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 Remember 09-09-2011, 10:08 PM #1 (permalink) Master EcoModder   Join Date: Feb 2011 Location: USA Posts: 346 Canyon - '07 GMC Canyon 2wd regular cab 90 day: 24.95 mpg (US) Thanks: 41 Thanked 39 Times in 24 Posts Tough (for me) math question Alright, let me set the picture: You have a bar loaded with an equal amount of weight on each side. It is evenly balanced on a stand placed directly at its center point. Now, somebody comes along and nudges the bar a distance of x units to the side - not enough to make it tip over, but enough to make the bar tilt. The system is now out of balance, despite the fact that the bar is still loaded with equal weight on each side. How can I figure out the downward force of the bar on the 'light' side and the 'heavy' side mathematically? I am not allowed to add weight to the light side until it becomes even - I must solve this on paper. Can someone send me in the right direction with this? __________________ EcoDriving: Turning more fuel into usable forward motion.  Today Popular topics Other popular topics in this forum... View the most popular topics in this subforum by views or by posts 09-09-2011, 10:26 PM #2 (permalink) Master EcoModder   Join Date: Apr 2008 Location: Northern Florida, USA Posts: 510 Hot Tamale - '10 Toyota Prius III Thanks: 27 Thanked 96 Times in 70 Posts Well the downward force on each side is the same, but the torques are different. The torque is the weight multiplied by the distance from the fulcrum (the pivot point). To figure the torques, multiply the weight that was on each end by the current distance from the pivot point. So you need to know the distance that the bar was moved. Subtract that from the midpoint for the short side and add it to the midpoint for the long side. Then multiply by the weight. To determine how much weight you need to add to the short side, divide the torque for the long side by the distance of the short side, then subtract the weight on the short side. This will make the torques equal and balance the system.  09-09-2011, 10:57 PM #3 (permalink) Master EcoModder   Join Date: Apr 2008 Location: Northern Florida, USA Posts: 510 Hot Tamale - '10 Toyota Prius III Thanks: 27 Thanked 96 Times in 70 Posts It's been a long time, but I think this is how you solve it algebraically, IIRC. W1 = initial weight on each side. W2 = weight to add on light side. D1 = initial distance to weight on each side. D2 = distance to weight on light side. Summation of torques must equal zero. W1(D1) = W1(D1) Weight is moved X distance. X = distance weight was moved. The equation then becomes W1(D1+X) = (W1 + W2)(D1-X). Call this Equation 1. Solve for W2: W1(D1+X)/(D1-X) = (W1 + W2), W1(D1+X)/(D1-X) - W1 = W2. Call this equation 2. Test Equation 2: Let W1 = 5 lbs, D1 = 4 feet, X = 2 feet. W2 = 5(4+2)/(4-2) - 5 = 5(6)/2 - 5 = 30/2 - 5 = 15 - 5 = 10. W2 = 10 lbs. Plugging those numbers into Equation 1: 5(4+2) = (5+10)(4-2), 5(6) = (15)(2), 30 = 30. Check.   09-09-2011, 11:54 PM   #4 (permalink)
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Quote:
 Originally Posted by Kodak ...How can I figure out the downward force of the bar on the 'light' side and the 'heavy' side mathematically?
It may be a trick question, as patrick notes, the force is the same, the moments (torque) have changed, and that is simply the weight times the arm. Figure out the moment on the long side and subtract the moment from the short side.

If you need to convert the result back to a force then divide the resulting moment by the long arm, though that is a bit arbitrary from the information given.
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Last edited by dcb; 09-10-2011 at 12:01 AM..  09-10-2011, 10:38 AM #5 (permalink) Smeghead   Join Date: Oct 2009 Location: South Central AK Posts: 933 escort - '99 ford escort sport 90 day: 42.38 mpg (US) scoobaru - '02 Subaru Forester s 90 day: 28.65 mpg (US) Thanks: 32 Thanked 146 Times in 97 Posts It is still in balance. If the bar is not moving the moment is the same on each side. (Edit: or the difference in moment is not enough to overcome the friction in the system) W___________________W .................A................. Now we slide the bar W____________________W ..........................A....... The bar will drop on the left and rise on the right. Slightly rotating on the fulcrum. until fulcrum is directly under the cg of the tilted bar or the bar falls over. This can farther be demonstrated with a "stickier" fulcrum with a larger radius. __________________ Learn from the mistakes of others, that way when you mess up you can do so in new and interesting ways. One mile of road will take you one mile, one mile of runway can take you around the world.  09-11-2011, 01:51 PM #6 (permalink) ...beats walking...   Join Date: Jul 2009 Location: . Posts: 6,190 Thanks: 179 Thanked 1,522 Times in 1,123 Posts ...look-up the terms: "MOMENT ARM" and "TORQUE"  09-12-2011, 10:18 AM #7 (permalink) EcoModding Lurker   Join Date: May 2011 Location: Michigan Posts: 4 Thanks: 0 Thanked 0 Times in 0 Posts Well....I believe the correct answer is they cannot be in balance and therefore the side with the longer arm has to drop to the surface. If it does not drop there is resistance at the pivot point that is upsetting the equation. With a friction free pivot it has to drop.  09-12-2011, 12:18 PM   #8 (permalink)
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Quote:
 Originally Posted by Patrick It's been a long time, but I think this is how you solve it algebraically, IIRC. W1 = initial weight on each side. W2 = weight to add on light side. D1 = initial distance to weight on each side. D2 = distance to weight on light side. Summation of torques must equal zero. W1(D1) = W1(D1)
...Patrick is correct, although I believe he meant the "first" basic equation to be:

W1(D1) = W2(D2)

...a simple mnemonic is: "...both strings-lengths have to be equal..." meaning a string from the center out the length of D1, then hanging down (representing weight W1) has to be the same length as D2×W2, ...ie: D2 can be ANY length, as long as the remaining W2 'length' totals up to the same 'length' of the W1(D1) side.

...the "other" way to express this "inverse-proportion" relationship is:

W1/W2 = D2/D1

Last edited by gone-ot; 09-12-2011 at 12:26 PM.. Thread Tools Show Printable Version Email this Page

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