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On 11 Dec 2003 04:35:05 -0800, (basskisser) wrote:
(Steven Shelikoff) wrote in message ... On 10 Dec 2003 09:00:31 -0800, (basskisser) wrote: K Smith wrote in message ... Snafu wrote: Gary, I'm not understanding the part about "the longer the tongue, the furthur back we can put the wheels." To maintain the ~10% of the weight on the hitch rule, increasing the tongue length works the opposite way. Absolutely the longer the tongue (drawbar?) the further "forward" the wheels will by to keep the balance right. K Once again, you've proven you stick your nose where it doesn't belong. You certainly don't understand simple physics. If you make the tongue longer, you will need to move the wheels BACK to keep the balance the same. Say you have the tongue 10' from the wheels, and the tongue weight is 100 pounds, to say, at that 10', you need to resist 100 pounds to lift the tongue. Now, we extend the tongue, to 15'. Now you have a longer lever, thereby LESS tongue weight. To compensate, you need to move the wheels BACK. Tell ya what I'll do, I'm not going to give you the answer, but I'll define a problem and let basskisser and Karen come up with solutions and see if they come up with the same one, and then hack out their differences. Let's go with the situation above. Assume you have a trailer where the distance from hitch to axle is 10' and the tongue weight is 100lbs. Also, assume that the trailer is level because if it isn't, it will affect the answer. Also assume that the tongue extension has a constant weight per unit length. Now increase the length of the tongue to 15' from axle to hitch. Just from the difference in leverage, what is the new, lower weight on the hitch? Since we assumed that the extension has a linear weight/length, the center of mass of the extension is 2.5' from the hitch and 12.5' from the axle. So what percentage of the weight of the extension is supported by the hitch? Knowing the percentage of the weight of the extension that's supported by the hitch, how much does the extension have to weigh to exactly counteract the decrease in weight due to leverage and keep the weight on the hitch at 100lbs? If the extension weighs any more than the answer you come up with, then you have to move the wheels forward to keep the weight 100lbs. Any less, then you have to move the wheels back. Bonus question: assume that the original level trailer with the 10' from axle to hitch weighs 1000lbs and the hitch weight is 100lbs, or 10% of the trailer weight. You extend the tongue by 5' using an extension that weighs 10 lbs/ft. How far and in what direction do you have to move the axle to keep the hitch weight 10% of the new trailer weight? Have fun! Steve I know damn well what the outcome is. But, I don't have to show/prove to you a damned thing. I solve baby **** like that on a daily basis. Apparently not if you think that the extension has to weigh the same per unit length as the boat and trailer to balance it out. Instead of thinking you know what the outcome is, why don't you solve the problem and see if you're right? Don't even worry about the bonus question if that's too hard. The main question is fairly simple to solve even with the most basic high school level of physics. Hell, I pretty much set the whole thing out for you on how to solve it. Steve |
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#4
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(Steven Shelikoff) wrote in message ...
On 11 Dec 2003 04:35:05 -0800, (basskisser) wrote: (Steven Shelikoff) wrote in message ... On 10 Dec 2003 09:00:31 -0800, (basskisser) wrote: K Smith wrote in message ... Snafu wrote: Gary, I'm not understanding the part about "the longer the tongue, the furthur back we can put the wheels." To maintain the ~10% of the weight on the hitch rule, increasing the tongue length works the opposite way. Absolutely the longer the tongue (drawbar?) the further "forward" the wheels will by to keep the balance right. K Once again, you've proven you stick your nose where it doesn't belong. You certainly don't understand simple physics. If you make the tongue longer, you will need to move the wheels BACK to keep the balance the same. Say you have the tongue 10' from the wheels, and the tongue weight is 100 pounds, to say, at that 10', you need to resist 100 pounds to lift the tongue. Now, we extend the tongue, to 15'. Now you have a longer lever, thereby LESS tongue weight. To compensate, you need to move the wheels BACK. Tell ya what I'll do, I'm not going to give you the answer, but I'll define a problem and let basskisser and Karen come up with solutions and see if they come up with the same one, and then hack out their differences. Let's go with the situation above. Assume you have a trailer where the distance from hitch to axle is 10' and the tongue weight is 100lbs. Also, assume that the trailer is level because if it isn't, it will affect the answer. Also assume that the tongue extension has a constant weight per unit length. Now increase the length of the tongue to 15' from axle to hitch. Just from the difference in leverage, what is the new, lower weight on the hitch? Since we assumed that the extension has a linear weight/length, the center of mass of the extension is 2.5' from the hitch and 12.5' from the axle. So what percentage of the weight of the extension is supported by the hitch? Knowing the percentage of the weight of the extension that's supported by the hitch, how much does the extension have to weigh to exactly counteract the decrease in weight due to leverage and keep the weight on the hitch at 100lbs? If the extension weighs any more than the answer you come up with, then you have to move the wheels forward to keep the weight 100lbs. Any less, then you have to move the wheels back. Bonus question: assume that the original level