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?