Virtual Experiment for every new member

[clu001]

There is a cup half full of water on a scale. If you put your finger into the water without touching the sides of the cup, and the bottom of cup, will the scale read a different weight?


Do not do experiment and just think it about and post back your answer.

[clu001]

There is a cup half full of water on a scale. If you put your finger into the water without touching the sides of the cup, and the bottom of cup, will the scale read a different weight?


Do not do experiment and just think it about and post back your answer.

[clu001]

There is a cup half full of water on a scale. If you put your finger into the water without touching the sides of the cup, and the bottom of cup, will the scale read a different weight?


Do not do experiment and just think it about and post back your answer.

[Fernanda]

I\'m assuming there are a lot of wet fingers...either that or no one bothered to type the answer...:P

:D

[Fernanda]

I\'m assuming there are a lot of wet fingers...either that or no one bothered to type the answer...:P

:D

[Fernanda]

I\'m assuming there are a lot of wet fingers...either that or no one bothered to type the answer...:P

:D

[cassieaw]

While you are exerting a force on the system that force simply displaces the water which is a fluid, just as air is a fluid. The effect is the same as if you were to hold you finger in the air a short distance above the scale.

The scale\'s reading will not change.

[cassieaw]

While you are exerting a force on the system that force simply displaces the water which is a fluid, just as air is a fluid. The effect is the same as if you were to hold you finger in the air a short distance above the scale.

The scale\'s reading will not change.

[cassieaw]

While you are exerting a force on the system that force simply displaces the water which is a fluid, just as air is a fluid. The effect is the same as if you were to hold you finger in the air a short distance above the scale.

The scale\'s reading will not change.

[Fernanda]

the water will displace...but now the system being weighed is water+cup+finger... up to the level of the water.

It would be no different than adding ice to it :D

[Fernanda]

the water will displace...but now the system being weighed is water+cup+finger... up to the level of the water.

It would be no different than adding ice to it :D

[Fernanda]

the water will displace...but now the system being weighed is water+cup+finger... up to the level of the water.

It would be no different than adding ice to it :D

[JR]

NO, the water will displace

[JR]

NO, the water will displace

[JR]

NO, the water will displace

[Fernanda]

I\'m not sure how many scales you think clu001 is talking about...

see http://physics.usc.edu/Classes/100/Bars/chapters/ch13/ch13next.pdf

question 10

[Fernanda]

I\'m not sure how many scales you think clu001 is talking about...

see http://physics.usc.edu/Classes/100/Bars/chapters/ch13/ch13next.pdf

question 10

[Fernanda]

I\'m not sure how many scales you think clu001 is talking about...

see http://physics.usc.edu/Classes/100/Bars/chapters/ch13/ch13next.pdf

question 10

[Fernanda]

hmm...but unlike a gas that has plenty of room to expand and compress...water has to be displaced. And it only displaces if there is some downward force displacing it...

So my assumption is that the scale will read more just like the ice cube I mentioned before. To maintain your finger in the water, you must apply a small downward force.
To me, it helps to think of something more boyant...say cork.

[Fernanda]

hmm...but unlike a gas that has plenty of room to expand and compress...water has to be displaced. And it only displaces if there is some downward force displacing it...

So my assumption is that the scale will read more just like the ice cube I mentioned before. To maintain your finger in the water, you must apply a small downward force.
To me, it helps to think of something more boyant...say cork.

[Fernanda]

hmm...but unlike a gas that has plenty of room to expand and compress...water has to be displaced. And it only displaces if there is some downward force displacing it...

So my assumption is that the scale will read more just like the ice cube I mentioned before. To maintain your finger in the water, you must apply a small downward force.
To me, it helps to think of something more boyant...say cork.

[Terry]

I would say that the weight on the scale would increase. The increase would be proportional to the amount of water displaced.

For example:

If I took a hollow ball that was filled with air and I would have to push down on it to partially submerge it. The magnitude of the downward force would be proportional to the amount of water displaced.

If I took a solid lead ball and held it while partially submerging it into the cup of water, the magnitude of buoyant force would be the same as for hollow ball as long as the displaced volume was the same, and the scale would read an increase in weight equivalent to the displaced volume of the water.

If I were to drop the hollow ball into the cup then the weight increase of the water and cup system would be increased by the weight of the hollow ball.

