## MCAT Physics and Math Review

**Chapter 4: Fluids**

### 4.2 Hydrostatics

**Hydrostatics** is the study of fluids at rest and the forces and pressures associated with standing fluids. A proper understanding of hydrostatics is important for the MCAT because the testmakers frequently include passages and questions on hydraulics and buoyancy.

PASCAL’S PRINCIPLE

For fluids that are incompressible—that is, fluids with volumes that cannot be reduced by any significant degree through application of pressure—a change in pressure will be transmitted undiminished to every portion of the fluid and to the walls of the containing vessel. This is **Pascal’s principle**. For example, an unopened carton of milk could be considered an incompressible fluid in a closed container. If one were to squeeze the container, exerting an increased pressure on the sides of the milk carton, the applied pressure would be transmitted through the entire volume of milk. If the cap were to suddenly pop off, the resulting geyser of milk would be evidence of this increased pressure.

**REAL WORLD**

When the air pressure changes above a large body of water, the water level rises or falls to re-establish pressure equilibrium between the air and the water. The surface of a water body directly below a high-pressure air pocket forms a very small but measurable valley of water. A low-pressure air system has the opposite effect, creating a hill of water.

One application of Pascal’s principle can be seen in **hydraulic systems**. These systems take advantage of the near-incompressibility of liquids to generate mechanical advantage, which, as we’ve seen in our discussion of inclined planes and pulleys in Chapter 2 of *MCAT Physics and Math Review*, allows us to accomplish a certain amount of work more easily by applying reduced forces. Many heavy machines use hydraulics, including car brakes, bulldozers, cranes, and lifts.

Figure 4.1 shows a simple diagram of a hydraulic lift. Let’s determine how such a lift could allow an auto mechanic to raise a heavy car with far less force than the weight of the car. We have a closed container that is filled with an incompressible liquid. On the left side of the lift, there is a piston of cross-sectional area *A*_{1}. When this piston is pushed down the column, it exerts a force with a magnitude equal to *F*_{1} and generates a pressure equal to *P*_{1}. The piston displaces a volume of liquid equal to *A*_{1}*d*_{1} (the cross-sectional area times the distance gives a volume). Because the liquid inside is incompressible, the same volume of fluid must be displaced on the right side of the hydraulic lift, where we find a second piston with a much larger surface area, *A*_{2}. The pressure generated by piston 1 is transmitted undiminished to all points within the system, including to*A*_{2}. As *A*_{2} is larger than *A*_{1} by some factor, the magnitude of the force, *F*_{2}, exerted against *A*_{2} must be greater than *F*_{1} by the same factor so that *P*_{1} = *P*_{2}, according to Pascal’s principle.

**Equation 4.7**

**Figure** **4.1.** **Hydraulic Lift**

What this series of equations shows us is that hydraulic machines generate output force by magnifying an input force by a factor equal to the ratio of the cross-sectional area of the larger piston to that of the smaller piston. This does not violate the law of energy conservation; an analysis of the input and output work in a frictionless system reveals that there indeed is conservation of energy. As mentioned above, the volume of fluid displaced by piston 1 is equal to the volume of fluid displaced at piston 2.

Combining the equations for pressure and volume, we can generate an equation for work as the product of constant pressure and volume change, as this is an isobaric process.

This shows us the familiar form of work as the product of the magnitude of force and displacement (times the cosine of the angle between them, which is 0° in this case). Because the factor by which *d*_{1} is larger than *d*_{2} is equal to the factor by which *F*_{2} is larger than *F*_{1}, we see that no additional work has been done or unaccounted for; the greater force *F*_{2} is moving through a smaller distance *d*_{2}. Therefore, an auto mechanic needs only to exert a small force over a small area through a large distance to generate a much larger force over a larger area through a smaller distance.

**KEY CONCEPT**

Remember when applying Pascal’s principle that the larger the area, the larger the force, although this force will be exerted through a smaller distance.

