Balance reading when new object immersed in water

[songoku]

Homework Statement

The weight of a glass full of water is being measured using balance.
a. An object is fully immersed so that some of water spilled out. How is the reading compared to before the object is immersed?

b. If the experiment is repeated and now the object immersed has same volume as (a) but twice the mass of (a), what will be the reading now?

Homework Equations

Archimedes principle

The Attempt at a Solution

a. The reading will be the same because the weight of spilled water = weight of object immersed

b. The reading will also be the same because same reason as (a)?

Are my answers to (a) and (b) correct? Thanks

 

[stockzahn]

a. An object is fully immersed so that some of water spilled out. How is the reading compared to before the object is immersed?
a. The reading will be the same because the weight of spilled water = weight of object immersed

How would the reading change, if the object isn't immersed fully, but floats?

 

[songoku]

How would the reading change, if the object isn't immersed fully, but floats?

I think the reading will still be the same but the volume of water spilled out will be less?

Thanks

 

[stockzahn]

I think the reading will still be the same but the volume of water spilled out will be less?

Another question: Imagine the object is fully immersed, but no water spills out. How does the reading change?

 

[stockzahn]

Are my answers to (a) and (b) correct? Thanks

Just to be clear: if fully immersed means sinking to the ground then your answers are not correct. If fully immersed means. the object has the same density like water, then they are - but that would be kind of a strange question: You substitute water with water, what's the readings's difference...

 

[songoku]

Another question: Imagine the object is fully immersed, but no water spills out. How does the reading change?

The reading will be higher than original before the object is put inside.

Just to be clear: if fully immersed means sinking to the ground then your answers are not correct. If fully immersed means. the object has the same density like water, then they are - but that would be kind of a strange question: You substitute water with water, what's the readings's difference...

Yes it is not sinking but like floating in the middle of water. Maybe it is a trick question to test my understanding.

So do I get all correct or I miss something?

Thanks

 

[stockzahn]

The reading will be higher than original before the object is put inside.

Yes it is not sinking but like floating in the middle of water. Maybe it is a trick question to test my understanding.

So do I get all correct or I miss something?

Thanks

If the object and the water have the same density (it is floating fully immersed in the middle of the water), then a) is correct - but not b).

 

[songoku]

If the object and the water have the same density (it is floating fully immersed in the middle of the water), then a) is correct - but not b).

The reading will be higher? I think because they experience same upthrust then the reading will be the same?

 

[stockzahn]

The question b) says, that the 2nd object has the same volume, but twice the mass - what does that mean?
Archimedes principle says, that the buoyancy force corresponds to the weight of the displaced volume, multiplied by the displaced fluid's density (= displaced fluid's mass).

What's the difference of the two buoyancy forces acting on the two objects?

 

[songoku]

The question b) says, that the 2nd object has the same volume, but twice the mass - what does that mean?

The density of 2nd object twice of 1st?

Archimedes principle says, that the buoyancy force corresponds to the weight of the displaced volume, multiplied by the displaced fluid's density (= displaced fluid's mass).

I thought buoyancy force = weight of displaced volume because they already has same unit? Still need to multiply with density?

What's the difference of the two buoyancy forces acting on the two objects?

The formula I know is:
Buoyancy = density of fluid . gravity . volume of object immersed

Using that formula, the buoyancy of 1st and 2nd object is the same.

Or maybe it is like this:
Reading on balance = weight of container and water + apparent weight of object?

Apparent weight of object = weight in air - buoyancy

Weight in air of 2nd object is higher so higher apparent weight then higher balance reading?

Thanks

 

[stockzahn]

The density of 2nd object twice of 1st?

Therefore it cannot float anymore, but must sink to the ground.

I thought buoyancy force = weight of displaced volume because they already has same unit? Still need to multiply with density?

The formula I know is:
Buoyancy = density of fluid . gravity . volume of object immersed

Yes, you are right, sorry, my bad - there were two thoughts in one sentence. What I wanted to write was: The buoyancy force corresponds the displaced volume, multiplied by the displaced fluid's density (= displaced fluid's mass) and the gravitational acceleration = weight of the displaced volume of the fluid.

Using that formula, the buoyancy of 1st and 2nd object is the same.
Thanks

In the first case the buoyancy force is strong enough to keep the object floating: $m_1 g = \rho_{Fluid} V g$
In the second case the buoyancy force is to weak to keep the object floating: $m_2 g > \rho_{Fluid} V g$

What happens with the difference $\Delta F =m_2 g-m_1 g$

Or maybe it is like this:
Reading on balance = weight of container and water + apparent weight of object?
Apparent weight of object = weight in air - buoyancy
Weight in air of 2nd object is higher so higher apparent weight then higher balance reading?
Thanks

Yes, that should be the correct solution of the problem, just accuracy: Weight in air of 2nd object with the same volume (= higher density) is higher so higher apparent weight then higher balance reading.

 

[stockzahn]

By the way, there is nice "riddle" to check the understanding:

You are in a boat holding a stone in your hand. Now you drop the stone into the water. Does the water level rise, stay the same or decrease?

 

[songoku]

Therefore it cannot float anymore, but must sink to the ground.

