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I've just been discussing this question with my eldest son, who is an extremely intelligent man, as well as being an engineer, a sailor and a scuba-diver, and he believes that an object heavier than water, such as his anchor, or his diving weights actually becomes less heavy when submerged in water. I described a scenario, whereby a solid steel, or lead weight was suspended from a spring balance and slowly lowered into a body of water, the weight reading on the spring balance would not change. He insisted that it would... I think it would not, because an object that is denser than water would not posess any buoyancy once submerged, therefore the density and weight would remain unchanged. Any comments anyone??? There's a £5 bet riding on this :)

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  • $\begingroup$ Nice question, but it's possible that it is a duplicate. Please check this question and answer. You may find it interesting $\endgroup$ – Alex Doe May 19 at 20:19
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You lost the bet, sorry to bring this you.

Consider a volume of water in the shape of the lead surrounded by more water. Its weight pushes on the surrounding water and this in turn pushes back by the same force. If not water would sink in water. When the volume of water is replaced by the lead, the push back force is still there. So the weight of the lead is diminished by the weight of the water it replaces. Eureka, this is the idea be hind Archinedes' law.

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  • $\begingroup$ Yes. Absolutely! I realise that now. I was wrong. $\endgroup$ – Simon Broadhead May 20 at 23:19
  • $\begingroup$ It turns out that you, Joel, and Francois are right! Bradley and I were wrong. For every object you submerge in water, you displace an equal volume of water and the weight of the water displaced, reduces the weight of the object you submerged by the weight of the water it's displaced... So, if you submerged a condom containing 1 litre of water (which weighs 1 Kg in air) into water, it would weigh only as much as an empty condom... QED. I owe Joel £5.00 $\endgroup$ – Simon Broadhead May 20 at 23:32
  • $\begingroup$ Physics is great fun! $\endgroup$ – my2cts May 20 at 23:41
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    $\begingroup$ Yes it is, but only when everyone agrees with the laws does it become fun. Until then, it can provoke some serious argument, which is not always fun!!! $\endgroup$ – Simon Broadhead May 21 at 0:15
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Simple experiment

Hang a dense object from a beam on a long string balanced with another object on a short string. Low the long side into the water and observe how the balance reacts.

The short-string side will go down because the bouyant force is acting upward on the dense object and as a consequence less tension is needed to balance the force of gravity.

Definitions

I wrote "force of gravity" there instead of "weight" on purpose.

There are two common ways to define "weight" that are different and both used.

  1. The force of gravity on an object: $mg$. Under this definition the weight of the object does not change, but then neither does the weight of a floating object.

  2. The support forces applied to an object. This is identical to the former when, for instance, you have an object resting statically on a surface, but is reduced for an object sliding on a frictionless incline. In this case you have to decide if you are going to count buoyancy as a support force (I would).1

I tend to prefer the support-force definition even though it conflicts with the first definition that most students receives (the force of gravity), and under that regime the weight of a submerged object that is otherwise un-supported will be less than it's static weight.

But I would accept an argument from a student based on either the force-of-gravity definition or the excluding buoyancy from support forces if they were explicit about their meaning.


1 Because a boat (or other buoyant object) floating at rest on the water's surface is "supported", and the force that does it is buoynacy.

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  • $\begingroup$ It turns out that you, Joel, and Francois are right! Bradley and I were wrong. For every object you submerge in water, you displace an equal volume of water and the weight of the water displaced, reduces the weight of the object you submerged by the weight of the water it's displaced... So, if you submerged a condom containing 1 litre of water (which weighs 1 Kg in air) into water, it would weigh only as much as an empty condom... QED. I owe Joel £5.00 $\endgroup$ – Simon Broadhead May 20 at 23:33
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It turns out that you, Joel, and Francois are right! Bradley and I were wrong. For every object you submerge in water, you displace an equal volume of water and the weight of the water displaced, reduces the weight of the object you submerged by the weight of the water it's displaced... So, if you submerged a condom containing 1 litre of water (which weighs 1 Kg in air) into water, it would weigh only as much as an empty condom... QED. I owe Joel £5.00

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The spring balance will show a lower weight. Take a boat and put the boat on the balance in mid air; the boat will have some weight. Now let the boat in the water; no weight left.

It depends on your point of view on what mass means to state what answer is good or wrong.

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  • $\begingroup$ It turns out that you, Joel, and Francois are right! Bradley and I were wrong. For every object you submerge in water, you displace an equal volume of water and the weight of the water displaced, reduces the weight of the object you submerged by the weight of the water it's displaced... So, if you submerged a condom containing 1 litre of water (which weighs 1 Kg in air) into water, it would weigh only as much as an empty condom... QED. I owe Joel £5.00 $\endgroup$ – Simon Broadhead May 20 at 23:33

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