More 2-D vectors: force of 490 lbs needed to keep car on 10-

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karmsy

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I have 2 questions, one specific, the other more general.

The specific question concerns the answer(s) to the following problem:

"A car is on a driveway that is inclined 10 degrees to the horizontal. A force of 490 lb is required to keep the car from rolling down the driveway.

a) Find the weight of the car.
b) Find the force the car exerts on the driveway."

I know a bit about the "decomposition" of vectors, and about the calculation of "work." But I don't feel I'd done a good job, even in setting this problem up, much less in solving it. The examples in the book are sparse, or I wouldn't be bothering you. I'd really appreciate any help.

The second question is more general, about the "decomposition" of vectors. Given vectors A and B, originating at the same point, we "break B down" into 2 new vectors, one parallel to A, with "magnitude along A," and the other orthogonal to A. This "magnitude along A" puzzles me. Is this the WHOLE magnitude of "A"? Or does A, being a different vector from B, also have its own magnitude--distinct from that of the new vector we derived from B?

Thanks so much. This forum really came through for me last time, and hopefully it will again.

I hope you're having a great Labor Day--if you're in the U.S..
 
Re: More 2-D vectors!!

karmsy said:
I have 2 questions, one specific, the other more general.

The specific question concerns the answer(s) to the following problem:

"A car is on a driveway that is inclined 10 degrees to the horizontal. A force of 490 lb is required to keep the car from rolling down the driveway.

a) Find the weight of the car.
b) Find the force the car exerts on the driveway."

I know a bit about the "decomposition" of vectors, and about the calculation of "work." But I don't feel I'd done a good job, even in setting this problem up, much less in solving it. The examples in the book are sparse, or I wouldn't be bothering you. I'd really appreciate any help.

The second question is more general, about the "decomposition" of vectors. Given vectors A and B, originating at the same point, we "break B down" into 2 new vectors, one parallel to A, with "magnitude along A," and the other orthogonal to A. This "magnitude along A" puzzles me. Is this the WHOLE magnitude of "A"? Or does A, being a different vector from B, also have its own magnitude--distinct from that of the new vector we derived from B?

Thanks so much. This forum really came through for me last time, and hopefully it will again.

I hope you're having a great Labor Day--if you're in the U.S..
Click to expand...

Are you taking a course in statics - for engineering?

If you are - then which book are you following?
 
Re: More 2-D vectors!!

You ever hear of "off-road vehicles"? Well, I'm an "off-road student" :) I'm not enrolled in a course, I'm doing this on my own. I'm preparing for California's state examination for new math teachers (what I'd actually be teaching is much more elementary than what's on the test).

The problem I cited comes out of the book "Precalculus: 5th Edition," by Stewart/Redlin/Watson. I am also referring regularly to "Precalculus," by Sullivan. The latest edition, I think.
 
Re: More 2-D vectors!!

karmsy said:
I have 2 questions, one specific, the other more general.

The specific question concerns the answer(s) to the following problem:

"A car is on a driveway that is inclined 10 degrees to the horizontal. A force of 490 lb is required to keep the car from rolling down the driveway.

a) Find the weight of the car.
b) Find the force the car exerts on the driveway."

I know a bit about the "decomposition" of vectors, and about the calculation of "work." But I don't feel I'd done a good job, even in setting this problem up, much less in solving it. The examples in the book are sparse, or I wouldn't be bothering you. I'd really appreciate any help.

The second question is more general, about the "decomposition" of vectors. Given vectors A and B, originating at the same point, we "break B down" into 2 new vectors, one parallel to A, with "magnitude along A," and the other orthogonal to A. This "magnitude along A" puzzles me. Is this the WHOLE magnitude of "A"? Or does A, being a different vector from B, also have its own magnitude--distinct from that of the new vector we derived from B?

Thanks so much. This forum really came through for me last time, and hopefully it will again.

I hope you're having a great Labor Day--if you're in the U.S..
Click to expand...

Triangles, triangles, triangles...

The weight of the car is toward the center of the planet on which the car resides, usually Earth.

