How does the weight on a scale change when an object is submerged in water?

[anesthesiologist]

TL;DR Summary

Law of Archimedes problem

Dear,

I am a medical doctor and my physics background is limited to that of secondary school many years ago, but should want to ask for your help.

In attachment, you can find a picture of my question.

On a scale, we place a jar filled with water. We place in the water a platform which is attached to a construction which is resting on the table next to the scale. Then we zero the scale.
Now, we place in the water the red irregular object, which sinks in the water and which is resting on the platform.

Which weight does the scale now display?

Thanks a lot for your answer!

 

[Nugatory]

What do you think and why?

A good starting point is to consider at all the forces involved in supporting the irregular object - every one will have a Newton's third law pair.

 

[anesthesiologist]

See attachment.

-The upward force is equal to the weight of the fluid that the body displaces = m_{fluiddisplaced} * g = V_body * ρ * g.
-The gravitational force is F = m_body * g.
-The platform exerces a normal force.

-The object is not moving, so the the gravitational force is equal to the sum of the archimedes force and the normal force.

So m_body * g = V_body * ρ * g + normal force
If what the scale displays is the normal force, then it is m_body - V_body * ρ = m_body - m_body = 0

But an experiment showed that it is not 0, so I forgot something.

 

[berkeman]

But an experiment showed that it is not 0, so I forgot something.

If you zero the scale before putting the platform in the water, does the scale reading change when you put the platform in?

 

[anesthesiologist]

Yes, but we zero the scale again after putting the platform in the water and after that my question starts.

 

[A.T.]

If what the scale displays is the normal force...

Why would the scale show the normal force at the platform? The scale is not placed under the platform rig, is it?

 

[anesthesiologist]

That is true. It will not be the normal force.
But what will the scale measure then?

 

[kuruman]

The scale measures the net downward force acting on its platform. That is also true for a bathroom scale. It does not measure your weight. Think of what it measures when you push on it with your hand. So in your case the submerged scale measures the downward weight of the object minus the upward Archimedes force on the object.

It's good that you rezeroed it upon immersion because that eliminates the extraneous Archimedes force on the submerged parts of the assembly.

 

[A.T.]

That is true. It will not be the normal force.
But what will the scale measure then?

As you correctly stated:

...the gravitational force is equal to the sum of the archimedes force and the normal force.

Since the normal force is not supported by the scale, what is left?

 

[anesthesiologist]

So in your case the submerged scale measures the downward weight of the object minus the upward Archimedes force on the object.

But the downward weight is supported by the platform which is supported by the table under the scale. The scale cannot measure this force?

Since the normal force is not supported by the scale, what is left?

Gravitational force and Archimedes force

 

[kuruman]

The scale doesn't know anything about the table. All it knows is how much downward force acts on its platform and that is what it reports on its dial. If you put a 2 lbs weight on it, it will report that. With the weight on, if you put your finger on it and push down, it will report more than 2 lbs and if you push up with your finger, it will report less than 2 lbs. As long as the net downward force is non zero, it cannot distinguish the origin or direction of the individual forces that act on its platform.

 

[A.T.]

But the downward weight is supported by the platform which is supported by the table under the scale. The scale cannot measure this force?

How would a scale measure something placed next to it on the table, but not the scale itself?

- Ultimately, the total weight of the red object has to be supported by the table somehow.

- There are only two objects in contact with the table, which can provide that support: the platform base and the scale

- You have already identified how the weight is split up between the two:

...the gravitational force is equal to the sum of the archimedes force and the normal force.

 

[anesthesiologist]

I totally agree.
And what is then the downward force on the scale?

 

[kuruman]

I totally agree.
And what is then the downward force on the scale?

Before I say more I need a clarification. Is the black assembly that holds the weight up in your figure part of the scale sensor or is it just a support? I thought it was the former and I based my replies on that assumption. If it is the latter, the best way to analyze this is to consider Newton's 3rd law as @Nugatory already suggested in post #2. It's the same idea as just sticking your finger in the water. Will the scale read less, the same or more and why?

 

[A.T.]

I totally agree.
And what is then the downward force on the scale?

Consider having scales under both:
Scale A : under the platform base
Scale B : under the water tank

Both scales are zeroed, before the red object is introduced. So after the object is placed on the platform, their total reading must equal the objects weight:

A + B = weight

You also know that:

normal_force + buoyancy = weight

And obviously:

A = normal_force

Now solve for B.

 

[anesthesiologist]

Is the black assembly that holds the weight up in your figure part of the scale sensor or is it just a support?

