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PHY 120 - Lecture Notes by Alberto Rojo
Pressure and Pascal’s principle
1 Force vs. pressure
Pressure (P ) is simply the force (F ) experienced by an object divided by the area (A) of the surface on which the force acts:
P =
F
A
Note that the force here is the force acting perpendicular to the surface. Let’s say you take a 1-inch by 1-inch piece of wood that’s 3 feet long, and let’s say this piece of wood weighs 1 pound. If you were to stand that piece of wood on-end on your foot, it would place 1 pound of pressure on your toe. Since its cross-section is 1 square inch, it exerts 1 pound of force per square inch of pressure (1 psi) on your toe. If you were to take a 30-foot-long piece of the same wood and balance it on your foot, it would apply 10 psi of pressure. If it were 300 feet long, it would apply 100 psi, and so on. Water that is 1 foot deep exerts 0.43 psi, so if you are a mile underwater there’s about 2,270 psi being exerted. That is, a 1-inch-square column of water a mile high weighs 2,270 pounds.
Another frequently used unit for pressure is called Pascal,
1 Pascal = 0. 00014
Pounds of force square inch
and also other units are used (atmospheres, millibars, etc). When we talk about atmospheric pressure, we’re talking about the pressure exerted by the weight of the air above us. The air goes up a long way, so even though it has a low density it still exerts a lot of pressure:
atmospheric pressure at sea level = 103, 000 Pascals, Some examples of pressure found on our everyday life are indicated in the table below.
2 Pressure in gases and liquids
Two concepts are key to understand static properties of liquids and gases:
- A gas transmits pressure equally over the entire surface that contains it.
- Pressure is perpendicular to the walls of the container.
2.1 How do we exert pressure in a gas?
The way a gas like air exerts pressure inside a container like a tire or a balloon is through the action of the air atoms colliding with the sides of their container. Imagine that you have a single atom of nitrogen in a sealed container. That atom is in constant motion ricocheting off the sides of the container. The speed of the atom’s motion is controlled by the temperature, the higher the temperatures the higher the speed. By its collisions with the sides of the container, the atom exerts an outward pressure. So there are two ways to increase the pressure inside the container:
- Raise the temperature of the atoms inside the container - The hotter the atoms, the faster they move.
- Put more atoms in the container - The more gas atoms you put in the container, the more collisions you get and the greater the pressure they exert on the sides of the container. When you blow up a tire on a car or a bike, you use a pump to increase the pressure of the air inside the tire by increasing the number of atoms inside the tire. A car tire typically runs at 30 psi, and a bike tire might run at 60 to 100 psi. There is no magic here – the pump simply stuffs more air into a constant volume, so the pressure rises. If we cool the air inside a balloon (for example throwing liquid Nitrogen to it) the pressure will decrease and the balloon will contract. If we decrease the pressure of the air outside a balloon, the air will expand. This can be seen in two popular demos: a) put shaving cream inside a glass bell and connect it to a vacuum pump. The bubbles of the shaving cream are at atmospheric pressure. When the pressure inside the bell decreases the bubbles expand and the cream increases its volume. b) Put a glass of water inside the bell. When the pressure decreases the air bubbles inside the water expand and the water starts “boiling” at room temperature!
2.2 Pressure in liquids
The figure below shows two containers filled up with water up to a certain height H, which is the same for both containers. The base on the container of the right has a larger surface area than the one on the left. Let us call A (^) L and A (^) R the respective surface areas. The force on the bottom surface on the container on the right has two contributions: the force coming from the air pressure (represented as P (^) a ) and the force due to the weight of the liquid:
Total Force = density of water × Height × Area A (^) L + P (^) a × AreaA (^) L.
Since pressure is force per unit area, the pressure at the bottom of the container on the left is
Pressure = Total Force Area A (^) L
= density of water × Height + P (^) a = P (^) b ,
and we obtain that the pressure at the bottom is the same for both con- tainers, since the term AL has cancelled from the expression for P (^) b. This means that in order to know the pressure at a certain point of the liquid we need to know three things: a) the density of the liquid, b) the height of the point in question with respect to the liquid level and c) the pressure above the water level. As we mentioned above, the special feature about liquids is that they transmit the pressure and when it pushes perpendicularly to the surface con- taining it. Suppose we now add two tubes of equal diameter (and therefore
of that force? The force has to be large enough to equate the pressures at the bottom of the two containers. The new pressure at the bottom of the left container is P (^) b + F/L. Again, the water will be in equilibrium if the pressures are the same:
P (^) b +
F
A L
= P (^) b +
W
A R
which implies that with a force given by
F =
A L
AR
× Force due to weight W
we can maintain equilibrium. If the area AR is much larger than the area A (^) L we can maintain equilibrium with a force much smaller than the weight of the car. This is called mechanical advantage, which is defined as
mechanical advantage = Force out Force in
A L
AR
If we increase the force F , a little bit of water will flow from left to right and we can lift the car with a small force. This is the principle of the hydraulic press. Another application is the working of the break of your car. With a small force (and a the relatively large displacement of your foot) on the pedal you move fluid through a small tube with a large pressure that is converted to a large force that presses the calipers and slows down the wheels.
