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As you will see I know nothing about physics and after being asked to solve a physics problem in a recent interview wanted to ask it of professionals and see what the response would be:

I have a set of domestic scales (actually just one scale) that weigh up to the maximum weight of 5kg and a large package that weighs more than 5kgs but less than 10kgs. How can I tell the exact weight using the inadequate scales.

The package is a long item akin to a steal girder but not in weight obviously.

I know that stack exchange has rules about this type of question so please treat it as light entertainment from someone in awe of your intellect.

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Hint, think of lifting one end of a long lever. –  Martin Beckett May 3 '12 at 15:25
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The title of the question made me want to answer "so go on a diet!" ;-) –  Manishearth May 3 '12 at 17:13
    
Just edited the question, there is just one scale –  Mregan May 4 '12 at 4:11
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2 Answers

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You put the package horizontally across three scales and add up the weight you see. If the center scale registers more than the limit, move the box, or put some pieces of paper under the scales which are registering a small weight to redistribute the weight.

A more physics-y question would be how to determine the weight when you have only one scale. This can be done by tilting the package at a small angle, by placing one end on a scale (perhaps propped up on something light, like your shoe, so that the weight momentum flow goes through the scale to the floor). To weigh this way requires knowing where the CM of the box is, and this can be determined by balancing it precisely on the edge of the scale (or on your finger, or on something else narrow).

Then the weight registered by the scale (minus the weight of the shoe) is the ratio of the length of the box to the distance from the end which is on the floor to the center of mass. If this is still too heavy, you can tilt the box up to 45 degrees, which will cut down the weight registered by the scale further by a factor of .707, which can be undone by multiplying by 1.414.

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So the last part of your answer, using a single scale, how would you express that formulaically? –  Mregan May 4 '12 at 5:27
    
By the way the answer that was presented to me during the interview as being correct was to balance one end on the scale and double that? I expressed concern about that giving an accurate weight but was unable to give a reason for my concern. –  Mregan May 4 '12 at 5:30
    
@Mregan: The "formulaic" way is the weight is the mass times the ratio of the distance from the end of the box to the center of mass to the length of the box. Doubling the weight assumes the center of mass is right in the middle, which is usually not right. You should find the CM by balancing. –  Ron Maimon May 4 '12 at 15:40
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Find a plank of wood, or something similar and put it on the scales as shown below:

Weigh parcel

The right end of the plank is on the scales and the left end is balanced on some convenient support. Now zero the scales, and put your parcel as closely as possible to the middle of the plank. The scales will now show half the weight of the parcel, so just read the weight on the scales and multiply it by two.

This is basically what Martin Beckett suggested in his comment, so he deserves the credit!

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There is no plank of wood provided, and there are multiple scales. –  Ron Maimon May 4 '12 at 0:32
    
The package is a long item so I assumed you could just do this –  Martin Beckett May 4 '12 at 5:11
    
@MartinBeckett: when I saw your comment I agreed and wasn't going to answer. Then it occurred to me that the centre of gravity of the package isn't known - it may not be uniform along it's length. Using a plank avoids this or at least reduces the error. –  John Rennie May 4 '12 at 5:48
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