This question already has an answer here:

I attempted to modify a classic experiment to demonstrate Archimedes principle to my 5th graders and I created an error I can’t explain.

We started with a graduated cylinder containing 20 mL of water. We added 10 pennies to the cylinder forcing a rise of about 5 mL to the column, or a total of 25 mL We then placed the 10 pennies into one pan of a balance We the used a syringe to place 5 mL of water on the other side of the balance

At this point I was ready to scream Eureka and have them bask in the equivalence of the displaced water to the object(s) that caused the displacement.

Unfortunately, the two masses were no where near equal... no Eureka. The 5 mL of displaced water in the cylinder was not equal to the mass of the 10 pennies I had the kids keep adding water until it balanced and it was about 25 mL to match the mass of 10 pennies

I have seen this video experiment before using an overflow vessel and watched countless online professors show the mass of the displaced water equal to the mass of the object.

Yet for some reason that I can’t explain I could t get this to work when you measure the displacement in a cylinder

Anyone have any ideas on what kind of obvious error in thinking and / or procedure am I making here?



marked as duplicate by sammy gerbil, Community Nov 28 '17 at 2:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ Sure the videos you saw didn't involve floating objects? $\endgroup$ – DJohnM Nov 28 '17 at 0:35
  • $\begingroup$ Some of them were floating but some were steel cylinders submerged in an overflow vessel. I even saw a science teacher climb into a huge full tub of water and measure the overflow. I figured displacement is displacement whether object sinks or floats. $\endgroup$ – tyroneaphone Nov 28 '17 at 1:03
  • 1
    $\begingroup$ Your error is having an incorrect idea about what Archimedes' Principle actually states. Check your sources of information. Look at those videos again. $\endgroup$ – sammy gerbil Nov 28 '17 at 1:15

OK, first let's start with a reality check. You were expecting that the weight of the objects you placed in the water and which sank to the bottom (i.e., the ten pennies) would be equal to the weight of the volume of water which the objects displaced (i.e., 5 ml). But suppose that instead of the copper pennies you had dropped in much heavier coins of the same size as those pennies but made of a denser metal such as gold or lead? By your understanding of Archimedes' principle, these heavier coins should also weigh the same as 5 ml of water, which obviously can't be correct.

Read this statement of Archimedes' principle carefully: "Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces and acts in the upward direction at the center of mass of the displaced fluid."

Archimedes' principle does not say that the weight of an object dropped into a fluid is equal to the the weight of the fluid which is displaced by the object. What it says is that the buoyant force of the immersed object is equal to the weight of the fluid displaced by the object. When you put your pennies into the water there are effectively two forces acting on them: (1) the normal downward gravitational force and (2) the upward buoyant force which, by Archimedes' principle, has a magnitude equal to the weight of the water displaced by the pennies. So Archimedes' principle basically says that your pennies get lighter (i.e., weigh less) when you put them into the water by an amount equal to the weight of the water that those pennies displace.

Only for an object of neutral buoyancy will the gravitational force (or weight) of the object (i.e., force #1 above) be equal to the buoyant force of the object (i.e., force #2 above), but your pennies are not of neutral buoyancy. They're denser than water and they sink. That means that force #1 is greater than force #2, which is what you demonstrated when you found that the weight of the pennies is greater than the weight of the volume of water displaced by the pennies.

If you want an object that has a weight (in air) exactly equal to the weight of the water displaced by that object, then you need to figure out some way of making an object which is neutrally buoyant in water.

  • $\begingroup$ Thanks Samuel. I liked bit about the coins of the same size but heavier displacing the same amount of water as the pennies. So let me see if I have the jist of things. When I’m seeing this guy online dunk a steel cylinder hanging from a spring scale into an overflow vessel and then measuring the liquid displacement in the overflow cup with a spring scale as well.... the two values he is showing equivalence for is the buoyant force acting on the cylinder compared to the weight of the displaced water. It’s not a direct comparison of the mass of the cylinder to the mass of the displaced water. $\endgroup$ – tyroneaphone Nov 28 '17 at 2:26
  • $\begingroup$ That's right. The difference in the apparent weight of the cylinder between when it is measured in air and then when it is measured in water is equal to the buoyant force on the cylinder in water. And by Archimedes' principle, that buoyant force is equal to the weight of the amount of water displaced by the cylinder. So (weight of cylinder in air) - (weight of cylinder in water) = (buoyant force) = (weight of displaced water), with the last equality being Archimedes' principle. $\endgroup$ – Samuel Weir Nov 28 '17 at 3:15
  • $\begingroup$ Finally, note what happens in the special case that the object has neutral buoyancy: IF the cylinder were neutrally buoyant in water, then the weight of the cylinder in water would be zero and the equation I wrote in the above comment reduces to the simple equality: (weight of cylinder in air) = (weight of displaced water). That's the equation that you were mistakenly assuming, but this equality is only true in the special case that the object is neutrally buoyant. $\endgroup$ – Samuel Weir Nov 28 '17 at 3:20
  • $\begingroup$ Just FYI I feel like I had a much more successful lesson today so thank you for the insight. I constructed simple overflow vessels and had the pennies floating in a cup in the vessel. The overflow water was placed onto the balance pan and the pennies in the cup on the other side and sure enough the two things were equal... or close enough. I spoke about that volume of water’s buoyant force being just perfect to offset the downward pull of gravity. I ll look into this idea of neutral buoyancy..... sounds like a great next investigation. I’ll probably be back when I screw that up too. Cheers. $\endgroup$ – tyroneaphone Nov 29 '17 at 3:34

Not the answer you're looking for? Browse other questions tagged or ask your own question.