It is a support and is NOT part of the scale sensor

Now solve for B.

So, what is displayed on the scale under the water tank is the Archimedes force?

 

[A.T.]

So, what is displayed on the scale under the water tank is the Archimedes force?

Yes.

 

[jbriggs444]

See attachment.

-The upward force is equal to the weight of the fluid that the body displaces = m_{fluiddisplaced} * g = V_body * ρ * g.
-The gravitational force is F = m_body * g.
-The platform exerces a normal force.

-The object is not moving, so the the gravitational force is equal to the sum of the archimedes force and the normal force.

So m_body * g = V_body * ρ * g + normal force
If what the scale displays is the normal force, then it is m_body - V_body * ρ = m_body - m_body = 0

But an experiment showed that it is not 0, so I forgot something.

There is a complication that we may do well to explore. This is not an ivory tower thought experiment. This is a real physical experiment.

As drawn, the width of the vertical support post holding the platform up is similar to the diameter of the irregular object. That support post displaces water. More to the point, the section of that post between the original water level and the increased water level after insertion of the irregular object displaces water.

That displacement counts toward the measured change in the scale reading.

 

[sophiecentaur]

The upward force is equal to the weight of the fluid that the body displaces

Is there, seriously, any more to say about the problem than this? The difference between readings on the scale with and without the object immersed will be the weight of the displaced water (including the part of the rod that's under the water).

That is what Upthrust means in the modern statement of Archimedes' principle.

There are other non-ideal circs which can be discussed - or better still, discussed internally. The upthrust still follows Archimedes but the displaced volumes are not all the same.

Dense object resting on the bottom: the scale will show the weight of the object .
Object on a string which floats with the string slack: scale will show the weight of the object.
Object being forced onto the bottom by the rod: depends entirely on the strength of the support and the force being applied + object weight.

 

[anesthesiologist]

So, in conclusion the answer is: Vbody * ρ ?
And because the density of water is 1, the number of grams the scale displays, is the volume of the object in ml?
Is this correct?

 

[sophiecentaur]

So, in conclusion the answer is: Vbody * ρ ?
And because the density of water is 1, the number of grams the scale displays, is the volume of the object in ml?
Is this correct?

Yes - if it doesn't float.

 

[anesthesiologist]

Thanks a lot!

An analogue question for floating objects, see attachment.
We put a jar filled with water on scale B and place a standard in the water which is resting on scale A. We zero both scales, and then we put a FLOATING object under the standard in the water.

Are the following statements correct?
-The object exercises the Archimedes force on the standard. The object is not moving, so the Archimedes force is equal to the normal force excercised by the standard on the object.
-Scale A displays the negative of the Archimedes force, so the volume of the object.
-Scale B displays the gravitational force + the normal force (the latter is equal to the Archimedes force). So scale B displays the mass + volume of the object.

 

[kuruman]

Here is a sequence of actions that should clarify what's going on.

1. Place container on scale, zero the scale then pour in water.
What the scale reads = ρ V(water)

2. Insert the support assembly and re-zero the scale
What the scale reads = 0

3a. Stick your finger in the water (supported by your hand attached to it) to see what happens
What the scale reads = ρ V(part of finger under water)

3b. Instead of finger, allow irregular object to sink entirely under water (supported by the platform attached to table)
What the scale reads = ρ V(object)
Note that in this case the normal force that the platform must provide is the weight of the object less the Archimedes force.
N = m(object)*g - ρ V(object)*g

3c. Instead of using platform, allow the object to sink to the bottom of the container.
What the scale reads = m(object)
That's because the bottom of the container, hence the scale, must provide the additional force N that the platform provided in 3b. Thus,
What the scale reads in 3c = what the scale reads in 3b + N/g = ρ V(object) + m(object) - ρ V(object) = m(object).

See how it works?

 

[anesthesiologist]

And what about floating objects that are forced to be immerced in the water by the standard, like on my picture?

 

[kuruman]

And what about floating objects that are forced to be immerced in the water by the standard, like on my picture?

The assembly you show in post #10 is a bit complicated because torques, in addition to forces, need to be balanced for the system to be in equilibrium. For that one needs additional specifications such as the various dimensions of the standard. You might wish to simplify this and consider the following situation.

You fill the container with water, place it on the scale and zero it. Then you tie one end of of a string (of negligible volume and weight) to an object that floats in water. You then tie the other end of the string to the bottom of the container so that the object is completely submerged. What is the reading on the scale?

 

[A.T.]