1) Draw the weight vector - pointing downward.
2) Draw the equilibrium force - Parallel to and UP the ramp. 490 lbs
3) Draw the force 2) is equilibriuming (??) - parallel to and DOWN the ramp. 490 lbs
4) Draw the Normal force, perpendicular to the ramp and down through the ramp.
5) Vectors don't care where they start or stop, so make a copy of the DOWN equilibrium force, starting somewhere along the Normal Force and terminating at the end of the Weight vector. 490 lbs
6) You should have created a Right Triangle. Find the Right angle and label it.
7) You should have some similar triangles. You have the 10º slope of the ramp, this tells you also one of the angles of the Right Triangle.
8) You are SO CLOSE to done!!! What do you get? Were you expecting to use a little trigonometry?
 
Re: More 2-D vectors!!

Thanks for the reply.

Sorry--maybe you'd provide an illustration? That would help me a lot.
 
Re: More 2-D vectors!!

Hi karmsy:

I have a copy of Stewart's 5th edition.

karmsy said:
... The problem I cited comes out of the book "Precalculus: 5th Edition," by Stewart/Redlin/Watson ...
Click to expand...

Look at example 4 on page 621. It is the same problem with a heavier car. Figure 4 shows four diagrams; each illustrates how the component vectors change as the inclines goes from horizontal to about 75 degrees.

Do you "see" all of the possible triangles that can be formed?

Your problem gives you the magnitude for one side of such a triangle. It also gives you one of the triangle's angle measurements. Think about how you would use knowledge of trigonometry to answer the question.

If you're still stuck after reading page 621, then feel free to let us know where.

Cheers,

~ Mark :)

My edits: fixed ambiguous language
 
Re: More 2-D vectors!!

Karmsy:

Our previous posts passed each other in cyberspace! :)

karmsy said:
Sorry--maybe you'd provide an illustration?
Click to expand...

I'm not sure over what you're apologizing, but the illustration labeled Figure 5 on page 621 is exactly what you want. (Of course, the angle measurement in your problem is different.)

Vectors u and v are the component vectors of vector w. Vector w is pointing toward the center of the earth.

The second question is more general ... Given vectors A and B, originating at the same point, we "break B down" into 2 new vectors, one parallel to A, with "magnitude along A," and the other orthogonal to A. This "magnitude along A" puzzles me. Is this the WHOLE magnitude of "A"?
Click to expand...

No, they are not referring to the "WHOLE magnitude of A". They are not even referring to the magnitude itself. They are simply referring to the line that goes through A so that you know where to position the component. You are free to extend a line through a vector as far as you like, in either direction, in order to form triangles with other vectors.

Look at Figure 6 on page 622. I will paraphrase the same statement using the vector names in that diagram.

"Given vectors v and u, originating at the same point, we "break u down" into 2 new vectors, one parallel to v, with "magnitude along v", and the other orthogonal to v.

So, the red vector in the diagram is the vector to which I refer when I say, "one parallel to v with magnitude along v".

You can replace the phrase "with magnitude along A" with the phrase "and positioned on the line that goes through A".

Do you see the dotted line that goes from the head of the red vector to the head of vector u? That dotted line forms a triangle. We had to extend the line through vector v far enough to be able to form this triangle. The two legs are the components of vector u.

~ Mark
 
Looking at Figure 5 on p. 621 of Watson/Redlin/Stewart "Precalculus," it would help me very much to see somebody "set up" a problem, which is NOT spelled out in the book, but which I am going to interpret from the diagram:

"A van (weighing 4500 lb, we'll say) is parked on a hill, inclined 15 degrees to the horizon. Find the force needed to keep the van from rolling down the hill."

What are the critical vectors in this problem, and what are there angles?
 
karmsy said:
Looking at Figure 5 on p. 621 of Watson/Redlin/Stewart "Precalculus," it would help me very much to see somebody "set up" a problem, which is NOT spelled out in the book, but which I am going to interpret from the diagram:

"A van (weighing 4500 lb, we'll say) is parked on a hill, inclined 15 degrees to the horizon. Find the force needed to keep the van from rolling down the hill."

What are the critical vectors in this problem, and what are there angles?
Click to expand...
Excellent.

Take a copy of this drawing and change the following:

4500 lbs ==> x lbs

15 degrees ==> 10 degrees

Unknown Force ==> 490 lbs

Are we missing anything?
 
Thanks for the reply. It would help me to know why you've made the substitutions you have.

As I understood it, in the wording of the question, we already had the weight of the van--presumably, the magnitude of vector w, all by itself. Why are we replacing that weight with "x" now?

