### Check Your Understanding

The pressure found in part (a) of the example is completely independent of the width and length of the lake; it depends only on its average depth at the dam. Thus, the force depends only on the water’s average depth and the dimensions of the dam, not on the horizontal extent of the reservoir. In the diagram, note that the thickness of the dam increases with depth to balance the increasing force due to the increasing pressure.

The density of mercury is 13.6 times greater than the density of water. It takes approximately 76 cm (29.9 in.) of mercury to measure the pressure of the atmosphere, whereas it would take approximately 10 m (34 ft.) of water.

Yes, it would still work, but since a gas is compressible, it would not operate as efficiently. When the force is applied, the gas would first compress and warm. Hence, the air in the brake lines must be bled out in order for the brakes to work properly.

### Conceptual Questions

Mercury and water are liquid at room temperature and atmospheric pressure. Air is a gas at room temperature and atmospheric pressure. Glass is an amorphous solid (non-crystalline) material at room temperature and atmospheric pressure. At one time, it was thought that glass flowed, but flowed very slowly. This theory came from the observation that old glass planes were thicker at the bottom. It is now thought unlikely that this theory is accurate.

The density of air decreases with altitude. For a column of air of a constant temperature, the density decreases exponentially with altitude. This is a fair approximation, but since the temperature does change with altitude, it is only an approximation.

Pressure is force divided by area. If a knife is sharp, the force applied to the cutting surface is divided over a smaller area than the same force applied with a dull knife. This means that the pressure would be greater for the sharper knife, increasing its ability to cut.

If the two chunks of ice had the same volume, they would produce the same volume of water. The glacier would cause the greatest rise in the lake, however, because part of the floating chunk of ice is already submerged in the lake, and is thus already contributing to the lake’s level.

Because the river level is very high, it has started to leak under the levee. Sandbags are placed around the leak, and the water held by them rises until it is the same level as the river, at which point the water there stops rising. The sandbags will absorb water until the water reaches the height of the water in the levee.

Atmospheric pressure does not affect the gas pressure in a rigid tank, but it does affect the pressure inside a balloon. In general, atmospheric pressure affects fluid pressure unless the fluid is enclosed in a rigid container.

The pressure of the atmosphere is due to the weight of the air above. The pressure, force per area, on the manometer will be the same at the same depth of the atmosphere.

Not at all. Pascal’s principle says that the change in the pressure is exerted through the fluid. The reason that the full tub requires more force to pull the plug is because of the weight of the water above the plug.

The buoyant force is equal to the weight of the fluid displaced. The greater the density of the fluid, the less fluid that is needed to be displaced to have the weight of the object be supported and to float. Since the density of salt water is higher than that of fresh water, less salt water will be displaced, and the ship will float higher.

Consider two different pipes connected to a single pipe of a smaller diameter, with fluid flowing from the two pipes into the smaller pipe. Since the fluid is forced through a smaller cross-sectional area, it must move faster as the flow lines become closer together. Likewise, if a pipe with a large radius feeds into a pipe with a small radius, the stream lines will become closer together and the fluid will move faster.

The mass of water that enters a cross-sectional area must equal the amount that leaves. From the continuity equation, we know that the density times the area times the velocity must remain constant. Since the density of the water does not change, the velocity times the cross-sectional area entering a region must equal the cross-sectional area times the velocity leaving the region. Since the velocity of the fountain stream decreases as it rises due to gravity, the area must increase. Since the velocity of the faucet stream speeds up as it falls, the area must decrease.

When the tube narrows, the fluid is forced to speed up, thanks to the continuity equation and the work done on the fluid. Where the tube is narrow, the pressure decreases. This means that the entrained fluid will be pushed into the narrow area.

The work done by pressure can be used to increase the kinetic energy and to gain potential energy. As the height becomes larger, there is less energy left to give to kinetic energy. Eventually, there will be a maximum height that cannot be overcome.

Because of the speed of the air outside the building, the pressure outside the house decreases. The greater pressure inside the building can essentially blow off the roof or cause the building to explode.

