fluid mechanics hydraulic lift experiment

Lift a Large Load Using Liquids

A hydraulic-powered activity from Science Buddies

By Science Buddies & Sabine De Brabandere

fluid mechanics hydraulic lift experiment

Liquid power: Build a lift that uses water to help!

George Retseck

Key Concepts Physics Engineering Simple Machines Force Pressure

Introduction Not even the strongest human could lift a truck into the air. Our brains, however, are smart enough to create a tool that can lift heavy objects for us: hydraulic lifts! You can find them at car repair shops, in wheelchair lifts, and even on skyscraper construction sites. They use water (or other liquids) to increase the force available to lift things. And you will get to build one in this activity! Try it out, and discover how much easier it is to lift with the power of water.

Background If you have ever used a wheelbarrow, you know it can help to lift heavy objects: lifting the handles of the wheelbarrow is easier than lifting the object itself. A hydraulic lift accomplishes the same task. It allows you to move a heavy object with much less effort.

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"Hydraulic" refers to tools that operate by moving liquids, such as water or oil. You cannot compress liquids (unless you have special equipment). This means that no matter how hard you press on a liquid you cannot noticeably change its volume. Try it out! Fill an oral medicine syringe with water, close its open tip with a finger, and try to press the plunger in. You will see that you cannot. Even if you could place an elephant on the plunger, it would not compress the fluid in the syringe. So what will you be able to do with this water power?

In the effort to move something, it is not only important to exert force but also to consider the full area over which the force is spread. Think of a time when you played with Play-Doh. Pushing a finger into the dough is easy, but pushing the palm of your hand into the dough is harder—you probably felt more resistance. Because your hand has a larger surface area, your effort is spread out over a larger area, and the dough does not receive as much force for each square inch as it does with a single-finger push. Now think about this problem in reverse: sometimes you want less force per square inch, such as when you are trying to lift something. Is there a way to place an object on a larger surface so it creates less pressure when you're trying to lift it? Try the activity to find out!

Two stainless steel wall plates, at least 4.5 by 2.75 inches in size. (These are available in hardware stores.)

Two small oral medicine syringes (these are usually given in metric measurements—25 milliliters or 25CC works well.) (Available at pharmacies.)

Large oral medicine or irrigation syringe (one that is approximately 60 milliliters or 60CC works well; one with a catheter tip attached will be easiest to work with). (Available at pharmacies.)

Extra-strong wall-mount double-sided tape

Four feet of silicone fuel tubing with a 5/32-inch inner diameter. (This is available in hobby shops, hardware stores and online.)

Hardware vise

Sturdy surface to which the vise can be attached

Jar full of water (with lid closed tightly) and/or other heavy object that will fit on the wall plate

Large bowl of water

Paper and pen or pencil

Adult helper

Workspace that can get messy

Plumber's putty (optional) (available at hardware stores)

Scale (optional)

Other sizes of syringes with catheter tip (optional)

Preparation

Put one small syringe away to the side. This syringe will be the "leading" syringe; you will use it to make the lift move. The other two syringes (one small one and the large one) will hold the weight that needs to be lifted. These will be the "helping" syringes.

Use the extra-strong double-sided tape to adhere the center of one wall plate to the flat top of each of the plungers on the two helping syringes.

Carefully cut the tubing in two pieces, each about 2-feet long.

For each helping syringe (the syringes with plates attached) connect a plastic tube to its tip.

Push the plunger in each helping syringe all the way down so that it is fully depressed into the cylinder.

One at a time, put the free end of each tube into the water, and pull the plungers smoothly up as far as they will go without falling out of the syringes. The syringes will suck up water.

Hold each syringe (one at a time) with its tip pointing up. Remove the air bubble near the tip of the syringe by pushing the plunger in slowly until all the air is pushed out of the tube.

Place the end of the small syringe's tube back in the water, and suck in some more water.

Very little air should be left in either of the tubes or helping syringes. The small syringe should be full, and the larger one can be half-full.

Attach the vice to a sturdy surface.

Insert the tip of the leading syringe into the free end of the tube connected to the small helping syringe. Push the plunger of this helping syringe in. What happens to the leading syringe? Why would this happen? Do you need to push hard to get the water to transfer from one syringe to the other? (If the tubing pops off, refill the tubing and helping syringe with water, and reattach the tubing to the leading syringe. If needed, seal the connection with plumber's putty.)

The leading syringe should now be full of liquid.

Use the vise to carefully hold the helping syringe vertically with the plate horizontal and facing up. Be careful to not squish the cylinder of the syringe as this will obstruct the movement of the plunger.

Fill the jar with water, and place it in the precise center of the platform attached to the helping syringe in the vice. Do you think you can easily lift this weight by pushing on the leading syringe?

Try it out . Push gently so the jar does not tumble off. You might need a helper to gently keep the construction upright . Was your prediction correct?

Take the jar off the platform, push the plunger of the helping syringe down until the leading syringe is filled with liquid, and then place the jar (or another heavy object) on the platform and lift it again. Use a ruler to measure how high your load (the jar with water) is moving. (You might need a helper for this step.) Write this distance down.

You just tested a design in which the helping and leading syringes are equal in size. Switch the helping syringe (the syringe that holds the weights) to one that is larger than the leading syringe. Do you think lifting the jar will be easier or harder with this design? Do you think the jar will move up more, less or about the same? Why do you think this?

Repeat the procedure with the larger helping syringe to find out! (Notice you will not have to press the helping syringe completely to fill the leading syringe.)

Use the ruler to measure how high the plate is being moved, and write down that distance.

Compare the measurements you made for each design . Why do you think they are different?

Did one design make it easier to move the jar? Why do you think this is so? Which design would you choose to lift a heavy truck—and why?

Extra: Try to lift heavier and heavier objects until you barely can lift them. Remember to place the objects centered on the plate (if a heavy weight is placed off center, the tape might give and your object might fall). Try this for each configuration and, if you have a scale available, weigh the different weights you were able to lift with each syringe combination. Which combination was able to lift more weight? Why do you think that is?

Extra : Measure the radius of the small and large helping syringe cylinders. Can you use these numbers to find the factor by which the weight you can lift gets multiplied when you move from a small to a large syringe?

Extra: Try other combinations of syringe sizes. Can you see a pattern in your findings?

Extra : Make a hydraulically operated wheelchair lift, dental office chair, or car lift for toys.

Observations and Results You probably found it easier to lift the jar with the large helping syringe. In that combination the jar probably also moved a much shorter distance.

When you press one plunger the liquid inside the closed system presses with an equal amount of pressure against the other plunger. Because pressure is created by applying force over an area, a small push on a small area creates the same pressure as a much larger push on a much larger area. That is why a hydraulic lift can multiply a force: you push on a small plunger, the pressure that is created moves through the liquid to a much larger plunger, and the load on the larger plunger feels a much greater push than the one you first made. For that reason it was easier and, if you tried, you could move more weight with a large helping syringe compared to a small one.

Did you see how liquid was transferred from one column to the other in the process? Because the volume of liquid does not change, the larger plunger will not move as much as the smaller plunger. But that is the tradeoff for being able to lift a heavy object!

More to Explore Jack it Up! Lift a Load Using Hydraulics , from Science Buddies The Multiplication of Force by a Hydraulic System , by Wisc Online Heavy Lifting with a Lever , from Scientific American STEM Activities for Kids , from Science Buddies

This activity brought to you in partnership with Science Buddies

Build a Hydraulic Lift

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( Physics for ages 8+)

Machines are all around us, and they can be very helpful for doing everyday tasks, like lifting or moving heavy objects. Have you ever used a wheelbarrow? Have you noticed how much easier it makes moving rocks or dirt than having to move them by hand?

Cardboard Box knife Ruler 2 small oral syringes (10-25 mL or CC) Flexible plastic tubing (can be found in hobby stores or medical supplies) Water Food coloring Hot glue gun Adult supervision (Adult supervision at all times please)

14.3 Pascal's Principle and Hydraulics

Learning objectives.

