Vector Components and Resultant Vectors: A Homework Challenge

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Homework Help Overview

The discussion revolves around vector components and resultant vectors, specifically involving a series of movements in different directions and their respective distances. The original poster describes a path consisting of movements east, south, and at angles relative to cardinal directions, leading to confusion regarding the resultant vector's magnitude and direction compared to an instructor's provided answer.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the components of each vector and derive the resultant vector using the Pythagorean theorem and inverse tangent functions. Some participants question the quadrant of the resultant vector and the implications of using inverse trigonometric functions, while others suggest potential errors in calculations or interpretations of angles.

Discussion Status

The discussion is active, with participants providing insights and questioning assumptions about the calculations. There is acknowledgment of potential errors in the original poster's approach, particularly regarding the quadrant of the resultant vector and the interpretation of calculator outputs. Guidance has been offered regarding the use of inverse trigonometric functions and their limitations.

Contextual Notes

Participants note the importance of accurately interpreting directional angles and the potential for misunderstanding in vector representation. The original poster mentions a lack of prior knowledge in trigonometry and geometry, which may influence their understanding of the problem.

scottzilla
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Homework Statement


A person follows the path outlined below. 100m east, then 300m south then 150m 30 degrees south of west, then 200m 60 degrees north of east.


Homework Equations


I used component vectoring and have not been coming up with the resultant answer that the instructor has outlined.


The Attempt at a Solution


I label each vector as A, B, C, and D. Then figure the components of each: Ax=100 Ay=0, Bx=0 By=-300, Cx-129.9 Cy=-75, and Dx=-100 and Dy 173.2. This gives me Rx- -129.9 and Ry= 201.8. I then use pythagorean theorum to do radicand (-129.9 squared)+(-201.8 squared) which gives me 154.4. Yet ther instructor shows his answer as 240. I also then do angle theta =Tan-1(-201.8/-129.9) which gives me tan-1(1.553) wihch gives me the answer 57.2, but the instructor gives me the answer 237 degrees.

Any help would be greatly appreciated as I feel I am getting hte hang of it except for this type of question.

Thanks in advance,

Scottzilla
 
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scottzilla said:
...then 200m 60 degrees north of east.

... Dx=-100 and Dy 173.2.

I think I see the problem. What quadrant is vector D in? (Incidentally, I don't see how your instructor gets their answer, either. I'm wondering if they misinterpreted one of the directions, for C or D...)
 
Crud, sorry it should read 60 degrees north of West. My fault, hope this can help you figure it out. Thanks for taking a look at it so far though:) Any help would be greatly appreciated.

Scottzilla
 
In that case, this is correct:

scottzilla said:
I label each vector as A, B, C, and D. Then figure the components of each: Ax=100 Ay=0, Bx=0 By=-300, Cx-129.9 Cy=-75, and Dx=-100 and Dy 173.2. This gives me Rx- -129.9 and Ry= -201.8.

But I will ask you how you got the hypotenuse to be smaller than one of the sides...

I then use pythagorean theorum to do radicand (-129.9 squared)+(-201.8 squared) which gives me 154.4.

Mayhap there was a calculator entry error: you should get your instructor's result now...

I also then do angle theta =Tan-1(-201.8/-129.9) which gives me tan-1(1.553) wihch gives me the answer 57.2, but the instructor gives me the answer 237 degrees.

Inverse trig functions constantly reveal the danger of trusting a calculator... Your tangent is calculated correctly. But what quadrant is your resultant vector in? How many angles can have the same tangent value? What other angle has a tangent of 1.553?
 
Thanks, I got it now. If I had just looked at my diagram it would have become clear to me. Being in quadrant III means it would be what my calculator got (57.2) but I should have added 180 degrees, correct? Sorry about not following your explanation completely but I have never had trig or geometry. I am learning them as I go through this class:)

Scottzilla
 
When you have picked up enough trigonometry, you'll learn that there are (generally) two angles in the cycle from 0º to 360º which have the same values for sine, cosine, and tangent. If you reverse the question to ask what angle has a particular value of one of these functions, you have two answers (usually) then. But a mathematical function can only give one "output" value for a chosen "input" value and a calculator can only work with functions.

So the "inverse trig functions" -- arcsine, arccosine, arctangent, etc. -- are created by restricting their "outputs": arcsine and arctangent only give values from -90º to +90º and arccosine only gives values from 0º to 180º. However, real problems can lead to angles in any of the four quadrants, so the user of calculators or software has to know how to interpret the "output" from a requested function. In your problem, the resultant vector points into the third quadrant, but arctangent can only provide results in the fourth or first. When that happens, knowing that the output is in the "wrong" quadrant, you have to take the value of the arctangent and add 180º to it. (There are similar, but different, manipulations for arcsine and arccosine.)
 
Thanks, I am getting the hang of it now.

Scottzilla
 

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