The air inside the hose has kinetic energy due to its motion. The kinetic energy can be used to do work against the pressure difference.

Potential energy due to position, kinetic energy due to velocity, and the work done by a pressure difference.

The water has kinetic energy due to its motion. This energy can be converted into work against the difference in pressure.

The water in the center of the stream is moving faster than the water near the shore due to resistance between the water and the shore and between the layers of fluid. There is also probably more turbulence near the shore, which will also slow the water down. When paddling up stream, the water pushes against the canoe, so it is better to stay near the shore to minimize the force pushing against the canoe. When moving downstream, the water pushes the canoe, increasing its velocity, so it is better to stay in the middle of the stream to maximize this effect.

You would expect the speed to be slower after the obstruction. Resistance is increased due to the reduction in size of the opening, and turbulence will be created because of the obstruction, both of which will clause the fluid to slow down.

### Problems

The mass is 2.58 g. The volume of your body increases by the volume of air you inhale. The average density of your body decreases when you take a deep breath because the density of air is substantially smaller than the average density of the body.

$0.760\phantom{\rule{0.2em}{0ex}}\text{m}=76.0\phantom{\rule{0.2em}{0ex}}\text{cm}=760\phantom{\rule{0.2em}{0ex}}\text{mm}$

a. Pressure at $h=7.06\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{6}\text{N}$;

b. The pressure increases as the depth increases, so the dam must be built thicker toward the bottom to withstand the greater pressure.

$5.76\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{3}\phantom{\rule{0.2em}{0ex}}\text{N}\phantom{\rule{0.2em}{0ex}}\text{extra force}$

If the system is not moving, the friction would not play a role. With friction, we know there are losses, so that ${W}_{\text{o}}={W}_{\text{i}}-{W}_{\text{f}};$ therefore, the work output is less than the work input. In other words, to account for friction, you would need to push harder on the input piston than was calculated.

a. 39.5 g; b. $50\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{3}$; c. $0.79\phantom{\rule{0.2em}{0ex}}{\text{g/cm}}^{3}$; ethyl alcohol

$\begin{array}{ccc}\text{net}\phantom{\rule{0.2em}{0ex}}F\hfill & =\hfill & {F}_{2}-{F}_{1}={p}_{2}A-{p}_{1}A=({p}_{2}-{p}_{1})A=\left({h}_{2}{\rho}_{\text{fl}}g-{h}_{1}{\rho}_{\text{fl}}g\right)A\hfill \\ & =\hfill & ({h}_{2}-{h}_{1}){\rho}_{\text{fl}}gA,\phantom{\rule{0.2em}{0ex}}\text{where}\phantom{\rule{0.2em}{0ex}}{\rho}_{\text{fl}}=\text{density of fluid}\text{.}\hfill \\ \text{net}\phantom{\rule{0.2em}{0ex}}F\hfill & =\hfill & ({h}_{2}-{h}_{1})A{\rho}_{\text{fl}}g={V}_{\text{fl}}{\rho}_{\text{fl}}g={m}_{\text{fl}}g={w}_{\text{fl}}\hfill \end{array}$

a. 12.6 m/s; b. $0.0800\phantom{\rule{0.2em}{0ex}}{\text{m}}^{3}\text{/s}$; c. No, the flow rate and the velocity are independent of the density of the fluid.

If the fluid is incompressible, the flow rate through both sides will be equal:

$Q={A}_{1}{\stackrel{\text{\u2013}}{v}}_{1}={A}_{2}{\stackrel{\text{\u2013}}{v}}_{2},$ or $\pi \frac{{d}_{1}^{2}}{4}{\stackrel{\text{\u2013}}{v}}_{1}=\pi \frac{{d}_{2}^{2}}{4}{\stackrel{\text{\u2013}}{v}}_{2}\Rightarrow {\stackrel{\text{\u2013}}{v}}_{2}={\stackrel{\text{\u2013}}{v}}_{1}({d}_{1}^{2}\text{/}{d}_{2}^{2})={\stackrel{\text{\u2013}}{v}}_{1}{({d}_{1}\text{/}{d}_{2})}^{2}$