By the end of this section, you will be able to:

State Pascal’s principle
Describe applications of Pascal’s principle
Derive relationships between forces in a hydraulic system

In 1653, the French philosopher and scientist Blaise Pascal published his Treatise on the Equilibrium of Liquids , in which he discussed principles of static fluids. A static fluid is a fluid that is not in motion. When a fluid is not flowing, we say that the fluid is in static equilibrium. If the fluid is water, we say it is in hydrostatic equilibrium . For a fluid in static equilibrium, the net force on any part of the fluid must be zero; otherwise the fluid will start to flow.

Pascal’s observations—since proven experimentally—provide the foundation for hydraulics , one of the most important developments in modern mechanical technology. Pascal observed that a change in pressure applied to an enclosed fluid is transmitted undiminished throughout the fluid and to the walls of its container. Because of this, we often know more about pressure than other physical quantities in fluids. Moreover, Pascal’s principle implies that the total pressure in a fluid is the sum of the pressures from different sources. A good example is the fluid at a depth depends on the depth of the fluid and the pressure of the atmosphere.

Pascal’s Principle

Pascal’s principle (also known as Pascal’s law) states that when a change in pressure is applied to an enclosed fluid, it is transmitted undiminished to all portions of the fluid and to the walls of its container. In an enclosed fluid, since atoms of the fluid are free to move about, they transmit pressure to all parts of the fluid and to the walls of the container. Any change in pressure is transmitted undiminished.

Note that this principle does not say that the pressure is the same at all points of a fluid—which is not true, since the pressure in a fluid near Earth varies with height. Rather, this principle applies to the change in pressure. Suppose you place some water in a cylindrical container of height H and cross-sectional area A that has a movable piston of mass m ( Figure 14.15 ). Adding weight Mg at the top of the piston increases the pressure at the top by Mg / A , since the additional weight also acts over area A of the lid:

According to Pascal’s principle, the pressure at all points in the water changes by the same amount, Mg / A . Thus, the pressure at the bottom also increases by Mg / A . The pressure at the bottom of the container is equal to the sum of the atmospheric pressure, the pressure due to the fluid, and the pressure supplied by the mass. The change in pressure at the bottom of the container due to the mass is

Since the pressure changes are the same everywhere in the fluid, we no longer need subscripts to designate the pressure change for top or bottom:

Interactive

Pascal’s Barrel is a great demonstration of Pascal’s principle. Watch a simulation of Pascal’s 1646 experiment, in which he demonstrated the effects of changing pressure in a fluid.

Applications of Pascal’s Principle and Hydraulic Systems

Hydraulic systems are used to operate automotive brakes, hydraulic jacks, and numerous other mechanical systems ( Figure 14.16 ).

We can derive a relationship between the forces in this simple hydraulic system by applying Pascal’s principle. Note first that the two pistons in the system are at the same height, so there is no difference in pressure due to a difference in depth. The pressure due to F 1 F 1 acting on area A 1 A 1 is simply

According to Pascal’s principle, this pressure is transmitted undiminished throughout the fluid and to all walls of the container. Thus, a pressure p 2 p 2 is felt at the other piston that is equal to p 1 p 1 . That is, p 1 = p 2 . p 1 = p 2 . However, since p 2 = F 2 / A 2 , p 2 = F 2 / A 2 , we see that

This equation relates the ratios of force to area in any hydraulic system, provided that the pistons are at the same vertical height and that friction in the system is negligible.

Hydraulic systems can increase or decrease the force applied to them. To make the force larger, the pressure is applied to a larger area. For example, if a 100-N force is applied to the left cylinder in Figure 14.16 and the right cylinder has an area five times greater, then the output force is 500 N. Hydraulic systems are analogous to simple levers, but they have the advantage that pressure can be sent through tortuously curved lines to several places at once.

The hydraulic jack is such a hydraulic system. A hydraulic jack is used to lift heavy loads, such as the ones used by auto mechanics to raise an automobile. It consists of an incompressible fluid in a U-tube fitted with a movable piston on each side. One side of the U-tube is narrower than the other. A small force applied over a small area can balance a much larger force on the other side over a larger area ( Figure 14.17 ).

From Pascal’s principle, it can be shown that the force needed to lift the car is less than the weight of the car:

where F 1 F 1 is the force applied to lift the car, A 1 A 1 is the cross-sectional area of the smaller piston, A 2 A 2 is the cross sectional area of the larger piston, and F 2 F 2 is the weight of the car.

Example 14.3

Calculating force on wheel cylinders: pascal puts on the brakes.

Manipulate this algebraically to get F 2 F 2 on one side and substitute known values.

Significance

Check your understanding 14.3.

Would a hydraulic press still operate properly if a gas is used instead of a liquid?

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Simple Hydraulic Lifter WIP

Meta Description

Learning Objectives

Understanding the relationship between pressure, area and force.

Observing that fluids transmit pressure.

Awareness of the importance of hydraulics to lift heavy loads.

Cross Sectional Area The area of the shape produced when an object is cut in half and it’s internal shape viewed.

Force A pull or a push which acts on an object.

Hydraulics The science associated with the flow of liquids and their use to produce forces.

Hydraulic Lifter A machine used to lift objects using liquids under pressure.

Pressure The force per unit area.

Pressure in Liquids Pressure is distributed equally throughout a trapped liquid.

Weight The downward force on a mass due to the Earth’s gravitational pull.

Step 1 Pre-stretch the balloon by inflating and then deflating it.

Step 2 Insert the end of the tubing into the balloon and secure the joint with tape. Ensure that the joint is watertight by using the funnel to fill the balloon through the tube. Empty the balloon through the tube.

Step 3 Cut off the top of the bottle, so that it is slightly taller than the can.

Step 4 Use scissors and a pencil to make a hole in the side of the bottle, close to the bottom.

Step 5 Pass the free end of the tube through the hole in the bottle, from the inside of the bottle to the outside. Leave the balloon inside the bottle.

Step 6 Place the can on top of the balloon.

Step 7 Place the heavy book on top of the bottle.

Step 8 Fit the funnel at the free end of the tube and pour the water into the pipe. Holding the funnel high allows air bubbles to escape.

Step 9 As the balloon fills, what do you observe?

Step 10 Once the balloon is full, remove the book and can and use your hand to gently press down on the balloon – hold the tube facing upwards when doing this step. What do you notice?

Ensure that the edges of the can are blunt and do not pose a cutting hazard.
Use scissors responsibly. Do not place your hand in the way of the blade and place on a table when not in use.
Protect any surfaces which could be damaged by the water.

When you’re swimming underwater, you can feel the pressure of the liquid on your skin. All liquids exert a pressure on their container and on objects submerged in them.

The water filling the balloon creates a pressure which causes the balloon to expand. This pressure acts on the can sitting on top of the balloon, pushing against it. This pushing effect is called a force. The can is moved upwards until it starts pushing against the book.

So the book has two forces acting on it. It’s weight, which acts downwards, and the force from the balloon, which pushes upwards. The book only moves when the upward force is greater than it’s own weight.

Why does the water not come out of the pipe? Due to the pressure of the water in the pipe and the angle of the pipe.

Why can the balloon move the book? Upward resultant force due to pressure in the water > weight of book.

Why does the balloon not burst? If stretched beyond the capabilities of the rubber it would burst.

Is the pressure even across the surface of the balloon? Yes.

Why do we have to hold the pipe in the air? To release air bubbles and give the fluid more potential energy, to create greater pressure at the outlet.

Hydraulic machines use liquids under pressure. They rely on two important characteristics of fluids: firstly that liquids are incompressible (you can’t squash them) and secondly that when pressure is applied to a trapped liquid, the pressure is transferred to all parts of the liquid.

In general, pressure can be expressed by a simple equation:

P = F/ A [1]

where P is the applied pressure (in Pa or N/m 2 ), F is the resulting force (in N) and A is the cross sectional area (in m 2 ).