trailer with the 10' from axle to hitch weighs 1000lbs and the hitch weight is 100lbs, or 10% of the trailer weight. You extend the tongue by 5' using an extension that weighs 10 lbs/ft. How far and in what direction do you have to move the axle to keep the hitch weight 10% of the new trailer weight? Have fun! Steve I know damn well what the outcome is. But, I don't have to show/prove to you a damned thing. I solve baby **** like that on a daily basis. Apparently not if you think that the extension has to weigh the same per unit length as the boat and trailer to balance it out. didn't say that. Read for context, please. Instead of thinking you know what the outcome is, why don't you solve the problem and see if you're right? Don't even worry about the bonus question if that's too hard. The main question is fairly simple to solve even with the most basic high school level of physics. Hell, I pretty much set the whole thing out for you on how to solve it. Steve I KNOW I'm right. And, yes, as I've stated, it IS basic high school physics. Again, I KNOW the answer, but alas, I'm not about to play your idiotic games. I have nothing that I need to prove to an ignoramus like you. Explaining and teaching something to a dimwit is all but impossible. |
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#5
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On 11 Dec 2003 10:42:43 -0800, (basskisser) wrote:
(Steven Shelikoff) wrote in message ... On 11 Dec 2003 04:35:05 -0800, (basskisser) wrote: (Steven Shelikoff) wrote in message ... On 10 Dec 2003 09:00:31 -0800, (basskisser) wrote: K Smith wrote in message ... Snafu wrote: Gary, I'm not understanding the part about "the longer the tongue, the furthur back we can put the wheels." To maintain the ~10% of the weight on the hitch rule, increasing the tongue length works the opposite way. Absolutely the longer the tongue (drawbar?) the further "forward" the wheels will by to keep the balance right. K Once again, you've proven you stick your nose where it doesn't belong. You certainly don't understand simple physics. If you make the tongue longer, you will need to move the wheels BACK to keep the balance the same. Say you have the tongue 10' from the wheels, and the tongue weight is 100 pounds, to say, at that 10', you need to resist 100 pounds to lift the tongue. Now, we extend the tongue, to 15'. Now you have a longer lever, thereby LESS tongue weight. To compensate, you need to move the wheels BACK. Tell ya what I'll do, I'm not going to give you the answer, but I'll define a problem and let basskisser and Karen come up with solutions and see if they come up with the same one, and then hack out their differences. Let's go with the situation above. Assume you have a trailer where the distance from hitch to axle is 10' and the tongue weight is 100lbs. Also, assume that the trailer is level because if it isn't, it will affect the answer. Also assume that the tongue extension has a constant weight per unit length. Now increase the length of the tongue to 15' from axle to hitch. Just from the difference in leverage, what is the new, lower weight on the hitch? Since we assumed that the extension has a linear weight/length, the center of mass of the extension is 2.5' from the hitch and 12.5' from the axle. So what percentage of the weight of the extension is supported by the hitch? Knowing the percentage of the weight of the extension that's supported by the hitch, how much does the extension have to weigh to exactly counteract the decrease in weight due to leverage and keep the weight on the hitch at 100lbs? If the extension weighs any more than the answer you come up with, then you have to move the wheels forward to keep the weight 100lbs. Any less, then you have to move the wheels back. Bonus question: assume that the original level trailer with the 10' from axle to hitch weighs 1000lbs and the hitch weight is 100lbs, or 10% of the trailer weight. You extend the tongue by 5' using an extension that weighs 10 lbs/ft. How far and in what direction do you have to move the axle to keep the hitch weight 10% of the new trailer weight? Have fun! Steve I know damn well what the outcome is. But, I don't have to show/prove to you a damned thing. I solve baby **** like that on a daily basis. Apparently not if you think that the extension has to weigh the same per unit length as the boat and trailer to balance it out. didn't say that. Read for context, please. When I said that it's possible that you could be wrong, that the wheels would have to move forward to balance out the extra weight of the hitch, you said: The tongue would have to be made out of something VERY heavy, lead perhaps, to to make up for the additional lever arm from the fulcrum point. Basically, a foot of tongue would have to weigh the same as a foot of the boat and trailer. If you didn't mean that a foot of the extension would have to weight the same as a foot of the boat and trailer with your statement above, then what did you mean? Instead of thinking you know what the outcome is, why don't you solve the problem and see if you're right? Don't even worry about the bonus question if that's too hard. The main question is fairly simple to solve even with the most basic high school level of physics. Hell, I pretty much set the whole thing out for you on how to solve it. Steve I KNOW I'm right. And, yes, as I've stated, it IS basic high school physics. Again, I KNOW the answer, but alas, I'm not about to play You were wrong once, so apparently you don't KNOW the answer. your idiotic games. I have nothing that I need to prove to an ignoramus like you. Explaining and teaching something to a dimwit is all but impossible. You said it, not me. Steve |
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#6
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(Steven Shelikoff) wrote in message ...
