How can the weight of a container be affected by an object's buoyancy?

Question

Suppose a table tennis ball is immersed in a fluid and held down by a string. The container is placed on a scales. What will happen to the reading of the scales if the string breaks? How can you correctly explain this using Newton's Third Law?

I am assuming when the string is cut, the ball experiences only its weight and buoyancy. If buoyancy exceeds the weight, the ball will experience a net force upwards (Fb - W). So, according to Newton's Third Law (every action has an equal and opposite reaction), will the container experience the same magnitude of the force (Fb - W) but act downwards? If so, will the reading on the scales be greater?

Answer 1

The flaw in your logic comes from neglecting the water's acceleration after cutting the string, and forces on the water before cutting the string. The ball is accelerating upward, yes, and that would seem to indicate that more force needs to be supplied from the scale. But the center of mass of the system (ball, water, and container) is accelerating down. Of course it is, that's the whole point of the buoyant force - there is less potential energy when the ball is replaced with water and the ball moves up because the water is heavier. Like I said in a comment, this is analogous to what is shown in this video. This is a bit paradoxical - it visually looks like something is accelerating up, but rest assured more mass is accelerating down than is accelerating up. The buoyant force is not in the business of increasing the gravitational potential of the whole system. We expect the reading on the scale to be lower in the moments after the string is cut, and that's exactly what we'll find after calculating the size of the effect properly.

Before the string is cut the weight measured on the scale is definitely the entire mass of the system times $g$, as it is for all static systems (that's what scales are for). One might think the weight on the scale is reduced by the tension in the string but it definitely is not. To see why, consider this more careful accounting of the forces on the container, the ball, and the water. Ultimately, the buoyant force that would seem to pull up on the container is cancelled by extra force in the water that is also transferred to the container.

Likewise if a person stands on a scale and pulls up really hard on a handle connected to the scale, the scale still measures exactly the person's mass times $g$. The force that the person pulls up is countered by additional force in his feet, which is also directed into the scale.

So now what about the situation after cutting the string? Like I said we need to think about the water accelerating down, not just the ball moving up. To make the geometry more trivial my ball is a square, but rest assured the answer in the end does not depend on the shape of the ball.

In a short time $t$ the "ball" (side length $h$) moves up by a distance $(1/2)at^2$. As you correctly calculated $a=(F_B-W)/m$. Water is transferred from above the ball to below, through height $h$. A total volume $(1/2)ah^2t^2$, or a mass of $(1/2)at^2h^2\rho_W$ ($\rho_W$ is the density of water), moves down by $h$. Equivalently, a mass of water $\rho_W h^3$ accelerates downward at a rate $a$.

So finally the answer... the container pushes up on the water enough that the rest of the water not accelerating downward remains still. The forces on the water are $gM_w$ pulling down, and $B$ (the buoyancy of the ball) pushing down (the reaction force of the buoyant force on the ball). So the total force that the container exerts on the water, and thus the reading on the scale, is $$ gM_w+B-\rho_W h^3 a $$ But this can be made nicer by inserting, for example, $B=\rho_W Vg$ ($V$ is now the generic volume of a ball) and $a=(B-M_b g)/M_b$, $M_b=\rho_B V$ (mass of the ball is the density of the ball times volume) $$ gM_w+\rho_W V g-\rho_W V(\rho_W Vg-\rho_B V g)/\rho_B V $$ $$ gM_w+Vg(2\rho_W- \rho_W^2/\rho_B)<gM_w+\rho_B Vg $$ This is always lower than the initial reading $gM_w+\rho_B Vg$. $2-1/x<x$ except when $x=1$. I.e. obviously except when the ball is the same density as water there is no change. This answer even shows that when the ball is heavier (and presumably pushed up by a stiff rod that is removed), the weight goes down as well, but that's more obvious.

Interesting to see that after the string is cut, whether the weight is higher or lower than the mass of the water alone depends on if the density of the ball is less than half of the density of water.

Another amazing feature of this solution is that the mass drop goes to infinity as $\rho_B$ goes to zero. This is somewhat real - when the mass of the ball goes to zero it accelerates up infinitely fast. A lot of water is transported down really quickly to compensate. You might think it's crazy for water to accelerate down faster than $g$, but the acceleration of a small volume of water near the ball is driven by the pressure of all the water above it more than it's driven by the effect of gravity on that specific fluid volume. Probably this accelearation is very short-lived; it's limited by viscosity. Of course we're more familiar with things floating up at roughly constant speed, because they reach terminal velocity really fast.

