Re: Does the weight of an hourgalss depend on whether the sand is flowing?

This would be a really cool experiment to try if you could make a big hour glass (people size), use a large digital scale, and very heavy sand or gravel.

There are a couple things to keep in mind to make solving this problem a little easier.

First of all, a scale measures force applied which is usually weight but not always. The scale really measures the normal force required to keep the thing being weighed from falling. It is fairly accurate to think of a scale as a spring that is being compressed between the floor and the top face of the scale by the mass of the object being weighed.

Secondly, the amount of sand doesn't change so I expect that the weight of the hourglass and sand is going to remain constant if nothing is moving.
This means that in a regular situation on earth, I expect that if I could keep the sand from falling from top to bottom that it will weight the same as if all the sand is in the bottom. Things will get more complicated in a complex gravitational field or if the hourglass is tall enough that the weight is different enough to measure. On the surface of the earth this would require the height to something on the order of a few percent of the radius of the earth (6.38 x 10⁶ m).

Thirdly, the scale will take some time to stabilize -- a change in force applied will produce some oscillation of the reading on the scale. Typically there is an overshoot, but we suppose that we have a perfect scale that measures forces perfectly and instantaneously. We expect there to be changes in forces as our objects move around. Rapid changes in force are usually described in Impulses (change in momentum divided by time describing the change). Therefore, the force we read on our scale is impulse force when the configuration of stuff is changing.

Let me change the problem to a more obvious case and then we can apply what we learn to the case of the sand in the hourglass. Let's start with a large perfect scale that is fairly sensitive, a step ladder, and a bright young physicist (BYP). Place the ladder on the scale and the BYP on top of the ladder. The scale will then read the weight (force) of the ladder and the BYP.

Now suppose the BYP steps off the ladder. While the BYP is falling the scale is not supporting his weight, therefore it will read only the weight of the ladder.

What happens when the BYP lands? After the impulse has passed, her/his weight is supported by the scale, so the reading on the scale is the weight of the ladder and the BYP.

Here is a summary of what I have concluded.

• Force Measured by Scale Before Falling = (mass of ladder + mass of BYP) * acceleration from gravity
• Force Measured while BYP in Air = (mass of ladder ) * acceleration from gravity
• Force During Landing = Changing as body and scale absorb landing.
• Force Measured by Scale After Landing = (mass of ladder + mass of BYP) * acceleration from gravity

So now, let's try to come up with a reasonable estimate for the magnitude of the impulse force when the BYP lands on the scale. I will set my axis point upward and set up the equations of motion verticle velocity V_y (speed and sign), and height Y:

V_y = V_y0 - g * t

Y = Y₀ + V_y0 * t - 1/2 * g * t²

where V_y0 is initial speed in the vertical direction, Y₀ is the initial height, g is the acceleration from gravity, and t is the time. The initial velocity is zero, and the initial height is the height of the ladder and the height of the scale is zero. Solving the height equation to solve for the time of the fall:

0 = h_ladder + 0 * t - 1/2 * g * t_fall²

or

t_fall = SQRT (2 * h_ladder / g).

V_{y @ landing} = -g * t_fall

or

V_{y @ landing} = -SQRT( 2 * g * h_ladder).

The momentum of the BYP when she/he lands is then given by:

m_byp * V_{y @ landing} = -m_byp * SQRT( 2 * g * h_ladder).

Now here comes the approximation, Let's suppose the landing can be defined by a small time dt. The impulse to stop the landing is given by

Change in momentum / dt = m_byp * V_{y @ landing} / dt.

Notice the change in sign! This the impulse force to stop the landing. This represents force as read by the scale. This is not the force from the weight of the BYP or the ladder. That force is separate, so the total force as measured by the scale:

Force _{@ landing} = m_byp * ( g + V_{y @ landing} / dt) + m_ladder * g.

Even with this simple model, if I let dt grow, it can be seen what happens if the BYP is lands over a lower time (less impulse) or if V_{y @ landing} changes to zero -- gently floating onto the scale (less impulse). Both of these extremes seem to make sense.

Let's put some real numbers into it.

Take our ladder to be about 10 kg and 1 meter tall, and our BYP to be 60 kg. Let's suppose it takes our BYP 0.20 seconds to "land." The vertical velocity of the BYP when she/he lands is 4.4 m/s. I could call the ratio V_{y @ landing} / dt an effective acceleration boost, which with these numbers gives 22 m/s² which is slightly more than 2 x gravity.

With the numbers above:

• Force Measured by Scale Before Falling = 70 kg * 9.8 m/s²= 686 Newtons =154 pounds
• Force Measured while BYP in Air = 10 kg * 9.8 m/s² = 98 Newtons = 22 pounds
• Force During Landing = 686 Newtons + 60 kg * 4.4 m/s / 0.20 s = 686 + 1320 = 2006 Newtons = 450 pounds!
• Force Measured by Scale After Landing = 686 Newtons = 154 pounds

(Note: This points the effectiveness of impulse forces. An average size guy can bust down a door simply by running into it shoulder first. )

Now let's consider the case of sand in an hourglass. It should be pretty similar to this setup. While the sand is falling, we need to consider small pieces of mass (sand) that are dropping to the bottom and consider the distance over which it falls. The sand that is currently not moving is like the ladder of my example, except that it makes up most of the mass. Since the falling sand is probably a small fraction of the total mass and it is falling a short distance, I would think that the impulse force would be very small. However, because the sand is falling for a long time (new sand falling all the time), you might be able to see a slight increase in weight (force) read by the scale because the impulse force is probably more than the loss in weight from the falling sand. But this can depend on the sand density, distance of fall, and amount that is falling. Try the experiment! It would be cool.

Sincerely,

Tom "Impulsive" Cull