# Ball in beaker

1. Oct 4, 2015

### SataSata

1. The problem statement, all variables and given/known data
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?

2. Relevant equations

3. 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?

2. Oct 4, 2015

### andrewkirk

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.

3. Oct 4, 2015

### SataSata

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).
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?
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.

Last edited: Oct 4, 2015
4. Oct 4, 2015

### haruspex

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?
Yes, and yes.

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

5. Oct 4, 2015

### SataSata

The contact force from beaker on ball? Since the ball is in contact with it?
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?

6. Oct 4, 2015

### haruspex

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.
Yes.

7. Oct 4, 2015

### SataSata

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?

8. Oct 4, 2015

### haruspex

yes. So how does this affect the balance reading?

9. Oct 4, 2015

### SataSata

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?

10. Oct 4, 2015

### haruspex

Yes. But you can simplify a).

11. Oct 4, 2015

### SataSata

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.

12. Oct 4, 2015

### haruspex

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.

13. Oct 4, 2015

### andrewkirk

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.