If I were to drop the lead ball into the cup then the weight would increase by the amount of displaced fluid until the lead ball came to rest at the bottom of the cup, at which time the weight would increase by the weight of the ball.

[Terry]

I would say that the weight on the scale would increase. The increase would be proportional to the amount of water displaced.

For example:

If I took a hollow ball that was filled with air and I would have to push down on it to partially submerge it. The magnitude of the downward force would be proportional to the amount of water displaced.

If I took a solid lead ball and held it while partially submerging it into the cup of water, the magnitude of buoyant force would be the same as for hollow ball as long as the displaced volume was the same, and the scale would read an increase in weight equivalent to the displaced volume of the water.

If I were to drop the hollow ball into the cup then the weight increase of the water and cup system would be increased by the weight of the hollow ball.

If I were to drop the lead ball into the cup then the weight would increase by the amount of displaced fluid until the lead ball came to rest at the bottom of the cup, at which time the weight would increase by the weight of the ball.

[Terry]

I would say that the weight on the scale would increase. The increase would be proportional to the amount of water displaced.

For example:

If I took a hollow ball that was filled with air and I would have to push down on it to partially submerge it. The magnitude of the downward force would be proportional to the amount of water displaced.

If I took a solid lead ball and held it while partially submerging it into the cup of water, the magnitude of buoyant force would be the same as for hollow ball as long as the displaced volume was the same, and the scale would read an increase in weight equivalent to the displaced volume of the water.

If I were to drop the hollow ball into the cup then the weight increase of the water and cup system would be increased by the weight of the hollow ball.

If I were to drop the lead ball into the cup then the weight would increase by the amount of displaced fluid until the lead ball came to rest at the bottom of the cup, at which time the weight would increase by the weight of the ball.

[Beowulf]

The amount of water displacement is depended on the amount of force exerted by the finger. Therefore my guess is that the force of the finger is being negated by water displacement.

So the answer is no:P

[Beowulf]

The amount of water displacement is depended on the amount of force exerted by the finger. Therefore my guess is that the force of the finger is being negated by water displacement.

So the answer is no:P

[Beowulf]

The amount of water displacement is depended on the amount of force exerted by the finger. Therefore my guess is that the force of the finger is being negated by water displacement.

So the answer is no:P

[dealo]

if you had a cork in the glass, the water would be holding it up, which would make the corks wight count. But with our finger. We are holding the weight so that shouldnt alter the messuremnet on the scale.

My question is, now that there is more water touching the glass(in surface area), would that make any more reaction either in energy or chemical. And if so, eneogh to alter the weight?

[dealo]

if you had a cork in the glass, the water would be holding it up, which would make the corks wight count. But with our finger. We are holding the weight so that shouldnt alter the messuremnet on the scale.

My question is, now that there is more water touching the glass(in surface area), would that make any more reaction either in energy or chemical. And if so, eneogh to alter the weight?

[dealo]

if you had a cork in the glass, the water would be holding it up, which would make the corks wight count. But with our finger. We are holding the weight so that shouldnt alter the messuremnet on the scale.

My question is, now that there is more water touching the glass(in surface area), would that make any more reaction either in energy or chemical. And if so, eneogh to alter the weight?

[Fernanda]

I\'m not sure what reaction you expect water to have with the glass...
I hope at your house that reaction isn\'t a violent one...

:lol:

[Fernanda]

I\'m not sure what reaction you expect water to have with the glass...
I hope at your house that reaction isn\'t a violent one...

:lol:

[Fernanda]

I\'m not sure what reaction you expect water to have with the glass...
I hope at your house that reaction isn\'t a violent one...

:lol:

[dealo]

let me put it this way. If you had an empty bucket on the scales. And put you hand in there without touching. Would the weight change. No but your still displacing the air just like you are the water

[dealo]

let me put it this way. If you had an empty bucket on the scales. And put you hand in there without touching. Would the weight change. No but your still displacing the air just like you are the water

[dealo]

let me put it this way. If you had an empty bucket on the scales. And put you hand in there without touching. Would the weight change. No but your still displacing the air just like you are the water

[Fernanda]

but water has surface tension...and when you place you finger in there it doesn\'t spill over the cup.... unlike air which is unrestricted to move in and out of the bucket...

[Fernanda]

but water has surface tension...and when you place you finger in there it doesn\'t spill over the cup.... unlike air which is unrestricted to move in and out of the bucket...

