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An hourglass H weighs h. When it's placed on a scale with all the sand rested in the lower portion, the scale reads weight x where x = h.

Now, if you turn the hourglass upside down to let the sand start to flow down, what does the scale read?

I imagine initially, when the sand starts to fall but before the first batch of grains touch the bottom of the hourglass, these grains of sand effectively are in a state of free fall, so their weight would not register onto the scale. The weight at this point has to be less than h. However, what about the steady state when there is always some sand falling and some sand hitting the bottom of the hourglass? In the steady state, although we are having some sands in the free fall state and thus decrease the weight of H, there are also sands that are hitting (decelerating) the bottom of the hourglass. This deceleration should translate increase the reading on the scale more than the actual weight of those impacting sands. To illustrate the last point, imagine a ball weighing 500g rested on a scale. If you drop this ball from a mile high onto the same scale, on impact, the scale would read higher than 500g. in the same way, in our hourglass question, will the decreasing effect of weight due to free-fall cancel out exactly the increasing effect of weight due to sand impacting? does it depend on the diameter of the opening? does it depend on the height of the free-fall? Does it depend on the air pressure inside the hourglass?

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You need to consider the entire mass of sand. It can be shown that its center of mass does indeed decelerate, leading to an increased apparent weight. See my response below. –  Doru Constantin Dec 5 '13 at 16:53

7 Answers 7

up vote 12 down vote accepted

Analyzing the acceleration of the center of mass of the system might be the easiest way to go since we could avoid worrying about internal interactions.

Let's use Newton's second law: $\sum F=N-Mg=Ma_\text{cm}$, where $M$ is the total mass of the hourglass enclosure and sand, $N$ is what you read on the scale (normal force), and $a_\text{cm}$ is the center of mass acceleration. I have written the forces such that upward is positive

The center of mass of the enclosure+sand moves downward during process, but what matters is the acceleration. If the acceleration is upward, $N>Mg$. If it is downward, $N<Mg$. Zero acceleration means $N=Mg$. Thus, if we figure out the direction of the acceleration, we know how the scale reading compares to the gravitational force $Mg$.

The sand that is still in the top and already in the bottom, as well as the enclosure, undergoes no acceleration. Thus, the direction of $a_\text{cm}$ is the same as the direction of $a_\text{falling sand}$ . Let's just focus on a bit of sand as it begins to fall (initial) and then comes to rest at the bottom (final). $v_\text{i, falling}=v_\text{f, falling}=0$, so $a_\text{avg, falling}=0$. Thus, the (average) acceleration of the entire system is zero. The scale reads the weight of the system.

The paragraph above assumed the steady state condition that the OP sought. During this process, the center of mass apparently moves downward at constant velocity. But during the initial "flip" of the hour glass, as well as the final bit where the last grains are dropping, the acceleration must be non-zero to "start" and "stop" this center of mass movement.

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Imagine an hourglass with just one stone inside. When the stone start to falling a scale would stop to measure it's weight, but it will measure a spike corresponding to the moment when it hits the bottom. The bigger the airtime, the bigger the spike. It is like concentrating the weight of the stone in a very specific time interval: when it hits. However the average weight over the falling time is exactly the total weight with the stone at the bottom.

Going back to sand the only thing that change is that instead of a big spike, you have a lot of small spikes in the weight. So you don't have to wait a falling time to get the static weight as an average and a scale will average by itself inertia showing always the same weight. However if you find a scale with an outstanding resolution both in mass and time according to the size of the grains, you may be able to see these spikes.


Well, how to prove that we are just concentrating the weight in time. I think a quite simple, still effective argument is found in the high-school-level relation:

$$m \Delta v = F \Delta t$$

The momentum acquired by the rock during the free falling is:

$$M_f = m g T$$

during the impact the velocity is killed in a time $t$ and so the momentum, calling $a$ the involved acceleration we have (modulo any signs which are trivial to fix):

$$ m g T = m a t $$

It's now clear that

$$ m a = m g \frac{T}{t} $$

This means that, if for a time $T$ we are not measuring the stone weight, then for a time $t$ we measure a weight $\frac{T}{t}$ times bigger. The average in time equals to:

$$\frac{0\cdot T+mg\frac{T}{t} \cdot t+mg\cdot t}{T+t} = mg$$

The first term is related to the time of flight (zero force), the second is the force that kills the momentum over a time $t$ and for the same time $t$ also the natural weight of the rock acts.

If there is some air resistance, it will transmit some force to the scale while the stone's falling, in the average expression it will move some force from the second term to the first. The idea is that during the falling time the scale measures the drag, but then the velocity is a bit smaller when the stone hits. This would be proven in a similar way as above.


Actually sources of weight change has to be searched in the famous equation: $E=mc^2$. The energy of the sand down, will be a little less and so it's mass.

At the same time one could consider that going closer to the Earth surface the gravity is a little bigger, so sand down weight a little more.