The water flowing into the balloon increases the volume of water in the balloon and contributes to the pressure (Video) transmitted through the can to the book. The pressure in the fluid is transmitted to the book via the can. The can has a fixed cross-sectional area. Thus, the pressure in the balloon exerts a pressure on the balloon via the can. When this force is large enough, it pushes the book upwards. This happens when the upward force due to the pressure in the balloon is greater than the weight of the book, which acts downwards.

This problem is more complicated than than a simple hydraulic jack problem. In a simple hydraulic jack pressure is transmitted through a trapped liquid. In this example, some of the fluid is not trapped, but flows into a reservoir (the balloon).

From the law of conservation of energy, the energy of the water at any point in the pipe in constant. Bernoulli’s equations were based on this concept (Video) and can be used to understand the energy of the water in the pipe:

Energy of water at the top of the pipe = Energy of the water exiting the pipe

Top pressure + Top kinetic energy density + Top potential energy = Outlet pressure + Outlet kinetic energy density + Outlet potential energy

Ρ top + ½ v 2 top + ρgγ top = Ρ out + ½ρ v 2 out + ρgγ out

As the water flows down the pipe it’s kinetic energy can be expected to increase slightly (because of acceleration due to gravity), while it’s potential energy decreases appreciably (because of a decrease in height). Therefore, the pressure at the output of the pipe can be expected to be greater than the pressure at the top.

The water flowing into the balloon increases the volume of water in the balloon and creates a pressure within the balloon. Recall that pressure = Force/ Cross-sectional area . The pressure in the balloon acts on the book via the can, which has a fixed cross sectional area. Thus, the water pressure exerts a force on the book. When the force on the book exceeds the weight of the book, there is a resultant force upwards, which causes the book to be pushed higher. Since fluids are virtually incompressible, and the weight of the water in the pipe exerts enough pressure to prevent water being pushed back up the pipe, the balloon continues to grow as more water is added.

Applications Hydraulic lifts are used widely in industry to lift personnel and heavy loads, particularly in docks, construction sites, car repair, warehouses and factories. They can be used to raise heavy products, vehicles or machinery. In car repair mechanics use a hydraulic lift to raise vehicles. Many industries use huge lifters to load and unload goods from vehicles or ships. They can be used to lift personnel to, for example, service street lights. Hydraulic systems are also used in lifts within buildings. Hydraulic lifts can be controlled remotely or manually.

Hydraulic lifts are very powerful: safety checks must be carried out to make sure they are stable and operators must adhere to various precautions.

Research Heavy hydraulic machines including lifters, diggers and bulldozers use oil-based hydraulic fluids. Over a machine’s lifetime, about 85% of these fluids leak away, posing an immediate fire risk and causing long-term harm to wildlife and the environment. Disposal of the fluid is also costly. Recent research has been focused on developing hydraulic systems which use water instead of oil. This makes them environmentally friendly. However, water creates problems of rusting and friction in the power transmission, which must be overcome.

Add a smaller book on top of the first. This increases the downward force due to the weight of the books. How does this affect the system?
Experiment with lifting the pipe to different heights and angles and observe how this affects the outcome.

Preparation: 15 minutes

Conducting: 15 minutes

Clean Up: 10 minutes

Number of People

1 participant

Large empty plastic bottle

Short piece of tubing

Contributors

Complete Physics, Stephen Pople, Pg 56-57; 60-61 (Book)

Force, Mass and Acceleration

Hands-On Hydraulics – Science Fun for Kids

More about Hydraulic Lifts

The Working of a Hydraulic Lift

Water-friendly Hydraulic Technology to make Heavy Machinery Greener

Additional Content

Applications of Hydraulics (Beginner)

Easy Hydraulic Machine (Beginner)

Rise of the Human Exoskeleton (Beginner)

Archimedes (c.287 – 212 BC) (Intermediate)

Easy Hydraulic Machines (Intermediate)

Leaking Hydraulic Fluid could Damage Aircraft Rudders. (Intermediate)

Fracking Can Contaminate Drinking Water (Advanced)

Korean Shipbuilder Testing Industrial Exoskeleton for Future Cybernetic Workforce (Advanced)

What’s the Difference Between Pneumatic, Hydraulic, and Electrical Actuators? (Advanced)

Cite this Experiment

Padfield, N., & Fenech Salerno, B. (2017, September 29). Simple Hydraulic Lifter. Retrieved from http://steamexperiments.com/experiment/simple-hydraulic-lifter/

First published: September 29, 2017 Last modified: October 29, 2019

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14.3 Pascal’s Principle and Hydraulics

Learning objectives.

By the end of this section, you will be able to:

State Pascal’s principle
Describe applications of Pascal’s principle
Derive relationships between forces in a hydraulic system

Pascal’s Principle

Note that this principle does not say that the pressure is the same at all points of a fluid—which is not true, since the pressure in a fluid near Earth varies with height. Rather, this principle applies to the change in pressure. Suppose you place some water in a cylindrical container of height H and cross-sectional area A that has a movable piston of mass m ( Figure ). Adding weight Mg at the top of the piston increases the pressure at the top by Mg / A , since the additional weight also acts over area A of the lid:

Figure A is a schematic drawing of a cylinder filled with fluid and opened to the atmosphere on one side. A disk of mass m and surface area A identical to the surface area of the cylinder is placed in the container. Distance between the disk and the bottom of the cylinder is h. Figure B is a schematic drawing of the cylinder with an additional disk of mass Mg placed atop mass m causing mass m to move lower.

According to Pascal’s principle, the pressure at all points in the water changes by the same amount, Mg / A . Thus, the pressure at the bottom also increases by Mg / A . The pressure at the bottom of the container is equal to the sum of the atmospheric pressure, the pressure due the fluid, and the pressure supplied by the mass. The change in pressure at the bottom of the container due to the mass is

Since the pressure changes are the same everywhere in the fluid, we no longer need subscripts to designate the pressure change for top or bottom:

Pascal’s Barrel is a great demonstration of Pascal’s principle. Watch a simulation of Pascal’s 1646 experiment, in which he demonstrated the effects of changing pressure in a fluid.

Applications of Pascal’s Principle and Hydraulic Systems

Hydraulic systems are used to operate automotive brakes, hydraulic jacks, and numerous other mechanical systems ( Figure ).

A schematic drawing of a hydraulic system with two fluid-filled cylinders, capped with pistons and connected by a tube. A downward force F1 on the left piston with the surface area A1 creates a change in pressure that results in an upward force F2on the right piston with the surface area A2. Surface area A2 is larger than the surface area A1.

According to Pascal’s principle, this pressure is transmitted undiminished throughout the fluid and to all walls of the container. Thus, a pressure [latex]{p}_{2}[/latex] is felt at the other piston that is equal to [latex]{p}_{1}[/latex]. That is, [latex]{p}_{1}={p}_{2}.[/latex] However, since[latex]{p}_{2}={F}_{2}\text{/}{A}_{2},[/latex] we see that

This equation relates the ratios of force to area in any hydraulic system, provided that the pistons are at the same vertical height and that friction in the system is negligible.

Hydraulic systems can increase or decrease the force applied to them. To make the force larger, the pressure is applied to a larger area. For example, if a 100-N force is applied to the left cylinder in Figure and the right cylinder has an area five times greater, then the output force is 500 N. Hydraulic systems are analogous to simple levers, but they have the advantage that pressure can be sent through tortuously curved lines to several places at once.

Figure A is a schematic drawing of a U tube filled with a fluid. A downward force F1 is applied at the left side with the surface area A1. A downward force F2 is applied on the right side with the surface area A2. Surface area A2 is larger than the surface area A1.Figure B is a photo of passenger car placed on the hydraulic jack.

