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Does anyone know anything about math?

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Patrick N.

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Then the base is two times x because x is only half of the base. x has to be only half of the base because f(x) is symmetrical over the y-axis and so you can only get a rectangle if the two points are an equal distance from the y-axis.

Is this right?

Thanks again for your help.  ^_^

Yep, exactly.

Thanks to Felipe for generating that equation. I knew what I was looking for, but was not smart enough to put two and two together. Good job! :)

I also forgot that you could put a graph in a post. I just graphed it on my calculator and tried to convey what you should be seeing. It didn't work especially well. I'm one of those people who needs a pencil in hand no matter what. (It helps me think. ^_^ )

Zak

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Yep, exactly.

...

I also forgot that you could put a graph in a post. I just graphed it on my calculator and tried to convey what you should be seeing. It didn't work especially well. I'm one of those people who needs a pencil in hand no matter what. (It helps me think.  ^_^ )

Zak

Hooray! :D

Yeah, my TI-89 is my best friend. :) However, when I was trying to do the problem, I must have been kind of out of it, because the graph went too high to see the curve, and I didn't notice that that was because the window was really tiny. I knew it was supposed to be a curve, but it didn't look like one, so it just confused me even more. But obviously I should have known the y-intercept and realized that the graph needed to be bigger. Oh, well. I always need a pencil and paper as well. And I need to stop trying to do my problem sets at midnight. It doesn't work very well. ^_^

Thanks again to everyone for your help. ^_^

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Heh, you must not have read the first of my "Questions of the Week." I check them every Monday and post answers. But since you asked... The answer to the second one is correct, I do not remember the answer to the first one off the top of my head (and I am having a little trouble remebering how I did that one this late at night ^_^, but since you got the second one right, which is harder, I'd be willing to bet that you got the correct answer for that one as well. ^_^ Anyway, I will have both answers and a new question on Monday. :)

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  • 2 weeks later...

Okay, I need help again. :P

I need help on two problems:

1. Find the angle between the forces given the magnitude of their resultant.

Force 1= 45 pounds

Force 2= 60 pounds

Resultant Force= 90 pounds

2. There is an upside down triangle of cables, with angles A (upper left), B (upper right), and C (bottom). A and B are attached to a bridge, and there is a 2000 lb weight hanging off C.

Angle A= 50 deg

Angle B= 30 deg.

What is the tension in each cable supporting the load?

Note: The answers are in the back of the book; I need to know how to do them, not what the answers are.

Thanks guys! All help is greatly appreciated. :D

Zak

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Okay, I need help again.  :worry:

I need help on two problems:

1. Find the angle between the forces given the magnitude of their resultant.

Force 1= 45 pounds

Force 2= 60 pounds

Resultant Force= 90 pounds

2. There is an upside down triangle of cables, with angles A (upper left), B (upper right), and C (bottom). A and B are attached to a bridge, and there is a 2000 lb weight hanging off C.

Angle A= 50 deg

Angle B= 30 deg.

What is the tension in each cable supporting the load?

Note: The answers are in the back of the book; I need to know how to do them, not what the answers are.

Thanks guys! All help is greatly appreciated.  :)

Zak

Well, math happens to be a huge interest in my life - I think it even contributed to my interest in Objectivism - so I'll take a crack at this! I didn't major in this branch of Engineering, and it's been a few years, but let's see what we can do... B)

First, I got: acos(11/24) = 62.72deg (to two decimal places) for the first one. How am I doing so far?

For the second one - the way I remember it, it's a matter of the forces summing to zero (unless you're in Japan or San Francisco and something REAL BAD is going down... :( ):

-cos(50deg)*A + cos(30deg)*B = 0 (the components in the x-direction)

sin(50deg)*A + sin(30deg)*B -2000 = 0 (the components in the y-direction)

Then you just have to solve the two equations for A and B. Start with

B = [cos(50deg)/cos(30deg)]*A

directly from the 1st equation. Put this in the second equation, and you get:

[sin(50deg)+sin(30deg)*cos(50deg)/cos(30deg)]*A = 2000 lbs

So, to two decimal places, A = 1758.77 lbs. Use this result in the preceeding equation, and B = 1305.41 lbs.

Of course, this assumes that the cables are the quintessential "rigid bodies". If they're real cables that actually FLEX, that comes out to some sort of hyperbolic trig function, and you've got a long day ahead of you. But this sounds like mechanics class - not calculus... :)

Ain't Reason beautiful!?! :P

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First of all, welcome to the forum. This is quite a first post!

Well, math happens to be a huge interest in my life - I think it even contributed to my interest in Objectivism - so I'll take a crack at this!  I didn't major in this branch of Engineering, and it's been a few years, but let's see what we can do...  B)
Cool; I love math too! What did you major in, by the way?

