Water Filling a Cylindrical Container with a Scale Under It

[yellowcakepie]

Homework Statement

A scale is adjusted so that when a large, shallow pan is placed on it, it reads zero. A water faucet at height h = 4.0 m above is turned on and water falls into the pan at a rate R = 0.20 kg/s.

A) Determine a formula for the scale reading as a function of time t if at the moment t=0 first drops of water reach the pan.

B) Determine the reading for t = 9.4 s.

C) Repeat part B, but replace the shallow pan with a tall, narrow cylindrical container of area A = 18 cm^2 (the level rises in this case).

Homework Equations

Δx = 0.5gt^2

v = gt

I = R*v

F(t) = I + Rgt

The Attempt at a Solution

I have already solved parts A and B, but I am having trouble with part C.

A) F(t) = R√(2h/g)*g + Rgt (force due to change of momentum of the water plus weight of water in the container)

B) F(9.4 s) = 20 N

C) I have A = 0.0018 m^2 area for the cylindrical container, and density of water (D) is 1000 kg/m^3, so I got the fill rate by dividing D/R, and then divided A by D/R to get the change in height per second, which is 0.11 m/s.

I have no clue how to incorporate this height change rate into my answer.

 

[mfb]

The height difference between faucet and water level will change, but at the same time, you have to consider that the container fills a bit faster than you estimated because the stream of water represents a volume that is filled already. There is an argument that avoids explicit calculations of these effects.
It looks like the teacher didn’t think of the second effect when making this problem.

 

[kuruman]

The pan is said to be "large and shallow". I think that means that the linear parameter(s) defining the area (the radius if the pan is circular) are much larger than its height. I would interpret that to mean "ignore the height change of the level in the pan."

 

[mfb]

The question is about part C, where the container is not shallow any more.

 

[yellowcakepie]

The height difference between faucet and water level will change, but at the same time, you have to consider that the container fills a bit faster than you estimated because the stream of water represents a volume that is filled already. There is an argument that avoids explicit calculations of these effects.
It looks like the teacher didn’t think of the second effect when making this problem.

I didn't quite understand what you meant when you said "the stream of water represents a volume that is filled already".

I just need to calculate the force reading of the scale at 9.4 seconds after the first drop of water has hit the bottom of the container and started filling up.

As it fills up, I'd imagine the water would hit the surface quicker and quicker. But I don't know how to represent that in a mathematical equation.

 

[mfb]

I didn't quite understand what you meant when you said "the stream of water represents a volume that is filled already".

The shaded area in this very shady sketch:

R√(2h/g)*g

(from A)
Can you interpret this physically in relation to the amount of water falling down?

 

[yellowcakepie]

(from A)
Can you interpret this physically in relation to the amount of water falling down?

Yes, that is the force exerted on the container by the falling water when it hits the container. It is dependent on mass flow rate R, and the velocity of water upon impact, which depends on the time it takes to fall the distance between the faucet and surface of the water. I got the rate of change in height already (0.111 m/s). The time (t) it takes to hit the surface of water is then:

h - 0.111t = 1/2gt^2

t = 0.892 s

This is slightly different from what I got in part A (0.904 s) though...

 

[mfb]

Looks like the difference betweeen 9.81 and 10 m/s^2.

No, that's not what I meant. If you multiply the mass flow rate by the time it takes the water to fall down, you get a quantity of water. Where can you find this quantity of water in the setup?
How can you relate this quantity to the force you found for (A)?

 

[yellowcakepie]

Looks like the difference betweeen 9.81 and 10 m/s^2.

No, that's not what I meant. If you multiply the mass flow rate by the time it takes the water to fall down, you get a quantity of water. Where can you find this quantity of water in the setup?
How can you relate this quantity to the force you found for (A)?

That quantity of water would be the column of water falling down to hit the surface, which is just anywhere inside the container, above the rising water level.

That quantity of water is R*t, so with an empty container, would be 0.20 kg/s * 0.904 s = 0.1808 kg (using the time for water to hit in part A). The weight of water would be 1.77 N.

But this force would decrease as the water level in the container rises, since less time is needed to travel a smaller distance. Distance (d) is d = h - 0.111t.

h - 0.111t = 1/2gt^2 (to calculate time, right?)

 

[mfb]

That quantity of water would be the column of water falling down to hit the surface, which is just anywhere inside the container, above the rising water level.

The key point here: It is the total amount of water falling at a given point. And its force is the weight of this water.

For the problem here it doesn't matter if water arrived at the bottom already or not, as long as the stream started hitting the container.

 

[yellowcakepie]

The key point here: It is the total amount of water falling at a given point. And its force is the weight of this water.

For the problem here it doesn't matter if water arrived at the bottom already or not, as long as the stream started hitting the container.

Why is that the case? This concept seems physics-defying...

 

[mfb]

It only applies to a steady flow rate. If you would suddenly stop the water flow it wouldn’t be true any more of course.

Imagine a larger system where the faucet and a second tank are on the scale as well. While the water is flowing down at a constant rate, the center of mass moves downwards at a constant velocity. Based on conservation of momentum, whst the scale shows has to be the total mass of the system - including the water that is in free fall. The additional force at the bottom to stop the water is precisely the weight of the water in free fall.

 

[yellowcakepie]

It only applies to a steady flow rate. If you would suddenly stop the water flow it wouldn’t be true any more of course.

Imagine a larger system where the faucet and a second tank are on the scale as well. While the water is flowing down at a constant rate, the center of mass moves downwards at a constant velocity. Based on conservation of momentum, whst the scale shows has to be the total mass of the system - including the water that is in free fall. The additional force at the bottom to stop the water is precisely the weight of the water in free fall.

Oh, so turns out I just had to plug in the changing height (h - 0.111t) into the h term inside the square root and plug in t = 9.4s. And the resulting force is exactly the same as if the container was large and shallow, which is 20 N, because of the effect that you described. Thank you for the help.