Question

A tin can has a radius of 0.0410m and is 185 m high if the air is removed from inside the can how much forces will the outside air exert on the bottle ?

Answer 1

The force exerted on the bottle is 532 N

Explanation:

When the air is removed from inside the bottle, vacuum is created, therefore the external pressure (the atmospheric pressure) is no longer balanced, and it creates a net force downward on the can.

The pressure is related to the force by the equation:

$p=\frac{F}{A}$

where

p is the pressure

F is the force

A is the area of the can

Here we have:

$p=1.01\cdot 10^5 Pa$ is the atmospheric pressure

r = 0.0410 m is the radius of the can, so the area is

$A=\pi r^2 = \pi (0.0410)^2=5.27\cdot 10^{-3}m^2$

And solving for F, we can find the force on the can:

$F=pA=(1.01\cdot 10^5)(5.27\cdot 10^{-3})=532 N$

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