Vector Math

General Forums Off-topic Chat
#1

I like to think that I understand this vector math stuff fairly well, but this problem has me totally stumped… I’ve had a few “eureka!” moments, but none turned out to be right.

A collar that can slide on a vertical rod is subjected to the three forces shown. Determine (a) the value of the angle beta for which the resultant of the three forces is horizontal, (b) the corresponding magnitude of the resultant.

http://img127.imageshack.us/img127/4355/polekk3.jpg

I’ve changed the numbers on the vectors so just giving me an answer doesn’t help at all. If someone could explain what the process is that’d be great. Any ideas?

I think we can assume that since the two non-vertical vectors are the same distance from the nearest axis (angle beta) that they are 90 degrees apart, but I can’t seem to get a hold of where to go after that.

#2

Well I’m no good at math but if you add them all up you get… er… 284 lbs. (Addition is my forte :slight_smile: )

#3

Well if it was that easy, I’d be finished :). You can’t simply add them up because they are not all going in the same direction. You use a method I’ve always called tail to tip which is where you align one vector’s tip up with another vectors tail and then the resultant (the final vector) goes from the tail of the first to the tip of the last.

Any other ideas?

#4

I think we can assume that since the two non-vertical vectors are the same distance from the nearest axis (angle beta) that they are 90 degrees apart, but I can’t seem to get a hold of where to go after that.

Yup, so work out the magnitude of the force 45 degrees from each (between the two arrows) and treat it as one force. (edit - you don’t even have to do this, it just means you have more beta elements in the equation)

Then you need to work out what angle gives you a vertical force of 110lbs, pythag will show you the light here, because it’s a right angled triangle.

That should work in my mind.

#5

Hmm, the angle between the two non-vertical vectors is 90 degrees, but it doesn’t necessarily mean the beta angle is 45. It could be anything. Am I misunderstanding what you are saying?

#6

No, he’s saying that because the two vectors are at 90 degrees, you can take components of each at 45 degrees and treat it as one vector.

You could also sum the forces in each direction. You’ll get a cyclic equation (sine and cosines), but any graphing calculator should be able to let you solve for beta.

Something like:

0 = 110 + 50sinB - 124cosB

#7

82.950227741358958 degrees or 1.4477545893658532 radians?

#8

Well, it’s obviously a homework problem (tsk…tsk…) but at the same time I will cheerfully admit that it is a slightly-engaging one.

The essential nature of this problem is that you have three forces at work: one of these is independent (110lb straight-up), while the other two are dependent (an identical “theta” is a factor in both, applied in related(!!) but differing ways).

Since the goal-state is horizontal, you know that this is another way of saying “equivalent to 110-lbs of force straight-up.” So, in a perfectly-ordinary algebraic equation, “x = 110.” Now, x is obviously going to be an equation, and that equation is obviously going to be based on “theta,” and your chore will simply be to deduce the nature of that relationship.

We observe that the angle “theta” is being used in two different ways: in the first, it’s being measured from a horizontal line; in the second, it’s measured from the vertical. (Or: one is “theta” and the other is “90-degrees minus theta.”) Thus, for any value at all of “theta,” we automatically know the value of the remaining angle. In other words, “x” is going to be an equation of the nature “f(x) + f(90-x) …”

If you ponder this problem just a little bit more, it quickly devolves into a simple algebraic problem … albeit a problem in which the familiar trigonometric functions sin(theta), cos(theta) and/or tan(theta) will no-doubt play a part. The mere presence of … a constant, a variable “x”, and a factor “(n - x)” … is a dead-giveaway to the experienced eye: truly a bellweather of all such problems of its kind.

#9

Um yeah that went over my head. Not something I’ve learned in eight grade yet. But if you have additions problems I’d be glad to help.

#10

sundialsvc4: It is from a homework assignment, but the problem I posted wasn’t the same numbers as what was assigned. Fortunately, I like to learn so I prefer to learn the process rather than just writing down numbers. It helps in the long run too as cheating doesn’t hurt anyone but yourself.

Thanks for your help everyone, I just was making it much much harder than it really was. After lining the vectors up tail tip (actually, lining them up in a different organization that I was before) I realized that calculating the resultant for the 50lb and the 124lb and then using that as the hypotenuse for the right triangle generated by the resultant and the 110lb vector will allow me to calculate the final resultant. A lot easier than I thought.

@M3ta: Don’t worry about it, You probably won’t learn this stuff until sometime in your high school career. Depending on what you want to study in college, you may never even see this stuff again (in school that is).

#11

Hello tango!

First draw the x and y axes. Then take these axes as your reference points where the +x axis refer to positive magnitude and positive angles. From here, measure and take the components – x,y values – of each vector. If you’re still in doubt, please check out COMPONENT METHOD in solving vectors. I’m sure you’ll be able to solve the angle beta and the magnitude.

#12

just to see if my math are still correct,
is F=sin(B).50-sin(90-B).124+110?

#13

The key raison d’entre of any problem of this kind, of course, is to oblige you to think through it … with the purpose that “problems such as this one” will one-day become quite commonplace for you “on the job.”

This particular problem has a very important characteristic that (aww, shucks, it’s the community-college instructor in me…) I would hasten to point out. It’s subtle…

The essential nature of this problem is that … it devolves to an equation of the form: f(x) + f(90-x) + c, for an identical f(x) in both terms.

And so, given that the function f([/i]x[/i]) is identical in both terms, it necessarily becomes less-important to determine exactly what f(x) is, than to realize that one term is (x) and the other term is (90-x). Provided only that f(x) is a non-exponential function (as all of the trigonometric functions are…), their total influence on the outcome of the equation is inferior … no matter what the exact definition of f() may be.

#14

Hmmm… nope they dont teach this in the 10th grade. these forums either make me feel stupid or immature. oh well, i doubt ill need this when i join the military anyway. :stuck_out_tongue: (thats my excuse to be lazy, do you like it?)

#15

No…I don’t. The military kept trying to sign me up.

I laugh at you guys debating over a simple vector problem. You should know that if he can’t figure it out himself, he won’t understand what you are explaining. Be a little better in your explanations. Imagine you are teaching someone who… isn’t in community college?

I also laugh because the other guy in my apartment is studying physics for graduate school. He showed me some of his problems that he couldn’t get.

#16

lol, samf, I definitely tacked it, a while ago in fact. And, to your friend, that’s great, but he’s still a few years ahead of me (We all have to start somewhere…). Sorry I wasn’t more clear that I had solved the problem (and picked up a few different methods in the process)…

And, I really don’t mind when people give me hints as to where to approach it, what to think about, because actually, approaching it a bit differently is what helped me solve it. What I really don’t like is people feeding me the answer. So I appreciate the… somewhat vague (but helpful) responses.

Don’t assume that if I can’t figure it out right away (or in a given amount of time) that I can’t understand what they’re saying. That’s an arrogant and incorrect assumption.

Admin: feel free to remove this thread.

#17

I know. I just felt like posting.