[Fernanda]

but water has surface tension...and when you place you finger in there it doesn\'t spill over the cup.... unlike air which is unrestricted to move in and out of the bucket...

[dealo]

but its still the same amount of water

[dealo]

but its still the same amount of water

[dealo]

but its still the same amount of water

[WhiteKnight]

Of course it\'ll increase, for the reasons already mentioned by others. :)

[WhiteKnight]

Of course it\'ll increase, for the reasons already mentioned by others. :)

[WhiteKnight]

Of course it\'ll increase, for the reasons already mentioned by others. :)

[khallow]

Same as Terry. Your finger would displace water, and the force required to displace that water would be added to the observed weight of the glass of water.

[khallow]

Same as Terry. Your finger would displace water, and the force required to displace that water would be added to the observed weight of the glass of water.

[khallow]

Same as Terry. Your finger would displace water, and the force required to displace that water would be added to the observed weight of the glass of water.

[Plusunim]

the more you dip your finger, the more the overall mass increases (just upto the bottom of the cup is the maximum mass). and since we live in a relatively constant gravitational field, by Newton\'s secon law the weight increases.

the weight doesn\'t increase because of the \'push\' in the water, that is counteracted by the displacement/buoyancy. the weight increases bcause of the added mass (the dipped portion of the finger).

yes water is a fluid like air, but the air is not bounded as the water in a cup is.

saying that water is displaced...it must displace upwards, thus increasing in height relative to the bottom of the cup. doesn\'t mean more pressure? since the area of the cup is the same, mass also, and pressure increased, i can say that the force increased.

displacement occurs to counter the \'alien\' volume and the fact that no two atoms (or more) can be in the same place at the same time.

In all, I say that there is an increase in the scale\'s reading.

[Plusunim]

the more you dip your finger, the more the overall mass increases (just upto the bottom of the cup is the maximum mass). and since we live in a relatively constant gravitational field, by Newton\'s secon law the weight increases.

the weight doesn\'t increase because of the \'push\' in the water, that is counteracted by the displacement/buoyancy. the weight increases bcause of the added mass (the dipped portion of the finger).

yes water is a fluid like air, but the air is not bounded as the water in a cup is.

saying that water is displaced...it must displace upwards, thus increasing in height relative to the bottom of the cup. doesn\'t mean more pressure? since the area of the cup is the same, mass also, and pressure increased, i can say that the force increased.

displacement occurs to counter the \'alien\' volume and the fact that no two atoms (or more) can be in the same place at the same time.

In all, I say that there is an increase in the scale\'s reading.

[Plusunim]

the more you dip your finger, the more the overall mass increases (just upto the bottom of the cup is the maximum mass). and since we live in a relatively constant gravitational field, by Newton\'s secon law the weight increases.

the weight doesn\'t increase because of the \'push\' in the water, that is counteracted by the displacement/buoyancy. the weight increases bcause of the added mass (the dipped portion of the finger).

yes water is a fluid like air, but the air is not bounded as the water in a cup is.

saying that water is displaced...it must displace upwards, thus increasing in height relative to the bottom of the cup. doesn\'t mean more pressure? since the area of the cup is the same, mass also, and pressure increased, i can say that the force increased.

displacement occurs to counter the \'alien\' volume and the fact that no two atoms (or more) can be in the same place at the same time.

In all, I say that there is an increase in the scale\'s reading.

[errandir]

might it depend on the relative density of the finger to the density of water?

[errandir]

might it depend on the relative density of the finger to the density of water?

[errandir]

might it depend on the relative density of the finger to the density of water?

[errandir]

The force at the surface of the liquid on the liquid below is (~1 atm)*(A of surface).

The force at any depth in the liquid on all the liquid below is (~1 atm)*(A of surface) + integral of {(~density)*(x-sectional A at depth)*(~g)*dy} from y = 0 to y = depth. (The density, g, and A are approximated as constants.)

The force at the very bottom of the cup will basically be (~1 atm)*A + K*y, where A is the assumed constant x-sectional area, K is a constant of proportionality, and y is the level of the liquid with respect to the bottom of the cup.

Since this analysis is based on fluid pressure (not force directly), the only thing that matters is the height of the liquid (with respect to the bottom). In this light, I believe that the weight should increase. The narrower the cup, the more the weight should increase. I.e. sticking your finger in a shot glass will have a much more profound effect than sticking your finger in a bowl.

With that said, I\'m now going to get my finger wet.

