11.2.1.1: Illustrations

Illustration 1: Pressure in a Liquid

With fluids, instead of discussing forces, we usually talk about pressure, which is defined as the force per unit area or \(P = F/A\). This is because the direction of the force a liquid exerts on its container depends on the shape of the container (force is normal to the surface of the container) and the size of the container. Pressure is not a vector (no direction) and does not depend on the size of the container (position is given in meters and pressure is given in pascals). Restart.

Move the pressure indicator in the tube and note the pressure readings (the pressure is only measuring the effect of the liquid asdescribed below). Let's discuss why pressure increases as a function of depth. Assume the blue liquid is water (density \(1000\text{ kg/m}^{3}\)). Pick a point to measure the pressure somewhere in the upper tube. If the dimension of the container into the screen (the dimension you cannot see) is \(1\text{ m}\), what is the volume of water above the point you picked? What is the mass and thus the weight of the water at that point? For example, consider a depth of \(3\text{ m}\). The pressure is \(29,400\text{ N/m}^{2}\). The volume of water above this point is a cylinder of volume \(9.4\text{ m}^{3}\). The mass of the water is the volume times the water's density, or \(9,400\text{ kg}\), and therefore the weight of the water is \(92,120\text{ N}\).

What is the force downward at that point? Well take the weight and divide by the cross-sectional area of the column of water at that point, which is \(3.14\text{ m}^{2}\). This pressure should be equal to the pressure reading. The units of pressure are \(\text{N/m}^{2} =\) pascals (abbreviated Pa).

Strictly speaking, this is the gauge pressure, not the absolute pressure, because we assumed \(P = 0\) at the top of the water column when the pressure (due to the atmosphere) is actually around \(1\times 10^{5}\text{ Pa}\). The absolute pressure then would be the pressure at the top due to the atmosphere added to the pressure due to the weight of the water. All of this comes together in the equation:

\[P=P_{0}+\rho gy\nonumber\]

where \(P_{0}\) is the pressure at the top, \(\rho\) is the density of the liquid, \(g\) is acceleration due to gravity and \(y\) is the depth of the liquid.

What will be the pressure at point A? Add a second pressure indicator to check.

Illustration authored by Anne J. Cox.

Illustration 2: Pascal's Principle

The animation shows a model of a hydraulic lift (position is given in centimeters, force is given in newtons, and time is given in seconds). A gas or liquid, usually oil (yellow in this animation), fills a container with pistons. Restart.

Notice that because the pressure is the same at the same height in the fluid, the force required for the hand to support the mass is much less than the weight of the mass (\(10\) times less). This is because the areas of the gray circular "lids" on the top of the oil are different by a factor of \(10\). Now calculate the pressure exerted by the green mass and the pressure exerted by the force arrow. They should be the same. The ratio of the areas is the ratio of the forces required for equilibrium. Enter a new value for the mass (between \(100\) and \(300\text{ kg}\)) and try it.

Now, play the animation. As the arrow moves down, the mass moves up (but the motion is too small to accurately measure how far it moves up). How far up should the mass move and why? The volume of fluid pushed out of the left tube must be the same amount that goes into the right cylinder.

What is the work done by the force arrow? What is the work done on the mass? The work done by the force arrow is exactly the work done on the mass. So what is the advantage of this lift? The advantage is that it allows a small force to raise something much more massive (e.g., lifting up a car in a repair garage). One can think of this as a small force over a large displacement doing the same amount of work as a large force over a small displacement. The force arrow could not directly lift the green mass, but using the hydraulic lift, it can.

In this setup we neglected the change in pressure as a function of depth in the liquid. We assumed the pressure was still the same at the two gray lids on top of the oil at the end of the animation as it was at the beginning of the animation. At the end of the animation, however, the force arrow should actually be larger than that listed because it is some distance below the liquid level of the right side at the end of the animation.

Illustration authored by Anne J. Cox.

Illustration 3: Buoyant Force

The buoyant force on an object (whether or not it floats) is due to the pressure difference between the bottom of the object and the top of the object. If the object is going to be buoyed up, the bottom of the object is subjected to a greater pressure than the top of the object. Remember that pressure is force/area and so, if the pressure at the bottom of an object (submerged under a liquid) is greater than the pressure on top (still in air), there is a net upward force.