Answer 2

The buoyant force is an internal force, and does not affect the scale reading. The scale tells you the force exerted by the floor on the tank, which (by Newton's third law) is equal in magnitude to the force exerted by the tank on the floor. If the tank is not accelerating, and the floor is level, then this normal force is equal to the tank's weight. But if the tank is accelerating, then we have

$$ (\text{weight}) - (\text{scale reading}) = ma \tag{third law} $$

You can demonstrate this in your bathroom by standing on your scale and bouncing on the balls of your feet; the scale reading bounces around as you do.

While the buoyant object is rising through the water, an equivalent volume of denser water is flowing beneath it to take its place. The rising of the float corresponds to a net downward motion of the tank's center of mass, even though the outside of the tank may be stationary. But it's the center of mass whose acceleration we consider in the third-law equation. We can divide the motion into five stages:

1. The float is at rest under the surface of the water; the center of mass is at rest.
2. The velocity of the float is changing.
3. The float is rising with constant velocity; the center of mass is moving downwards with constant velocity as water fills the space under the float.
4. The velocity of the float is changing.
5. The float is at rest on top of the water; the center of mass is at rest.

Most students are comfortable with the prediction that the scale reading at the beginning and end, (1) and (5), will be equal to the weight of the tank plus the weight of the water plus the weight of the float. Most students are surprised that, during the constant-velocity phase (3), the scale reading is also equal to the weight. Remember that "constant velocity" is a synonym for "zero acceleration." My experience with table-tennis balls and water is that the ball will reach its terminal velocity very quickly, and nearly all of the motion will be in the constant-velocity phase (3).

When the ball is first released (2), the velocity of the center of mass goes from "zero" to "downwards." A downwards acceleration happens when the upward forces are less than the downward forces, so the scale reading will dip lower than the weight during the initial acceleration phase (2). Likewise the scale reading will nudge higher during the final acceleration (4) when the ball reaches the surface of the water. This is another one that you can demonstrate on your bathroom scale, by suddenly bending your knees.

A commenter links to a video by The Action Lab where a scale consistently reads less than the mass of an object while an internal part of that object is accelerating downwards. When the internal part reverses direction at the bottom of its motion, there should be a moment where the scale reads more than the weight as well — but the equipment in the video doesn't have the time resolution to display the effect.

Answer 3

The buoyant force pushing up on the ball does indeed result in an equivalent force pushing down on the liquid as the ball rises. Note that this would be a difficult experiment to perform, but there is a related experiment that I have done that verifies that Newton's 3rd law is indeed involved. Take a 100 ml beaker and put approximately 90 ml of water in it. Put that beaker on a scale and note the scale's reading (i.e., assume 150 g). Now carefully place your index finger into the water, being careful to avoid touching the beaker. Your finger will be displacing several ml of water, so there will be an upward buoyant force on your finger. Despite the fact that you are only touching the water and not the beaker, you will note that the scale reading increases, depending on how deeply you push your finger into the water, indicating that there is an additional force pushing down on the beaker. You can rest assured that kind of downward force will also exist in your proposed experiment.

The following equations apply to the tennis ball in your experiment:

$F_B-T-mg = 0$ before the string is cut

A rearrangement of this equation yields $F_B-mg=T$

Immediately after the string is cut, the equation that applies is $F_B-mg=ma$, indicating that without tension force $T$ from the string, the tennis ball will accelerate upward. Comparing the equations before and after the string is cut, this indicates that $ma=T$ after the string is cut. Due to Newton's 3rd law, the equivalent and opposite force from this acceleration will push down on the beaker, indicating (in my opinion) that the scale reading will not change as the tennis ball floats towards the surface.

Answer 4

What will happen to the reading of the scales if the string breaks? How can you correctly explain this using Newton's Third Law?

The flaw in the argument is that to apply Newton's third law, each (one) force on body one due to body two must be paired the other (one) force of the same type on body two due to body one and $Fb-W$ is not one force acting on one body due to another body.

The other thing which is important is to define the system.

Before the string is cut the system, container and contents, is in static equilibrium, which means that the force on the system due to the balance is equal to force on the balance due to the system.

The string is cut and this removes some internal forces due to the interaction of the string with the container and the ball.
These "removed" forces all are Newton third law pairs: force on string due to ball and force on ball due to string; force on base of container due to string and force on string due to base of container.

The centre of mass of the system will accelerate downwards which means that there is a net force downwards on the system.

Hence the force on the system due to the balance decreases as all other forces on the system stay the same which in turn means that the reading on the balance decreases.