Hourglass on a balance

You are right. Perhaps my wording is not clear, but I did not intend to ask the question about the initial period, when the sand starts to flow. Once the flow is established fully, the weight will be approximately the same as the "static" weight, since the total momentum of the system is nearly steady, as long as we neglect the gradual shortening of the length of the free falling sand column.

The center of mass does not move downward with constant velocity before $t_1$ .

If $\mu$ is the mass flow rate of the sand (in gram per meter, e.g.), $y_1$ the position of the top of the sand in the top container and $y_2$ the height of the top of the sand in the bottom container, then the velocity of the center of mass is $v_{CM} = \mu(y_2 - y_1)$ . This would be constant if the difference $y_2 - y_1$ were constant; however, this is not the case because the cross sections of top and bottom are not equal, and even vary in different ways.

Because of that, you would expect the graph left of $t_1$ to show some curvature.

"At any given moment, the downward force exerted by the hourglass is the weight of the sand that isn't in freefall (plus the glass)." Accordingly, the red graph should be correct. I do not agree with this. The impact of the sand hitting the bottom will give an extra force, that is approximately equal to the missing weight.

Assuming the rate of flow (mass per unit time) is $f$ , the force of the impacting sand is $F=fv$ , where $v=\sqrt{2gh}$ is the velocity at the impact. The mass of the free-falling sand is $m=f t$ , where $t=\sqrt{2h/g}$ is the time-of-flight of a given grain of sand. The weight of the free-falling sand is $F'=mg=gf\sqrt{2h/g}=f\sqrt{2gh}$ . It is clear that $F'=F$ . When the sand is in steady flow the balance will show the same weight as it shows when the hourglass have run out of sand.

As Steven Chase suggested, another way of getting the same conclusion is by realizing that the hourglass with the moving sand has a downward linear momentum that is constant in time. The hourglass with the sand stopped has a linear momentum of zero - also constant in time. Since the force relates to the change of the momentum, it has to be the same in both cases.

Imagine that we have a small valve that can start the hourglass without applying external force. The complete story is this: When the valve opens there will be a gradual decrease of the weight, as the sand grains start free-falling. When the first sand grain hits the bottom, the weight jumps back to the original, and it remains there until the sand starts to run out (or the valve is closed). There will be a gradual increase of the weight, ending at the moment the last grain of sand hits the bottom. This last part is depicted in the green graph.

All of this is valid is we assume that the length of the free fall $h$ does not change i.e. we neglect the pile-up of sand in the bottom container.

That is not true. Imagine that you put a big, closed box on the scale and a person in that box throws a big rock to the air. As long as the rock is in the air, the scale will show a lower weight.

The movement of sand does matter.

Overall, the falling sand is being slowed in its downward motion--i.e. it is accelerating upward (!)

The acceleration of the sand is due to a net force acting on the sand. This force must come from the glass container around it and is directed upward. (Newton's Second Law)

If the glass container pushes upward on the sand, the sand pushes downward on the glass container. (Newton's Third Law)

When the sand is flowing, the weight is slightly more that the weight after the sand stopped. This is discussed in detail in this problem: https://brilliant.org/weekly-problems/2018-05-21/advanced/?p=2

There is also a serious scientific study of this, with experiment, a https://aapt.scitation.org/doi/full/10.1119/1.4973527

I agree with you, the hourglass is a closed environment, the weight before t1 and after t2 should remain the same. No new grains were added or grains being removed. All lines in the solution that have the same weight t1 and after t2 is the blue line.

Thanks. There is also a third problem, made up by Blake Farrow, in the "Advanced" section.

my guess is yes, because the height of sand in the upper part of the sand glass is getting smaller. By Torricelli's law the flow of sand leaving the upper part of the hourglass gets smaller. If you follow the link https://aapt.scitation.org/doi/full/10.1119/1.4973527 then you will see that the force starts to increases before $t_1$ .

The "excess weight" is due to the sand grains hitting the top of the sand pile in the bottom container. The sand grains that are already falling do not know if the supply of sand runs out in the upper container, and therefore the excess weight does not change.

No. The stable, time-independent "missing weight" is a feature of the continuous sand flow. It is compensated by the force of the impacting sand.

When the sand runs out of the top container, the missing weight starts to decrease. As long as sand is hitting the bottom, the "compensation" is unchanged. That appears as an increase of the weight.

The referenced link explains why the horizontal portions of the green line are at different levels. In essence, it is due to the fact that the length of the free falling sand column slowly decreases as the bottom container is filling up.

I think the green curve is pretty similar to the referenced link. They have more complicated model, and some details are different, but the essential features are the same. In the experiment some of the sharp jumps are rounded off. BTW, when I posted the problem I did not know about the referenced link, it was uncovered later by someone else.

Could you please tell why?

The problem in basic is in steady state, since the water is falling all the time, and the height of the water column does not change. This one however, is changing state.

The situation in Basic is in a steady state , i.e. the state of the system (waterfall), including its center of mass, does not change, and the waterfall could run forever, in principle.

The situation here is not in a steady state; the state of the system changes, its center of mass moves downward; and at some time the sand stops flowing.

When the last grain starts to fall there are also lots of other grains in the air. Some of these are just hitting the lower pile of sound. As these grains keep hitting the ground, and there is no more grains coming after the last one, the weight will go up temporarily.

the weight is minimally increased at the time of impact but remains constant, does not decrease