My Math Forum Unequal Weights

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 January 26th, 2013, 06:43 PM #1 Senior Member   Joined: Dec 2012 Posts: 450 Thanks: 0 Unequal Weights I have a weghing balance.But when nothing is placed at both ends it droops down to one end. I weighed something using this.It weighed 100 grams on one side and 144 grams on the other side. Can you figure out the real weight of the object? This is a real world problem that i observed recently.And i solved it. Can anybody do what i did?Give me an explanation too.
 January 26th, 2013, 07:22 PM #2 Member     Joined: Jul 2012 Posts: 60 Thanks: 0 Math Focus: Calculus Re: UNEQUAL WEIGHTS Did you mean the [color=#8000BF]same object [/color]weighs 100 grams on one side and 144 on the other? Since you said it's a real world problem, a bit more information is needed like: what's wrong with the weighing balance? Is one plate heavier than the other, is it not leveled, or is one of the bars longer than the other? I'll just assume you meant that one plate is heavier than the other. Then the mass of the object you are weighing is 100 grams. Why? Let's first see how a weighing balance works: You place an object on one side, and then you have to place a metal weight on the other side until it's balanced. When it's balanced, you read the number the scale is showing. Well let's just say plate 1 is heavier than plate 2, and the difference between them is $m_0$ If you place your object on plate 1, then you will have to place a weight of $m_0+m_1$ onto the other side, so you will be measuring the mass of your object + the additional mass of the heavier plate = 144 If you place your object on plate 2, (assuming your object is heavier than $m_0$) you will have to place a weight of $m_1-m_0$ onto the other side. So you will successfully be weighing your object and getting a reading of 100 grams. $m_1=100g m_0=44g$
 January 26th, 2013, 09:32 PM #3 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,260 Thanks: 198 Re: Unequal Weights I have a different approach. Assuming 1 end is heavier, the fulcrum exactly in the middle, and everything else about the balance is uniform, homogeneous and isotropic... you can say let m = mass of your object, let x = additional weight at one end, then, m + x = 144 m - x = 100 m = 122
January 30th, 2013, 05:24 PM   #4
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Re: Unequal Weights

Quote:
 Originally Posted by agentredlum I have a different approach. Assuming 1 end is heavier, the fulcrum exactly in the middle, and everything else about the balance is uniform, homogeneous and isotropic... you can say let m = mass of your object, let x = additional weight at one end, then, m + x = 144 m - x = 100 m = 122
But what if there is no extra weight on any side but the bars are of unequal length,can you find the real weight?

But your solution is very clear and simple,i appreciate that.

January 30th, 2013, 05:25 PM   #5
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Re: UNEQUAL WEIGHTS

Quote:
 Originally Posted by chocolatesheep Did you mean the [color=#8000BF]same object [/color]weighs 100 grams on one side and 144 on the other? Since you said it's a real world problem, a bit more information is needed like: what's wrong with the weighing balance? Is one plate heavier than the other, is it not leveled, or is one of the bars longer than the other? I'll just assume you meant that one plate is heavier than the other.
What if one of the bars is longer,try it?

 January 30th, 2013, 09:08 PM #6 Math Team     Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,260 Thanks: 198 Re: Unequal Weights We can use archimede's law of the lever and center of gravity idea, explained wonderfully by julius sumner miller in the links below. http://www.youtube.com/watch?v=vYbBakL9Cj8 http://www.youtube.com/watch?v=3CBuoe9iCQI The weight of the object in question is the same for both weighings. Whithout loss of generality, let the length of the ballance be 1 and the distance of the fulcrum from the short end be D. Then according to the law of the lever. $WD= 100(1-D) \$ (1) $W(1-D)= 144D \$ (2) Expand (1) and (2) $WD= 100 - 100D \$ (3) $W - WD= 144D \$ (4) add (3) and (4) $W= 100 + 44D \$ (5) Substitute (5) in (1) $(100 + 44D)D= 100 - 100D \$ (6) re-arrange (6) $11D^2 + 50D - 25= 0 \$ (7) the positive solution of (7) is D = 5/11 , plug this into (2) $W(1 - \frac{5}{11})= 144 \frac{5}{11}$ $W= 144 \frac{5}{11} \frac{11}{6}$ W = 120 P.S. I hope you have an easier way.  We could have plugged 5/11 into (5) but it's also good to plug into (2) in order to come full circle so to speak. [endedit]
 January 31st, 2013, 05:44 AM #7 Member     Joined: Jul 2012 Posts: 60 Thanks: 0 Math Focus: Calculus Re: Unequal Weights Well did I do it right or what? Is that how you solved it?
January 31st, 2013, 07:09 AM   #8
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Re: Unequal Weights

