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When right triangles go wrong.

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.

shannon-mcleod
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When right triangles go wrong.

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  1. 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?

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

  3. CONSIDER THE FOLLOWING... 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. dt 4.00 m 2.00 m This can’t be solved using our right-triangle math because IT ISN’T A RIGHT TRIANGLE!

  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. We need to break the two individual vectors into pieces, called x- & y-components This can’t be solved using our right-triangle math because it isn’t a RIGHT TRIANGLE!

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

  6. Adding these component vectors together produces the original vector.

  7. 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!

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

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

  10. 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!

  11. 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

  12. R2 = 4.00 m 70°N of E 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

  13. x y R1 R2 +1.81262 m +1.36808 m +0.84524 m +3.75877 m Now we have the following information:

  14. 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

  15. +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

  16. 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

  17. 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

  18. 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)

  19. 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”

  20. So, to summarize what we just did…

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

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

  23. YES, IT'S THAT EASY!!!

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

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

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

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

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

  29. In the compass direction, 55° N of E, the 55° angle is referenced to the EAST; therefore, it is a horizontal angle. 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°.)

  30. Yeah, baby! Let’s give it a try!

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