I think that this is the same principle used in hydraulic lifts.

[errandir]

The force at the surface of the liquid on the liquid below is (~1 atm)*(A of surface).

The force at any depth in the liquid on all the liquid below is (~1 atm)*(A of surface) + integral of {(~density)*(x-sectional A at depth)*(~g)*dy} from y = 0 to y = depth. (The density, g, and A are approximated as constants.)

The force at the very bottom of the cup will basically be (~1 atm)*A + K*y, where A is the assumed constant x-sectional area, K is a constant of proportionality, and y is the level of the liquid with respect to the bottom of the cup.

Since this analysis is based on fluid pressure (not force directly), the only thing that matters is the height of the liquid (with respect to the bottom). In this light, I believe that the weight should increase. The narrower the cup, the more the weight should increase. I.e. sticking your finger in a shot glass will have a much more profound effect than sticking your finger in a bowl.

With that said, I\'m now going to get my finger wet.

I think that this is the same principle used in hydraulic lifts.

[grahamc]

This is easier to understand if we assume that the water container is full to the brim. Placing fingers in the water will cause water displacement equal in mass to the same volume of water taken by the fingers. Assuming that all the displaced water falls from the container (no drips left) then the scales will stay the same.
However, in our case, the container is not full to the brim and is able to contain the displaced water so increasing the containers weight reading on the scale.
Try it.

[Fernanda]

yuppers! Let\'s try it with beer :beer:

[spirit_wasa]

HA HA HA everyone this is a well known experiment.

It was on my final exam paper (QCST) in queensland australia. What happens is that everyone knows the force due to boyancy that all objects feel in water however since your finger is not moving not only are you pulling against gravity but pushing slightly against the boyancy force and this must go through the water into the table

[jchristophm]

I tryed it with beer :beer:, but then I drank the beer, and then I forgot what happened. :puzzled:

Is this a group of theorists?! Jeez, the expiment just took me 2 minutes. The weight goes up because you are adding mass.

[Fernanda]

lol...it\'s the old joke about theorists.

we have a super smart theorist here is the department...nuclear guy. I mean, the guy KNOWS his stuff inside, out, sideways.

So there\'s an oral defense on detectors and they are showing some angle dependent data with a range of say 90deg to about 130deg

so he asks \"where are the other angles?\"

student \"what do you mean?\"

him \"where\'s 0 thru 89?\"

student \"we don\'t have it\"

him \"why not?\"

student \"cause we can\'t move the detectors that far\"

him confused \"move? you mean you don\'t have detectors all around?\"

student \"uh no.... we have to physically move the detectors and then recalibrate and do it all over again\"

him \"how much do they weigh?\"

finally, students advisor in an annoyed voice \"a hell of a lot\" turning to student \"please continue.\"

[physics137]

First consider some limiting cases (always a good idea when looking at a problem involving multiple subtleties or interactions):

Suppose that I drop a piece of lead into the water - it sinks to the bottom. The lead pushes downward on the bottom of the glass with a force mg, and the bottom of the glass responds (a la the 3rd law) with an upward reaction force on the lead. Similarly, the glass pushes down on the scale with some force Mg (where M = the mass of the container PLUS the mass of the water (specifically the hydrostatic pressure exerted on the bottom of the container times its surface area times g, which is the total downward hydrostatic force, which happens to be equal to the total weight of the water) PLUS that of the added lead), and the scale returns in kind, and that mass M is what is read.

Now suppose I drop an ice cube on the water. The ice cube is pulled downward by gravity with a force mg, but does not sink because of a buoyancy force (which is equal in magnitude and opposite in direction). The water level rises, thus the hydrostatic pressure at the bottom of the container increases, and so the downward force exerted by the container on the scale increases, and the reaction force to this (the upward force exerted by the scale\'s mechanism to hold the container in mechanical equilibrium) and the weight recorded by the scale increases.

Now suppose I HANG a piece of lead, suspended in the water by a string. True, the water level rises, so one might think that the scale\'s reading goes up (since hydrostatic pressure is p*g*h, and h increases). But consider this:

Look at the case where the ice is floating on top of the water. Look at the water IMMEDIATELY below the ice cube. What is its pressure? It\'s exactly the same as the pressure at another point at the same height, exactly the same as if the part of the ice cube submerged were replaced by an equivalent volume of water (which by Archimedes principle is of the same mass as the total ice cube). Thus the hydrostatic pressure at the bottom of the glass is indeed p*g*h.

