Liquid pressure problems
Now let us put P = hρg to work on the five problems of the exercise. Throughout, take g = 10 m s−2.
Problem (1)
The problem
The pressure produced on the base of a vessel by the liquid it contains is 1500 Pa. What does "the pressure is 1500 Pa" mean here?
The solution
That on every 1 m2 of area of the base of the vessel, the liquid exerts a normal force of 1500 N.
Problem (2)
The problem
Find the pressure produced by a mercury column 50 cm high. (the density of mercury is 13 600 kg m−3)
The solution
h = 50 cm = 0.5 m
P = h ρ g = 0.5 m × 13 600 kg m−3 × 10 m s−2
= 68 000 Pa
Problem (3)
The problem
The depth from the water surface to the bottom of a pond is 1.5 m. Calculate the pressure produced by the water at the bottom of the pond. (the density of water is 1000 kg m−3)
The solution
P = 1.5 m × 1000 kg m−3 × 10 m s−2
= 15 000 Pa
Problem (4)
The problem
The depth at a certain place in the sea is 1 km. Find the pressure produced by the sea water on the sea bed at that place. (the density of sea water is 1050 kg m−3)
The solution
h = 1 km = 1000 m
P = 1000 m × 1050 kg m−3 × 10 m s−2
= 10 500 000 Pa
That is more than a hundred times atmospheric pressure — which is why deep-sea vessels need thick walls.
Problem (5)
The problem
A tank 5 m long, 3 m wide and 2 m deep is filled with a liquid of density 800 kg m−3.
(a) What is the pressure at the base of the tank due to that liquid?
(b) What force does that pressure produce on the base of the tank?
The solution
(a) P = h ρ g = 2 m × 800 kg m−3 × 10 m s−2 = 16 000 Pa
area of the base = 5 m × 3 m = 15 m2
(b) force = pressure × area = 16 000 N m−2 × 15 m2 = 240 000 N
Notice that the length and the width were not used in (a) — the pressure depends on the depth alone. They are needed only to work out the area for the force in (b).