How Does Buoyancy Affect Balance Scale Readings?

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Buoyancy significantly affects balance scale readings in a beaker filled with water, as demonstrated through various scenarios involving submerged objects. When a ping pong ball is submerged, the reading increases due to the buoyant force displacing water, resulting in R2 being greater than R1. Similarly, a submerged steel ball tied to a thread also leads to an increased reading (R3 = R2) due to the same buoyant principles. However, when the steel ball sinks to the bottom without the thread, it adds its weight to the scale, making R4 greater than R3. If the steel ball accelerates upward while submerged, the reading R5 will exceed R3 due to the additional force exerted on the scale.
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Homework Statement


A beaker is filled with water and its weight is measured by a balance scale. The reading is recorded as R1.

1. A ping pong ball is now submerged in the water without touching the wall and the bottom of the beaker. What is the reading, R2 of the balance scale?

2. A steel ball with the same volume as the ping pong ball tied to a light thread is now submerged in the water without touching the wall and the bottom of the beaker. What is the reading,R3 of the balance scale?

3. The same steel ball is put into the beaker without the thread, it sink to the bottom of the beaker. What is the reading,R4 of the balance scale?

4. Back to question 2, but now instead of a stationary steel ball submerged in the water, we make the steel ball moves upward with an acceleration a. Should the reading R5 be R5=R3, R5>R3 or R5<R3?

Homework Equations


Newton Second Law F=ma, Newton Third Law F1 = -F2 , Weight = mg , Pressure exerted by the liquid= Force/ Area = density x gravity x height of liquid

The Attempt at a Solution


1. When a ping pong ball is submerged in the water, some water is displaced by the ball. The weight of the ping pong ball is balanced by the buoyant force of the water. Since the water is now displaced to a higher level, the bottom of the beaker experienced more pressure than a beaker filled with water only, thus R2 > R1. I am uncertain if my concept is right. If so, how do i explain this in terms of Newton's Law?

2. I think the reading for this one, R3, should be the same as R2 with the same explanation.

3. When the steel ball is sunk to the bottom of the beaker, it exerts its own weight to the bottom of the beaker. Therefore, the balance scale has 3 forces acting on it, the weight of the beaker, the weight of the water and the weight of the steel ball. R4 > R3 > R1

4. I believe this can be solved by using Newton's Law of Motions. My intuition believes the reading R5 should be bigger than R3 because by Newton 2nd law, the normal reaction of the scale, N = Resultant force,F + Weight of beaker and water, W. But I don't really understand why does the scale displays the magnitude of normal reaction(which is an upward force), instead of the weight(which is downward force).

I do not know the answer for these questions, so my answer might be wrong.
 
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Check with free body diagrams, and by considering small changes to the setup.

1, you draw the free body diagrams ... consider: if the ball were just floating on the surface, would the scale read more? How much by? What has to happen for the ball to be submerged?

2+3. free body diagram - what are the forces on the ball?

4... you are correct, again a free body diagram is your freind.

If you are unclear how to draw these, perhaps practise by drawing them for a regular floating object and for a neutral buoyancy object and a sunk object.
 
honlin said:
But I don't really understand why does the scale displays the magnitude of normal reaction(which is an upward force), instead of the weight(which is downward force).
A scale does not care what exerts a force on its surface. It can be the a hunk of meat at the deli counter. Or it can be the salesman's finger.
 
Simon Bridge said:
Check with free body diagrams, and by considering small changes to the setup.

1, you draw the free body diagrams ... consider: if the ball were just floating on the surface, would the scale read more? How much by? What has to happen for the ball to be submerged?

I think it would. For the ball to float on the water, the ball has 2 forces acting on it, an upward buoyant force, and a downward weight. Buoyant force must be bigger than the weight for the ball to float. Since the water exerts an upward buoyant force, by Newton Third Law, the ball exerts a same magnitude but opposite direction force on the water, thus the total downward force exerted on the scale would be the weight of the water + beaker and the buoyant force reaction. Am i right?

Simon Bridge said:
2+3. free body diagram - what are the forces on the ball?
In 2, there is a total of 3 forces acting on the ball: the steel ball's weight(downward) , buoyant force and tension(upward). Same principle, the total downward force exerted on the scale would be the weight of the water + beaker and the buoyant force.

jbriggs444 said:
A scale does not care what exerts a force on its surface. It can be the a hunk of meat at the deli counter. Or it can be the salesman's finger.
I think i understand now. As long as there are downward forces acting on it, the scale will simply display the total of the magnitude of that force, which can be expressed as the normal reaction by the scale.
 
honlin said:
1. A ping pong ball is now submerged in the water without touching the wall and the bottom of the beaker. What is the reading, R2 of the balance scale?
This is awkward. It says submerged, implying it is completely covered by the water, yet we are not told how this is achieved. We are left to guess that something is holding it down. If so, this part of the answer is not correct, since it refers to the weight of the ping pong ball:
honlin said:
The weight of the ping pong ball is balanced by the buoyant force of the water.
If submerged, but not on the bottom, what matters is the volume of the ball.
honlin said:
2. I think the reading for this one, R3, should be the same as R2 with the same explanation
Yes, but only in view of the point I make above.
honlin said:
4. Back to question 2, but now instead of a stationary steel ball submerged in the water, we make the steel ball moves upward with an acceleration a.
Again, we are not told how this is achieved, but presumably some other applied force. What direction is that force? What will that do to the reading on the scale?
 
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