Question

A container has a large cylindrical lower part with a long thin cylindrical neck open at the top. The lower part of the container holds 22.8 m^3 of water and the surface area of the bottom of the container is 9.10 m^2. The height of the lower part of the container is 2.50 m, and the neck contains a column of water 8.50 m high. The total volume of the column of water in the neck is 0.200 m^3.

(a) What is the magnitude of the force exerted by the water on the bottom of the container? (b) Explain why it is not equal to the weight of the water.

Answer 1

(a) 1045.5 KN

(b) 225.63 KN

Step-by-step explanation:

Since Pressure,
`P=\frac {F}{A}` where F is force exerted and A is area of the bottom of container

Making F the subject then

F=PA

Height of container=8.5m+2.5m=11.0 m

Density of water
`\rho_(water)=1000 Kg/m^(3)`

Surface area of the bottom of the container is
`9.10 m^(2)`

Pressure at the bottom of container

`P=P_(atm)+h\rho_(water) g` where
`P_(atm)` is atmospheric pressure taken as
`101.3*10^(3) Kg/m.s^(2)`, h is height which is 11 m,
`\rho_(water)` is density of water and g is acceleration due to gravity which is taken as
`9.81 m/s^(2)`

`P=101.3*10^(3) Kg/m.s^(2) +11m*1000 Kg/m^(3) *9.81 m/s^(2)=209.1*10^(3) Kg/m.s^(2)`

Force exerted is then found by

`F=PA=209.1*10^(3) Kg/m.s^(2)*5 m^(2)=104.55*10^(4) N`

Therefore, force at the bottom is 1045.5 KN

(b)

Volume of container at lower part is given as 22.8 cubic meters hence mass of water =volume*density of water

Mass=22.8*1000=22800 Kg

Volume of water confined in the column is 0.2 cubic meters hence the mass of water confined in the column is 0.2*1000=200 Kg

Total mass=200+22800=23000 Kg

Weight of water, W=mg=23000*9.81=225630 N=225.63 KN

Therefore, the weight of water is less than force applied at the bottom of container since pressure exerted by atmosphere on the surface of water is considered during calculation of force exerted at the bottom of the container