9
$\begingroup$

If I'm carrying a bucket of water in one hand and a piece of plastic in the other, and then I decide to keep the plastic in the bucket of water (it floats). Will it feel less heavy in the second case?

I think it will feel the same because it's mass adds up to the bucket's mass and will be pulled by gravity with the same extent. But somehow I can't get my mind off from the fact that it's weight is already balanced by the up-thrust.

Is there a simple way to explain how this works? It would be clearer if you helped me with some free body diagrams or an analogy or something simple.

$\endgroup$
  • $\begingroup$ I tried to convince myself by considering a solid object as an extremely dense fluid which gives normal force when something with weight is placed in it and the weight balances normal force. I felt up-thrust is more or less the same thing. Just because a weight is balanced by normal force doesn't mean that it doesn't exert force. Something like that. Is that near? $\endgroup$ – Bikash Adhikari Sep 11 at 13:01
  • $\begingroup$ On bathroom scale put your bucket and you too stand next to the bucket. Next hold the same bucket on your head. Would there be any difference in the reading? $\endgroup$ – across Sep 11 at 13:11
  • $\begingroup$ related: physics.stackexchange.com/questions/461149/… $\endgroup$ – Andrew Steane Sep 11 at 17:12
  • 1
    $\begingroup$ Note: in a situation where you don't have to carry the water with you (for example, if you're wading through a pond and carrying something submerged), the answers might be somewhat different. $\endgroup$ – probably_someone Sep 12 at 0:10
  • 1
    $\begingroup$ If you pick up an empty chair, then you pick up the same chair with someone sitting in it, the person "floats" on the chair. Does it weight more with the person sitting in it? $\endgroup$ – J... Sep 12 at 12:20
13
$\begingroup$

Fluids, like all objects, obey Newton's Third Law. This means that any upward buoyant force exerted on the plastic by the fluid has a counterpart downward force exerted on the fluid by the plastic. The force you must exert on the bucket & the fluid to keep it from falling is therefore increased by the amount of the buoyant force exerted on the plastic. If the plastic is in equilibrium, this is just equal to the weight of the plastic. Thus, the total weight doesn't change when you put the plastic in the water.

Your idea in the comment is basically correct. The situation is not fundamentally different from putting an object on top of another object that's sitting on a scale. When you do this, the reading of the scale will increase by the weight of the top object. The explanation is essentially the same as given above; just replace "buoyant force" with "normal force".

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for the answer. It cleared much of my confusion. I wanted to ask how does that change when the body has sunk? Because it sinks when the up-thrust is less than its weight. So, " the force increased by the amount of buoyant force " as you said will be less than the plastic's weight, won't it? How does that work? $\endgroup$ – Bikash Adhikari Sep 11 at 15:47
  • 6
    $\begingroup$ @BikashAdhikari: In that case, it’ll sink to the bottom. Once it’s there, there will be a buoyant force from the fluid and a normal force from the bucket, both of which will have to be canceled out by the force you exert on the bucket & fluid. It all works out the same in the end. $\endgroup$ – Michael Seifert Sep 11 at 17:32
  • 2
    $\begingroup$ As there object is added to the water the level of the water will increase also... $\endgroup$ – Lamar Latrell Sep 12 at 5:26
4
$\begingroup$

Assuming the densities of the water in the bucket and the density of the plastic are constant, and that no water spills out of the bucket when the plastic is placed in the bucket, then

$$W_{Tot}=ρ_{w}V_{w}+ρ_{p}V_{p}$$

Where $W_{Tot}$ = Total weight of the combination of the water and plastic in the bucket (ignoring the weight of the empty bucket, i.e., the container), in terms of the volumes and densities of the water and plastic, denoted by subscripts $w$ and $p$ respectively.

The first term on the right side is the weight of the plastic outside the bucket and the second term is the weight of the bucket of water without the plastic floating in it. Although the weight of the plastic decreases when placed in bucket due to the buoyant force acting upward, at the same time the weight of the bucket that now includes the plastic increases by an equal amount, so that the combined weight is still the same as the sum of the weights of each by themselves.

The above assumes that no water spills out of the bucket when the plastic is placed in the water. If before putting the plastic in the bucket the bucket was filled to the rim with water, then the water displaced by the plastic will spill out of the bucket, reducing the total weight. In this case, the reduction in total weight exactly equals the reduction in the weight of the plastic floating in the water.

