precalculus question: 2-D vectors

karmsy

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First time I have ever used this forum. Sorry if I didn't assign the question to quite the right category, I tried my best.

The question concerns 2-D vectors. It's on p. 615, from "Precalculus 5th Edition," Stewart/Lothlin/Redlin, if that makes a difference. Here it is:

"A man pushes a lawn mower with a force of 30 lb exerted at an angle of 30 degrees to the ground. Find the horizontal and vertical components of the force."

OK, given that v = |v|(cosine theta)i + |v|(sine theta)j; |v| in this case equals 30 lb, and as best I can figure, theta is 30 degrees, so: v = 30(sqare root of 3)/2i + 30(1/2)j. As far as I can tell, there should be no negative signs in here anywhere. The solution should be: v = 15(square root of 3)i + 15j.

This isn't the book's answer, though. The book has - 15j instead of + 15j, and I'm not sure why. I am having the same difficulty with other problems in the section, and I've been all through the text and examples trying to find clarification. Would somebody enlighten me? Thanks.
 
karmsy said:
... exerted at an angle of 30 degrees to the ground ...

The book has - 15j instead of + 15j, and I'm not sure why.

They are using a system where the positive direction vertically is defined to be UP into the sky; thus, the negative direction vertically must be in the opposite direction (i.e., DOWN into the ground).

It's natural to define the positive and negative VERTICAL directions this way because it matches the standard xy-coordinate system where the positive y-axis is pointing UP, and this also matches our tendency as humans to think of "up" as positive and "down" as negative. However, you are free to make your own definitions, and, if you choose to define the positive direction (vertically) to be DOWN, then you would use +15j for the vertical component of vector v. You would also need to explicitly state this definition.

So, using this natural system, the vertical component of vector v is negative because the man is pushing the lawn mower forward and downward. If the man were pushing the lawn mower forward and upward into the sky, then the vertical component would be positive.

Here's a sloppy picture.

lawnmower.JPG


Here's two more points to ponder.

1) Did they tell you that the man is pushing the lawnmower from left to right? I drew it that way in my picture, so the horizontal component of vector v is positive. But what if the man were pushing the lawnmower from right to left instead? In this case, since our system defines the positive direction horizontally to be pointing to the right, we would have to write:

v = -15 sqrt(3) i - 15 j

2) Sometimes, it's more convenient to define the positive direction vertically to be DOWN. For example, if you're dropping a camera at the end of a cable into the ocean to conduct experiments, then it might be natural to think of the sinking camera as advancing through positive depths (eg: +200 feet, +400 feet, +600 feet) as it moves closer to the ocean floor. This is because you might be thinking in terms of how much cable has gone into the water. Within this definition, an instruction to the winch operator to "give me plus 50" would mean drop the camera 50 feet and "give me minus 25" would mean to raise the camera 25 feet.

The main thing to remember with vectors is the definition of positive direction both horizontally and vertically in whatever system you choose to use. If everybody understands the definition, then everybody is on the same page, and "positive" and "negative" are simply relative.

Please let us know if you need further explanations. :D

~ Mark
 
Thanks so much for the answer. This helps, but I do have one more question.

You seem to imply in your response that we're definitely still talking about a 30-degree angle, NOT an angle of 210 or 330 (both of which have NEGATIVE cosine values of (square root of 3)/2. So, actually, the angle in this case doesn't determine the sign (+ or -) of the component answer, the supposed direction of the force does.

What a nightmare. There is at least one other problem in this section involving 4 directions. I always think of a compass with the right-hand arm facing the West, but in the examples in the book, the compass's right-hand arm faces East. I guess I just make due, and figure this is one more variable I take into account, solving each problem :)
 
I always think of a compass with the right-hand arm facing the West, but in the examples in the book, the compass's right-hand arm faces East.

when you look at a map, which way (right or left) is East and West?
 
karmsy said:
You seem to imply in your response that we're definitely still talking about a 30-degree angle ...

I am definitely talking about a 30º angle because the problem you posted says, "... at an angle of 30 degrees to the ground."

Maybe you are stuck in your thinking because your mind keeps defaulting to placing the tail (i.e., the initial point) of any vector at the origin of some coordinate system. In other words, maybe you are thinking in terms of the polar coordinate system.

I'll illustrate the difference with the following pictures.

vectorgrid.JPG


In the middle picture, I've imposed a polar coordinate system such that the tail of the vector is at the origin. But if we use the polar coordinate system, then there is no need for horizontal and vertical components to describe the direction of the vector. The direction is entirely described by the angle measurement. In other words, the direction of any vector pointing in any direction in such a system could be described with a single angle measurement.

This is not how we describe the direction of vectors!