Oh my...I did not realize this...

Yes, you are right, sorry, my bad - there were two thoughts in one sentence. What I wanted to write was: The buoyancy force corresponds the displaced volume, multiplied by the displaced fluid's density (= displaced fluid's mass) and the gravitational acceleration = weight of the displaced volume of the fluid.

Just to clarify, based on Archimedes principle:
1. weight of displaced fluid = upthrust

2. volume of displaced fluid = volume of object immersed

Correct?

Weight of displaced fluid ≠ weight of object immersed? I think weight of displaced fluid = weight of object when fully immersed but weight of displaced fluid ≠ weight of object when partially immersed?

Yes, that should be the correct solution of the problem, just accuracy: Weight in air of 2nd object with the same volume (= higher density) is higher so higher apparent weight then higher balance reading.

Another simple question: if the original reading of balance (container + water) = A, weight of object in air = B, weight of displaced water when object immersed = C, what will be the reading when the object is completely immersed without sinking?

So, the reading on balance = original reading + apparent weight of object = A + B - C? The teacher told me the answer is A + C

By the way, there is nice "riddle" to check the understanding:

You are in a boat holding a stone in your hand. Now you drop the stone into the water. Does the water level rise, stay the same or decrease?

Before dropping: weight = buoyancy

After dropping: new weight = new buoyancy and new weight < original weight so new buoyancy < original buoyancy. This means that the boat will float more (displaced less water). Stone will displace some water and if we add new volume displaced by boat and volume displaced by stone, it will be equal to original volume of water displaced before stone is dropped, then water level does not rise?

Thanks

 

[stockzahn]

Just to clarify, based on Archimedes principle:
1. weight of displaced fluid = upthrust
2. volume of displaced fluid = volume of object immersed

1. weight of displaced fluid = upthrust $\rightarrow$ I think that's correct, it's also called "buoyancy"
2. volume of displaced fluid = volume of object fully immersed $\rightarrow$ if the object floats on the surface, the weight of the displaced water volume corresponds to the weight of the floating object. And it is obvious that an object floating at the surface does not displace its entire volume...

I think weight of displaced fluid = weight of object when fully immersed but weight of displaced fluid ≠ weight of object when partially immersed?

Exactely the opposite:

Weight of the displaced fluid = weight of the floating object
Weight of displaced fluid ≠ weight of object sank to the ground

Another simple question: if the original reading of balance (container + water) = A, weight of object in air = B, weight of displaced water when object immersed = C, what will be the reading when the object is completely immersed without sinking?

So, the reading on balance = original reading + apparent weight of object = A + B - C? The teacher told me the answer is A + C

Just to be clear: Totally immersed, but not sank to the ground is only possible, if the object has exactely the same density as water. This is a very rare case (except you fill water into the container until water spills out) and it also is confusing and doesn't help to understand buoyancy. Therefore let's go with the two statements above:

1) Weight of the displaced fluid = weight of the floating object (partly or fully (very rare and special case) immersed)
2) Weight of the displaced fluid < weight of object sank to the ground

In the first case the displaced water exerts a force on the object exactely the same as its weight ($B=C$). All the water displaced by the body therefore has the same weight as the object you put into the container. If the displaced water spills out, there will be no difference in the reading of the balance ($A+C=A-C+B$).

In the second case the displaced water exerts a smaller force on the object than its weight ($B>C$). All the water displaced by the body therefore has a smaller weight than the object you put into the container. If the displaced water spills out, there will be a difference in the reading of the balance ($A+C<A-C+B$).

In the second case the body sinks to the ground. Why? Because the buoyancy (upthrust) isn't sufficiently high to keep it floating. The difference of the weights of the object and displaced water is then supported by the ground. This difference will give you a different reading. Imagine you put your bath tub filled with water on a balance. Now you put in a football made of gold (and the displaced water spills out). Do you think the balance's reading will increase or stay the same?

In a next step you put as much gold into the bath tub to displace all of the water. What do you think now about the balance's reading?

Before dropping: weight = buoyancy

After dropping: new weight = new buoyancy and new weight < original weight so new buoyancy < original buoyancy. This means that the boat will float more (displaced less water). Stone will displace some water and if we add new volume displaced by boat and volume displaced by stone, it will be equal to original volume of water displaced before stone is dropped, then water level does not rise?

The water level decreases: If the stone floats with the boat it displaces water corresponding to its entire weight - it doesn't sink to the ground, does it? If you throw the stone into the water, it only displaces water corresponding to its volume, which is less water, since $\rho_{water}<\rho_{stone}$. Less displaced water = lower water level.

 

[songoku]

Sorry for late reply.

1. weight of displaced fluid = upthrust $\rightarrow$ I think that's correct, it's also called "buoyancy"
2. volume of displaced fluid = volume of object fully immersed $\rightarrow$ if the object floats on the surface, the weight of the displaced water volume corresponds to the weight of the floating object. And it is obvious that an object floating at the surface does not displace its entire volume...
Exactly the opposite:

Weight of the displaced fluid = weight of the floating object
Weight of displaced fluid ≠ weight of object sank to the ground

Let's try with number:
1. Total volume of object X = 100 cm³. This object has same density as water and is put into water so it is totally immersed but not sank. Then the volume of water displaced = 100 cm³?

