Understanding Vectors: Exploring an Interesting Physics Problem

[tiffanysnsd]

So I have constructed this vector for a physics problem:

http://img23.imageshack.us/img23/1753/vectorx.jpg

Why is it when I take the tangent of the labelled angle, I get an angle of 77*, which is the complementary pair for the correct angle (which is 13*).

I can rationalize this by adding 90* to the calculated angle of 77*, the sum for which is 167*. 180*-167* = 13*, so I get the correct answer by accounting for the quadrant the vector is located in. But in my (admittedly limited) experience with vectors, I have never had to adjust the calculated angles for the quadrant (on a cartesian plane) where the vector is located; the trigonometric ratio between the two sides has always given the correct angle.

 

[kuruman]

Can you state the problem the way it is phrased and show your work? It seems that the angle that you have interpreted as θ is the complementary angle of what the problem thinks is the correct θ.

 

[tiffanysnsd]

Sure.

Three charges are arranged as shown:

http://img8.imageshack.us/img8/1841/82829031.jpg

where,
q1 = 2.5x10E-17
q2 = 3.0x10E-17
q3 = 3.5x10E-17
(all are positively charged)

Calculate the magnitude and direction of the force acting on q1.

Let up and right be positive.

₂₁F = 2.7x10E-23 N
₃₁F = 8.8x10E-23 N
(Both of these were calculated using Coloumb's Law).

So, I then proceeded to resolve these forces into the horizontal and vertical components:
F_x = ₂₁Fcos37* + ₃₁Fcos90*
= -2.2x10E-23 N + 0N
= -2.2x10E-23 N (negative because left has been defined as negative)
F_y = ₂₁Fsin37* + ₃₁Fsin90*
= 1.0x10^-22 N

So using the Pythagorean Theorem, I calculated the net force to be: 1.0x10E-22 N

This is where I had to now calculate the angle. I also performed a solution in terms of cosine line and got an angle of approx. 12-13* as I should get.

 

[kuruman]

Your drawing shows two charges labeled q2 and no q1. Which one of the two q2 charges is q1? You don't need to repost the drawing, just describe it.

 

[tiffanysnsd]

Oh, I apologize for that. The top point is q1.

 

[kuruman]

Your method is fine. The correct answer could be many things depending on what your reference for measuring the angle is.

It is 13_o from the positive y-axis.
It is 103^o from the positive x-axis.
It is 77^o from the negative x-axis.

If this is a web-based problem graded by the machine using a stored formula, then the reference line should have been made clear in the statement of the problem. If this is a homework problem graded by a sentient human being, any of the above three answers would be correct. BTW, I didn't redo your numbers, I assume you did them correctly.

 

[tiffanysnsd]

Thank you, that is reassuring.

Do you suppose I could rationalize it the following way:

Instead of using the angle of 37* for my calculations, I would use 143* (considering the vector for q2 points northwest and would lie in the second quadrant on a cartesian plane, so 180-37=143). Therefore, when I calculate the tan ratio as I did, it makes sense to add 90* in order to validate the answer for the second quadrant .

 

[kuruman]

No, don't bother. You understand how to do this problem and that's what counts. The line of reference of the angle is irrelevant. Personally, I prefer to express vectors in terms of their x and y components and forget about angles and magnitudes.

 

[tiffanysnsd]

Does my rationalization not make sense (or is inapplicable) or do you just think explanations along these lines are useless, provided I know how to approach the problems?

It isn't really so much for this problem that I made this inquiry, but I don't want to be caught with my pants down again in a similar way in the future. For example, if I'm writing an exam and have limited time to check my answers, I think it might be useful to find a working approach - which, in this case, involves using angles from 0-360 (as in a cartesian plane) and adjusting the calculated angle (obtained from trig ratios) depending on the quadrant it would be located in a cartesian plane. Otherwise, I fear answering with the wrong angle again, since my initial answer for this question was the complementary angle to the correct one.

 

[kuruman]

As long as you specify with respect to what line you measure your angles and your calculations are correct, nobody in his/her right mind will mark you wrong. If you want to put your mind at ease, before you take an exam, talk to your instructor and ask "Is there a preferred line with respect to which angles are measured or can I specify it myself?" and see what (s)he says. If you asked me that, I would say "As long as you specify the reference line, and your answer is correct, you get full credit."

 

[tiffanysnsd]

I believe I should clarify myself a bit more.

As you can see from my diagram in my original post, the calculated angle clearly uses the x-axis as the reference. So one would expect that using the tan trig ratio to calculate the angle, dividing the y-component by the x-component, the correct angle would be obtained. However, in my original solution, for some reason, that gave the complementary angle (with the reference axis clearly defined).

So, I got the question marked wrong, since when I made the calculation, I thought my method would provide the correct angle. I mean, I did establish the x-axis as the reference, I had all the magnitudes correctly defined; yet, the trig ratios did not yield the correct answer and needed to be adjusted. Presumably, I think, because I have to account for the vectors' position on a cartesian plane.

 

[kuruman]

I redid your calculations and the original diagram you posted is correct. You say 77 degrees with respect to the negative x-axis, I say 76.13 degrees with respect to the negative x-axis. Your answer is correct to within roundoff errors.

Who marked your answer wrong? You need to take the matter to your instructor and have him/her explain why your answer was marked wrong. I don't see anything wrong with it.