Quote:
Originally Posted by mathmaniac
Quote:
 Originally Posted by chocolatesheep Did you mean the [color=#8000BF]same object [/color]weighs 100 grams on one side and 144 on the other? Since you said it's a real world problem, a bit more information is needed like: what's wrong with the weighing balance? Is one plate heavier than the other, is it not leveled, or is one of the bars longer than the other? I'll just assume you meant that one plate is heavier than the other.
What is one of the bars is londer,try it?
LaTeX is such a pain. I did it with pen and paper if you don't mind.
Here you go:

M is moment of force btw
In order to stay leveled, the moment of force of the left side and the right side need to be equal.
Attached Images

January 31st, 2013, 05:12 PM   #9
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Re: UNEQUAL WEIGHTS

Quote:
 Originally Posted by chocolatesheep Did you mean the [color=#8000BF]same object [/color]weighs 100 grams on one side and 144 on the other? Since you said it's a real world problem, a bit more information is needed like: what's wrong with the weighing balance? Is one plate heavier than the other, is it not leveled, or is one of the bars longer than the other? I'll just assume you meant that one plate is heavier than the other. Then the mass of the object you are weighing is 100 grams. Why? Let's first see how a weighing balance works: You place an object on one side, and then you have to place a metal weight on the other side until it's balanced. When it's balanced, you read the number the scale is showing. Well let's just say plate 1 is heavier than plate 2, and the difference between them is $m_0$ If you place your object on plate 1, then you will have to place a weight of $m_0+m_1$ onto the other side, so you will be measuring the mass of your object + the additional mass of the heavier plate = 144 If you place your object on plate 2, (assuming your object is heavier than $m_0$) you will have to place a weight of $m_1-m_0$ onto the other side. So you will successfully be weighing your object and getting a reading of 100 grams. $m_1=100g m_0=44g$
m1+m0=144
m1-m0=100

so you get m1=122

You have done it wrong.....i think....

January 31st, 2013, 05:51 PM   #10
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Re: Unequal Weights

Quote:
 Originally Posted by agentredlum We can use archimede's law of the lever and center of gravity idea, explained wonderfully by julius sumner miller in the links below. http://www.youtube.com/watch?v=vYbBakL9Cj8 http://www.youtube.com/watch?v=3CBuoe9iCQI The weight of the object in question is the same for both weighings. Whithout loss of generality, let the length of the ballance be 1 and the distance of the fulcrum from the short end be D. Then according to the law of the lever. $WD= 100(1-D) \$ (1) $W(1-D)= 144D \$ (2) Expand (1) and (2) $WD= 100 - 100D \$ (3) $W - WD= 144D \$ (4) add (3) and (4) $W= 100 + 44D \$ (5) Substitute (5) in (1) $(100 + 44D)D= 100 - 100D \$ (6) re-arrange (6) $11D^2 + 50D - 25= 0 \$ (7) the positive solution of (7) is D = 5/11 , plug this into (2) $W(1 - \frac{5}{11})= 144 \frac{5}{11}$ $W= 144 \frac{5}{11} \frac{11}{6}$ W = 120 P.S. I hope you have an easier way.  We could have plugged 5/11 into (5) but it's also good to plug into (2) in order to come full circle so to speak. [endedit]

Yes i have an easier way i think....But i doubt if it is the right way after seeing how you guys do it.

Suppose the lengths of the arms are a and b units and the weight of the object X grams.
a*X=b*100[* is the symbol for multiplication]------------>eq (1)[i dont have a good explanation for this]
a*144=b*X------------->eq (2)
eq(1)/eq(2)gives X/144=100/X
So X^2=100*144
which gives X=120 grams.

Am i right?

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