Pressure of air inside a glass

In summary, when a plastic card is placed on top of an uncovered glass, the pressure inside the glass is reduced due to the interaction between the air particles and the card. However, this reduction in pressure may be countered by the weight and density of the card resting on the glass. The resulting pressure in the glass may be either higher or lower than atmospheric pressure, depending on the thickness and weight of the card.
  • #1
Pushoam
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TL;DR Summary
How does putting a cover affect the pressure inside a glass filled with air?
Let's consider an uncovered glass. Air particles are present in the glass.
1679724724893.png

$$ P_1 = P_a$$ $$P_2 =P_1 +\rho gh = P_a +\rho g h$$where ##P_A## is atmospheric pressuere and ##\rho ## is air density.
1679725186723.png

Now, if I cover the glass with a plastic card, then what is ## P_1##?
$$P_2 =P_1 +\rho gh $$
1) ## P_1 ## is pressure due to motion of air particles and the air particles near the cover interact with the cover and its speed may change and hence ##P_1## may be less or more than ##P_a##.

2) Following three forces are acting on the cover:
1679725733426.png

a) force due to pressure ##P_1## of air particles
b) normal force N due to glass walls
c) cover's weight W

Applying Newton's first law gives,
$$ P_1 A+ N = W$$ $$P_1 A = W - N$$
Now, since normal force is self-adjustable, let's take a light plastic card such that N = 0. Hence, in this case ##P_1 A = W ##.
For a plastic card with mass 20g and area 20cm2, ## P_1 = 10 Pa## which is lower than the atmospheric pressure.
So, the conclusion is: putting a cover reduces the pressure of air inside the glass. Is this correct?
 
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  • #2
Pushoam said:
So, the conclusion is: putting a cover reduces the pressure of air inside the glass. Is this correct?
No.
There will be a hydrostatic difference in pressure due to the thickness of the card, but that difference will be overcome by the density of the card, which is resting on the glass.
 
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  • #3
Baluncore said:
No.
There will be a hydrostatic difference in pressure due to the thickness of the card, but that difference will be overcome by the density of the card, which is resting on the glass.
And/or the card will bend down/inward due to its own weight and the resulting pressure in the glass will be higher than atmospheric.
 
  • #4
Define Pa as the atmospheric pressure at the top edge, inside the glass.
The following four forces are acting on a cover of thickness; t
a) force due to pressure of air from below; A·Pa
b) force due to pressure of air from above; A·(Pa - ρ·g·t)
c) cover's weight; W
d) normal force upwards due to glass wall; N
N + A·Pa = W + A·(Pa - ρ·g·t)
N = W - ρ·g·t
ρ·g·t
is the buoyancy of the card
 

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