From Pascal’s principle, it can be shown that the force needed to lift the car is less than the weight of the car:

where [latex]{F}_{1}[/latex] is the force applied to lift the car, [latex]{A}_{1}[/latex] is the cross-sectional area of the smaller piston, [latex]{A}_{2}[/latex] is the cross sectional area of the larger piston, and [latex]{F}_{2}[/latex] is the weight of the car.

Calculating Force on Wheel Cylinders: Pascal Puts on the Brakes

Consider the automobile hydraulic system shown in Figure . Suppose a force of 100 N is applied to the brake pedal, which acts on the pedal cylinder (acting as a “master” cylinder) through a lever. A force of 500 N is exerted on the pedal cylinder. Pressure created in the pedal cylinder is transmitted to the four wheel cylinders. The pedal cylinder has a diameter of 0.500 cm and each wheel cylinder has a diameter of 2.50 cm. Calculate the magnitude of the force [latex]{F}_{2}[/latex] created at each of the wheel cylinders.

Figure is a schematic drawing of a hydraulic brake system. A foot is applied to the brake pedal with the force F1. It is transmitted to the pedal cylinder with the area A1 of 0.5 cm. The Pedal cylinder is connected to the hydraulic system with two wheel cylinders with an area A2 of 2.5 cm. Wheel cylinders create output force F2.

We are given the force [latex]{F}_{1}[/latex] applied to the pedal cylinder. The cross-sectional areas [latex]{A}_{1}[/latex] and [latex]{A}_{2}[/latex] can be calculated from their given diameters. Then we can use the following relationship to find the force [latex]{F}_{2}[/latex]:

Manipulate this algebraically to get [latex]{F}_{2}[/latex] on one side and substitute known values.

Pascal’s principle applied to hydraulic systems is given by [latex]\frac{{F}_{1}}{{A}_{1}}=\frac{{F}_{2}}{{A}_{2}}:[/latex]

Significance

This value is the force exerted by each of the four wheel cylinders. Note that we can add as many wheel cylinders as we wish. If each has a 2.50-cm diameter, each will exert [latex]1.25\times {10}^{4}\,\text{N}\text{.}[/latex] A simple hydraulic system, as an example of a simple machine, can increase force but cannot do more work than is done on it. Work is force times distance moved, and the wheel cylinder moves through a smaller distance than the pedal cylinder. Furthermore, the more wheels added, the smaller the distance each one moves. Many hydraulic systems—such as power brakes and those in bulldozers—have a motorized pump that actually does most of the work in the system.

Check Your Understanding

Would a hydraulic press still operate properly if a gas is used instead of a liquid?

Yes, it would still work, but since a gas is compressible, it would not operate as efficiently. When the force is applied, the gas would first compress and warm. Hence, the air in the brake lines must be bled out in order for the brakes to work properly.

Pressure is force per unit area.
A change in pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container.
A hydraulic system is an enclosed fluid system used to exert forces.

Conceptual Questions

Suppose the master cylinder in a hydraulic system is at a greater height than the cylinder it is controlling. Explain how this will affect the force produced at the cylinder that is being controlled.

How much pressure is transmitted in the hydraulic system considered in Figure ? Express your answer in atmospheres.

What force must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2000-kg car (a large car) resting on a second cylinder? The master cylinder has a 2.00-cm diameter and the second cylinder has a 24.0-cm diameter.

A host pours the remnants of several bottles of wine into a jug after a party. The host then inserts a cork with a 2.00-cm diameter into the bottle, placing it in direct contact with the wine. The host is amazed when the host pounds the cork into place and the bottom of the jug (with a 14.0-cm diameter) breaks away. Calculate the extra force exerted against the bottom if he pounded the cork with a 120-N force.

[latex]5.76\times {10}^{3}\,\text{N}\,\text{extra force}[/latex]

A certain hydraulic system is designed to exert a force 100 times as large as the one put into it. (a) What must be the ratio of the area of the cylinder that is being controlled to the area of the master cylinder? (b) What must be the ratio of their diameters? (c) By what factor is the distance through which the output force moves reduced relative to the distance through which the input force moves? Assume no losses due to friction.

Verify that work input equals work output for a hydraulic system assuming no losses due to friction. Do this by showing that the distance the output force moves is reduced by the same factor that the output force is increased. Assume the volume of the fluid is constant. What effect would friction within the fluid and between components in the system have on the output force? How would this depend on whether or not the fluid is moving?

If the system is not moving, the friction would not play a role. With friction, we know there are losses, so that [latex]{W}_{\text{o}}={W}_{\text{i}}-{W}_{\text{f}};[/latex] therefore, the work output is less than the work input. In other words, to account for friction, you would need to push harder on the input piston than was calculated.

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14 Fluid Mechanics

14.3 pascal’s principle and hydraulics, learning objectives.

By the end of this section, you will be able to:

State Pascal’s principle
Describe applications of Pascal’s principle
Derive relationships between forces in a hydraulic system

Pascal’s Principle

Note that this principle does not say that the pressure is the same at all points of a fluid—which is not true, since the pressure in a fluid near Earth varies with height. Rather, this principle applies to the change in pressure. Suppose you place some water in a cylindrical container of height H and cross-sectional area A that has a movable piston of mass m ( (Figure) ). Adding weight Mg at the top of the piston increases the pressure at the top by Mg / A , since the additional weight also acts over area A of the lid:

Figure 14.15 Pressure in a fluid changes when the fluid is compressed. (a) The pressure at the top layer of the fluid is different from pressure at the bottom layer. (b) The increase in pressure by adding weight to the piston is the same everywhere, for example, [latex] {p}_{\text{top new}}-{p}_{\text{top}}={p}_{\text{bottom new}}-{p}_{\text{bottom}} [/latex].

According to Pascal’s principle, the pressure at all points in the water changes by the same amount, Mg / A . Thus, the pressure at the bottom also increases by Mg / A . The pressure at the bottom of the container is equal to the sum of the atmospheric pressure, the pressure due the fluid, and the pressure supplied by the mass. The change in pressure at the bottom of the container due to the mass is

Since the pressure changes are the same everywhere in the fluid, we no longer need subscripts to designate the pressure change for top or bottom:

Pascal’s Barrel is a great demonstration of Pascal’s principle. Watch a simulation of Pascal’s 1646 experiment, in which he demonstrated the effects of changing pressure in a fluid.

Applications of Pascal’s Principle and Hydraulic Systems

Hydraulic systems are used to operate automotive brakes, hydraulic jacks, and numerous other mechanical systems ( (Figure) ).

Figure 14.16 A typical hydraulic system with two fluid-filled cylinders, capped with pistons and connected by a tube called a hydraulic line. A downward force [latex] {\overset{\to }{F}}_{1} [/latex] on the left piston creates a change in pressure that is transmitted undiminished to all parts of the enclosed fluid. This results in an upward force [latex] {\overset{\to }{F}}_{2} [/latex] on the right piston that is larger than [latex] {\overset{\to }{F}}_{1} [/latex] because the right piston has a larger surface area.

According to Pascal’s principle, this pressure is transmitted undiminished throughout the fluid and to all walls of the container. Thus, a pressure [latex] {p}_{2} [/latex] is felt at the other piston that is equal to [latex] {p}_{1} [/latex]. That is, [latex] {p}_{1}={p}_{2}. [/latex] However, since[latex] {p}_{2}={F}_{2}\text{/}{A}_{2}, [/latex] we see that

This equation relates the ratios of force to area in any hydraulic system, provided that the pistons are at the same vertical height and that friction in the system is negligible.

Hydraulic systems can increase or decrease the force applied to them. To make the force larger, the pressure is applied to a larger area. For example, if a 100-N force is applied to the left cylinder in (Figure) and the right cylinder has an area five times greater, then the output force is 500 N. Hydraulic systems are analogous to simple levers, but they have the advantage that pressure can be sent through tortuously curved lines to several places at once.

