MAS2803/PHY2033 Assignment 2 Semester 2, 2019/2020

The deadline is 15:00 Thursday 7th MaySpecify any forces as vectors.

Question 1 10Consider a reservoir filled with water of (uniform) density ρw, and a dam wall of height h andwidth W , as shown below. The atmospheric (air) pressure in the vicinity of the dam can be takento be constant, Patm. You can assume that the fluids are in equilibrium.

(a) Determine an expression for the pressure force exerted by the water on the dam wall.

Hint: For this question you should use the general result derived in the lectures for a fluidof uniform density ρ0 under gravity that the pressure varies as P (z) = P0 − ρ0gz, where P0

is the pressure at z = 0 and g is the acceleration due to gravity.

The pressure in the water is P (~x) = P0 − ρwgz. We calculate the pressure force via thesurface integral,

~F = −∫∫

P (~x)n dS = −∫∫

(P0 − ρwgz)(−x)dS = x

∫∫(P0 − ρwgz) dy dz

= x

∫dy

0∫−h

(P0 − ρwgz)dz = xW

[P0z −

ρwgz2

2

]0−h

= xW

(P0h−

ρwgh2

2

)

The direction of this force (in the positive x direction) represents the fact that the waterexerts a force on the dam wall.

(b) Determine an expression for the pressure force exerted by the air on the other side of thewall.

Now n = x and P (~x) = P0, so the pressure force is,

~F = −∫∫

P (~x)n dS = −xP0

∫∫S

dy dz = −P0xhW.

Here, for a shortcut, we used the general result that∫∫

SdS = area of S.

The force acts in the negative x direction, as we would expect.

(c) Hence determine the net pressure force on the dam wall.

The net force is just the sum of these two contributions:

~F =ρwgWh2

2x

Question 2 10

(a) A cube with sides of length 10 cm and density ρc = 900 kg m−3 floats in water of densityρw = 1000 kg m−3. What distance does the cube extend into the water? You may ignorethe buoyancy effect on the cube from air.

The weight of the cube is −ρcgL3z, where L is the side of the cube. The buoyancy force isρwgdL

2z, where d is the depth of the cube below the surface. Balancing these gives,

ρcgL3 = ρwgdL

2.

Rearranging for d gives,

d =ρcρwL =

9

10× 10cm = 9cm

.

(b) What mass M must be placed on the cube so that it just becomes submerged? State youranswer in grams.

With the additional mass, the weight of the cube+mass is −(Mg + ρcgL3)z. If the cube is

fully submerged the buoyancy force will be ρwgL3z. Balancing these gives,

Mg + ρcgL3 = ρwgL

3.

Rearranging for M gives,

M = (ρw − ρc)L3 = 100× 0.13 = 0.1kg

Thus, 100 g must be added to the cube.

(c) An identical cube (without the additional mass) is placed in oil of density 850 kg m−3. Itis weighed by a weighing scales. What mass does the scales register? Give your answer ingrams.

The weight of the cube is −ρcgL3z. The buoyancy force is ρwgL3. So the net downward

force, detected by the weighing scales, is,

−ρcgL3z + ρwgL3z = (ρw − ρc)gL3z.

This is the effective weight, −M ′gz, and so we arrive at,

M ′ = (ρc − ρw)L3 = (900− 850)× 0.13 = 50g

Question 3 10On a planet far far away the gravitational body force (per unit mass) varies with altitude z,

with the form ~f = (0, 0,−z). The air density ρ(~x) and pressure P (~x) vary in space, and arerelated by the expression ρ = 2P . Consider the fluid to be at equilibrium such that it satisfies

the equilibrium condition ∇P = ρ~f.

(a) By solving the equilibrium condition, obtain an expression for how the pressure varies withaltitude, P (z), subject to the condition P (z = 0) = P0.

(b) Sketch P (z). Indicate P0 on your sketch.

We have ∇P = ρ~f with ~f = (0, 0,−z). This gives rise to 3 partial differential equations,

∂P

∂x= 0,

∂P

∂y= 0,

∂P

∂z= −ρz.

These equations tell us that P depends on z only, and so we can write the third equation in termsof conventional derivatives and proceed to solve it as an ordinary differential equation,

dP

dz= −gρ = −2gP.

Separating and integrating,

dP

P= −2z dz → ln |P | = A− z2 → P (z) = Ae−z

2

= P0e−z2 .

Question 4 10Consider a cubic balloon∗ of side L and of density ρb. At ground level the density of air is ρ0 = 1.2kg m−3.

(a) Calculate the net force (of buoyancy and weight) on the balloon at ground level. Oneballoon has a net force of magnitude Fb, which we take to be up (in the positive z direction).This net force is a combination of buoyancy force (up) and the balloon weight (down), i.e.

Fb = ρ0gV − ρbgV = (ρ0 − ρb)gV = (1.2− 0.9)× 9.81× (0.3)3 = 0.0795N.

(b) Estimate the minimum number of balloons required to lift an apple of mass 0.1 kg off theground, if L = 30 cm and ρb = 0.9 kg m−3. Ignore the buoyancy force of the apple itselfand the weight of any string, etc.

N such balloons have a net upward force of NFb.

To lift the apple off the ground requires that this upward force overcomes the weight of theapple, mag, i.e.

NFb ≥ mag.

N ≥ mag

Fb

=≥ 0.1× 9.81

0.0795= 12.34.

Thereforce, 13 balloons are required.

(c) In fact, air density is not constant. Consider it to vary with altitude z as ρa(z) = ρ0

(1− z

h

),

where h is a constant. Taking into account that the density is not constant, obtain anexpression for the buoyancy force on a balloon whose centre is at altitude z0.

The buoyancy force is ~FB = M0gz. M0 is the mass of the displaced fluid, which we calculateas,

M0 =

∫∫∫ρ dV =

∫∫ρa(z) dx dy dz =

∫dx

∫dy

∫ρ0

(1− z

h

)dz

= L2ρ0gz

z0+L/2∫z0−L/2

(1− z

h

)dz = L2ρ0gz

[z − z2

2h2

]z0+L/2

z0−L/2= L3ρ0gz

(1− z0

h

)

(d) Hence determine an expression for the height to which the balloon will rise, assuming thatit maintains the same shape and size. How high would the balloon reach, for the parametersgiven above and taking h = 10km?

The balloon rises because the buoyancy force (acting up) overcomes the balloon weight(acting down). The balloon will stop when the magnitude of these forces are equal, i.e.

gρ0 L3(

1− z0h

)= ρbgL

3.

Cancelling terms and rearranging gives the maximum height to be,

z0 = h

(1− ρb

ρ0

).

Using h = 10 × 103m and the parameters from part (a), i.e. that L = 0.3m, ρb = 0.9 kgm−3 and ρ0 = 1.2 kg m−3, we obtain,

z0 = 2.5× 103m = 2.5km

∗Cubic balloons might not be in a shop near you but they are much easier to deal with mathematically, and give similarresults to the usual balloons.