# Exploring Weight Measurements in Water Experiments

• SataSata
In summary, we discussed the use of an electronic balance when a water filled beaker is placed on it and zeroed. We also explored the different scenarios of a ball resting at the bottom of the beaker, a ball suspended by a string in the water, and a ball being lifted up from below the water. In each case, we determined that the balance would measure the normal force, which is equal to the weight of the ball minus its buoyancy. Additionally, we considered the effects of tension, buoyancy, and fluid resistance on the system and how they would affect the balance readings.
SataSata

## Homework Statement

I place a water filled beaker on an electronic balance and zero the reading. Meaning the balance will not be measuring the weight of the water and the beaker.

a.) If a ball is resting at the bottom of the beaker. What would the balance be measuring?

b.) If I immersed and suspend a ball with a string in the water and not touching any sides. What will the electronic balance be measuring?

c.) This time I lift the ball up from below, but not getting it out of the water. What would the balance be measuring?

## The Attempt at a Solution

a.) Looking at the free body diagram, I guess the balance shows the normal force. Since N = mg - B where B is buoyancy.

b.) Since it is suspended, T + B = mg whereby T is tension. It is in equilibrium so what would the balance be measuring? Is it the mass of the displaced water according to Archimedes Principle? If it is so, does part a include the mass of the displaced water too?

c.) When I lift the ball up, fluid resistance f is pushing it down. Hence F = T + B - mg - f. The resultant force is upwards, but what is the balance measuring?

SataSata said:
Looking at the free body diagram, I guess the balance shows the normal force. Since N = mg - B where B is buoyanc
What do you mean by the 'normal force'? Do you mean that the scale will read (mg - B), where m is the mass of the ball and B is its buoyancy?

Think about the buoyancy. That's the water net pushing the ball up. In that case, in which direction is the ball net pushing the water? And what does that push ultimately transmit to?

In (c), how is the ball being lifted? How does this differ from (b)? (c) seems to be specified with insufficient clarity to me.

SataSata
andrewkirk said:
What do you mean by the 'normal force'. Do you mean that the scale will read (mg - B), where m is the mass of the ball and B is its buoyancy?
That is what I meant. When I say the normal force is because the ball come into contact with the bottom of the beaker. And this contact force is (mg - B).
andrewkirk said:
Think about the buoyancy. That's the water net pushing the ball up. In that case, in which direction is the ball net pushing the water? And what does that push ultimately transmit to?
The ball net is pushing downwards? And this push is transmitted to the bottom of the beaker which is on the electronic balance? This force would be (mg - B) yes?
andrewkirk said:
In (c), how is the ball being lifted? How does this differ from (b)? (c) seems to be specified with insufficient clarity to me.
Part (b) is when the ball is suspended and not in motion. Part (c) is when the ball is being lifted up so fluid resistance has to be taken into account. The reading for part (c) is taken when the ball is moving upwards. While part (b) is when it has already been lifted up and is not in motion.

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SataSata said:
The ball net is pushing downwards? And this push is transmitted to the bottom of the beaker which is on the electronic balance? This force would be (mg - B) yes?
Andrew means the buoyancy force. Just as the water exerts a buoyancy force on the ball, the principle of action and reaction tells you that the ball exerts an equal and opposite force on the water. What force is in turn affected by that?
SataSata said:
b.) Since it is suspended, T + B = mg whereby T is tension. It is in equilibrium so what would the balance be measuring? Is it the mass of the displaced water according to Archimedes Principle? If it is so, does part a include the mass of the displaced water too?
Yes, and yes.

For c, what is the change in the tension? What does that do to the beaker+water+ball system?

SataSata
haruspex said:
Andrew means the buoyancy force. Just as the water exerts a buoyancy force on the ball, the principle of action and reaction tells you that the ball exerts an equal and opposite force on the water. What force is in turn affected by that?
The contact force from beaker on ball? Since the ball is in contact with it?
haruspex said:
For c, what is the change in the tension? What does that do to the beaker+water+ball system?
The tension will increase because the ball is experiencing fluid resistance. The system become lighter? It is as though the string is pulling the entire system up?

