# What is the magnitude of the force exerted by the water on the bottom of a bottle?

## Homework Statement

A container has a large cylindrical lower part with a long thin cylindrical neck. The lower part of the container holds 12.5 m3 of water and the surface area of the bottom of the container is 4.68 m2. The height of the lower part of the container is h1 = 3.30 m and the neck contains a column of water h2 = 8.20 m high. The total volume of the column of water in the neck is 0.200 m3. What is the magnitude of the force exerted by the water on the bottom of the container?

## Homework Equations

F = PA
= (ρgh + P0)(4.68 m²)
ρ = density of water=1000kg/m^3
g = acceleration due to gravity = 9.8 m/sec²
h = height of water column = 3.30m + 8.20m = 11.50 m
P0 = air pressure at top of column = 1 atmosphere.

## The Attempt at a Solution

F = PA
= (ρgh + P0)(4.68 m²)
= (1000kg/m^3)(9.8m/s^2)(11.50m) + 1 atm.)(4.65 m²)=527440.68N need to convert to MN

Is my formula correct?
Are my units for force correct that it comes out in Newtons?

ideasrule
Homework Helper

It's correct, but the question asks for the pressure the water exerts. The pressure exerted by the air above is not usually taken into consideration.

I think the contribution of force exerting on bottom of container includes force by neck and force by lower part.

F = F_neck + F_low