The assembly you show in post #10 is a bit complicated because torques, in addition to forces, need to be balanced for the system to be in equilibrium. For that one needs additional specifications such as the various dimensions of the standard. You might wish to simplify this and consider the following situation.

You fill the container with water, place it on the scale and zero it. Then you tie one end of of a string (of negligible volume and weight) to an object that floats in water. You then tie the other end of the string to the bottom of the container so that the object is completely submerged. What is the reading on the scale?

Tying it to the tank bottom is quite different to external support for the forced submerging. This is not really a mere simplificarion of the OP`'s 2nd scenario.

 

[kuruman]

Tying it to the tank bottom is quite different to external support for the forced submerging. Not really a mere simplificarion of the OP 2nd scenario.

True. I wanted to have something more self-contained than a disembodied hand to keep the object submerged. The buoyant force is the same in all cases.

 

[A.T.]

And what about floating objects that are forced to be immersed in the water by the standard, like on my picture?

The density of the submerged object is irrelevant for the buoyant force. Only its (submerged) volume and the density of the fluid matter. If the only interaction of the submerged object with the tank & water is the buoyant force, the (previously zeroed) scale under the tank will measure the buoyant force.

But if you tie a lighter than water fully submerged object to the bottom of the tank (as @kuruman suggested), the string tension (second interaction) will be buoyant force minus weight, so the (previously zeroed) scale under the tank will show the weight of the object, not the buoyant force.

 

[A.T.]

True. I wanted to have something more self-contained than a disembodied hand to keep the object submerged. The buoyant force is the same in all cases.

The buoyant force is the same, but not the reading on the scale below the tank. See my post above.

 

[anesthesiologist]

So, scale B also displays the volume of the object?

 

[berkeman]

So, scale B also displays the volume of the object?

3a. Stick your finger in the water (supported by your hand attached to it) to see what happens
What the scale reads = ρ V(part of finger under water)

I like the finger stick method. Think about what happens to the height of the water in the tank when you stick something in it and support the weight of that object externally. If the level of the water rises, what does that do to the pressure on the bottom surface of the tank (and how does that pressure relate to the force on the scale)?

 

[Dale]

But an experiment showed that it is not 0, so I forgot something.

Excellent problem, and good reasoning for this not being your area of expertise!

As you correctly noted, the buoyancy force acting upward on the object is equal to the weight of the fluid displaced. By Newton’s 3rd law there is an equal and opposite force from the object acting downward on the fluid. This is what the scale measured.

The forces on the platform and the table are important for keeping everything in static equilibrium, but by design they are not measured by the scale.

 

[A.T.]

So, scale B also displays the volume of the object?

In principle, you can use the scale reading to directly compute the object's volume, if you know the density of the fluid.

As for the accuracy in practice, you have to consider what @jbriggs444 wrote earlier, about the variable displacement of the platform rig. The density of the fluid and the ratio of object volume to tank volume can also affect the signal to noise ratio.

 

[A.T.]

Here is a related puzzle:

- The upper balance scale has two identical balls hanging from it, so it is initially balanced.

- The lower balance scale has two buckets on it, with two different liquids in them:
left bucket : water
right bucket: ethanol (less dense than water)

The lower scale is zeroed, so it is initially also balanced. The balls are denser than both liquids.

What happens to the balance of the scales, when you lower the upper scale, so the balls are fully submerged but don't touch the bottom of the buckets?

 

[Arjan82]

It doesn't seem this was answered explicitly, so let me do so (although all bits of information are in the replies already :)

Are the following statements correct?
-The object exercises the Archimedes force on the standard. The object is not moving, so the Archimedes force is equal to the normal force excercised by the standard on the object.

Nope, the object exercises the vector sum of the Archimedes force and the gravitational force on the standard. I say vector sum because the vectors point opposite. Therefore, numerically, the upward force is equal to the Archimedes force minus the gravitational force.

-Scale A displays the negative of the Archimedes force, so the volume of the object.

Nope, scale A reads the negative of the Archimedes force minus the gravitational force (for this case the Archimedes force is larger than the gravitational force, so scale A will read a negative value).

-Scale B displays the gravitational force + the normal force (the latter is equal to the Archimedes force). So scale B displays the mass + volume of the object.

Nope, this will read solely the Archimedes force, since that is the only force exerted on the fluid (which is equal but opposite to the force the fluid exerts on the object). The gravitational force of the object is not exerted on the fluid and therefore also not on the scale.

Note that your first problem and this problem will have the same effect on scale B, i.e. $V_{object} \cdot \rho_{water}$. The density of the object is irrelevant here.