First, I got:  acos(11/24) = 62.72deg (to two decimal places) for the first one.  How am I doing so far?
Awesome, except I have no idea how you got that. The answer is right, but I'm stumped as to how you got 11/24. I did it by setting the first vector at {45,0}. The second vector is above by x degrees (it has an hypotenuse of 60). Add the vectors together and figure out where the resultant (90) goes. Now you have a parallelagram with one diagonal (90). Just solve for the base angle using law of cosines. Subtract that from 180.

For the second one - the way I remember it, it's a matter of the forces summing to zero (unless you're in Japan or San Francisco and something REAL BAD is going down...  :(  ):

  -cos(50deg)*A + cos(30deg)*B = 0             (the components in the x-direction)

  sin(50deg)*A + sin(30deg)*B -2000 = 0      (the components in the y-direction)

Then you just have to solve the two equations for A and B.  Start with

  B = [cos(50deg)/cos(30deg)]*A

directly from the 1st equation.  Put this in the second equation, and you get:

  [sin(50deg)+sin(30deg)*cos(50deg)/cos(30deg)]*A = 2000 lbs

So, to two decimal places, A = 1758.77 lbs.  Use this result in the preceeding equation, and B = 1305.41 lbs.

Yeah, I'll have to get back to you later on that one. I don't have the time to think about it right now. Plus, the teacher just told us the angle would be the same on the test which make things a heck of a lot easier. :P

Of course, this assumes that the cables are the quintessential "rigid bodies".  If they're real cables that actually FLEX, that comes out to some sort of hyperbolic trig function, and you've got a long day ahead of you.  But this sounds like mechanics class - not calculus... :)
Actually, its a high school pre-calculus class, but the cables are rigid. I would very much like to take a mechanics course, though. B)

Thanks very much for all of your help. My test is on Wednesday. :worry: I need an A; the quarter ends in two weeks. :o

Zak

Edited by realitycheck44
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First of all, welcome to the forum. This is quite a first post!

Cool; I love math too! What did you major in, by the way?

Thanks. I figured I'd wait to register until I had time to put together an intro, but then I spotted this "math emergency"... :)

To answer your question, I majored in Electrical Engineering, both as an undergrad & grad.

Awesome, except I have no idea how you got that. The answer is right, but I'm stumped as to how you got 11/24. I did it by setting the first vector at {45,0}. The second vector is above by x degrees (it has an hypotenuse of 60). Add the vectors together and figure out where the resultant (90) goes. Now you have a parallelagram with one diagonal (90). Just solve for the base angle using law of cosines. Subtract that from 180.

I think it just comes out to the equivalent of the Law of Cosines. If you drop a line segment down from the diagonal of your parallelogram to the x-axis, you have a new side. From the Pythagorean Theorem,

[90]^2 = [45+60*cos(x)]^2 + [60*sin(x)]^2

8100 = [45]^2 + 2*45*60*cos(x) + [60*cos(x)]^2 + [60*sin(x)]^2

8100 = 2025 + 5400*cos(x) + 3600*cos(x)^2 + 3600*sin(x)^2

6075 = 5400*cos(x) + 3600*[cos(x)^2 + sin(x)^2]

6075 = 5400*cos(x) + 3600

2475 = 5400*cos(x)

cos(x) = 2475/5400

Divide top and bottom by 225, and cos(x) = 11/24. Somewhere inbetween the second and third lines is the Law of Cosines, using cos(180-x) = -cos(x).

Yeah, I'll have to get back to you later on that one. I don't have the time to think about it right now. Plus, the teacher just told us the angle would be the same on the test which make things a heck of a lot easier.  :P

While we're at it, the equation for A simplifies a bit:

[sin(50deg)+sin(30deg)*cos(50deg)/cos(30deg)]*A = 2000 lbs

{[sin(50deg)*cos(30deg)+cos(50deg)*sin(30deg)]/cos(30deg)}*A = 2000 lbs

[sin(50deg+30deg)/cos(30deg)]*A = 2000 lbs

[sin(80deg)/cos(30deg)]*A = 2000 lbs

so

A = [cos(30deg)/sin(80deg)] * 2000 lbs

There's probably a way to do it with the Law of Sines, considering how simple it comes out.

Then

B = [cos(50deg)/cos(30deg)]*A

= [cos(50deg)/cos(30deg)]*[cos(30deg)/sin(80deg)]* 2000 lbs, or

B = [cos(50deg)/sin(80deg)]* 2000 lbs

Actually, its a high school pre-calculus class, but the cables are rigid. I would very much like to take a mechanics course, though.  B)

Thanks very much for all of your help. My test is on Wednesday. :worry: I need an A; the quarter ends in two weeks.  :o

Zak

I skipped pre-calc, myself, in High School, but good luck on your test! :) (No mysticism intended or endorsed by the term "luck", BTW ... B) ) The more successful Objectivists are at useful things, the safer a place the world becomes for Reason! But - hey - "no pressure"...

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