 An ant walks 2.00 m 25°N of E, then turns and walks 4.00 m 70°N of E. Find the total displacement of the ant. RIGHT TRIANGLE This can’t be solved using our right-triangle math because IT ISN’T A RIGHT TRIANGLE! 4.00 m 2.00 m dtdt CONSIDER THE FOLLOWING...

5 An ant walks 2.00 m 25°N of E, then turns and walks 4.00 m 70°N of E. Find the total displacement of the ant. This can’t be solved using our right-triangle math because it isn’t a RIGHT TRIANGLE! We need to break the two individual vectors into pieces, called x- & y-components

6 Adding these component vectors together produces the original vector. Any vector can be broken into two or more COMPONENT vectors.

7 Adding these component vectors together produces the original vector.

8 If we break the original vector into one component that follows the x-axis and one that follows the y-axis... …we get a right triangle!

9 If we break the original vector into one component that follows the x-axis and one that follows the y-axis... …we get a right triangle! This is REALLY cool because we can calculate the magnitudes of these component vectors using trig functions. (YEA!!!!!!)

10 …then: y = R sinθ and x = R cosθ θ R x y If we label the triangle as so…

11 We can use these relationships to find the x- and y-components of each individual vector. Once we have those 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 TRY IT!

12 R 1 = 2.00 m 25°N of E 25° x = R cosθ = (2.00 m) cos 25° = +1.81262 m y = R sinθ = (2.00 m) sin 25° = +0.84524 m 1.81262 m 0.84524 m

13 R 2 = 4.00 m 70°N of E x = R cosθ = (4.00 m) cos 70° = +1.36808 m y = R sinθ = (4.00 m) sin 70° = +3.75877 m 1.36808 m 3.75877 m

14 Now we have the following information: xy R1R2R1R2 +1.81262 m +1.36808 m +0.84524 m +3.75877 m

15 Now we have the following information: xy R1R2R1R2 +1.81262 m +1.36808 m +0.84524 m +3.75877 m 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.

16 xy R1R2R1R2 +1.81262 m +1.36808 m +0.84524 m +3.75877 m +3.18070 m +4.60401 m x-component of resultanty-component of resultant

17 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. 3.18070 m 4.60401 m dTdT θ

18 The Pythagorean theorem will produce the magnitude of d T : c 2 = a 2 + b 2 (d T ) 2 = (3.18070 m) 2 + (4.60401 m) 2 d T = 5.59587 m  5.60 m A trig function will produce the angle, θ: tan θ = (y/x) θ = tan -1 (4.60401 m / 3.18070 m) = 55º

19 Of course, ‘55º’ is an ambiguous direction. Since there are 4 axes on the Cartesian coordinate system, there are 8 possible 55º angles. 55º 55° …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)

20 3.18070 m 4.60401 m dTdT θ From the diagram we can see that the angle is referenced to the +x axis, which we refer to as EAST. The vector d T is 55° north of the east line; therefore, the direction of the d T vector would be “55° north of east”

21 So, to summarize what we just did…

22 We started with the following vector addition situation… 4.00 m 2.00 m dtdt …which did NOT make a right triangle.

23 dtdt Then we broke each of the individual d vectors ( the black ones) into x- and y-components… …and added them together to get x- and y- components for the total displacement vector.

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25 Just a few things to keep in mind... X-component vectors can point either EAST or WEST. EAST is considered positive. WEST is considered negative.

26 Just a few things to keep in mind... Y-component vectors can point either NORTH or SOUTH. NORTH is positive. SOUTH is negative.

27 This vector has a POSITIVE x-component... …and a NEGATIVE y-component.

28 This vector has a NEGATIVE x-component... …and a POSITIVE y-component.

29 And another thing... In order for the component equations, y = R sinθ and x = R cosθ to give correct values, θ must be a horizontal angle. θ R x y

30 In the compass direction, 55° N of E, the 55° angle is referenced to the EAST; therefore, it is a horizontal angle. ° W of S, the 38° angle is referenced to the SOUTH -- If the direction of a vector is 38° W of S, the 38° angle is referenced to the SOUTH -- it is a vertical angle. ( In order to obtain correct values from the component equations, you must use its complementary angle, 52°.)

31 Yeah, baby! Let’s give it a try!


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