This animation shows a block lowered into a liquid and then floating. The density of the block can be changed by click-dragging in the block on the upper left of the animation (click-drag at the white-gray interface of the box). The graph in the lower right of the animation shows the pressure difference (\(P\)) as a function of depth (\(Y\)). When the net upward force, the buoyant force, is equal to the weight of an object (the white and gray striped block), it floats in the gray liquid. In addition, the amount of fluid displaced when the block is in the fluid is shown in the block to the left.

Assume that the gray liquid is water (density \(1000\text{ kg/m}^{3}\)). The density of the block is a fraction of the density of water. So, to start with, the block has a density of \(0.4\times 1000\text{ kg/m}^{3} = 400\text{ kg/m}^{3}\). Note that approximately \(40\%\) of the block is submerged when it is floating. Assume the dimensions of the block are \(1\text{ m}\) (and it is a cube). First, let's find the net force on the block when it is floating by finding the pressure on the block.

Since pressure as a function of depth in a liquid is \(\rho_{\text{liquid}}gy\) (where \(\rho\) is the density of liquid, \(g\) is the acceleration due to gravity, and \(y\) is the depth in the liquid), what is the pressure at the bottom of the block? It is \(P_{\text{atm}} + \rho_{\text{liquid}}gy\) where \(y = 0.4\text{ m}\). What is the pressure at the top of the block? It is just \(P_{\text{atm}}\). Therefore, the liquid exerts a total force of \(\Delta P\) \(A = \rho_{\text{liquid}}gy\) \(A = 400\text{ N}\), which must also be the block's weight if it is in equilibrium. Why do we neglect the force on the sides of the block due to water pressure?

The animation also shows the spillover of water into a second container (to the left). What is the volume of water in this second container? Using the density of water, what is the weight of the water? Note that this is equal to the buoyant force because, if the block were removed and the water in the spillover tray (to the left) were put back into the main water container, that water would be supported, so the pressure difference supports that weight of water. This means that although the buoyant force is due to the pressure difference at different depths, it is also equal to the weight of the water displaced by the object, as expressed in the equation below:

\[F_{B}=\rho_{\text{liquid}}gV_{\text{liquid displaced}}=\rho_{\text{liquid}}gV_{\text{submerged part of object}}\nonumber\]

If the buoyant force is equal to the weight of the object, the object floats. Change the density of the object by click-dragging the mouse on the base of the red arrow in the top left-hand box and then let the animation run. Repeat the calculations to show that the buoyant force is equal to the weight of the object.

What happens if the object is pushed into the water below where it would naturally float (away from equilibrium)? Try using the mouse to drag the floating block farther down in the water. What type of motion do you observe, and why? Think about the forces acting on the block.

Illustration authored by Anne J. Cox.
Applet authored by Fu-Kwun Hwang, National Taiwan Normal University.

Illustration 4: Pumping Water up from a Well

Why can't you pump water up out of a well deeper than \(10.3\text{ m}\)? You may not have even known of this limitation! We can answer this question by considering the change in pressure in a liquid as a function of depth (position is given in meters). Restart.

First, what is the pressure at the top of the water in the well (the dark blue line)? It is just atmospheric pressure. Measure the pressure of the water in the tube at the pump by dragging the pressure gauge there. Change the well depth by dragging down the dark blue line and see what happens. Varying this level increases the height to which the pump must pump the water. When the well is deeper than \(10.3\text{ m}\) (i.e., when the height of the water column up to the pump is greater than \(10.3\text{ m}\)), what is the problem? Can the pump pressure ever be less than zero? No. The best a pump could do is to create a vacuum, which would be \(P = 0\text{ Pa}\). In reality, the pump could never reach \(P = 0\text{ Pa}\) since there would be vapor pressure in the top of the tube.

Note that this animation is exactly how you use a straw to get liquid out of a glass. You reduce the pressure in your mouth (from atmospheric pressure) and the liquid goes up the straw and into your mouth.

What would happen if you tried to pump out a less dense material (like oil)? (Remember that \(P = P_{0}+\rho gy\)). Could you have a deeper well or would it require a shallower well? Try it.

Illustration authored by Anne J. Cox.

Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College.