SAT Subject Test Physics (2012)
PART III. PHYSICS TOPIC REVIEW
Chapter 6. FLUID MECHANICS
Substances that can flow easily are known as fluids. Gases and liquids are fluids. Unlike solids, fluids do not have fixed shapes. Both liquids and gases adjust to the shape of their container. In addition, a gas expands to fill its container. While liquids have a definite volume, gases do not. You may encounter a few questions on SAT Physics that involve the properties of fluids, so it is best to understand them and recognize changes in them.
Pressure in a Liquid
The particles in a fluid are in constant motion. They collide with each other and the walls of their container. Pressure is the force they exert per unit area.
The pressure of a fluid depends on depth. Pressure increases with depth. Therefore, an object placed in a fluid experiences greater pressure as it is moved deeper within the fluid because the weight of the fluid above it increases. The weight of the fluid is related to the density of the fluid. Density is the mass per unit volume.
The denser the fluid above a submerged object is, the more pressure there is exerted on the object. The pressure can therefore be determined by finding the product of the height of the liquid (h), the density (d), and the acceleration due to gravity (g). [Note: Density is often represented by the Greek letter rho,ρ, instead of d.]
If a container of liquid is open, the atmosphere above it exerts additional pressure on the liquid. This added pressure must be included when considering the total pressure of the fluid.
The SI unit of pressure is the pascal (Pa). This unit is equivalent to 1 N/m2.
Example:
A biologist is studying a school of fish 11.0 meters below the surface of the ocean. If the density of sea water is 1.03 × 103 kg/m3 and atmospheric pressure is 1.01 × 105 N/m2, what is the pressure on the biologist?
Gas Pressure
Like the particles in a liquid, the particles of a gas are also in constant motion. In fact, they move at greater speeds and travel farther apart. As a result, gases also exert pressure. As with fluids, gas pressure depends on the weight of the gas above a given point. The higher you go in the atmosphere, for example, the less air there is above you. Therefore, atmospheric pressure decreases with altitude. Air is densest near Earth’s surface, which explains why air pressure is greatest there.
Atmospheric pressure can be measured with a mercury barometer. This device is made by inverting a glass tube from which the air has been removed into a reservoir of mercury. Air will push downward on the mercury in the reservoir, forcing some mercury into the tube. The height of the mercury in the tube will match the pressure of the atmosphere that supports it.
Atmospheric pressure varies with location, altitude, and weather conditions. Standard pressure is the pressure that supports a column of mercury to a height of 760 millimeters. You can use the pressure equation to calculate standard atmospheric pressure. The calculation becomes easier if you convert the height to 76.0 centimeters.
Note that when you are working with grams and centimeters, the resulting unit of power involves dynes. The dyne is an older unit of force, which can be converted as shown.
Pascal’s Principle
Suppose you apply pressure to a confined fluid. That pressure will be transmitted undiminished to every part of the fluid. This phenomenon is known as Pascal’s Principle, which, stated more formally, explains that an increase in pressure to any point of a confined fluid is transmitted equally to every other point in the fluid.
Consider the fluid in the tube shown in the following diagram. If the pressure exerted on the top of the fluid is increased by 3 units, that increase will be observed on all three gauges. No part of the fluid will experience a greater increase than any other part.
Pascal’s Principle is commonly utilized in hydraulic devices similar to the one shown in the following diagram. A small force F1 is applied to the piston with the small area A1. The increase in pressure is transmitted throughout the fluid and therefore acts on the bottom of the larger piston. Because the pressure is exerted over a larger area A2, the resulting force F2is greater than F1. If the device is designed properly, it can be used to magnify a force in such a way that a heavy load, such as a car, can be lifted.
Notice that d1 is greater than d2. The reason is that the hydraulic device can be considered to be a machine that does work. The smaller force must be exerted over a longer distance to produce a greater force exerted over a shorter distance. Think of F1as the effort and F2as the resistance. The mechanical advantage is then the ratio of the resistance to the effort.
Example:
A hydraulic press has an input cylinder with an area of 4 square centimeters and an output cylinder with an area of 40 square centimeters. If a force of 5 newtons is applied to the piston in the small cylinder, how large a weight can be supported by the piston in the larger cylinder?
Archimedes’ Principle
An object submerged in a fluid appears to experience a loss in weight. The reason is that an upward force known as the buoyant force acts on the object. According to Archimedes’ Principle, the magnitude of the buoyant force is equal to the weight of the fluid displaced by the object. Keep in mind that the buoyant force does not depend on the weight of the submerged object or even its shape. It depends only on the weight of the displaced fluid.
According to the following diagram, the object weighs less in the liquid than in air because of the buoyant force. As you can see, the difference in weight equals the weight of the liquid that spills out of the container when the object is submerged.
The buoyant force results from the pressure of the fluid. Because pressure increases with depth, the pressure acting on the top of a submerged object is less than the pressure acting on the bottom of the object. Therefore, the force pushing downward on the top of the object is less than the force pushing upward on the bottom of the object. The net force is the buoyant force.
The buoyant force depends on the volume of the object submerged because this is the volume of fluid that will be displaced. Therefore, spheres of different substances that are submerged in water will all displace the same volume of water and therefore experience the same buoyant force. Suppose, for example, spheres of lead, aluminum, and cork are all submerged in water. The spheres have identical volumes. The buoyant force exerted on each sphere will be the same. The difference is that once set free, the lead will quickly sink to the bottom. The aluminum will sink as well, and the cork will rise to the surface. The buoyant force acts opposite to the weight of the sphere. If the weight of the sphere is greater than the buoyant force, as in the cases of lead and aluminum, the sphere will sink. If the weight of the sphere is less than the buoyant force, as in the case of cork, the sphere will float.