3 Buoyancy
If one places a copper ball in a pail of water it will sink, whereas a wooden ball will float. Whether or not a given object will sink or float in a fluid is determined by the buoyant force on the object. The buoyant force is essentially caused by the difference between the pressure at the top of the object, which pushes it downward, and the pressure at the bottom, which pushes it upward. Since the pressure at the bottom is always greater than at the top, every object submerged in a fluid necessarily feels an upward buoyant force. Of course, objects also feel a downward force due to gravity, and the difference between the gravitational force and buoyant force on a submerged object determines whether that object will sink, or rise to the surface. If the weight is greater than the buoyant force, the object sinks, and vice versa. It was Archimedes (supposedly while in his bath), who realized that submerged objects always displace fluid upwards (the level of water in the bathtub rose when Archimedes got in). Thus, he reasoned that the buoyant force on an object must be equal to the weight of fluid that object displaces. If the weight of an object is greater than the weight of displaced fluid, it will sink, whereas if the weight of the object is less than the weight of displaced fluid, it will rise. Moreover, it is evident that the volume of displaced fluid is precisely equal to the volume of the submerged part of the object, so that the difference between the buoyant force and the weight is determined by the relative density of the object and the fluid. In particular, we come to Archimedes’s principle, which implies that
A solid object will sink in a fluid if its density is larger than the fluid’s density, and will float if its density is smaller.
This explains why wood and styrofoam float on water, whereas concrete and steel sink. It also explains why it is nonetheless possible to make boats out of steel or even concrete. As long as there are portions of the boat below the surface of the water that are hollow (i.e. contain air), the effective density of the boat can be less than that of water even though the real density of the material is greater. For a fluid that floats:
The buoyant force is equal to the weight of the fluid displaced.
Example: If you place a 1 cubic foot object that weighs 63 lbs into fresh water, the object is displacing 62.4 lbs of water, but weighs 63 lbs.
is squeezed upward it pivots around b, raising the part f, which engages the tooth on part d and pushes it upward. That pushes the valve e up, letting the gasoline flow. At the same time the vacuum pump is sucking air through a small tube that ends in a small hole (a few millimeters in diameter) outside the nozzle a centimeter of two from the end. When gasoline enters the small tube the pressure inside the little tube decreases and causes the diaphragm to pull in and, through the link c pull f to the left, unlatching d, letting the spring g pull the valve closed.
4.2 Why your ears “pop” in an airplane?
The picture below shows a diagram of the ear. For our purpose, it is im- portant to understand the role of the auditory (or Eustachian) tube of the middle ear. The middle ear consists of bones within an air-filled space. The outer ear has air in it, too, air that is continuous with the atmospheric air. The inner ear is not filled with air– it is filled with fluid. The Eustachian tube connects your middle ear with your nose-throat cavity. It is a limp tube with no muscles of its own and it is therefore normally collapsed like a toy balloon unless causes it to open. Different altitudes have different natural air pressures. The higher you ascend, the lower the air pressure, and the lower you descend, the higher the
air pressure. Therefore, if you hike up a mountain, when you near the top, the atmospheric air pressure is lower than what it had been at the bottom. Before you started your hike, the air pressures on either side of the tym- panic membrane (both outer and middle ear sides) were the same. As you hiked up to the top, the outer ear air pressure dropped. With less pressure inside, the tympanic membrane should buckle out, from the excess pressure in the middle ear. Any buckling of the tympanic membrane decreases your hearing ability and can be painful. The balance in pressures between the outer and middle ears must be restored. That’s what the auditory tube is for. When a discrepancy in pressure arises, the auditory tube can open for a short interval to allow air at the atmospheric pressure in your throat to enter and replace the middle ear air. This is when your ears “pop”
When you swallow or yawn, your muscles pull on the end of the Eu- stachian tube, straightening it out so that it can be forced open more easily than when it’s crumpled.
4.3 The physics of breathing
The process of inhaling and exhaling air is controlled by the diaphragm, a muscle below our lungs that, when contracted, displaces downwards and expands the lungs. This decreases the pressure in the lungs and air is forced in. When the diaphragm comes back to relaxation the lungs contract and we exhale.
4.5 Siphons
You have a small swimming pool on the elevated deck behind your house. How can you drain this pool using only a hose, without making a hole in the bottom of the pool?
4.6 Drinking straws
If you fill a drinking straw with water and seal the top with your finger, what will happen as you lift the straw out of the water glass? Answer: The water begins to descend but as it does, it creates an empty region near your finger. The pressure inside the straw near your finger drops below the atmospheric pressure so the water in the straw experiences a pressure imbalance, with lower pressure on top. This pressure imbalance creates an upward force that prevents the water from descending further. The longer the straw, the larger the pressure imbalance needed to support the increasing weight of the water. If the straw were about 10 meters long, the pressure near your finger would reach zero.