Now try to suspend the ice cube from a string so it\'s at the same location as before. You might think (as I first did) that now there is no need for a buoyancy force to hold the ice cube in position, so the scale would be unchanged.

But what happenswhen you try to suspend the cube by applying a force mg upward to suspend the cube in air, then slowly lower it into the water? Well, you can\'t. As soon as you pull the string taut the ice cube starts rising out of the water, and in fact in the ice cube rests so that its bottom surface is level with the water surface (so that there is no water displaced and thus no buoyancy force). The only way to submerge it is to apply a force smaller than the mg of the ice (i.e. so that if the water weren\'t there the ice cube would accelerate downwards) the reason being that the force you apply PLUS an upwards buoyancy force equal to the weight of the displaced water equals the ice\'s mg).

Long story short, the free body diagram for the suspended object shows three forces:

1. The upwards force you apply, Fa

2. The downwards gravitational force, mg

3. Possibly the buoyancy force Fb (which will ALWAYS be nonzero if the object is at least partially immersed, which is transmitted via hydrostatic pressure to the bottom and will increse the scale reading by that amount)

If Fb is not zero, then manifestly Fa will be less than mg (i.e. you\'re not supporting the whole weight of the suspended object; if you were you couldn\'t get it to submerge)

Therefore, the reading on the scale will be due to the weight of anything on the bottom of the water, anything floating on top, and the unsupported weight of anything hanging (i.e. mg - Fa, which is equal to the upwards buoyancy force applied by the water, which is equal to the weight of the water displaced by the hanging object).

So the answer is yes, the scale reading goes up, by the weight of the displaced water. In the special case of the unsupported floating object, the buoyancy force is equal to the total mass, so the scale reading goes up by the total weight of the floating object. (The same applies for something resting against the bottom.)

[physics137]

Another way to think about it:

Suppose I start with the container empty and have a lead weight hanging from a spring, suspended somewhere in the volume of the container. The spring has some original undeflected length, and stretches a distance D when I hang the lead from it; manifestly the stretching distance D is equal to the lead\'s weight divided by the Hooke\'s law constant of the spring:

D = mg/k

Now begin filling the container. As the water level reaches the lead weight, the buoyancy force begins to lift it, and when all is said and done the lead weight will be resting at a slightly higher position with deflection D\' from the original length: (neglect the mass and volume of the spring). Thus the force suspending the lead weight will be somewhat less:

amount shifted upward d = D-D\'

change in upward force provided by spring = kd\'

thus the upwards buoyancy force provided by the water is kd\', and this buoyancy force can be determined from the density of lead:

Fb = m_lead*g*(rho_lead-rho_water)*volume_lead = kd\'

so d\' = m_lead*g*volume_lead*(rho_lead-rho_water)/k

Since the buoyancy force is nonzero (and equal to the weight of whatever water happens to be displaced by the submerged volume of the suspended object), the situation reduces to that of an ice cube floating in water, and the scale reading goes up.

In the case that the density of the suspended object is significantly less than that of water (such as a baloon) the suspending force might actually be DOWNWARD (if you want to suspend it at a position lower than Archimedes dictates), and the result should be obvious in that case.

[PritamDamania]

The weight in the scale will increase as our finger will displace water a bouyant force will act on our finger and the reaction of this force acts on the bottom of the vessel thus incresing the force actting on the vessel

[pratikkotkar]

ofcourse reading will change!! Buoyancy is a two way force..Buoyancy acts on the finger simlarly it acts on the water..Hence reading will increase due to extra force downward on the cup

[arivero]

The spring is illuminating. Let me see

1)If density of \"(f)inger\" is greater than (w)ater.
Spring force is g * M_f - d_w * vol_f
Gravity force is -g* (M_w + M _f)
Weight (reaction force in the floor) is unknown, but the system being static, forces will equilibrate, W+G+S=0, then

-W=G+S= -g Mass_w- density_w * vol_f

So if signs are right, the weight of the system will increase.

2) if d_f < d_w.

The spring now must exert a downward force to keep the finger under water. But equations are the same:

Spring force is g * M_f - d_w * vol_f
Gravity force is -g* (M_w + M _f)

Thus the \"weight\" still will increase.

The funny thing in both cases in that the increment of weight does not depend on the density of the finger, only in the summerged part. God Save Archimedes :cool: