realitycheck44 Posted March 6, 2005 Report Share Posted March 6, 2005 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 Quote Link to comment Share on other sites More sharing options...
non-contradictor Posted March 6, 2005 Report Share Posted March 6, 2005 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! 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. Quote Link to comment Share on other sites More sharing options...
HaloNoble6 Posted March 6, 2005 Report Share Posted March 6, 2005 You're welcome. BTW, non-contradictor, check if I submitted correct answers to your blog post about that 3rd degree polynomial. Quote Link to comment Share on other sites More sharing options...
non-contradictor Posted March 6, 2005 Report Share Posted March 6, 2005 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. Quote Link to comment Share on other sites More sharing options...
realitycheck44 Posted March 21, 2005 Report Share Posted March 21, 2005 Okay, I need help again. 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 Quote Link to comment Share on other sites More sharing options...
realitycheck44 Posted March 21, 2005 Report Share Posted March 21, 2005 Okay, I understand how to do the first one, but if anybody knows how to do the second one, I would really appreciate it. Zak Quote Link to comment Share on other sites More sharing options...
Optimizer Posted March 22, 2005 Report Share Posted March 22, 2005 Okay, I need help again. 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... 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!?! Quote Link to comment Share on other sites More sharing options...
realitycheck44 Posted March 22, 2005 Report Share Posted March 22, 2005 (edited) 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... 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. 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. Thanks very much for all of your help. My test is on Wednesday. I need an A; the quarter ends in two weeks. Zak Edited March 22, 2005 by realitycheck44 Quote Link to comment Share on other sites More sharing options...
Optimizer Posted March 23, 2005 Report Share Posted March 23, 2005 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. 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. Thanks very much for all of your help. My test is on Wednesday. I need an A; the quarter ends in two weeks. 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 ... ) The more successful Objectivists are at useful things, the safer a place the world becomes for Reason! But - hey - "no pressure"... Quote Link to comment Share on other sites More sharing options...
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