Figure 14.17 (a) A hydraulic jack operates by applying forces [latex] ({F}_{1}\text{,}\,{F}_{2}) [/latex] to an incompressible fluid in a U-tube, using a movable piston [latex] ({A}_{1},{A}_{2}) [/latex] on each side of the tube. (b) Hydraulic jacks are commonly used by car mechanics to lift vehicles so that repairs and maintenance can be performed.

From Pascal’s principle, it can be shown that the force needed to lift the car is less than the weight of the car:

where [latex] {F}_{1} [/latex] is the force applied to lift the car, [latex] {A}_{1} [/latex] is the cross-sectional area of the smaller piston, [latex] {A}_{2} [/latex] is the cross sectional area of the larger piston, and [latex] {F}_{2} [/latex] is the weight of the car.

Calculating Force on Wheel Cylinders: Pascal Puts on the Brakes

Consider the automobile hydraulic system shown in (Figure) . Suppose a force of 100 N is applied to the brake pedal, which acts on the pedal cylinder (acting as a “master” cylinder) through a lever. A force of 500 N is exerted on the pedal cylinder. Pressure created in the pedal cylinder is transmitted to the four wheel cylinders. The pedal cylinder has a diameter of 0.500 cm and each wheel cylinder has a diameter of 2.50 cm. Calculate the magnitude of the force [latex] {F}_{2} [/latex] created at each of the wheel cylinders.

Figure 14.18 Hydraulic brakes use Pascal’s principle. The driver pushes the brake pedal, exerting a force that is increased by the simple lever and again by the hydraulic system. Each of the identical wheel cylinders receives the same pressure and, therefore, creates the same force output [latex] {F}_{2} [/latex]. The circular cross-sectional areas of the pedal and wheel cylinders are represented by [latex] {A}_{1} [/latex] and [latex] {A}_{2} [/latex], respectively.

We are given the force [latex] {F}_{1} [/latex] applied to the pedal cylinder. The cross-sectional areas [latex] {A}_{1} [/latex] and [latex] {A}_{2} [/latex] can be calculated from their given diameters. Then we can use the following relationship to find the force [latex] {F}_{2} [/latex]:

Manipulate this algebraically to get [latex] {F}_{2} [/latex] on one side and substitute known values.

Pascal’s principle applied to hydraulic systems is given by [latex] \frac{{F}_{1}}{{A}_{1}}=\frac{{F}_{2}}{{A}_{2}}: [/latex]

Significance

This value is the force exerted by each of the four wheel cylinders. Note that we can add as many wheel cylinders as we wish. If each has a 2.50-cm diameter, each will exert [latex] 1.25\,×\,{10}^{4}\,\text{N}\text{.} [/latex] A simple hydraulic system, as an example of a simple machine, can increase force but cannot do more work than is done on it. Work is force times distance moved, and the wheel cylinder moves through a smaller distance than the pedal cylinder. Furthermore, the more wheels added, the smaller the distance each one moves. Many hydraulic systems—such as power brakes and those in bulldozers—have a motorized pump that actually does most of the work in the system.

Check Your Understanding

Would a hydraulic press still operate properly if a gas is used instead of a liquid?

Pressure is force per unit area.
A change in pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container.
A hydraulic system is an enclosed fluid system used to exert forces.

Conceptual Questions

Suppose the master cylinder in a hydraulic system is at a greater height than the cylinder it is controlling. Explain how this will affect the force produced at the cylinder that is being controlled.

How much pressure is transmitted in the hydraulic system considered in (Figure) ? Express your answer in atmospheres.

[latex] 5.76\,×\,{10}^{3}\,\text{N}\,\text{extra force} [/latex]

If the system is not moving, the friction would not play a role. With friction, we know there are losses, so that [latex] {W}_{\text{o}}={W}_{\text{i}}-{W}_{\text{f}}; [/latex] therefore, the work output is less than the work input. In other words, to account for friction, you would need to push harder on the input piston than was calculated.

OpenStax University Physics. Authored by : OpenStax CNX. Located at : https://cnx.org/contents/[email protected]:Gofkr9Oy@15 . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]

FREE K-12 standards-aligned STEM

curriculum for educators everywhere!

Find more at TeachEngineering.org .

TeachEngineering
Hydraulic Arm Challenge

Hands-on Activity Hydraulic Arm Challenge

Grade Level: 7 (6-8)

(4.5 40-minute class periods; see the Procedure section for details)

(between $20 and $30, depending on the number of groups)

Group Size: 2

Activity Dependency: None

Subject Areas: Science and Technology

NGSS Performance Expectations:

TE Newsletter

Engineering connection, learning objectives, materials list, worksheets and attachments, more curriculum like this, introduction/motivation, vocabulary/definitions, investigating questions, additional multimedia support, user comments & tips.

Just like engineers, students in this activity work in teams and follow the steps of the engineering design process . Engineers develop hydraulic arms for a variety of reasons. Hydraulic arms can be used in situations that are too difficult or dangerous for people to deal with directly or in automated systems. Examples include arms that lift heavy weights and arms that hold a load and unload them into a specific position.

After this activity, students should be able to:

Identify the steps of the engineering design process .
Recognize the steps of the engineering design process as they design and build.
Represent solutions to a design process in multiple ways.
Describe and explain features and purpose of a design.
Explain the basic concepts of hydraulic and pneumatics.

Educational Standards Each TeachEngineering lesson or activity is correlated to one or more K-12 science, technology, engineering or math (STEM) educational standards. All 100,000+ K-12 STEM standards covered in TeachEngineering are collected, maintained and packaged by the Achievement Standards Network (ASN) , a project of D2L (www.achievementstandards.org). In the ASN, standards are hierarchically structured: first by source; e.g. , by state; within source by type; e.g. , science or mathematics; within type by subtype, then by grade, etc .

Ngss: next generation science standards - science.

NGSS Performance Expectation
MS-ETS1-2. Evaluate competing design solutions using a systematic process to determine how well they meet the criteria and constraints of the problem. (Grades 6 - 8) Do you agree with this alignment? Thanks for your feedback!


This activity focuses on the following aspects of NGSS:
Science & Engineering Practices	Disciplinary Core Ideas	Crosscutting Concepts
Evaluate competing design solutions based on jointly developed and agreed-upon design criteria. Alignment agreement: Thanks for your feedback!	There are systematic processes for evaluating solutions with respect to how well they meet the criteria and constraints of a problem. Alignment agreement: Thanks for your feedback!

NGSS Performance Expectation
MS-ETS1-4. Develop a model to generate data for iterative testing and modification of a proposed object, tool, or process such that an optimal design can be achieved. (Grades 6 - 8) Do you agree with this alignment? Thanks for your feedback!


This activity focuses on the following aspects of NGSS:
Science & Engineering Practices	Disciplinary Core Ideas	Crosscutting Concepts
Develop a model to generate data to test ideas about designed systems, including those representing inputs and outputs. Alignment agreement: Thanks for your feedback!	Models of all kinds are important for testing solutions. Alignment agreement: Thanks for your feedback! The iterative process of testing the most promising solutions and modifying what is proposed on the basis of the test results leads to greater refinement and ultimately to an optimal solution. Alignment agreement: Thanks for your feedback!

International Technology and Engineering Educators Association - Technology

View aligned curriculum

Do you agree with this alignment? Thanks for your feedback!

State Standards

Massachusetts - science.

Each group needs:

plastic syringes, such as from McMaster-Carr for ~$1.27 per syringe; see note below
plastic tubing, such as from hardware and pet supply stores for ~$20; see note below
various wood scraps
bolts, screws, nuts, washers
other APPROVED materials
1 empty soda can

Note about syringes and tubing : In this activity, all the syringes must be exactly the same and it is important that the plastic tubing fits snugly on the tip of the syringe. A good option is a 50 or 60 cc plastic manual syringe with a tapered tip, such one available from McMaster-Carr —the "60 cc plastic manual syringe with taper tip" for $1.27 per syringe. It may be helpful to purchase the syringes first and then bring one to a hardware or pet supply store to find appropriate sized plastic tubing.

