Unit 13: Liquids

The pressure on an object at a specific point depends on its depth below the surface. The horizontal forces on a submerged object cancel out because every horizontal force that acts on one side of the object is countered by a force of equal magnitude that pushes on the opposite side at the same depth.

60 kPa

Water has a mass density of $1000\,\text{kg}/\text{m}^3$. In terms of weight, the density of water is expressed as \begin{align} \text{Weight density} &= \text{mass density}\times g \nonumber\\[6pt] &= 1000\,\text{kg}/\text{m}^3\times 10\,\text{m}/\text{s}^2 \nonumber\\[6pt] &= 10{,}000\,\text{N}/\text{m}^3 \nonumber\\[6pt] &= 10\,\text{kN}/\text{m}^3 \end{align} For the barrel filled with water, you can calculate the pressure near the bottom (at 1 m depth) as \begin{align} \text{Water pressure}\, (1\,\text{m}) &= \text{weight density} \times \text{depth} \nonumber\\[6pt] &= 10\,\text{kN}/\text{m}^3 \times 1\,\text{m} \nonumber\\[6pt] &= 10\,\text{kN}/\text{m}^2 \nonumber\\[6pt] &= 10\,\text{kPa} \end{align} When the thin pipe extending from the barrel is filled with water (see Figure Q13.38), the pressure at the bottom of the barrel increases: \begin{align} \text{Water pressure}\, (6\,\text{m}) &= \text{weight density} \times \text{depth} \nonumber\\[6pt] &= 10\,\text{kN}/\text{m}^3 \times (5\,\text{m} + 1\,\text{m}) \nonumber\\[6pt] &= 10\,\text{kN}/\text{m}^3 \times 6\,\text{m} \nonumber\\[6pt] &= 60\,\text{kPa} \end{align} Since liquid pressure depends on vertical location only, all points at the bottom of the barrel experience a pressure of 60 kPa.

Nine times the force.

When comparing the area of the larger piston (radius 3 cm) with that of the smaller piston (radius 1 cm), you obtain the following ratio: \begin{align} \frac{\text{Area of larger piston}}{\text{Area of smaller piston}} &= \frac{\pi\, (3\,\text{cm})^2}{\pi\, (1\,\text{cm})^2} \nonumber\\[6pt] &= \frac{9}{1} \nonumber\\[6pt] &= 9 \end{align} Since the area of the larger piston is nine times the area of the smaller piston, a 1-N input force applied to the smaller piston results in an output force of 9 N exerted by the larger piston.

c, a, b

The pressure at a certain depth below the surface of a liquid is calculated using the formula \begin{equation} \text{Liquid pressure} = \text{weight density} \times \text{depth} \end{equation} where the weight density refers to the weight per unit volume of the liquid. Table 12.1 in the eText gives the density (i.e., mass per unit volume) for each of the three liquids. The corresponding weight densities are \begin{align} 1025\,\text{kg}/\text{m}^3 \times g &= 10{,}250\,\text{N}/\text{m}^3 \quad \text{for saltwater} \nonumber\\[6pt] 1000\,\text{kg}/\text{m}^3 \times g &= 10{,}000\,\text{N}/\text{m}^3 \quad\text{for freshwater}\nonumber\\[6pt] 13{,}600\,\text{kg}/\text{m}^3 \times g &= 136{,}000\,\text{N}/\text{m}^3 \quad\text{for mercury} \\[6pt] \end{align} where the acceleration due to gravity $g = 10\,\text{m}/\text{s}^2$.

1. At a 20-cm depth in saltwater, \begin{align} \text{Liquid pressure} &= 10{,}250\,\text{N}/\text{m}^3 \times 0.2\,\text{m} \nonumber\\[6pt] &= 2050\,\text{N}/\text{m}^2 \end{align}
2. At a 20-cm depth in freshwater, \begin{align} \text{Liquid pressure} &= 10{,}000\,\text{N}/\text{m}^3 \times 0.2\,\text{m} \nonumber\\[6pt] &= 2000\,\text{N}/\text{m}^2 \end{align}
3. At a 10-cm depth in mercury, \begin{align} \text{Liquid pressure} &= 136{,}000\,\text{N}/\text{m}^3 \times 0.1\,\text{m} \nonumber\\[6pt] &= 13{,}600\,\text{N}/\text{m}^2 \end{align}

The block of lead displaces the same amount of water as the block of aluminum of equal volume.

Aluminum and lead have densities greater than that of water (see Table 12.1 in the eText). Therefore, each block sinks to the bottom of the beaker and becomes entirely submerged. Since both blocks have the same volume, they displace equal amounts of water (10 cm³).

