[Guest]

There is a glass of water, standing on a scale, showing a digital weigth of 300.00 grams.

You hold a toothpick in your hand and submerge it in to the water. You don't touch the glass, and no water is overflowed from the glass.

Does the number on the scale changes?

[Guest]

Hi!

I'm the OP of this question.

Before giving the solution, I want to share my unhappiness with this: I had posted this question a month ago. But a genious moderator banned it beacuse of not being suitable to brain teasers. I insisted and wrote him that it is an interesting subject, but nothing helped him to convince. This time, I think that he was sleeping when I posted, so he missed to ban me.

As neide nicely explained, the scale will show a greater value.

I used toothpick to make the question harder. Suppose that you have a large stripor foam in your hand and submerge it to water. You will continiously have to apply a force to it to provide the foam to stay in water. Because you are making some water (equal to volume of foam) to be raised. The force that you must apply is the weight of that water. And you will always see in the scale:300.00 + weight of water in volume of foam (or toothpick)

[bonanova]

There is a glass of water, standing on a scale, showing a digital weigth of 300.00 grams.

You hold a toothpick in your hand and submerge it in to the water. You don't touch the glass, and no water is overflowed from the glass.

Does the number on the scale changes?

If the toothpick has a density less than that of water, then it would float.

If you submerge it, you exert a downward force that registers as increased weight on the scale.

The force needed to submerge it is the difference in weight of a toothpick's volume of water to the toothpick's weight.

[Guest]

If the toothpick has a density less than that of water, then it would float.

If you submerge it, you exert a downward force that registers as increased weight on the scale.

The force needed to submerge it is the difference in weight of a toothpick's volume of water to the toothpick's weight.

I don't think that toothpicks weight or density has a role. You hold it, thus its weight is a matter of you. Maybe you've misunderstood as leave the toothpick free on water. In question, you don't leave it, instead hold it. Only the volume of toothpick should be taken in account.

This is my own decision, the original source only says that the scales value raises. As a master, maybe you explain where I'm wrong.

[Guest]

Neida, you have quite a talent for illustrating your argument effectively and concisely, without making one feel lectured to. It is rare and appreciated.

Thanks! I don't really see any point in trying to make enemies!

I don't think that toothpicks weight or density has a role. You hold it, thus its weight is a matter of you. Maybe you've misunderstood as leave the toothpick free on water. In question, you don't leave it, instead hold it. Only the volume of toothpick should be taken in account.

This is my own decision, the original source only says that the scales value raises. As a master, maybe you explain where I'm wrong.

Bonanova is correct in that density and weight do play a role in determining the answer, and you are also correct that only the volume of the toothpick should be taken into account!

If the density is greater than water, then the maximum increase the scales could measure is the weight of the toothpick, which would happen if you were to let go entirely (provided that your fingers aren't in the water). As, in this case, the toothpick would sink, this is greater than the weight of a volume of water equal to the weight of the toothpick.

If the density is less than water, then the scales can measure a greater increase than the weight of the toothpick because, in order to submerge it, you must also apply an external downward force which is in addition to the toothpicks own weight. You can measure the total of this increase on the scales as weight of toothpick + weight applied by external downward force = weight of water equalling volume of the submerged part of the toothpick.

Note though, that if the toothpick is less dense than water and is allowed to float, the scales measure an increase equal to the weight of the toothpick. The above equation still applies though (weight of toothpick = weight of water equalling volume of the submerged part of the toothpick = increase on scales).

Note also though that we are not told how much we submerge the toothpick in the water. If we were only to put the tip in then we would actually need to support some of the weight ourselves, which equates to us applying an external upward force on the toothpick (or alternatively a negative downward force). Again the equation still stands and the answer is found by looking at the volume of the submerged part of the toothpick.

d3k3 has absolutely the right maths in his answer - I just always like thinking it out more in words as I sometimes find bare equations can be difficult to comprehend.

[Guest]

Bonanova is correct in that density and weight do play a role in determining the answer,

If the density is greater than water, then the maximum increase the scales could measure is the weight of the toothpick.

Thanks, now I see that if density of toothpick is greater, then I will not apply a force downward.

But in this case, you say the scale will measure the weight of toothpick. Then, I shouldn't apply a force to hold the toothpick, but I have to apply a force in order the toothpick not to sink completely.

I suppose the measurement increase will be = Weight of toothpick - weight of water in volume of toothpick in water. Not so?

[bonanova]

I don't think that toothpicks weight or density has a role. You hold it, thus its weight is a matter of you. Maybe you've misunderstood as leave the toothpick free on water. In question, you don't leave it, instead hold it. Only the volume of toothpick should be taken in account.

This is my own decision, the original source only says that the scales value raises. As a master, maybe you explain where I'm wrong.

[1] drop the pick onto the water's surface.

The scale's reading increases by the weight of the pick, but it's floating; it's not submerged.

[2] submerge the pick.

In order to submerge the pick you exert a downward force on it.

The scale reading increases once more, by an amount equal to the force you exert to submerge it.

That force is the difference in weight of the pick with that of an imaginary pick [same size and shape] made out of water.

Why? because a "water" pick would have neutral buoyancy and not require any force to submerge it.

The total increase of the scale reading is the sum of these two: weight of pick + [weight of water pick - weight of pick] = weight of water pick.

That is, the scale's reading will increase [from 300 g] by an amount equal to the weight of water equal in volume to the volume of the pick.

Edited January 9, 2009 by bonanova
put my reply into spoiler ;)

[Guest]

Thanks, now I see that if density of toothpick is greater, then I will not apply a force downward.

But in this case, you say the scale will measure the weight of toothpick.

Not quite - the "maximum" the scale "could" increase by will be the weight of the toothpick, but this will only happen if you let go.

Then, I shouldn't apply a force to hold the toothpick, but I have to apply a force in order the toothpick not to sink completely.

Exactly. This is actually the same case where I have said about applying a negative downward force.

I suppose the measurement increase will be = Weight of toothpick - weight of water in volume of toothpick in water. Not so?

Not quite. The equation (as I have given it) is that the increase measured in the scales equals weight of toothpick + weight applied by downward force = weight of volume of water equal to submerged part of toothpick. This still applies even if the external downward force is negative (as in this case). By rearranging this equation you see that the answer that you have given is actually the force that needs to be applied to hold the toothpick out of the water or, alternatively, the amount of weight of the toothpick that you are supporting.