A drinking straw 20 cm long and 3.0 mm in diameter stands vertically in a cup of juice 8.0 cm in

A drinking straw 20 cm long and 3.0 mm in diameter stands...

A drinking straw $20 \mathrm{cm}$ long and $3.0 \mathrm{mm}$ in diameter stands vertically in a cup of juice $8.0 \mathrm{cm}$ in diameter. A section of straw $6.5 \mathrm{cm}$ long extends above the juice. A child sucks on the straw, and the juice level begins dropping at $2.0 \mathrm{mm} / \mathrm{s}$. (a) By how much does the pressure in the child's mouth differ from atmospheric pressure? (b) What's the greatest height above the water surface from which the child could drink, assuming this same mouth pressure?

A drinking straw $20 \mathrm{cm}$ long and $3.0 \mathrm{mm}$ in diameter stands vertically in a cup of juice $8.0 \mathrm{cm}$ in diameter. A section of straw
$6.5 \mathrm{cm}$ long extends above the juice. A child sucks on the straw, and the juice level begins dropping at $2.0 \mathrm{mm} / \mathrm{s}$. (a) By how much does the pressure in the child's mouth differ from atmospheric pressure?
(b) What's the greatest height above the water surface from which the child could drink, assuming this same mouth pressure?

Key Concepts

Pressure Differential and Suction

Pressure differential is the difference in pressure between two regions. In a drinking straw, when a person sucks, they lower the pressure in their mouth relative to the atmospheric pressure acting on the fluid's surface. This difference creates a force that pushes the liquid upward into the straw. Understanding how this pressure difference relates to the height of the fluid column is critical in determining both the pressures involved and the maximum height from which fluid can be drawn.

Hydrostatic Pressure

Hydrostatic pressure refers to the pressure exerted by a fluid at equilibrium due to the force of gravity. It is directly proportional to the density of the fluid, the gravitational acceleration, and the height of the fluid column above the point of measurement. This concept is essential for calculating pressure differences in static fluids and is used to determine the pressure difference between the atmospheric pressure and the lower pressure created by suction in a straw.

Atmospheric Pressure

Atmospheric pressure is the pressure exerted by the weight of the Earth’s atmosphere on the surface of the Earth. It provides the baseline or maximum pressure available to push a fluid up a column when suction reduces the pressure at one end. In the context of drinking straws, atmospheric pressure limits the maximum height a liquid can be raised because the maximum possible pressure difference is essentially set by the ambient atmospheric pressure.

Transcript

00:01 Hi, everybody.

00:01 So we need to find how much pressure the child's mouth differ from atmospheric pressure, and what's the greatest height above the water's surface from which the child could drink, assuming the same mouth pressure.

00:15 So for this, we have a length of 20 centimeters, or we can do times if my pen lets me.

00:32 Come on.

00:35 There we go um times 10 to negative 2 meters and we have the diameter of 8 centimeters which is going to be 0 .08 meters then we have a diameter 2 which is again given to us in millimeters so it's going to be 3 times 10 to negative 3 meters mit looks kind of weird.

01:07 It's okay.

01:09 We have a velocity of 1 of 0 .2 centimeters squared, and that's going to be 0 .2 times 10 to negative 2 meters per second.

01:22 And here we have a area velocity 1 equals area 2, velocity 2, and here we know that your areas are going to, equal pi and i'm just going to insert it here so here we have pie and then we have our direction and we have divided by four velocity one equals pi d2 for a diameter four times v2 okay and if we put v2 onto its side, it's going to be v1 equals d2 squared.

02:13 Okay.

02:14 And now we're going to plug this into this equation.

02:19 So we got p1 plus one half times the density times velocity squared plus the density gravity times height 1 equals pressure 2 plus.

02:34 One half and we got density velocity two squared plus density grams height okay and it's height two and now we have p1 plus one half times density v1 squared plus density times gravity times height one equals pressure of two plus one half density velocity 1 squared times d1 d2 squared and it's going to be cubed in this case plus density gh2.

03:22 Okay.

03:23 And i want to keep going down.

03:25 So now we have p1 minus p2 equals one half density v1 squared.

03:37 Times your d1, d2, to the fourth, minus v1, square, plus your density of gravity, height 2 minus height 1.

03:55 And this is going to continue to equal 1 half, density v1 squared times d1, d2.

04:11 To the four minus one plus density gravity times height minus height.

04:24 And now we're just going to plug in the information that we have.

04:28 So we have one half times actually we can just a little bit differently because we have the densities all going to be the same.

04:40 So let's take the density out...