$\begin{array}{}\\ \\ \hfill F& =\hfill & pA\Rightarrow p=\frac{F}{A},\hfill \\ \hfill \left[p\right]& =\hfill & {\text{N/m}}^{2}=\text{N}\xb7{\text{m/m}}^{3}={\text{J/m}}^{3}=\text{energy/volume}\hfill \end{array}$

a. $1.58\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}{\text{N/m}}^{2}$; b. 163 m

a. ${v}_{2}=3.28\frac{\text{m}}{\text{s}}$;

b. $t=0.55\phantom{\rule{0.2em}{0ex}}\text{s}$

$x=vt=1.81\phantom{\rule{0.2em}{0ex}}\text{m}$

a. $3.02\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\text{\u2212}3}\phantom{\rule{0.2em}{0ex}}\text{N}$; b. $1.03\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\text{\u2212}3}$

a. $2.40\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{9}\phantom{\rule{0.2em}{0ex}}\text{N}\cdot {\text{s/m}}^{5}$; b. $48.3\phantom{\rule{0.2em}{0ex}}({\text{N/m}}^{2})\cdot \text{s}$; c. $2.67\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{W}$

a. Nozzle: $v=25.5\frac{\text{m}}{\text{s}}$

${N}_{\text{R}}=1.27\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{5}>2000\Rightarrow $

Flow is not laminar.

b. Hose: $v=1.96\frac{\text{m}}{\text{s}}$

${N}_{\text{R}}=\mathrm{35,100}>2000\Rightarrow $

Flow is not laminar.

$3.16\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\text{\u2212}4}\phantom{\rule{0.2em}{0ex}}{\text{m}}^{3}\text{/s}$

### Additional Problems

a. $\begin{array}{}\\ \\ {p}_{120}=1.60\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}{\text{N/m}}^{2}\hfill \\ {p}_{80}=1.07\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}{\text{N/m}}^{2}\hfill \end{array}$;

b. Since an infant is only approximately 20 inches tall, while an adult is approximately 70 inches tall, the blood pressure for an infant would be expected to be smaller than that of an adult. The blood only feels a pressure of 20 inches rather than 70 inches, so the pressure should be smaller.

a. 41.4 g; b. 41.4 cm^{3}; c. 1.09 g/cm^{3}. This is clearly not the density of the bone everywhere. The air pockets will have a density of approximately $1.29\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-3}}\phantom{\rule{0.2em}{0ex}}{\text{g/cm}}^{3}$, while the bone will be substantially denser.

a. $3.02\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-2}}\phantom{\rule{0.2em}{0ex}}\text{cm/s}$. (This small speed allows time for diffusion of materials to and from the blood.) b. $2.37\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{10}$ capillaries. (This large number is an overestimate, but it is still reasonable.)

a. $2.76\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}{\text{N/m}}^{2}$; b. ${P}_{2}=2.81\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}{\text{N/m}}^{2}$

$8.7\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{\mathrm{-2}}\phantom{\rule{0.2em}{0ex}}{\text{mm}}^{3}\text{/s}$

a. 1.52; b. Turbulence would decrease the flow rate of the blood, which would require an even larger increase in the pressure difference, leading to higher blood pressure.

### Challenge Problems

$p=0.99\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}\text{Pa}$

a. $150\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{3}\text{/s}$; b. $33.3\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{3}\text{/s}$; c. $25.0\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{3}\text{/s}$; d. $0.0100\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{3}\text{/s}$; e. $0.0300\phantom{\rule{0.2em}{0ex}}{\text{cm}}^{3}\text{/s}$

a. $1.20\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{5}\phantom{\rule{0.2em}{0ex}}{\text{N/m}}^{2}$; b. The flow rate in the main increases by 90%. c. There are approximately 38 more users in the afternoon.