- The weight is constant
- The weight decreases and then increases
- The weight increases
- The weight increases and then decreases

Solution :

Consider an hourglass on a scale pictured below at times t=0, 0.001, and 1 hour. What happens to the scale's measure of weight of the hourglass plus the sand combination as the sand falls?

- The weight is constant
- The weight decreases and then increases
- The weight increases
- The weight increases and then decreases

Solution :

Initially the hourglass has all its sand on the bottom and so its at a weight **W**.

When we invert the hourglass, the weight decreases as the falling sand do not contribute their weight to the weight of the hourglass (so at 0.001 s, the weight of the hourglass is less than**W**).

When the sand hits the bottom of the hourglass, the momentary impulse delivered by the sand makes the hourglass weigh heavier than it actually is, so at exactly one hour, the hourglass is heavier than**W**. Thus the weight of the hourglass **decreases and then increases**.

When we invert the hourglass, the weight decreases as the falling sand do not contribute their weight to the weight of the hourglass (so at 0.001 s, the weight of the hourglass is less than

When the sand hits the bottom of the hourglass, the momentary impulse delivered by the sand makes the hourglass weigh heavier than it actually is, so at exactly one hour, the hourglass is heavier than

Subscribe to:
Post Comments (Atom)

## 12 comments:

## Post a Comment

This webpage is LaTeX enabled. To type in-line formulae, type your stuff between two '$'. To type centred formulae, type '\[' at the beginning of your formula and '\]' at the end.

Wouldn't the answer also depend on how much sand is in free-fall at once? If the impulse of the sand is not enough to overcome the loss of the weight from the sand in free-fall behind it, then I agree it will decrease, then increase to W presumably.

This also doesn't consider the loss of weight when you pick up the system to flip it, plus the increase in weight when you put it back down.

Seems like a better way to write this problem would be to have a solid mass falling in a cylindrical container.

I was assuming that the last speck of sand hits the bottom of the hourglass at exactly one hour. This would mean that the scale reads more than its supposed to, since to increase and then decrease to W, you need some split seconds of adjustment time (however you are right in that it'll decrease, increase, and then decrease to W).

As for whether it depends on if how much sand is in free-fall, at 0.001 seconds, no sand has reached the bottom of the hourglass, so we don't need to worry about that if we're just evaluating the 3 specific times they gave us.

And yes, it would be much better to rewrite this problem as a solid mass falling thorough a trap door that opens at some time (maybe I'll do this in the future once you guys have forgotten all about this problem :) ).

What kind of interaction do you mean?

Also, think about this:

If you were standing on a scale, and you bent down a little, how does the scale reading change?

Air is somewhere in between, but at drag forces would counter-act some of the missing weight from sand in free fall. I just don't know how much. I need a scale and an hourglass. :P

Hmm. Ok. I'll check it out.

Geez! You're right! The comment box just disappeared! Thanks for the heads up!

I've switched it back to the old system for now.

Comments for the new posts should be working.

Post a Comment