**Example:**

A hydraulic press has a piston of radius 5 cm, which pushes down on an enclosed fluid. A 45 kg weight rests on this piston. Another piston in contact with this system has a radius of 20 cm. Taking what force is needed on the larger piston to keep the press in equilibrium?

**Solution:**

Use Pascal’s principle:

ARCHIMEDES’ PRINCIPLE

You’ve probably heard some version of this story before: Archimedes, a physicist in ancient Greece, was tasked by his king to determine the metallic composition of a certain crown given to the king as a gift. Archimedes knew that he could do this by finding the crown’s volume and mass, which would allow him to find its density and compare that density to those of known metals. Weighing the crown would be easy enough, but he was having trouble finding a way to measure its volume without melting it down and ruining its workmanship. Then one day, while getting into his bath, the water that overflowed from the tub gave him the idea to submerge the crown in water and measure the volume of the displaced liquid.

The principle that derives from the story is one of Archimedes’ lasting contributions to the field of physics. **Archimedes’ principle** deals with the **buoyancy** of objects when placed in a fluid. It helps us understand how ships stay afloat and why we feel lighter when we’re swimming. The principle states that a body wholly or partially immersed in a fluid will be buoyed upwards by a force equal to the weight of the fluid that it displaces.

Just as Archimedes’ body and his crown caused the water level to rise in the tub, any object placed in a fluid will cause a volume of fluid to be displaced equal to the volume of the object that is submerged. Because all fluids have density, the volume of fluid displaced will correspond to a certain mass of that fluid. The mass of the fluid displaced exerts a force equal to its weight against the submerged object. This force, which is always directed upward, is called the buoyant force, and its magnitude is given by:

*F*_{buoy} = *ρ*_{fluid}*V*_{fluid displaced}g = *ρ*_{fluid}*V*_{submerged}g

**Equation 4.8**

**MCAT EXPERTISE**

The most common mistake students make using the buoyancy equation is to use the density of the object rather than the density of the fluid. Remember always to use the density of the fluid itself.

When an object is placed in a fluid, it will sink into the fluid only to the point at which the volume of displaced fluid exerts a force that is equal to the weight of the object. If the object becomes completely submerged and the volume of displaced fluid still does not exert a buoyant force equal to the weight of the object, the object will accelerate downwards and sink to the bottom. This will be the case if an object is more dense than the fluid it’s in—a gold crown will sink to the bottom of the bathtub because it is more dense than water. On the other hand, an object that is less dense than water, such as a block of wood or an ice cube, will stop sinking (and start floating) because it is less dense than water. These objects will submerge enough of their volume to displace a volume of water equal to the object’s weight.

**KEY CONCEPT**

An object will float if its average density is less than the average density of the fluid it is immersed in. It will sink if its average density is greater than that of the fluid.

One way to conceptualize the buoyant force is that it is the force of the liquid trying to return to the space from which it was displaced, thus trying to push the object up and out of the water. This is an important concept because the buoyant force is due to the liquid itself, not the object. If two objects placed in a fluid displace the same volume of fluid, they will experience the same magnitude of buoyant force even if the objects themselves have different masses.

How can one determine how much of a floating object lies below the surface? To do this, one can make comparisons of density or specific gravity. Remember that an object will float, no matter what it is made of and no matter how much mass it has, if its average density is less than or equal to the density of the fluid into which it is placed. If we express the object’s specific gravity as a percent, this directly indicates the percent of the object’s volume that is submerged (when the fluid is pure water). For instance, the density of ice is so its specific gravity is 0.92. An ice cube floating in a glass of water has 92 percent of its volume submerged in the water—only 8 percent is sitting above the surface. Therefore, any object with a specific gravity less than or equal to 1 will float in water and any object with a specific gravity greater than 1 will sink in water. A specific gravity of exactly 1 indicates that 100 percent of the object will be submerged but it will not sink.

