CIVIL , ENVIRONMENTAL & GEOMATIC ENGINEERING DEPARTMENT
1st YEAR FLUID MECHANICS
CEGE1009 MECHANISMS
SOLUTIONS for TUTORIAL SHEET 1
3
Density of water ρ = 1000 kg / m
Q1.
A steel sphere for which σ = 7.8 has a diameter of 30 mm. Find its mass, its weight in air
and its weight in water.
i) Mass of object = volume x density
4
4
= /3 π r3 ρ = /3 π (0.015)3 [m3] x 7.8 [ ] x 1000 [kg m-3] = 0.110 kg
ii) Weight of object in air = mass x gravitational force/unit mass = m g = 0.110 x 9.81 = 1.082 N
(- neglecting buoyancy effect of air)
iii) Weight of object in water = m g - (buoyancy force) = m g - weight of water displaced
4
3
= V g (ρsteel - ρwater ) = /3 π (0.015) x 9.81 x (7800 - 1000) = 0.943 N
Q2.
A circular disc of glass (σ = 2.2) is 3 mm thick, it has a diameter of 50 mm and a uniform
layer of cork (σ = 0.15) covers one face. Find the minimum thickness of the cork layer
which will prevent the disc from sinking in water. If the cork thickness is increased to 10
mm, how much cork will protrude above the water surface?
i) If the disc is just able to avoid sinking, then its upper face is level with the water surface.
Take the thickness of the cork = t, and thickness of glass disc = s
If in equilibrium, then upward force (acting on bottom face of disc) must equal the total weight.
That is: hydrostatic pressure on bottom face x area of that face = m g
2
2
2
Hence: ρwater g (t + s) . π r = ρcork g π r . t + ρglass g π r . s
and t = {(ρglass - ρwater ) s} / ( ρwater - ρcork ) = (2200 - 1000) x 0.003 / (1000 - 150) = 0.00424m
ii) When the cork thickness becomes 10 mm, it is still necessary for the hydrostatic force on the bottom face
to balance the total weight. Letting the depth of immersion = h:
then: ρwater g h . π r2 = ρcork g π r2 . t + ρglass g π r2 . s
giving: h = {ρcork . t + ρglass . s } / ρwater = {150 x 0.01 + 2200 x 0.003} / 1000 = 0.0081 m
and height of cork protruding above water surface = (0.010 + 0.003 - 0.0081) = 0.0049 m = 4.9 mm
Q3.
Determine the position adopted by a square slab of material of relative density σ = 0.9 when
floating completely immersed in a vessel containing two liquids which do not mix and
whose relative densities are 1.2 and 0.8
Let the slab have a total thickness t, a depth s of which is immersed into the lower, denser fluid. Also, let
the fluid pressure on the top face of the slab be p, and the area of the slab be A.
Then: upthrust on bottom face of slab must equal the weight of the slab + downthrust on the top face
{p + ρupper fluid g (t - s) + ρlower fluid g s} . A = ρslab t A g + {p A}
Hence: s = {(ρslab - ρupper ) t} / ( ρlower - ρupper ) = (900 - 800) x t / (1200 - 800) = t / 4
In other words, the slab lies with a quarter of its thickness immersed in the lower fluid.
[Note that it is also legitimate to approach this problem by considering the buoyancy of the slab - in terms of
the weight of the fluids displaced. The same answer follows!]
Q4.
An object weighs 10 N in air and 8 N when immersed in water. What is its relative density?
Let the volume of the object = V, and its density = ρ b
Upthrust from buoyancy in air = ρ air g V and upthrust from buoyancy in water = ρ water g V
Hence, weight in air = g V ( ρ b - ρ air ) and weight in water = g V ( ρ b - ρ water )
Dividing weight in air by the weight in water - and neglecting the small term induced by air upthrust:
ρ b / ( ρ b - ρ water ) = 10 / 8
in other words: 8 x ρ b = 10 x ( ρ b - ρ water ) ; or
and the relative density ρb / ρwater = 10 / 2 = 5
2 x ρ b = 10 x ρ water
Q5.
A closed rectangular pontoon of dimensions 12 m x 8 m x 4 m high has a mass of 84,000
kg. Find the depth of immersion when launched in sea water (σ = 1.029). The pontoon is
moored to the seabed by four vertical cables of equal length, one at each corner. Find the
load in the cables required to produce a depth of immersion of 3 m. What is the maximum
volume of oil (σ = 0.9) which can be stored in the pontoon without influencing the depth of
immersion?
i) If the pontoon floats when launched into the sea, its weight downwards must be balanced by the upthrust
on its bottom face. If the unloaded depth of immersion is h:
m g = ρsea water g . h . (area of bottom face):
hence, h = m / { ρsea water (area of bottom face)}
= 84000 / (1029 x 12 x 8) = 0.85 m
ii) If the cables hold the pontoon down such that its depth of immersion is 3 m, then by equilibrium:
downward forces = upthrust, or: m g + 4 T = ρsea water g . h . (area of bottom face)
84000 x 9.81 + 4 T = 1029 x 9.81 x 3 x 12 x 8:
hence: T = {1029 x 9.81 x 3 x 12 x 8 - 84000 x 9.81} / 4 = 0.52 x 106 N
iii) If oil is stored in the pontoon until it is about to alter the depth of immersion, then its weight will at that
stage exactly have replaced the tensile force downwards previously provided in the cables.
That is: Volume of oil . g . ρoil = 4 T:
6
3
Volume of oil = 4 T / ( ρoil g) = 4 x 0.52 x 10 / {900 x 9.81} = 235.6 m
Q6.
Water in an 8 m wide canal is 2.5 m deep where it passes over a 12 m span aquaduct bridge.
Estimate the hydrostatic force on the bridge i) when a 30 Mg barge is at centre-span on the
aquaduct, and ii) after the barge has moved off the aquaduct.
i) When the barge is at mid-span, it is supported by the hydrostatic pressure under its hull (equivalent to its
buoyancy - which is equal to the weight of water displaced). The water displaced makes a negligible
difference to the water depth in the canal - assuming that the aquaduct is not contained between two locks!
Hence, the load on the aquaduct itself is caused simply by the hydrostatic force on the canal bed:
F = pressure at bed x area of canal bed in the aquaduct = ρwater g . h . A
= 1000 x 9.81 x 2.5 x 12 x 8 = 2.35 x 106 N = 2.35 MN
ii) As the barge moves off the aquaduct, so water returns to the space previously displaced by the barge leaving the water level still as 2.5m. Hence the load on the aquaduct will remain unaltered.