Another way to state this relationship is that an object will float in a fluid denser than itself and sink in a fluid that is less dense than itself. The volume of the object is equal to the volume of the displaced fluid. If the object floats, its weight must be less than the weight of the displaced fluid. Weight equals the product of the mass of a sample and the acceleration due to gravity. Because the acceleration due to gravity is the same for the object and the displaced fluid, the mass of the displaced fluid must be greater than the mass of the object. Density is mass divided by volume, so a greater mass divided by the same volume results in a greater density. The fluid must be denser than the object. The same logic explains why an object sinks.
Bernoulli’s Principle
An airplane wing experiences lift because the air moving across the top of the wing exerts less pressure than the air moving across the bottom of the wing. The design of a wing is such that air moves faster across the top than across the bottom. Bernoulli’s Principle states that the pressure of a fluid decreases as the velocity of the fluid increases. In other words, a slow-moving fluid exerts more pressure than a fast-moving fluid.
In addition to airplanes, Bernoulli’s Principle explains why a shower curtain billows inward. The flow of water causes air around it to move faster than the air outside the shower curtain. The faster-moving air creates a region of lower pressure, allowing the higher pressure outside the shower curtain to push it inward.
Bernoulli’s Principle also explains how pitchers can cause baseballs to curve. Suppose a ball is thrown from right to left with a clockwise spin. As the ball spins, it drags air with it. As a result, the air flowing over the ball is opposite to the spin, whereas the air flowing under the ball is in the same direction as the spin. Because the air below the ball moves faster than the air above it, the net pressure is downward. Depending on how the pitcher spins the ball, curves in different directions can be achieved.
Test-Taking Hint
Be sure to differentiate between individual quantities and net quantities. You might quickly identify a force on an object, such as the buoyant force on an object submerged in a fluid, but you cannot ignore other forces acting on the same object, such as its downward weight. Only the net force can be used to determine the resulting motion of the object. You can often compare individual quantities by combining arrows used in diagrams.
REVIEW QUESTIONS
Select the choice that best answers the question or completes the statement.
1. A scientist is investigating conditions along the bottom of a lake. The lake is 100.0 meters deep. What is the pressure due to the water at the bottom of the lake? 
(A) 1.14 × 103Pa
(B) 1.03 × 105Pa
(C) 9.80 × 105Pa
(D) 1.18 × 106Pa
(E) 9.80 × 106Pa
2. What is the total pressure at the bottom of the lake described in question 1? 
(A) 2.04 × 103Pa
(B) 8.79 × 105Pa
(C) 9.80 × 105Pa
(D) 9.89 × 105Pa
(E) 1.08 × 103kPa
3. What is the total pressure on a diver 42.0 meters below sea level? [The density of sea water is 1.025 × 103 kg/m3.]
(A) 422 kPa
(B) 523 kPa
(C) 565 kPa
(D) 621 kPa
(E) 693 kPa
4. The input piston of a hydraulic system has an area of 0.10 m2 and the output piston has an area of 0.40 m2. If the input force is 1500 N, what is the output force?
(A) 2500 N
(B) 3000 N
(C) 3750 N
(D) 4500 N
(E) 6000 N
5. The input piston of a hydraulic device moves a distance of 0.6 m when a force of 1200 N is exerted on it. The result is a force of 3600 N on the output piston. What distance does the output piston move?
(A) 0.2 m
(B) 0.3 m
(C) 0.4 m
(D) 0.6 m
(E) 0.8 m
6. A force of 8 N is exerted on the stopper of a container as shown. If the area of the stopper is 4 cm2 and the area of the bottom of the jug is 400 cm2, what is the increase in force on the bottom of the jug?
(A) 100 N
(B) 200 N
(C) 600 N
(D) 800 N
(E) 1600 N
7. Due to buoyant forces, an object experiences an apparent loss of 28.0 grams when submerged in water. What is the volume of the object?
(A) 2.6 cm3
(B) 14.0 cm3
(C) 18.2 cm3
(D) 28.0 cm3
(E) 37.8 cm3
8. Which conclusion is reasonable based on the illustration?
(A) Warm water is denser than cold water.
(B) Cold sea water is denser than warm sea water.
(C) Fresh water is denser than sea water.
(D) A boat is denser in warm sea water than in cold sea water.
(E) A boat is less dense in cold fresh water than in warm fresh water.
Questions 9 and 10 relate to the diagram below, which shows a hydraulic device along with the input force and area of the input and output cylinders.
9. What pressure is transmitted throughout the liquid in the device?
(A) 0.1 kPa
(B) 0.2 kPa
(C) 0.4 kPa
(D) 2 kPa
(E) 4 kPa
10. What force is produced on the output piston as a result of the input force?
(A) 60 N
(B) 80 N
(C) 100 N
(D) 120 N
(E) 160 N
QUESTION ANSWERS AND EXPLANATIONS
1. 
2. 
3.
4. 
5. 
, which becomes 
. so 
6. D First, find the pressure added to the fluid in the jug.
Second, find the force that results when the pressure is transmitted to the bottom of the jug. Note that the pressure is transmitted undiminished to all parts of the liquid so the increase in pressure P2 equals P1.
7. D The mass of the water displaced by the object must be 28 g. The density of water is 1 g/cm3. Therefore, the volume of water displaced must be 28 cm3. The volume of water displaced by the object must be equal to the volume of the object. The volume of the object is 28 cm3.
8. B The buoyant force pushing the boat upward increases from left to right in the illustration. Therefore, the weight of the water displaced by the boat must be increasing. The weight depends on the density of the displaced water, which therefore suggests that cold sea water is the densest liquid in which the boat is placed.
9. A The pressure applied to the input piston is transmitted undiminished throughout the fluid.
10. 