ecoli Posted October 6, 2005 Share Posted October 6, 2005 You're given a cable, weighing 2lb/fot, that is used to lift 800 lb of coal up a mineshaft 500ft deep. find the work Help me finish this problem, please! [math] \int_0^500 [(800+1000)(9.8) ]X \ , dx [/math] Is this right, is it worth it to solve? Link to comment Share on other sites More sharing options...
ecoli Posted October 6, 2005 Author Share Posted October 6, 2005 sorry, I don't know what happened. It's supposed to be integrated from 0 to 500. A latex error I don't know how to fix, put the "00" with the integral, for some reason. Link to comment Share on other sites More sharing options...
timo Posted October 6, 2005 Share Posted October 6, 2005 sorry, I don't know what happened. It's supposed to be integrated from 0 to 500. A latex error I don't know how to fix, put the "00" with the integral, for some reason. You forgot the {} around the 500. This is how it probably should look like: [math] \int_{0}^{500} [(800+1000)(9.8) ]x dx [/math] Actually, this isn´t correct. The total work you have to do is the length-integral over the applied force. The nessecary force F for a remaining cable-length h is given by the weight of the coal (800lb) and the weight of the cable ( h*2lb/ft ). Integrate the force over the whole distance and you have your answer. As a sidenote: At the level of physics that you seem to be you are absolutely not doing yourself a favor by ommiting the units. Keeping track of the units might well be the very best way of finding out that a certain result simply can´t be right. EDIT: You are really working with units of foot and ... well, whatever "lb" is ? Link to comment Share on other sites More sharing options...
ecoli Posted October 6, 2005 Author Share Posted October 6, 2005 W = Fd and F = mg, correct? so w = mgd this is basically what I have... mg (1800* 9.8) is a constant so pull it outside the integral and integrate x from 0 to 500 and you get the work. What's wrong with that? Link to comment Share on other sites More sharing options...
timo Posted October 6, 2005 Share Posted October 6, 2005 I do not have the slightest clue what 1800 is. I can only guess what 9.8 is. But the solution you presented is definitely -and regardless of what the numbers stand for- wrong. The pull is not constant. Depending on your current height there is more or less weight of the cable that pulls down. You should really do yourself the favor of plugging the units in. Link to comment Share on other sites More sharing options...
ecoli Posted October 6, 2005 Author Share Posted October 6, 2005 1800 is the weight (800 pounds + 1000 pounds (rope is 2 pounds per foot)). 9.8 is the acceleration due to gravity (m/s^2) otherwise known as g. Link to comment Share on other sites More sharing options...
timo Posted October 6, 2005 Share Posted October 6, 2005 You seem to be missing my point that the weight of the cable (to be more correct: The weight of the part of the cable that you still have to pull up) does not remain constant. It is 1000 pounds (assuming "lb" is pounds) when you start to pull but it gets less with the distance you pulled. Therefore, your integrand cannot be constant. You attempt is ok in general, you are just missing the two points that 1) The force you need decreases over the distance pulled 2) You are calling out loud for errors if you don´t plug in the units - especially if there are some non-SI units involved. Link to comment Share on other sites More sharing options...
ecoli Posted October 6, 2005 Author Share Posted October 6, 2005 right... I'm crossing units too, oops. So this would be correct? [math] \int_0^{500} 800lb * h * 2lb/ft\ dh [/math] Is that right, or I'm I still missing distance? Link to comment Share on other sites More sharing options...
timo Posted October 6, 2005 Share Posted October 6, 2005 You should take a few minutes more before you make a reply. No, it´s not correct for several reasons, most of them probably being caused by the fact that you typed that in so fast. - the units at the interval ends are missing. - h*2lb/ft is a weight. So you are integrating over weight squared (800lb * a weight). That certainly contradicts what you already found out yourself, namely that you have to integrate over a force. Link to comment Share on other sites More sharing options...
ecoli Posted October 7, 2005 Author Share Posted October 7, 2005 Alright, first, I didn't type that in "so fast" I just really have no idea what you're trying to tell me. It seems like you're telling me I need two variables, one for weight and one for force... but I'm taking single variable calculus, so I'm sure that's not the case. ... I figured that 0 to 500 was implied to be the height in feet. Link to comment Share on other sites More sharing options...
ecoli Posted October 7, 2005 Author Share Posted October 7, 2005 Ok, I found a few things out since the last post. I looked up the answer, which 650,000 ft-lbs. The weight is the force, so there is no need for the acceleration due to gravity. I tried something that I thought was right, but it didn't yield the correct answer. Am I not integrating correctly or is my set-up wrong? [math] \int_{0 ft}^{500 ft} (800lbs + 2x) x\ dx [/math] I distributed the x and integrated, ending up with [math] ({800/2})x^2 + ({2/3})x^3 [/math] from 0 to 500. That doesn't come up with the rgith answer, though. Any ideas? Link to comment Share on other sites More sharing options...
ecoli Posted October 7, 2005 Author Share Posted October 7, 2005 [math] \int_{0 ft}^{500 ft} (800lbs + 2x) \ dx [/math] Link to comment Share on other sites More sharing options...
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