On 11 Dec 2003 10:42:43 -0800, (basskisser) wrote: (Steven Shelikoff) wrote in message ... On 11 Dec 2003 04:35:05 -0800, (basskisser) wrote: (Steven Shelikoff) wrote in message ... On 10 Dec 2003 09:00:31 -0800, (basskisser) wrote: K Smith wrote in message ... Snafu wrote: Gary, I'm not understanding the part about "the longer the tongue, the furthur back we can put the wheels." To maintain the ~10% of the weight on the hitch rule, increasing the tongue length works the opposite way. Absolutely the longer the tongue (drawbar?) the further "forward" the wheels will by to keep the balance right. K Once again, you've proven you stick your nose where it doesn't belong. You certainly don't understand simple physics. If you make the tongue longer, you will need to move the wheels BACK to keep the balance the same. Say you have the tongue 10' from the wheels, and the tongue weight is 100 pounds, to say, at that 10', you need to resist 100 pounds to lift the tongue. Now, we extend the tongue, to 15'. Now you have a longer lever, thereby LESS tongue weight. To compensate, you need to move the wheels BACK. Tell ya what I'll do, I'm not going to give you the answer, but I'll define a problem and let basskisser and Karen come up with solutions and see if they come up with the same one, and then hack out their differences. Let's go with the situation above. Assume you have a trailer where the distance from hitch to axle is 10' and the tongue weight is 100lbs. Also, assume that the trailer is level because if it isn't, it will affect the answer. Also assume that the tongue extension has a constant weight per unit length. Now increase the length of the tongue to 15' from axle to hitch. Just from the difference in leverage, what is the new, lower weight on the hitch? Since we assumed that the extension has a linear weight/length, the center of mass of the extension is 2.5' from the hitch and 12.5' from the axle. So what percentage of the weight of the extension is supported by the hitch? Knowing the percentage of the weight of the extension that's supported by the hitch, how much does the extension have to weigh to exactly counteract the decrease in weight due to leverage and keep the weight on the hitch at 100lbs? If the extension weighs any more than the answer you come up with, then you have to move the wheels forward to keep the weight 100lbs. Any less, then you have to move the wheels back. Bonus question: assume that the original level trailer with the 10' from axle to hitch weighs 1000lbs and the hitch weight is 100lbs, or 10% of the trailer weight. You extend the tongue by 5' using an extension that weighs 10 lbs/ft. How far and in what direction do you have to move the axle to keep the hitch weight 10% of the new trailer weight? Have fun! Steve I know damn well what the outcome is. But, I don't have to show/prove to you a damned thing. I solve baby **** like that on a daily basis. Apparently not if you think that the extension has to weigh the same per unit length as the boat and trailer to balance it out. didn't say that. Read for context, please. When I said that it's possible that you could be wrong, that the wheels would have to move forward to balance out the extra weight of the hitch, you said: The tongue would have to be made out of something VERY heavy, lead perhaps, to to make up for the additional lever arm from the fulcrum point. Basically, a foot of tongue would have to weigh the same as a foot of the boat and trailer. If you didn't mean that a foot of the extension would have to weight the same as a foot of the boat and trailer with your statement above, then what did you mean? It's called vector mechanics, there Steve. Instead of thinking you know what the outcome is, why don't you solve the problem and see if you're right? Don't even worry about the bonus question if that's too hard. The main question is fairly simple to solve even with the most basic high school level of physics. Hell, I pretty much set the whole thing out for you on how to solve it. Steve I KNOW I'm right. And, yes, as I've stated, it IS basic high school physics. Again, I KNOW the answer, but alas, I'm not about to play You were wrong once, so apparently you don't KNOW the answer. your idiotic games. I have nothing that I need to prove to an ignoramus like you. Explaining and teaching something to a dimwit is all but impossible. You said it, not me. Did you figure out where you are wrong in your analysis? Another hint... You completely left out the fact that the fulcrum is NOT at the CG in the Z direction. |