Both the effects are far from measurable.

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How can you be sure that the average weight of the stone is exactly the same as the weight of the stone in rest? That doesn't seem at all trivial to calculate. –  JSQuareD Dec 3 '13 at 13:23
    
@JSQuareD Whenever you talk about hits in a classical framework, the time involved is an unknown hard to determine. However, as often happens, you don't need a brute force computation to prove something. I hope that the argument that I added satisfies you. –  DarioP Dec 3 '13 at 14:14
    
how do you know "This means that, if for a time T we are not measuring the stone weight, then for a time t we measure a weight Tt times bigger." Also, how about air resistance in the hourglass that slows down the drop? Furthermore, the impact will cause heat which increase energy (convert potential to kinetic energy), so how does it affect the system? Does the added energy increase the weight of the hourglass? (Says our scale is absolute percise) –  Gary Chou Dec 3 '13 at 18:40
    
@DarioP Ah yes, that's a really nice way of proving it, thanks. –  JSQuareD Dec 3 '13 at 21:11
    
@user35167 1)I know because the momentum acquired during the falling is the same as the one dissipated during the hit, so I equal the impulses. 2)I added the point of air resistance. 3)Both the drag and the hits dissipate some energy, this is fine as the final energy (stone down) is less than the initial (stone up). However momentum and energy conservations are two well separated concepts in physics, forces are related to the first. You may object E=mc²: all the sand down should weight a little less, I would add that gravity closer to Earth surface is a little higher, isn't this going to far? –  DarioP Dec 4 '13 at 8:13

Suppose you are standing on a tower which is on a scale. Jump off. While you're in the air, what does the scale read? (Assume here that you will land on the scale, so the analogy to sand grains holds.)

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interesting way to separate my question into individual sand. i am more concerned about steady state though, so using your method, the question could be rephrased as, two people on top of a tower on top of a scale. one jumps first. second jumps at the moment the first hits the ground. i think overall at steady state, the scale will actually read higher then!! (do you agree?) –  Gary Chou Dec 3 '13 at 12:39
    
@user35167 You're getting into calculus :-) - try taking the limit as sand-grain mass --> 0 while grain rate (grains/sec falling) -->Inf such that the mass/sec hitting the bottom remains constant. –  Carl Witthoft Dec 3 '13 at 12:44
    
Sorry I didn't really get the calculus part maybe because Calculus was what I failed at 15 years ago in college...haha –  Gary Chou Dec 3 '13 at 18:41

There are quite a few things to consider here.

First, The "hourglass" if this vessel is filled with air the results will be much more complex to determine.

Second, the diameter of the grains and their uniformity will influence the measurements.

Third, The size of the opening will also impact grainflow.

Fourth, The sensitivity of the scale in relation to both time and mass.

In a perfect situation the scale would drop and spike for each individual grain as it leaves the opening and begins free-fall and eventually strikes the sand/bottom of the vessel.

Now to the meat of the question it would be extremely unlikely dare I say impossible for a measurement with enough sensitivity to be perfectly displaced at the exact rate as the sand flow.

On a side note thinking this experiment out as the pile of sand grows on the bottom you will lose some momentum when the grain strikes others and pushes them sideways which would eventually hit other grains or the bottom and cause further spikes.

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The apparent weight is indeed larger when the hourglass is running than at rest. See here for a detailed write-up. This effect has even been verified experimentally.

In a nutshell: the net effect of the flow is to move sand from the top surface (where it has a downwards velocity $v$) to the bottom pile, at rest. Thus, the sand is decelerating and the force on the balance is higher than at rest.

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brevity is appreciated but a link only answer is not very helpful. If you could possibly paraphrase or share some relavent points based on the op's question to expand on your answer it may be more helpful. –  Argus Dec 8 '13 at 1:42

I think as time goes on, the layer on the down part of the hourglass becomes thicker. So as the result of free fall, each next grain will have less velocity as it has come a lesser distance.So as we consider that each grain stops after hitting the bottom, the amount of impact of each grain decreases.So at first, the scale will show a great (comparing to others) loss in weight.As the time goes, it will show a lesser loss in weight, and at last it will show the complete weight of system in static term.

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I considered a similar question once.

Consider a hollow (perfectly rigid) container of mass $M$ in a homogeneous gravitational field filled with $N$ identical particles of mass $m$ whirling around inside the container as an ideal gas in equilibrium (only exposed to the external gravitational field). Let the container be in contact with a weighing machine. Even though most of the time most of the particles are not in contact with the container - and thus with the weighing machine - it can be shown that the averaged (over time) weight measured by the weighing machine of the container plus its "filling" is exactly

$$M + N\cdot m$$

This results from calculating the difference of the pressure at the top of the container and at the bottom of the container, exerted by the particles in the gravitational field.

To this day I find this astonishing and remarkable.

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