The floating Hourglass

Marilyn is Wrong Copyright © 1997-1998 Herb Weiner. All rights reserved.

Ask Marilyn ® by Marilyn vos Savant is a column in Parade Magazine, published by PARADE, 711 Third Avenue, New York, NY 10017, USA. According to Parade, Marilyn vos Savant is listed in the "Guinness Book of World Records Hall of Fame" for "Highest IQ."


Fred Moolten <> remembers the following Marilyn column. If you have this column, please send it to me.

An hour glass containing air and sand is in a cylinder of water. While sand is flowing from the top to the bottom compartment, the hour glass remains submerged at the bottom of the cylinder, but when the sand has completed its downward journey, the hour glass floats to the top. Why?


Marilyn's answer had something to do with the sand flow causing the hour glass to tilt and hit the walls of the cylinder, thereby being impeded from moving upward.

An Alternative

Fred say's it's been a while since he took elementary physics, but he suggests that while the sand is falling, the force pushing the glass downward is equal to the weight of all the stationary items in the hour glass plus the force provided by the kinetic energy of the falling sand. This force is greater than the force the sand would exert if it were stationary. This is the same principle that applied to Juggling in a Raft. When all the sand is stationary, the hour glass is just bouyant enough to rise to the top.

Herb Responds

It's also been a while since I've taken elementary physics, so I'll phrase my response as questions rather than facts. Assume that the sand falls at a constant rate until it has all fallen. While a grain of sand is in freefall, would it be exerting any force at all on the hourglass? As long as the sand is flowing, would the force exerted exerted by the impact of the sand on the bottom together with the lack of force being exerted by the sand in freefall average out to greater, equal, or less force than that exerted by the stationary sand? Prior to the assent of the hourglass, the sand ends up at a lower altitude than it started at, so by the same argument used in Juggling in a Raft, wouldn't the effective downward force exerted by the sand be less, not more, than its stationary weight?

Fred's Experiment

For the hourglass, you are certainly right in saying that sand in free fall won't contribute any weight, and I should probably have included that in my suggested explanation. As you point out, however, the question remains as to whether the net effect of the falling sand is positive or negative. I did the following experiment (repeating it several times):

On the pan of a top-loading balance, I placed a plastic cylinder (to represent the bottom of an hourglass). In its top, I seated a syringe barrel with its aperture facing downward (to represent the top of the hourglass). I put 10 ml of water in the syringe, keeping it from escaping by plugging the aperture with a thin stick. I then removed the stick so that the water could flow. At the beginning, the weight reading on the balance bounced around (due mainly to the removal of the stick), but as the flow continued downward, it quickly stabilized at a constant value. Then, when the flow ended, with all the water at the bottom of the cylinder, the weight shifted to a reading a few milligrams lighter. In other words, the apparatus appeared heavier to the balance during flow than after it ended. This seems analogous to the hourglass floating upwards after all the sand has fallen. What's going on?

I suspect that the kinetic energy of the falling sand is doing the work of moving the hourglass downward (until the motion is halted, e.g., when it hits bottom, from which point forward any additional kinetic energy can only be converted into heat). In trying to understand this in terms of what is actually happening with the sand particles, the following explanation occurred to me. When the first grain of sand starts to fall, its weightlessness during free fall removes downward force from the hourglass (i.e., a "one grain" reduction in the effective weight of the hourglass), but that same quantity is restored when the grain lands, in the form of the force needed to return its velocity to zero, and in addition, once at zero velocity it will exert a separate one grain of downward force due to its weight. The result is a net Plus of one grain. When the next grain starts to fall, its weightlessness brings the net value back to zero, but when it too lands, it reverts to One Plus, and so on, so that the process stabilizes with an average small net positive force. I visualize the distinction between the force needed to stop a falling object and the force of its weight as follows: consider a plywood fence that must be strong enough to resist the force of a baseball pitched at 95 miles per hour (traveling horizontally). It must have enough force in its structure to bring a baseball from 95 mph to zero velocity without breaking. Now consider how much force a plywood floor would need in order to resist a baseball dropped from a height so as to land at 95 mph. It would also need the same force to slow the baseball from 95 mph to zero, but would need additional force to withstand the weight of the stationary baseball.

I'm not sure this is right, but it's the best I can do at the moment. I'll keep working at it, and will ask around.

A Plausible Explanation suggests:
Imagine the hourglass starting with all the sand in the top half and exerting a certain amount of force on whatever supports it. One grain of sand starts to fall. While it falls, the weight of the hourglass is reduced by the weight of the grain of sand. The grain accelerates in free fall until it strikes the floor of the hourglass. This decelerates it to a stop in a very short time. Obviously this means the deceleration is at a much higher rate than the acceleration due to gravity. Which, by F=ma, means that the force exerted on the floor of the hourglass is much more than the weight of the grain. As this is a continuous process with lots of grains of sand, the average weight is greater than the static weight. In a small time interval, subtract the weight of the number of grains starting to fall, and add the bigger force of those landing.


Robert D. Mathews <> writes:
The trick is based on friction. While there is still sand in the upper half of the glass, the hourglass is topheavy and leans over against the cylinder. Friction keeps the glass from rising. When all the sand is in the lower half, the hourglass goes upright. No longer pressed to the side of the cylinder, it floats.

Fred's explanation for the larger weight observed by the falling sand is correct. It is a problem in Halliday and Resnick's Physics.

Martin Gardner's Explanation

Jud McCranie <> wrote to report that the "rising hourglass" is discussed by Martin Gardner in Mathematical Carnival, in chapter 14 entitled "The Rising Hourglass and other Physics Puzzles". Hardback: Knopf, 1975; softcover: Vintige, 1977. When the hourglass is top-heavy, it leans over and the friction keeps it in place. When enough sand falls to the bottom, the friction is reduced and the hourglass will move.

A Novelty Toy

Kevin W. Clark <> wrote to report that In regards to the Floating Hourglass puzzle, I read about this many years ago, possibly in Scientific American's Mathematical Games or Amateur Scientist column. It claimed that this was available in the form of a novelty toy made in Hong Kong or somewhere, but that it came in two variants. One of which would float to the top of the tube when the sand ran out, the other of which would sink to the bottom of the tube when the sand ran out. The mechanism for both was given as tilt and friction holding it in place, combined with either a net positive or negative bouyancy, to effect the transfer, although it was playfully suggested in the instructions, that the changing position of the sand changed the bouyancy of the hourglass. last updated June 30, 1998 by