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#7
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On 12 Dec 2003 04:21:15 -0800, (basskisser) wrote:
(Steven Shelikoff) wrote in message ... On 11 Dec 2003 10:42:43 -0800, (basskisser) wrote: (Steven Shelikoff) wrote in message ... On 11 Dec 2003 04:35:05 -0800, (basskisser) wrote: (Steven Shelikoff) wrote in message ... On 10 Dec 2003 09:00:31 -0800, (basskisser) wrote: K Smith wrote in message ... Snafu wrote: Gary, I'm not understanding the part about "the longer the tongue, the furthur back we can put the wheels." To maintain the ~10% of the weight on the hitch rule, increasing the tongue length works the opposite way. Absolutely the longer the tongue (drawbar?) the further "forward" the wheels will by to keep the balance right. K Once again, you've proven you stick your nose where it doesn't belong. You certainly don't understand simple physics. If you make the tongue longer, you will need to move the wheels BACK to keep the balance the same. Say you have the tongue 10' from the wheels, and the tongue weight is 100 pounds, to say, at that 10', you need to resist 100 pounds to lift the tongue. Now, we extend the tongue, to 15'. Now you have a longer lever, thereby LESS tongue weight. To compensate, you need to move the wheels BACK. Tell ya what I'll do, I'm not going to give you the answer, but I'll define a problem and let basskisser and Karen come up with solutions and see if they come up with the same one, and then hack out their differences. Let's go with the situation above. Assume you have a trailer where the distance from hitch to axle is 10' and the tongue weight is 100lbs. Also, assume that the trailer is level because if it isn't, it will affect the answer. Also assume that the tongue extension has a constant weight per unit length. Now increase the length of the tongue to 15' from axle to hitch. Just from the difference in leverage, what is the new, lower weight on the hitch? Since we assumed that the extension has a linear weight/length, the center of mass of the extension is 2.5' from the hitch and 12.5' from the axle. So what percentage of the weight of the extension is supported by the hitch? Knowing the percentage of the weight of the extension that's supported by the hitch, how much does the extension have to weigh to exactly counteract the decrease in weight due to leverage and keep the weight on the hitch at 100lbs? If the extension weighs any more than the answer you come up with, then you have to move the wheels forward to keep the weight 100lbs. Any less, then you have to move the wheels back. Bonus question: assume that the original level trailer with the 10' from axle to hitch weighs 1000lbs and the hitch weight is 100lbs, or 10% of the trailer weight. You extend the tongue by 5' using an extension that weighs 10 lbs/ft. How far and in what direction do you have to move the axle to keep the hitch weight 10% of the new trailer weight? Have fun! Steve I know damn well what the outcome is. But, I don't have to show/prove to you a damned thing. I solve baby **** like that on a daily basis. Apparently not if you think that the extension has to weigh the same per unit length as the boat and trailer to balance it out. didn't say that. Read for context, please. When I said that it's possible that you could be wrong, that the wheels would have to move forward to balance out the extra weight of the hitch, you said: The tongue would have to be made out of something VERY heavy, lead perhaps, to to make up for the additional lever arm from the fulcrum point. Basically, a foot of tongue would have to weigh the same as a foot of the boat and trailer. If you didn't mean that a foot of the extension would have to weight the same as a foot of the boat and trailer with your statement above, then what did you mean? It's called vector mechanics, there Steve. Is that what you meant when you said that to balance out the weight of adding length to the tongue that a foot of the tongue would have to weigh the same as a foot of the boat and trailer: "It's called vector mechanics"? You really don't comprehend english very well, do you? Instead of thinking you know what the outcome is, why don't you solve the problem and see if you're right? Don't even worry about the bonus question if that's too hard. The main question is fairly simple to solve even with the most basic high school level of physics. Hell, I pretty much set the whole thing out for you on how to solve it. Steve I KNOW I'm right. And, yes, as I've stated, it IS basic high school