Using the concepts of horizontal and vertical to describe the direction of vectors does not require an origin. It does not even require a coordinate system. IN FACT, once we have calculated the magnitude of both the horizontal and vertical components, we no longer NEED the angle measurement of 30º. We can throw it away! Describing the direction of a vector only requires that everybody understands what constitutes horizontal OR what constitutes vertical, and the positive direction of each, followed by stating how far we move horizontally and how far we move vertically in going from the tail of the vector to its head. In the problem you posted, we all understand that the surface of the ground constitutes horizontal. Therefore, an angle of 30º with respect to the ground is an angle that makes 30º with the horizontal.

This is illustrated by the red line in the bottom picture. I know that the ground constitutes horizontal, so I simply drew a line parallel to the ground positioned to make an angle with the vector. I could have drawn the red line higher, so that it touches the tail of the vector instead. It doesn't matter; there would still be 30º between the vector and the horizontal red line.

If I describe a different problem that deals only with an arrow in my mind, and I tell you that this vector makes a 30º angle with respect to the horizontal, then you have NO WAY to determine the direction because the head of the vector could be at either end. BUT, if I give you the horizontal and vertical components of the arrow, then you have everything that you need to know in order to draw that arrow on a piece of paper pointing in the same direction as the arrow in my mind. Of course, this assumes that you and I have already agreed on what constitutes horizontal or vertical and the positive direction of each.

karmsy said:
... the angle in this case doesn't determine the sign (+ or -) of the component answer, the supposed direction of the force does.

EXACTLY! And the "supposed direction" comes from our understanding of what it means to push a lawnmower. We all understand that the man is pushing foward and downward because we've all used a lawnmower and we realize that there is a downward direction because we are taller than the lawnmower. The angle determines the magnitude of the vertical and horizontal components. If the man were really tall, then the direction of a 30-lb force he exerts might make a 75º angle with respect to the ground; in this case, the magnitude of the horizontal component would be smaller and the magnitude of the vertical component would be larger.

karmsy said:
What a nightmare.

Oh, come on, now. :p

Maybe you said the same thing about negative and positive numbers when you first learned that "subtraction" is the same as "addition with opposites". Eventually, you gained a solid understanding -- and the nightmare passed. So, try not to become discouraged.

karmsy said:
There is at least one other problem in this section involving 4 directions. I always think of a compass with the right-hand arm facing the West, but in the examples in the book, the compass's right-hand arm faces East.

I'm not confident that I understand what you mean by "compass". When you say, "the right-hand arm", are you referring to some type of arrow pointing to the right?

When dealing with vectors, there are not simply "four directions". There are an infinite number of directions.

However, I think you are referring to the positive and negative directions of what we define to be horizontal and vertical for any given problem. This results in four directions.

On a map, the direction of East is entirely up to the publisher of the map. For sure, the vast majority of maps are printed such that North is pointing toward the top of the paper when you're holding it so that the words are not upside down or sideways. But not all maps are oriented this way.

If you always think of the direction pointing to the right as West, then you must also always think that North is down. Am I correct?

The point I would like to make is this: IT DOES NOT MATTER if moving West on a map takes you toward the left side of the paper, or the right side of the paper, or the top, or the bottom. What does matter is that everybody understands and agrees at the beginning which side of the paper is the North end of the map; the other three compass directions are then obvious.

If I am used to always thinking about the top of the map as being North, then I will have to adjust MY thinking when somebody hands me a map on which an arrow pointing to the right is labeled North.

Likewise, if your book shows the direction of due West to be pointing toward the left side of the page, and you are used to seeing due West pointing toward the right side of the page instead, then you will need to adjust YOUR thinking for that exercise.

Please feel free to post more questions and/or thoughts about concepts related to vectors; we can come up with lots more examples.

Cheers,

~ Mark :D
 
In general, for cartesian co-ordinate system the angle is:

\(\displaystyle 0 \, \le \, \theta \, \le \, \frac{\pi}{2}\)

So you take care of the sign by "words" like "pushing down" or "pulling to left" or "pulling toward positve x direction"(the last type of description is more helpful but not always practical).

If you have learned about 2-D polar-coordinate system, then the vector description in that system is:

\(\displaystyle (r,\theta)\)

and converted to cartesian system that would be,

\(\displaystyle (r\cdot\cos\theta\, , \, r\cdot\sin\theta)\)

here r is always positive

and:

\(\displaystyle 0 \, \le \, \theta \, \le \, 2\pi}\)

The range of the angle here takes care of the sign.
 
Subhotosh Khan said:
If you have learned about 2-D polar-coordinate system, then the vector description in that system is:

\(\displaystyle (r,\theta)\)

The following suggestion may mostly be a technicality, but I would alter the above statement to specify that the polar description is for a position vector.

~ Mark :)
 
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