The block of lead displaces less water than the block of aluminum of equal mass.

In Table 12.1 in the eText, you see the density of lead is approximately four times the density of aluminum. Therefore, an aluminum block is four times larger in size (or volume) than a block of lead of equal mass.

Note that the volume of a 1-kg block of lead is calculated as follows: \begin{align} \text{Volume of lead block} &= \frac{\text{mass}}{\text{density}} \nonumber\\[6pt] &= \frac{1\,\text{kg}}{11{,}340\,\text{kg}/\text{m}^3} \nonumber\\[6pt] &= \frac{1000\,\text{g}}{11.34\,\text{g}/\text{cm}^3} \nonumber\\[6pt] &= 88\,\text{cm}^3 \end{align} For a 1-kg block of aluminum, \begin{align} \text{Volume of aluminum block} &= \frac{1000\,\text{g}}{2.7\,\text{g}/\text{cm}^3} \nonumber\\[6pt] &= 370\,\text{cm}^3 \end{align}

Consider the effect of the sugar content in the regular soda.

Regular soda is denser than water due to its sugar content. A can of regular soda normally has 39 g of sugar dissolved in it, causing the density of the soda to be greater than that of pure water. Therefore, the can sinks in water.

To make diet soda, a relatively tiny amount of sweetener (e.g., aspartame) is dissolved in the liquid, which ends up being slightly less dense than water. Therefore, the can of diet soda floats on water.

You can also say that a can of regular soda sinks because its weight is slightly greater than the buoyant force, while the weight of a can of diet soda is slightly less than the buoyant force.

10 N

The buoyant force exerted on a 1-L carton submerged under water is equal to the weight of the displaced water. Since 1 L of water has a mass of 1 kg, you can write \begin{align} \text{Buoyant force} &= \text{weight of displaced water} \nonumber\\[6pt] &= 1\,\text{kg}\times 10\,\text{m}/\text{s}^2 \nonumber\\[6pt] &= 10\,\text{N} \end{align} Given that the carton has negligible weight, you will need to push down with a $10$-$\text{N}$ force to hold the carton beneath the water surface.

The weight of the displaced water is greater for the submerged ball.

The buoyant force that pushes upward on an object is equal to the weight of the displaced water. A volleyball held beneath the surface displaces more water than if it is floating on the surface (or partially submerged).

The buoyant force remains the same.

The buoyant force required to keep the ship floating is equal to its weight. The buoyant force is also equal to the weight of displaced water. Since freshwater is slightly less dense than seawater, the ship sinks slightly deeper, displacing more freshwater and maintaining the same buoyant force.

The weight of the half-filled aquarium increases when a fish is placed inside but remains unchanged if the aquarium is initially filled to the brim.

When the aquarium is half-filled with water, the added fish causes the reading of the scale to increase by an amount equal to the weight of the fish. Note that the amount of water in the aquarium does not change in this case.

The reading of the scale remains unchanged if the fish is placed in the aquarium when it is filled with water to the brim. This is because the density of the fish is equal to that of water and, therefore, the weight of the water that spills over the brim is equal to the weight of the fish.

He will not be successful.

Since the area of the larger piston is 50 times that of the smaller piston, the strong man needs to exert a force that is at least 50 times the weight of the 10-kg object. This is impossible in the arrangement shown because the strong man cannot apply a downward force on the piston of more than his weight, which is less than 500 kg.

Normally, a hydraulic system is arranged so the input force is applied against the smaller piston and the output force is exerted by the wider piston. The arrangement in this question is reversed.

Hot water has less surface tension.

The molecules in hot water move more rapidly and do not adhere to one another as well as the molecules in cold water. As a result, hot water has lower surface tension and leaks more easily through thin cracks and small openings in a radiator.

The wood would float higher.

Since the block of wood and the piece of iron have the same weight in both situations, equal amounts of water are displaced to support their combined weight. As shown in Figure Q13.106, when the piece of iron is placed on top of the wood block, water is displaced by only the submerged portion of the block. However, when the piece of iron is suspended underneath the wood, the submerged iron displaces an amount of water equal to its volume, leaving less water to be displaced by the wood block.

The water level would go down.

The amount of water displaced by a floating ship is equal to its weight. However, if a battleship sinks, the water displaced has the same volume as the material (predominantly steel) that makes up the ship. Since the density of steel is eight times that of water, the weight of the displaced water is one-eighth the weight of the submerged battleship.

Stay at the same depth.

If the gravitational field of Earth were to increase (i.e., $g$ becomes larger), the weight of the fish would increase. However, the weight density of water and the water pressure would also increase by equal factors. Therefore, the fish would remain at the same depth.