SataSata said:
The contact force from beaker on ball? Since the ball is in contact with it?
No. If something is somehow pushing down on the water (as the ball's reaction to the buoyancy force does) how will the water transfer that force? The water does not accelerate, so some other force on the water must counteract the reaction from the ball.
SataSata said:
The tension will increase because the ball is experiencing fluid resistance. The system become lighter? It is as though the string is pulling the entire system up?
Yes.

SataSata
haruspex said:
No. If something is somehow pushing down on the water (as the ball's reaction to the buoyancy force does) how will the water transfer that force? The water does not accelerate, so some other force on the water must counteract the reaction from the ball.
I don't quite understand this part. We are talking about part (a) right? When the ball is resting at the bottom of the beaker.
The ball's reaction is acting on the water. Some sort of force on the water must counteract it. It would be an upward force. The only upward force in the system other than buoyancy of the ball would simply be the normal force from the balance right?

SataSata said:
I don't quite understand this part. We are talking about part (a) right? When the ball is resting at the bottom of the beaker.
The ball's reaction is acting on the water. Some sort of force on the water must counteract it. It would be an upward force. The only upward force in the system other than buoyancy of the ball would simply be the normal force from the balance right?
yes. So how does this affect the balance reading?

SataSata
haruspex said:
yes. So how does this affect the balance reading?
The balance reading will increase. And what forces are each reading corresponding to?
a.) Normal force from balance on beaker(in terms of forces what would this be? mg-B?) + mass of displaced water?
b.) Mass of displaced water
c.) ?? This is the tension force pulling the entire system up right?

SataSata said:
The balance reading will increase. And what forces are each reading corresponding to?
a.) Normal force from balance on beaker(in terms of forces what would this be? mg-B?) + mass of displaced water?
b.) Mass of displaced water
c.) ?? This is the tension force pulling the entire system up right?
Yes. But you can simplify a).

SataSata
haruspex said:
Yes. But you can simplify a).
How do I simplify it?
And what about part (c)? I understood it but I don't know how to explain it in terms of forces.

SataSata said:
How do I simplify it?
If you think about the effect of the string tension on the rest of the system it should become clear. No time to answer your second qn right now.

SataSata
To agree with haruspex and take it just one step further: without even thinking about things like buoyancy in water, you can reason that adding the ball to a system on a zeroed scale will, if nothing else changes, make the scale show the weight mg of the ball, if the system is in a vacuum or, if not, the weight of the ball less the buoyancy of the ball in the air, which is minuscule and can reasonably be ignored.

It can't make any difference where on the scale the ball is placed, whether in the water or to the side of the beaker. How could it?

If we then add to the system a string that is attached in some way to the system and is pulling up, then the scale will read the weight of the ball minus the tension in the string. That is the case in (ii) and (iii).

That's all there is to it. The arguments about water buoyancy and water resistance to motion just allow one to dissect the net result and see how the various parts offset one another. But they are not necessary in order to know what the balance reading will be.

SataSata

## 1. What is the purpose of conducting weight measurements in water experiments?

The purpose of conducting weight measurements in water experiments is to understand the relationship between the weight of an object and its buoyancy in water. This can help scientists make predictions about the behavior of objects in water and also aid in the study of density and volume.

## 2. How do weight measurements in water experiments differ from weight measurements in air?

Weight measurements in water experiments differ from weight measurements in air because of the presence of buoyancy force. In air, the object's weight is equal to the force of gravity acting on it. In water, the buoyancy force opposes the weight of the object, causing it to appear lighter or even float.

## 3. What equipment is needed for weight measurements in water experiments?

The equipment needed for weight measurements in water experiments includes a scale, a container of water, and the object being measured. It is important to use a waterproof scale that is accurate and precise to get accurate measurements.

## 4. How do you calculate the weight of an object in water?

To calculate the weight of an object in water, you can use the following formula: Weight in air = Weight in water + Buoyant force. The buoyant force can be calculated by multiplying the volume of water displaced by the object by the density of water (1000 kg/m3) and the gravitational acceleration (9.8 m/s2).

## 5. How can weight measurements in water experiments be applied in real-life situations?

Weight measurements in water experiments can be applied in real-life situations such as determining the weight of objects that are partially or fully submerged in water, understanding the behavior of ships and submarines, and calculating the weight of underwater structures like oil rigs or pipelines.

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