To share with the entire class:

20 x 20 cm piece of wood or cardboard to serve as a wall
drill (for teacher use or with appropriate supervision)
saw (for teacher use)
empty soda can

Have you ever seen a car lifted into the air at an auto repair place? Have you ever wondered how an elevator can lift a load of people up into the air? Well, after our project today, you'll have a better understanding of how these work, because we're going to look at hydraulic systems.

Hydraulic systems use a liquid, usually oil, to transmit force. This system works on the same principles as other mechanical systems and trades force for distance. Hydraulic systems are used on construction sites and in elevators. They help users perform tasks that they would not have the strength to do without the help of hydraulic machinery. They are able to perform tasks that involve large amounts of weight with seemingly little effort.

Suggested Timing

This activity is comprised of two parts:

Part 1 - Investigating Pneumatics and Hydraulic Systems Handout: 1 ½ - 2 class periods at 40 minutes each.
Part 2 – Creating the hydraulic arm: three 40-minute classes (This activity can be done in fewer class periods, but giving students this amount of time enables them to test numerous design ideas and further understand the engineering design process and the underlying concepts.)

Hydraulic systems are used in many different types of machines: control surfaces on airplanes, elevators, automobile lifts, and backhoes. The idea behind a hydraulic system is that force is applied to one point and is transmitted to a second point using an incompressible fluid. You can find detailed background information oon how hydraulic machines work at http://science.howstuffworks.com/transport/engines-equipment/hydraulic1.htm .

Before the Activity

A sketch shows a tall barrier in the middle of the image. On the left side is a drawing of a can. On the right side of the image is a bulls eye that is the target for the can. There is an arrow beginning at the can. It goes over the barrier wall and points to the target.

Make copies of the journals and handouts.
Gather materials.

With the Students

Divide the class into groups of two students each. Have each design team:

Research the engineering design process and answer the questions on the Investigating Pneumatics and Hydraulics Systems Student Handout.
Research possible solutions to the challenge. Tips: Look for pictures of other mechanical arms (or parts of arms) that perform functions similar to the ones that they must perform. Think about the connection between their team's component and the components it is connect to. The connections are the most challenging part!
Develop a portfolio (a collection) of sketches that attempt to solve the problem. Share with the entire design team. Upon identifying a promising design, brainstorm with the next design team about attaching them together. Critique (be nice, constructive) the designs and make a short list of pros (+) and cons (-) for each idea. Identify the best ideas and vote to decide upon them.
Make final engineering sketches of the parts that are needed.
Construct the prototypes, noting changes, modifications, failures and successes. It is perfectly fine to mark up your engineering sketches. Show your work!
Test the prototype. TRY TO MAKE IT FAIL. What do you have to do to get it to fail? Can you redesign it to prevent that from happening? Make your design the best it can be. (Students like to make their designs fail. They understand that as an instruction and see it as a good mindset for testing prototypes.)
Write down information on how long it took for your device to fail.
Redesign and reconstruct.
Once satisfied, plot your found data to see how your device improved as you modified it.
Present the portfolio of marked-up drawings, the finished arm, and demonstrate the arm to the class.

hydraulics: Involving or moved by fluid under pressure.

pneumatics: Involving the mechanical properties of air and other gases. Safety Factor(N): A number used to describe how much more force your device should withstand past the max expected force based on a number of parameters such as material and dimensions (N=1 means only can withstand 100% of expected force, so it will fail at 101% of expected load).

prototype: A working model of a new product or new version of a product.

Activity Embedded Assessment : Administer the Arm Investigating Questions and Design Check List.

Post-Activity Assessment : Evaluate the student project using the attached Hydraulic Arm Rubric, with criteria on research, imagining-planning-improving, creativity, written or oral sharing, and how the mechanism meets the challenge.

Making Sense: Have students reflect about the science phenomena they explored and/or the science and engineering skills they used by completing the Making Sense Assessment .

See Researching the Engineering Design Process Handout

Safety Issues

Cut and drill the wood if students do not have experience.

Students learn about the fundamental concepts important to fluid power, which includes both pneumatic (gas) and hydraulic (liquid) systems.

Carpenito, K. and E. Chilton. Hydraulic Arms Challenge. Posted January 2006. Accessed November 7, 2011. (activity inspiration) https://docs.google.com/View.aspx?docid=ah7pxzjtrzfd_baddp39ndp3dv

Hydraulic Arm Research. Posted January 27, 2006. Beebe School of Engineering. Accessed November 7, 2011. (a list of references to support this activity, including info on the arm joint and the engineering design process) http://k12engineering.blogspot.com/2006/01/hydraulic-arm-research.html

Brain, Marshall. How Hydraulic Machines Work. How Stuff Works. Accessed November 7, 2011. http://science.howstuffworks.com/hydraulic.htm

Contributors

Supporting program.

Last modified: October 31, 2020

Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices.

Experiment #1: Hydrostatic Pressure

1. introduction.

Hydrostatic forces are the resultant force caused by the pressure loading of a liquid acting on submerged surfaces. Calculation of the hydrostatic force and the location of the center of pressure are fundamental subjects in fluid mechanics. The center of pressure is a point on the immersed surface at which the resultant hydrostatic pressure force acts.

2. Practical Application

The location and magnitude of water pressure force acting on water-control structures, such as dams, levees, and gates, are very important to their structural design. Hydrostatic force and its line of action is also required for the design of many parts of hydraulic equipment.

3. Objective

The objectives of this experiment are twofold:

To determine the hydrostatic force due to water acting on a partially or fully submerged surface;
To determine, both experimentally and theoretically, the center of pressure.

In this experiment, the hydrostatic force and center of pressure acting on a vertical surface will be determined by increasing the water depth in the apparatus water tank and by reaching an equilibrium condition between the moments acting on the balance arm of the test apparatus. The forces which create these moments are the weight applied to the balance arm and the hydrostatic force on the vertical surface.

5. Equipment

Equipment required to carry out this experiment is the following:

Armfield F1-12 Hydrostatic Pressure Apparatus,
Calipers or rulers, for measuring the actual dimensions of the quadrant.

6. Equipment Description

The equipment is comprised of a rectangular transparent water tank, a fabricated quadrant, a balance arm, an adjustable counter-balance weight, and a water-level measuring device (Figure 1.1).

The water tank has a drain valve at one end and three adjustable screwed-in feet on its base for leveling the apparatus. The quadrant is mounted on a balance arm that pivots on knife edges. The knife edges coincide with the center of the arc of the quadrant; therefore, the only hydrostatic force acting on the vertical surface of the quadrant creates moment about the pivot point. This moment can be counterbalanced by adding weight to the weight hanger, which is located at the left end of the balance arm, at a fixed distance from the pivot. Since the line of actions of hydrostatic forces applied on the curved surfaces passes through the pivot point, the forces have no effect on the moment. The hydrostatic force and its line of action (center of pressure) can be determined for different water depths, with the quadrant’s vertical face either partially or fully submerged.

A level indicator attached to the side of the tank shows when the balance arm is horizontal. Water is admitted to the top of the tank by a flexible tube and may be drained through a cock in the side of the tank. The water level is indicated on a scale on the side of the quadrant [1].

Diagram of the Armfield F1-12 Hydrostatic Pressure Apparatus. The top of the apparatus consists of a balance arm fixed with a counterbalance on the right end and a level indicator on the left end. The arm is held in place by a clamping screw and knife edge pivot. A weight hanger is clipped to the pointed end of the balance arm on the left hand side. In the center of the apparatus is a plastic container which holds the scale. The plastic container is held up off the table by three adjustable feet. There is a drain valve fixed to the bottom right hand side of the apparatus.