Also, you substituted "10 degrees" for "15." Was this all about ease of working with cosines and sines?

Finally, I thought the "unknown force," which you gave as 490 lbs, was exactly what the problem asks us to find!

Thanks in advance for your info.
 
My point was that your original problem is just a slight variation from the example. See the concepts in the example that are exactly the same as the concepts in the given problem.

Example vs. Problem
1) Given Vehicle Weight vs. Solve for Vehicle Weight
2) 15º angle vs 10º angle
3) Solve for equilibrium force vs. Given equilibrium force
 
I'm puzzled, because your example isn't even looking to me like the example with the diagram on p. 621. Degrees are different, for one. Be that as it may, let's take the diagram on p. 621, and set up a problem with it. It would be most helpful to me to see a problem whose solution involves a whole vector, NOT merely a component. So, something involving complete vectors, with "the projection of u onto v" formula from p. 622.
 
I see you were referring to MY example problem above, not to anything in the book. OK, my bad.

Still not happy. This one hasn't been "put to bed," and it isn't because I (and other people, to be fair) haven't been trying.

I don't know how to solve this problem. I don't know how to solve similar problems I might run into.

Using the example I started out with, which is analogous to the diagram on p. 621 in the Watson/Redlin/Stewart book, is the "projection of w onto u," the force pulling the van down the 10-degree incline, u=490(cos 10 degrees)i - 490(sin 10 degrees)j? Now, how do we get magnitude for the vector w, the vector of the van's weight, relative to the earth? Which I guess is exactly the question we're trying to answer.

A step-by-step, with vector equations written out in full, would be most helpful.
 
Sadly, that's where we started. See 8 steps dated 1 September.
 
Say what?

karmsy said:
... It would be most helpful to me to see a problem whose solution involves a whole vector, NOT merely a component.
Click to expand...

From these words, I am not sure that I understand what you're thinking. When I read them, I think of something like a trigonometry student saying, "I would like to see a problem whose solution involves a whole triangle, not merely an angle and two sides."

Solving these types of problems involves more than seeking a solution using different numbers for the same problem, followed by exchanging old numbers with new. Somebody has said something similar to the following. Spoon feeding, in the end, teaches us nothing more than the size of the spoon.

You told us that you are preparing on your own for an entrance exam. That's to be commended. What do you intend to do during the exam when there are no "templates" for you to use (this is a rhetorical question).

I'm concerned that you're not seeing the big picture. I do not know whether or not you understand vectors and how their attributes make them a good mechanism for solving questions about stuff involving directions in two or three dimensions. I don't even know whether or not vector problems will be on the exam.

~ Mark :)

 
Perhaps the fact that the last 2 replies strike me as condescending is a sign I'm deeply frustrated, and, with regard to this topic, really not going to get what I need in this venue, either.

I own two college precalculus textbooks. Neither has provided me with a full example of a worked-through vector problem, with steps explained. I have yet to see such a problem. I don't think that's a "cheating" request, as one of you implied, or a sign of stupidity, to want to see a vector problem, worked-through, with steps clearly explained. I'd defend it as an entirely reasonable thing to want.

I am seeing red, to tell you the truth. Not so much at the people who have tried to help here, but the level of math illiteracy in American culture, of which my troubles are a symptom, on many levels.

This is good-bye. Certainly with regard to this topic, and perhaps for quite a while.
 
The challenges of learning ...

karmsy said:
I own two college precalculus textbooks. Neither has provided me with a full example of a worked-through vector problem, with steps explained ...
Click to expand...

There is a very good reason why college text books do not provide complete worked-out examples for every problem. At the college level, you will be expected to think independently about basic concepts presented and you will be expected to reach some conclusions and discoveries without being lead by the hand.

High-school prepares young people how to think for themselves; college forces people to do it. You will not find many instructors at the college level willing to lead students by the hand.

When I have time after dinner, I will upload an image of the diagram from your book, after having crossed out the old numbers and having replaced them with the new numbers from the exercise you posted. I will also post all of the steps, so that you do not need to think or discover anything more about this exercise.

Cheers,

~ Mark :)

 
Please let us know if you have more questions ...

Please double-click the image below to expand it. The answers to your exercise are in green.

~ Mark :)
 

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