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Old 09-09-2011, 10:08 PM   #1 (permalink)
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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?

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Old 09-09-2011, 10:26 PM   #2 (permalink)
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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.
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Old 09-09-2011, 10:57 PM   #3 (permalink)
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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.
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Old 09-09-2011, 11:54 PM   #4 (permalink)
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Quote:
Originally Posted by Kodak View Post
...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..
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Old 09-10-2011, 10:38 AM   #5 (permalink)
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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.
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Old 09-11-2011, 01:51 PM   #6 (permalink)
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...look-up the terms: "MOMENT ARM" and "TORQUE"
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Old 09-12-2011, 10:18 AM   #7 (permalink)
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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.
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Old 09-12-2011, 12:18 PM   #8 (permalink)
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Quote:
Originally Posted by Patrick View Post
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 D2W2, ...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..
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