Chapter 9: Mechanical Properties of Fluids (Properties of Bulk Matter)

(a) Pressure in a fluid increases with depth ($p=p_0+\rho gh$). The column of blood above the feet is taller than that above the brain, so the pressure at the feet is greater.

(b) Air is compressible: its density is greatest near the ground and falls rapidly with height. Most of the mass of the atmosphere lies in the lowest few kilometres, so at about 6 km roughly half of the air is already below, and the pressure is about half.

(c) At a point in a fluid the pressure is the same in every direction, and the force it exerts on any surface is always normal to that surface. No particular direction is attached to pressure itself, so it is a scalar.

(a) Greater depth of blood above the feet (b) air density falls rapidly with height (c) pressure acts equally in all directions, so it has no direction.

(a) For water and glass, adhesion (water–glass) is stronger than cohesion (water–water), so water clings to glass and the angle of contact is acute. For mercury, cohesion is much stronger than adhesion, so the surface curves away and the angle is obtuse.

(b) For the same reason: $S_{sa}>S_{sl}$ for water, so spreading lowers the energy and water spreads; for mercury, forming drops (less contact with glass) lowers the energy.

(c) Surface tension is force per unit length (or energy per unit area) along any line on the surface; it depends on the nature of the liquid and its temperature, not on how big the surface is.

(d) Detergent lowers the surface tension of water against oil and grease, making $\cos \theta$ larger. A small angle of contact means water wets the fabric well and penetrates it, which helps washing.

(e) With no external force, surface tension makes the liquid take the shape of least surface energy, i.e. least area for a given volume, which is a sphere.

(a) Adhesion vs cohesion (b) water wets glass, mercury does not (c) T is a property per unit length (d) better wetting (e) a sphere has the least surface area.

(a) Intermolecular forces weaken as temperature rises. (b) In gases, faster molecules carry more momentum between layers; in liquids, cohesive forces weaken. (c) Solids resist deformation; fluids resist the rate of deformation. (d) $A_1{v}_1=A_2{v}_2$ follows from conservation of mass. (e) Turbulence sets in at a definite Reynolds number $\frac{\rho vd}{\eta }$; a smaller model (smaller d) needs a greater speed.

(a) decreases (b) increases; decreases (c) shear strain; rate of shear strain (d) conservation of mass (e) greater

(a) Blowing over the paper makes the air above move faster, so by Bernoulli's principle the pressure above falls. The higher pressure below holds the paper up.

(b) Closing the tap reduces the area of the outlet. By the equation of continuity ($Av$ = constant), the water must come out through the small gaps at a much higher speed.

(c) By Poiseuille's equation the flow rate depends on $r^4$ but only on the first power of the pressure. A small change in needle radius changes the flow far more than any change of thumb pressure.

(d) The fluid leaving the hole carries momentum forwards. By conservation of momentum (third law), the vessel receives equal momentum in the opposite direction: a backward thrust.

(e) A spinning ball drags air around with it, so the air speed is greater on one side than the other. The pressure difference (Magnus effect) gives a sideways force, so the path is not a simple parabola.

(a) Faster air above, lower pressure (b) continuity (c) flow ∝ r⁴ (d) momentum conservation (e) Magnus effect.

$F=50\times 9.8=490$ N; $A=\pi (0.5\times 10^{-2}{)}^2=7.85\times 10^{-5}{\text{m}}^2$.

$p=\frac{F}{A}=\frac{490}{7.85\times 10^{-5}}$

6.2 × 10⁶ Pa

$p_a=\rho gh\Rightarrow h=\frac{1.013\times 10^5}{984\times 9.8}=\frac{1.013\times 10^5}{9643}$

10.5 m

Pressure at the sea bed $=\rho gh=1.03\times 10^3\times 9.8\times 3000\approx 3\times 10^7$ Pa (plus 1 atm, negligible).

This is much less than $10^9$ Pa, so the structure can bear it.

Yes; the pressure (≈ 3 × 10⁷ Pa) is far below 10⁹ Pa.

$p=\frac{F}{A}=\frac{3000\times 9.8}{425\times 10^{-4}}=\frac{29400}{0.0425}$. By Pascal's law the same pressure acts on the smaller piston.

