Floating problems
Now let us work the four problems of exercise 15.2. Throughout, take g = 10 m s−2.
Problem (1)
The problem
(i) The depth of a certain reservoir is 1.2 m. Calculate the pressure arising at its bed due to the water. (density of water = 1000 kg m−3)
(ii) Find the force (thrust) the water produces on an area of 200 cm2 of that reservoir bed.
The solution
(i) P = h ρ g = 1.2 m × 1000 kg m−3 × 10 m s−2 = 12 000 Pa
200 cm2 = 0.02 m2
(ii) force = pressure × area = 12 000 N m−2 × 0.02 m2 = 240 N
Do not forget to convert the area to m2 — 200 cm2 is 0.02 m2, not 2 m2.
Problem (2)
The problem
(i) "The pressure in a liquid increases with depth." Write a simple experiment to demonstrate this.
(ii) Write a simple experiment to find out whether the pressure of the air inside a balloon is greater than atmospheric pressure or not.
The solution
(i) Make a row of evenly spaced holes down a plastic bottle from top to bottom and fill it with water. Hold the bottle above ground level and watch the jets. Water leaves the lower holes faster and travels further horizontally — which shows that the pressure is greater lower down.
(ii) Put water in a U-tube and check that the levels in the two arms are equal. Connect an inflated, tied balloon to one arm and undo the knot. If the level in that arm falls while the level in the other arm rises, the pressure in the balloon is greater than atmospheric pressure.
Problem (3)
The problem
(i) At sea level atmospheric pressure is 76 cm Hg. What is this pressure in pascals? (density of mercury 13 600 kg m−3)
(ii) What height of water column produces a pressure equal to the one above?
The solution
(i) P = 0.76 m × 13 600 kg m−3 × 10 m s−2 = 103 360 Pa
(ii) h × 1000 × 10 = 103 360
h = 103 360 ÷ 10 000 = 10.336 m
Problem (4)
The problem
(i) State Archimedes' principle.
(ii) The weight of a certain object in air is 20 N. When it is fully immersed in water its apparent weight is 5 N.
(a) What upthrust does the water produce on the object?
(b) What is the weight of the water displaced when the object is fully immersed?
The solution
(i) When an object is immersed partly or wholly in a fluid, the upthrust acting on it is equal to the weight of the fluid it displaces.
(ii) (a) upthrust = 20 N − 5 N = 15 N
(ii) (b) By Archimedes' principle the weight of the water displaced is also 15 N.
Part (b) needs no fresh calculation — the principle is precisely the statement that the two are equal. This object sinks because the upthrust (15 N) is less than its weight (20 N).