In this experiment, when the quadrant is immersed by adding water to the tank, the hydrostatic force applied to the vertical surface of the quadrant can be determined by considering the following [1]:

The hydrostatic force at any point on the curved surfaces is normal to the surface and resolves through the pivot point because it is located at the origin of the radii. Hydrostatic forces on the upper and lower curved surfaces, therefore, have no net effect – no torque to affect the equilibrium of the assembly because the forces pass through the pivot.
The forces on the sides of the quadrant are horizontal and cancel each other out (equal and opposite).
The hydrostatic force on the vertical submerged face is counteracted by the balance weight. The resultant hydrostatic force on the face can, therefore, be calculated from the value of the balance weight and the depth of the water.
The system is in equilibrium if the moments generated about the pivot points by the hydrostatic force and added weight (=mg) are equal, i.e.:

$mg\times L = F\times y \qquad (1)$

m : mass on the weight hanger,

L : length of the balance arm (Figure 1.2)

F : Hydrostatic force, and

y : distance between the pivot and the center of pressure (Figure 1.2).

Then, calculated hydrostatic force and center of pressure on the vertical face of the quadrant can be compared with the experimental results.

7.1 Hydrostatic Force

The magnitude of the resultant hydrostatic force (F) applied to an immersed surface is given by:

$F=P_c A=\rho g y_c A\qquad (2)$

P c : pressure at centroid of the immersed surface,

A : area of the immersed surface,

y c : centroid of the immersed surface measured from the water surface,

$\rho$

g : acceleration due to gravity.

The hydrostatic force acting on the vertical face of the quadrant can be calculated as:

Partially immersed vertical plane (Figure 1.2a):

$\displaystyle F=\frac{1}{2} \rho gBd^2\qquad (3a)$

Fully immersed vertical plane (Figure 1.2b):

$\displaystyle F=\rho gBD \left( d-\frac{D}{2} \right)\qquad (3b)$

B : width of the quadrant face,

d : depth of water from the base of the quadrant, and

D : height of the quadrant face.

7.2 Theoretical Determination of Center of Pressure

The center of pressure is calculated as:

$\displaystyle y_p=\frac{I_x}{A y_c}\qquad (4)$

Partially immersed vertical plane:

$\displaystyle I_x =\frac{Bd^3}{12}+Bd\left(\frac{d}{2}\right)^2=\frac{Bd^3}{3} \qquad (6a)$

Fully immersed vertical plane:

$\displaystyle I_x=BD \left[ \frac{D^2}{12}+\left(d-\frac{D}{2}\right)^2 \right] \qquad (6b)$

The depth of the center of pressure below the pivot point is given by:

$y=y_p+H-d \qquad (7)$

in which H is the vertical distance between the pivot and the base of the quadrant.

Substitution of Equation (6a and 6b) and into (4) and then into (7) yields the theoretical results, as follows:

$y=\displaystyle H-\frac{d}{3}\qquad (8a)$

Fully immersed vertical rectangular plane (Figure 1.2b):

$y =\displaystyle \frac{\frac{ D^2}{12}+(d-\frac{D}{2})^2}{d-\frac{D}{2}}+H-d\qquad (8b)$

7.3 Experimental Determination of Center of Pressure

For equilibrium of the experimental apparatus, moments about the pivot are given by Equation (1). By substitution of the derived hydrostatic force, F from Equation (3a and b), we have:

$y=\displaystyle \frac{m g L}{F}=\frac{2mL}{\rho Bd^{\\2} }\qquad (9a)$

8. Experimental Procedure

Begin the experiment by measuring the dimensions of the quadrant vertical endface ( B and D ) and the distances ( H and L ), and then perform the experiment by taking the following steps:

Wipe the quadrant with a wet rag to remove surface tension and prevent air bubbles from forming.
Place the apparatus on a level surface, and adjust the screwed-in feet until the built-in circular spirit level indicates that the base is horizontal. (The bubble should appear in the center of the spirit level.)
Position the balance arm on the knife edges and check that the arm swings freely.
Place the weight hanger on the end of the balance arm and level the arm, using the counter weight, so that the balance arm is horizontal.
Add 50 grams to the weight hanger.
Add water to the tank and allow time for the water to settle.
Close the drain valve at the end of the tank, then slowly add water until the hydrostatic force on the end surface of the quadrant is balanced. This can be judged by aligning the base of the balance arm with the top or bottom of the central marking on the balance rest.
Record the water height, which displayed on the side of the quadrant in mm. If the quadrant is partially submerged, record the reading in the partially submerged portion of the Raw Data Table.
Repeat the steps, adding 50 g weight each time, until the final weight of 500 g is reached. When the quadrant is fully submerged, record the readings in the fully submerged part of the Raw Data Table.
Repeat the procedure in reverse by progressively removing the weights.
Release the water valve, remove the weights, and clean up any spilled water.

9. Results and Calculations

Please visit this link for accessing the excel workbook for this experiment.

Record the following dimensions:

Height of quadrant endface, D (m) =
Width of submerged, B (m)=
Length of balance arm, L (m)=
Distance from base of quadrant to pivot, H (m)=

All mass and water depth readings should be recorded in the Raw Data Table:

Raw Data Table

Test No.		Mass, m (kg)	Depth of Immersion, d (m)
Partially submerged	1
	2
	3
	4
	5
Fully Submerged	6
	7
	8
	9
	10

9.2 Calculations

Calculate the following for the partially and fully submerged quadrants, and record them in the Result Table:

Hydrostatic force ( F )
Theoretical depth of center of pressure below the pivot (y)
Experimental depth of center of pressure below the pivot (y)

Result Table

Test No.	Mass m(kg)	Depth of immersion d(m)	Hydrostatic force F(N)	Theoretical depth of center of pressure (m)	Experimental depth of center of pressure (m)
1
2
3
4
5
6
7
8
9
10

Use the template provided to prepare your lab report for this experiment. Your report should include the following:

Table (s) of raw data
Table (s) of results
Hydrostatic force (y-axis) vs depth of immersion (y-axis),
Theoretical depth of center of pressure (y-axis) vs depth of immersion (x-axis),
Experimental depth of center of pressure (y-axis) vs depth of immersion (x-axis),
Theoretical depth of centre of pressure (y-axis) vs experimental depth of center of pressure (x-axis). Calculate and present value for this graph, and
Mass (y-axis) vs depth of immersion (x-axis) on a log-log scale graph.
Comment on the variations of hydrostatic force with depth of immersion.
Comment on the relationship between the depth of the center of pressure and the depth of immersion.
For both hydrostatic force and theoretical depth of center of pressure plotted vs depth of immersion, comment on what happens when the vertical endface of quadrant becomes fully submerged.
Comment on and explain the discrepancies between the experimental and theoretical results for the center of pressure.

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Hydraulic Lift: Working, Hydraulic System & Pascal's Law

Collegedunia Team

Content Curator

Hydraulic Lift is a device that lifts things by applying force to a liquid inside a cylinder, which drives a piston higher.

The piston is pushed higher by incompressible oil injected into the cylinder.
The piston descends due to gravity force when a valve opens to discharge the oil.
The hydraulic lift concept is based on Pascal’s Law for producing force or motion.
Pascal’s Law states that pressure change on an incompressible liquid in a confined area is distributed evenly in all directions.
A hydraulic system works when the applied force is at one point to an incompressible liquid, sending the force to a second point.
The method involves two pistons connected via an oil-filled pipe.

In a hydraulic lift, the two pistons are seen to be separated by the space filled with a liquid. A piston here has a small cross-section A 1 that aids in the exertion of a force, i.e. F 1 directly on the liquid. The pressure indicated by P =F/A is further transferred throughout the liquid to the larger cylinder that is attached to a bigger piston of A 2 area resulting in an upward force of P × A 2 .

What is a Hydraulic Lift?

[Click Here for Sample Questions]

A hydraulic lift is used for moving objects using the force created by pressure on a liquid inside a cylinder. This force moves a piston upward. The hydraulic Lift Principle is based on Pascal‘s law for generating force or motion. The principle states that pressure change on an incompressible liquid in a confined space is passed equally throughout the liquid in all directions.