**REAL WORLD**

At first it may seem strange that cruise ships, constructed of dense metals and weighing thousands of kilograms, can float on water. But remember that any object will float if its average density is less than that of water. The steel hull of the ship would sink by itself, but all the air submerged beneath the water level, between the ship’s lower decks, lowers the ship’s average density to be less than that of water.

**Example:**

A wooden block floats in the ocean with half its volume submerged. Find the density of the wood *ρ*_{b}. (Note: The density of sea water is )

**Solution:**

The magnitude of the weight of the block of total volume *V*_{b} is

*F*_{g,b} = *m*_{b}g = *ρ*_{b}*V*_{b}g

The weight of the displaced seawater is the buoyant force and is given by

*F*_{buoy} = *m*_{water}g = *ρ*_{water}*V*_{water}g

*V*_{water} is the volume of displaced water, which is also the volume of the part of the block that is submerged Because the block is floating, the buoyant force equals the block’s weight:

**REAL WORLD**

The Dead Sea, located between Israel and Jordan, is the deepest hypersaline lake in the world. Having a salt content of about 35 percent, it is almost nine times saltier than the ocean. All of this dissolved salt makes for some of the densest water on the surface of the Earth, with a specific gravity of 1.24. Humans have a specific gravity around 1.1; thus, in most bodies of water, we have a tendency to sink—but in the Dead Sea, one is unable to do anything but float.

MOLECULAR FORCES IN LIQUIDS

Water striders are insects in that have the ability to walk on water. Water striders are able to glide across the water’s surface without sinking, even though they are denser than water, because of a special physical property of liquids at the interface between a liquid and a gas. **Surface** **tension**causes the liquid to form a thin but strong layer like a “skin” at the liquid’s surface. Surface tension results from **cohesion**, which is the attractive force that a molecule of liquid feels toward other molecules of the same liquid. Consider the intermolecular forces between the separate molecules of liquid water. For those molecules below the surface, there are attractive intermolecular forces coming from all sides; these forces balance out. However, on the surface, the molecules only have these strong attractive forces from the molecules below them, which pulls the surface of the liquid toward the center. This establishes tension in the plane of the surface of the water; when there is an indentation on the surface (say, caused by a water strider’s foot) then the cohesion can lead to a net upward force.

**REAL WORLD**

Remember that cohesion occurs between molecules with the same properties. In a container of both water and oil, the water molecules will be cohesive with other water molecules, and the oil will be cohesive with other oil molecules.

Another force that liquid molecules experience is **adhesion**, which is the attractive force that a molecule of the liquid feels toward the molecules of some other substance. For example, adhesive forces cause water molecules to form droplets on the windshield of a car even though gravity is pulling them downward. When liquids are placed in containers, a **meniscus**, or curved surface in which the liquid “crawls” up the side of the container a small amount, will form when the adhesive forces are greater than the cohesive forces. A **backwards (convex)** **meniscus** (with the liquid level higher in the middle than at the edges) occurs when the cohesive forces are greater than the adhesive forces. Mercury, the only metal that is liquid at room temperature, forms a backward meniscus when placed in a container. Both types of menisci are shown in Figure 4.2.

**Figure** **4.2.** **Types of Menisci** *(A) A concave meniscus (more common); (B) A convex (backwards) meniscus. The dotted line indicates where measurements of depth or volume should be taken with each type of meniscus.*

**MCAT Concept Check 4.2:**

Before you move on, assess your understanding of the material with these questions.

1. Contrast cohesion and adhesion.

· Cohesion:

· Adhesion:

2. What would the meniscus of a liquid that experiences equal cohesive and adhesive forces look like?

3. A block is fully submerged three inches below the surface of a fluid, but is not experiencing any acceleration. What can be said about displaced volume of fluid and the buoyant force?

4. True or False: To determine the volume of an object by fluid displacement it must have a specific gravity greater than 1.

5. To which side of a hydraulic lift would the operator usually apply a force—the side with the larger cross-sectional area, or the side with the smaller cross-sectional area? Why?