2. Total volume of object Y = 100 cm³. This object has lower density than water and is put into water so it partially immersed. The volume immersed in water only half of total volume. Then the volume of water displaced = 50 cm³?

3. Weight of object P in air = 100 N. This object has same density as water and is put into water so it is totally immersed but not sank. Then the weight of water displaced = 100 N?

4. Total volume of object Y = 100 N. This object has lower density than water and is put into water so it partially immersed. The volume immersed in water only half of total volume. Then the weight of water displaced = 50 N?

Imagine you put your bath tub filled with water on a balance. Now you put in a football made of gold (and the displaced water spills out). Do you think the balance's reading will increase or stay the same?

Stay the same because the weight of water displaced = weight of football put in the water?

In a next step you put as much gold into the bath tub to displace all of the water. What do you think now about the balance's reading?

Since it is the same as measuring the weight of bath tub + football, the reading will increase because weight of football > weight of water?

Thanks

 

[stockzahn]

1. Total volume of object X = 100 cm³. This object has same density as water and is put into water so it is totally immersed but not sank. Then the volume of water displaced = 100 cm³?

Independent of the density: If an object is immersed entirely it displaces an amount of water corresponding to its volume. So your statement is correct.

2. Total volume of object Y = 100 cm³. This object has lower density than water and is put into water so it partially immersed. The volume immersed in water only half of total volume. Then the volume of water displaced = 50 cm³?

Of course, an object displaces an amount of water corresponding to its (partially) immersed volume. Since it doesn't sink to the ground:

- Its density must be lower than the water's density
- The fraction of the immersed volume corresponds to the ratio of the densities: $\frac{V_{displaced}}{V_{total}} = \frac{50\;cm^3}{100\;cm^3}=\frac{\rho_{body}}{\rho_{water}}$

3. Weight of object P in air = 100 N. This object has same density as water and is put into water so it is totally immersed but not sank. Then the weight of water displaced = 100 N?

The buoyancy force acting on a partially or fully immersed body corresponds the the weight force of the displced water - otherwise the weight force of the object would be higher than the buoyancy force and it would sink to the ground. So your statement is correct, but of course also applies to partially immersed objects (hence even if the density is lower than the one of water).

4. Total volume of object Y = 100 N. This object has lower density than water and is put into water so it partially immersed. The volume immersed in water only half of total volume. Then the weight of water displaced = 50 N?

Correct (apart from the unit $N$ in the first sentence): $\frac{G_{body}}{G_{displaced\;water}}=1 =\frac{gV\rho_{body}}{gV\rho_{displaced\;water}}=\frac{100\;cm^3\cdot\rho_{body}}{50\;cm^3\cdot\rho_{displaced\;water}}$ and therefore $\frac{V_{displaced}}{V_{total}} =\frac{\rho_{body}}{\rho_{water}}$.

Stay the same because the weight of water displaced = weight of football put in the water?

Since it is the same as measuring the weight of bath tub + football, the reading will increase because weight of football > weight of water?

I'm not sure about these two statements, they seem to contradict each other. However, a floating object does not increase the reading (as long as the displaced water spills out of the container). An object sank to the ground will increase the reading (even if the displaced water spills out of the container).

 

[songoku]

Correct (apart from the unit $N$ in the first sentence):

Sorry, I mean the weight of object Q in air = 100 N

Thank you very much for the help stockzahn

 

[stockzahn]

4. Total volume of object Y = 100 N. This object has lower density than water and is put into water so it partially immersed. The volume immersed in water only half of total volume. Then the weight of water displaced = 50 N?

Sorry, I mean the weight of object Q in air = 100 N

Then no. If the object's weight force is 100 N ($=m\cdot g$) and has a lower density, then it floats and therefore will displace water mass corresponding to the mass of the floating object ($m_{object} = m_{water}$ for $\rho_{object} \leq \rho_{water}$). If the volume of the displaced water is only half of the volume of the floating object then $\frac{V_{water}}{V_{object}}=\frac{1}{2}=\frac{\rho_{object}}{\rho_{water}} \rightarrow V_{water}\rho_{water}=V_{object}\rho_{object}$.

 

[songoku]

Then no. If the object's weight force is 100 N ($=m\cdot g$) and has a lower density, then it floats and therefore will displace water mass corresponding to the mass of the floating object ($m_{object} = m_{water}$ for $\rho_{object} \leq \rho_{water}$). If the volume of the displaced water is only half of the volume of the floating object then $\frac{V_{water}}{V_{object}}=\frac{1}{2}=\frac{\rho_{object}}{\rho_{water}} \rightarrow V_{water}\rho_{water}=V_{object}\rho_{object}$.

So for this case the weight of water displaced will also be 100 N? Is this right?

Thanks

 

[stockzahn]

So for this case the weight of water displaced will also be 100 N? Is this right?

Thanks

Right, 100 N. If floating, then mass of displaced water = mass of floating object, hence their weight (force) is identical.

 

[songoku]

Thank you very much