Hope this helps.

| cite | improve this answer | |
$\endgroup$
2
$\begingroup$

If I'm carrying a bucket of water in one hand and a piece of plastic in the other, and then I decide to keep the plastic in the bucket of water (it floats). Will it feel less heavy in the second case?

There are at least two ways to consider this question.

First of all, does the bucket feel any greater force due to the object placed in it. The answer is yes. If you have 1 kg water bucket and add a 10 g bottle, then on a scale the bucket will now weigh 1010 g. Yes, the weight of the bottle is balanced by the force from the water. But that (extra) force is transmitted down to the bucket and to the scale (or your hand).

It does not matter if the object floats or sinks in the water. As long as we are ignoring the buoyancy of the air, then the bucket is going to weigh exactly the same as the sum of the bucket and the bottle.

The other thing to ask is if you could feel the difference as you walk. Probably not. Our senses act in a logarithmic fashion rather than a linear one. The hand can easily tell the difference between 0 g and 10 g, but it cannot easily tell the difference between 1000 g and 1010 g. The absolute difference is the same, but the relative difference is tiny. Your senses may tell you that the bottle doesn't weigh anything in the bucket, but that's just because you're using an inaccurate scale.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Would buoyancy of the air change the result? In which direction? Wouldn't both systems have the same volume, assuming the piece of plastic doesn't get compressed by the water? $\endgroup$ – Eric Duminil Sep 12 at 12:30
1
$\begingroup$

I don't think you will feel less however the plastic (if could sense ) will of course feel less weight .

Let me explain.

Since the liquid applies buoyant force on the plastic (say $F_b$ ) so by Newton's third law the plastic also applies a force $ F_b$ on it in the downward direction . So the forces acting on the liquid are $Mg$ , $F_b$ and the normal force $N$ due to the bottom part of the bucket . Here's the fbd for the situation

enter image description here

Now what you feel is the normal force due to the bucket and this is equal to the normal force ($N$) applied by the liquid on the bucket in downward direction ( since bucket Also applied the same force on the liquid which is shown in the fbd).

So you feel more force .

Note : I have considered forces only in the vertical direction however there are forces on the liquid by the bucket in contact in other direction also.

Hope it helps 🙂.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks. This helped a lot. Will this change when the body is sunk and not afloat? The fbd will be the same but the magnitude of Buoyant force will change , won't it? $\endgroup$ – Bikash Adhikari Sep 11 at 15:50
  • $\begingroup$ @Bikash Adhikari no there will be no change neither in the fbd nor in the magnitude because magnitude of buoyant force depends on the weight of displaced fluid which will be affected only by changing the volume of the plastic or density of the liquid. $\endgroup$ – Ankit Sep 11 at 16:01
  • $\begingroup$ @Bikash Adhikari if volume of plastic remains the same and only density is increased( mass increased) then the body will sink($mg$ > $F_{buoyant}$) but the buoyant force will be the same as earlier one. $\endgroup$ – Ankit Sep 11 at 16:05
  • $\begingroup$ I got that. So, in that situation when mg > F(buoyant) , does it imply that the plastic feels lighter to me that it was outside ? $\endgroup$ – Bikash Adhikari Sep 11 at 16:14
  • $\begingroup$ @Bikash Adhikari considering only the effect of plastic on you then yes it will feel lighter. $\endgroup$ – Ankit Sep 11 at 16:20
1
$\begingroup$

As Bob D says, it will weigh the same unless some of the water spills out.

But it's likely to feel somewhat heavier because you do better with weights balanced in both hands than all the weight in one hand. And if you carry one bucket with both hands that will be awkward, unless it's an unusually small bucket.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

For an object to float it must displace its weight in water. This means that the water level in the bucket will rise when the plastic is in it. The level will rise the same as if you had added the same weight in water as the weight of the plastic. If you put a piece of iron in the bucket of water it would sink. In this case part of the iron's weight would displace water and the part heavier than the same volume of water would rest on the bottom of the bucket.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

I regularly carry my little daughter's bike and helmet to the park. If I carry the helmet in one hand and the bike in the other, I am aware of the weight of the helmet; if I put the helmet in the bike basket I don't notice the difference because the bike is much heavier.

So I would say, yes, it will feel less heavy in the second case, though as all the other answers point out it there will be the same total mass and total weight in both arrangements.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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