physics. Again, I KNOW the answer, but alas, I'm not about to play You were wrong once, so apparently you don't KNOW the answer. your idiotic games. I have nothing that I need to prove to an ignoramus like you. Explaining and teaching something to a dimwit is all but impossible. You said it, not me. Did you figure out where you are wrong in your analysis? Another hint... You completely left out the fact that the fulcrum is NOT at the CG in the Z direction. I guess you haven't figured out that, in the problem where I said "Also, assume that the trailer is level because if it isn't, it will affect the answer." I've taken into account the fact that the fulcrum is not at the CG in the Z direction. So you're wrong above where you claim I left out that fact. I would have thought someone as knowledgeable as you claim to be in high school physics would have picked up on the fact that if the trailer is level, the Z offset between the fulcrum and the CG contributes absolutely nothing to the weight/balance between the hitch and wheels and you can solve the problem without knowing what that offset is. You also obviously lied when you originally claimed to KNOW the answer... especially now that you've revealed that you (incorrectly) think the problem is flawed and has no answer. Keep digging yourself in deeper. lol Steve |
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#8
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"Steven Shelikoff" wrote in message ... I'll define a problem Can I play too?????? distance from hitch to axle is 10' and the tongue weight is 100lbs. Okay, that is 10' x 100 lbs = 1000 ft-lbs moment around the axel. Now increase the length of the tongue to 15' from axle to hitch. 1000 ft-lbs / 15 feet = 66.666 pounds. center of mass of the extension is 2.5' from the hitch and 12.5' from the axle. So what percentage of the weight of the extension is supported by the hitch? 12.5' / 15' x 100 = 83% Knowing the percentage of the weight of the extension that's supported by the hitch, how much does the extension have to weigh to exactly counteract the decrease in weight due to leverage and keep the weight on the hitch at 100lbs? 33 lbs / .83 = 40.1 lbs. Bonus question: assume that the original level trailer with the 10' from axle to hitch weighs 1000lbs and the hitch weight is 100lbs, or 10% of the trailer weight. 100 lbs x 10 feet = 1000 ft pounds. This means that the center of mass for the trailer is 1 foot forward of the axels. You extend the tongue by 5' using an extension that weighs 10 lbs/ft. We have added 50 pounds of material that has a center of mass 12.5 from the axel. The moment about the axels would be 50 lbs x 12.5 ft = 625 ft-lbs. Total moment is now 1625 ft-lbs. The new tongue weight will be 108.3 pounds. The new trailer weight is 1050 pounds. 1625 ft-lbs / 1050 pounds = 1.5 feet, thus the new center of mass is 1.5 feet in front of the axel, or 13.5 feet from the hitch. How far and in what direction do you have to move the axle to keep the hitch weight 10% of the new trailer weight? The desired of tongue weight of 10% would be 105 pounds. To get this, the distance from the hitch to the center of mass (13.5 feet) needs to be 10 times greater than the distance from the center of mass to the axel. Thus, the distance from the center of mass to the axel is 13.5/10 = 1.35 feet. Answer: The axel will have to move FORWARD 0.15 feet Did I get it right? Huh? Huh? Huh? Do I get a star on my paper????? Rod McInnis |
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#9
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On Thu, 11 Dec 2003 13:46:15 -0800, "Rod McInnis"
wrote: "Steven Shelikoff" wrote in message ... I'll define a problem Can I play too?????? Did you ever see the movie "My Cousin Vinnie" when Vinnie is questioning the old black woman with thick glasses and asks her how many fingers he's holding up, and the Judge gives her the answer? ![]() [answers snipped] Did I get it right? Huh? Huh? Huh? Do I get a star on my paper????? I'm not gonna tell you. Karen and Basskisser could always come up with something different. I will tell you something about the bonus question that may burn your britches a little: It's sort of a trick question because I didn't specify that the parts that move when you move the axle are balanced fore and aft. I *really* didn't think anyone would pick up on thatthough. ![]() But for Karen and Basskisser, when you work the problem out, you can assume that the axle assembly is balanced. Steve |
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