6.92 × 10⁵ Pa

The mercury levels are equal, so the pressures of the two columns above them are equal:

${\rho }_{w}g{h}_w={\rho }_{s}g{h}_s\Rightarrow \frac{{\rho }_s}{{\rho }_w}=\frac{h_w}{h_s}=\frac{10.0}{12.5}$

0.800

Now: water 25.0 cm, spirit 27.5 cm. In cm of water: water column $=25.0$, spirit column $=27.5\times 0.8=22.0$.

The water side presses harder by 3.0 cm of water, so mercury is pushed up in the spirit arm by h with $13.6h=3.0$.

$h=\frac{3.0}{13.6}=0.221$ cm.

0.221 cm, mercury higher in the spirit arm

No. Bernoulli's equation holds only for steady, streamline flow of a non-viscous fluid. The flow in a rapid is turbulent.

No; the flow is turbulent, not streamline.

No, as long as the atmospheric pressure is the same at the two points compared. Gauge pressure = absolute pressure − atmospheric pressure; the same constant is subtracted on both sides of Bernoulli's equation and cancels.

No, unless the atmospheric pressures at the two points differ significantly.

Volume flow $Q=\frac{4.0\times 10^{-3}}{1.3\times 10^3}=3.08\times 10^{-6}{\text{m}}^3{\text{s}}^{-1}$.

Poiseuille: $p=\frac{8\eta lQ}{\pi r^4}=\frac{8\times 0.83\times 1.5\times 3.08\times 10^{-6}}{3.14\times (10^{-2}{)}^4}=9.8\times 10^2$ Pa.

Check: $v=\frac{Q}{\pi r^2}=9.8\times 10^{-3}{\text{m s}}^{-1}$; $R_e=\frac{\rho vd}{\eta }=\frac{1.3\times 10^3\times 9.8\times 10^{-3}\times 0.02}{0.83}=0.3$, far below about 2000, so the flow is laminar.

9.8 × 10² Pa (Reynolds number ≈ 0.3: laminar)

Bernoulli (wing thin, same height): $p_{\text{below}}-p_{\text{above}}=\frac{1}{2}\rho (v_1^2-v_2^2)=\frac{1}{2}\times 1.3\times (4900-3969)=605$ Pa.

Lift $=605\times 2.5=1513$ N.

About 1.5 × 10³ N

At the constriction the area is smaller, so by continuity the speed is larger, and by Bernoulli's principle the pressure is smaller. The pressure column there must be lower, as in (b).

Fig (a) is incorrect: pressure must be lower at the constriction.

$A_1=8.0\times 10^{-4}{\text{m}}^2$, $v_1=\frac{1.5}{60}=0.025{\text{m s}}^{-1}$.

Total hole area $A_2=40\times \pi (0.5\times 10^{-3}{)}^2=3.14\times 10^{-5}{\text{m}}^2$.

$v_2=\frac{A_1{v}_1}{A_2}=\frac{8.0\times 10^{-4}\times 0.025}{3.14\times 10^{-5}}$

0.64 m s⁻¹

The film has two surfaces, so the upward force is $2Tl$: $2T\times 0.30=1.5\times 10^{-2}$.

T = 2.5 × 10⁻² N m⁻¹

The force of surface tension on the slider is $2Tl$; it depends only on the length l of the slider and on T, not on the area or shape of the film. All three sliders are 40 cm long and the liquid and temperature are the same.

4.5 × 10⁻² N in both (b) and (c)

Excess pressure $=\frac{2T}{r}=\frac{2\times 0.465}{3.00\times 10^{-3}}=310$ Pa.

Total $=1.01\times 10^5+310=1.0131\times 10^5$ Pa, i.e. $1.01\times 10^5$ Pa to three significant figures.

Excess 310 Pa; inside 1.01 × 10⁵ Pa

Soap bubble (two surfaces): $\frac{4T}{r}=\frac{4\times 0.025}{5.00\times 10^{-3}}=20.0$ Pa.

Air bubble in the solution (one surface): excess $=\frac{2T}{r}=10.0$ Pa.

Pressure outside it $=1.01\times 10^5+0.40\times 1200\times 9.8=1.01\times 10^5+4704=1.057\times 10^5$ Pa. Inside $=1.057\times 10^5+10\approx 1.06\times 10^5$ Pa.

Soap bubble: 20.0 Pa excess; air bubble: 10.0 Pa excess, pressure inside ≈ 1.06 × 10⁵ Pa