Question:

An hourglass timer is being weighed on a sensitive scale, first when all the sand is in the lower chamber and then after the timer is turned over and the sand is falling. Will the scale show the same weight in both cases?

.

.

C

A

P

T

A

I

N

I

N

T

E

R

V

I

E

W

.

.

Solution: There are actually two plausible responses. One response is that falling grains are essentially weightless and exert no force on the scale as long as they are falling. Hence, the hourglass will weigh less after it is turned over. The other response is that from the instant that the first grain of falling sand strikes the bottom of the hourglass to the instant the last grain falls, the force resulting from the impact of the sand on the bottom of the glass remains constant and helps make the total weight equal to the weight of the hourglass before being inverted. When the stream begins to fall, the freely falling sand does not contribute to the weight, and so there is slightly less weight registered for the first few hundredths of a second. As the last grains of falling sand strike, there is a short time interval when the weight exceeds the initial weight. For each grain of sand now striking the bottom, no longer is there a grain of sand leaving the upper chamber, and so the hourglass weighs more.

Answer: In reality, the inverted hourglass weighs more.