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

[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.

 

[Arjan82]

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?

Ill leave this one to @anesthesiologist :)