The hydraulic lift is used typically in automobiles. In it, there are two pistons that are separated by a space filled with a liquid.
A piston with a small cross-section A 1 has been used to exert a force F 1 directly on the liquid.
The pressure characterised by P = F/A is then transferred throughout the liquid to the larger cylinder attached to a larger piston of area A 2 , resulting in an upward force P × A 2 .
Therefore, it can be said that the piston can support a large force, which is the large weight of any car or truck set on the platform.
By varying the area A 1 , the platform can be shifted either up or down.
Therefore, the applied force is thus increased by a factor of A 2 /A 1 and this factor can be expressed as the mechanical advantage of the device.

Examples of Hydraulic Lift

Hydraulic lifts are used for moving goods or people vertically. Some common examples are:

Scissor lifts
Two-post lifts
Four-post lifts
Carousel lifts

Concept Related Topics

Pascal’s Law

[Click Here for Previous Year Questions]

Pascal’s Law states that:

Any surface in touch with the fluid is subjected to static pressure, which operates at right angles to it. Pascal also discovered that the pressure at a given location in a static fluid is the same in all planes travelling through that point. Pascal's law is also known as Pascal's principle or the principle of fluid-pressure transfer.

Hydraulic lift Pascal's Law states that hydraulic lift enables the usage of a relatively small input force to form a larger output force . It is because the pressure applied on one piston is equivalent to the pressure of the second piston.

Pascal's Law

Pascal Law Formula

The formula of Pascal law is:

F = Force applied
P = Pressure transmitted
A = Cross-sectional area

and area of B is 4,200 cm , determine the external input force of F.

Force of F calculated using the equation of Pascal’s principle:

F / A = F / A

F / 60 cm = 3500 N / 4200 cm

F / 60 = 35 N / 42

F = (60)(35) / 42

F = 2100 / 42

= 50 Newton

Hydraulic Lift Parts

Parts of Hydraulic Lift are elaborated below:

Hydraulic Circuit

The flow and pressure of the liquid in the system are controlled by hydraulic circuits. The diagram below depicts the many components of a hydraulic circuit:

Hydraulic Pump

The hydraulic pump is a device that transforms mechanical energy into hydraulic energy. At the pump inlet, hydraulic pumps generate a vacuum, which drives liquid from the reservoir into the intake line and out to the hydraulic system's outlet.

Hydraulic Motor

A hydraulic motor is a device that converts hydraulic pressure into spin and torque. It works similarly to a linear actuator in that it converts hydraulic energy's pressure and flow into rotating mechanical energy. The hydraulic motor is pushed by the pump, which provides hydraulic energy into the system.

Hydraulic Cylinder

The hydraulic cylinder transforms the hydraulic fluid's energy into force and starts the pressure in the fluid, which is regulated by the hydraulic motor.

The Fluid pressure moves hydraulic pistons in a straight path. Axial designs have a revolving housing with a number of pistons arranged in a circular arrangement.

Applications of Hydraulic Lift

Applications of Hydraulic Lift are:

Hydraulic Elevator

It is a type of elevator that works by using fluid pressure created by a suitable fluid. It's used to elevate cars in garages and service facilities.

Two pistons are separated by a liquid-filled gap in a hydraulic lift.
To apply a force F 1 directly on the liquid, a piston with a tiny cross-section A 1 is employed.
P =F/A is transferred through the liquid to the bigger cylinder linked to a larger piston of area A 2 , resulting in an upward force of P x A 2 .
As a result, the piston can withstand a significant amount of force (large weight of, say a car, or a truck, placed on the platform).
The platform may be moved up or down by adjusting the force at A 1 .
As a result, the applied force is enhanced by a factor of A2/A1, which is the device's mechanical advantage.

Hydraulic Brake

The hydraulic brake is a type of braking system in which appropriate brake fluid is utilised to transmit pressure from the control system to the braking system.

Automobile hydraulic brakes function on the same basis.
When we press down on the pedal, the master piston inside the master cylinder moves, and the resulting pressure is conveyed through the brake fluid to operate on a bigger piston.
The piston is forced down by a significant force, expanding the braking shoes against the brake lining.
A little push on the pedal creates a significant retarding force on the wheels in this manner.

Chapter Related Articles

Things to Remember

Hydraulic lift technology is used in a variety of industries, including construction and transportation.
It is frequently used to operate heavy machinery or to transport and carry huge and heavy things such as vehicles, dirt, and shipping containers.
"Pressure x area = force" is related to hydraulic lift technology. This aids in determining the amount of pressure that must be exerted on a liquid in a piston in order to create an adequate force to lift and move an item.
Hydraulics are used in vehicle brakes as well. When you use the brakes, a tiny piston in the brake master cylinder is pushed.
Hydraulic lift technology may be found in a variety of devices, including hydraulic jacks, forklifts, and vehicle lifts. It may be used by machines to give the necessary lift effort (force) to perform work, like moving another item.

Sample Questions

Ques: Is Pascal's Law valid for gases? (1 mark)

Ans: Pascal's Law holds true for gases. Pascal's concept is also known as the principle of fluid (water or gas) pressure transfer.

Ques: What is Hydraulic lift in Pascal's law? (1 mark)

Ans: Pascal's law claims that when there is a pressure increase at any point in a confined fluid, there is an equivalent increase at every other point in the container. For example, the hydraulic lift for automobiles shows a force multiplied by a hydraulic press and is based on Pascal's principle.

Ques: What is the formula for Hydraulic Lift? (1 mark)

Ans: The pressure P = F/A is conveyed throughout the liquid of the larger cylinder which has been attached to a larger piston of area A 2 , resulting in an upward force of P × A 2 .

Ques: In a hydraulic machine, two pistons of area of cross-section are in the ratio 1:10. Determine the force required on the narrow piston to withstand a force of 100N on the wider piston. (2 marks)

Ans: F 1 =100N

Since pressure = constant

$\frac{F_1}{A_1} = \frac{F_2}{A_2} = \frac{100}{10} = \frac{x}{1}$

Hence, x = 10 N

Ques: What is the meaning of the hydrostatic paradox? (2 marks)

Ans: The hydrostatic paradox in fluid dynamics refers to the liquid pressure at all locations at the same depth (horizontal level).

"The pressure at a given horizontal level in the fluid is proportional to the vertical distance to the fluid's surface," says the definition.

Ques: Explain the working of hydraulic lift. (2 marks)

Ans: There are two pistons in the system. The object to be raised is held on the bigger piston while the force is given to the smaller one. When we wish to raise a box or a weight, we provide an equivalent pressure to the smaller piston, which is transferred to all directions in the liquid, and therefore to the bigger piston as well. As a result, the weight is lifted. This is how a hydraulic lift works.

Ques: What is Pascal's principle? (2 marks)

Ans: Pascal's principle (also known as Pascal's law) says that when a change in pressure is made to an enclosed fluid, it is transferred unchanged to all parts of the fluid as well as the container's walls. Because the fluid's atoms are allowed to move about in an enclosed fluid, they transfer pressure to all areas of the fluid as well as the container's walls. Any change in pressure is unaffectedly communicated.

Ques: Would a hydraulic press still operate properly if a gas is used instead of a liquid? (2 marks)

Ans: Yes, it would function, but because gas is compressible, it would not perform very efficiently. When the force is applied, the gas compresses and warms up first. As a result, in order for the brakes to operate correctly, the air in the brake lines must be bled out.

Ques: How will you hydraulic lift an automobile? (3 marks)

Ans: If the lift cylinder has a diameter of 25 cm and the small cylinder has a diameter of 1.25 cm, the ratio of the areas is 400, and the hydraulic press arrangement produces a force multiplier of 400 times.

To raise a 6000-newton automobile, just 6000 N/400 = 15 N of force would be applied to the fluid in the tiny cylinder.
To elevate the automobile 10 centimetre, however, the oil would have to be moved 400 x 10cm = 40 metres.
This is accomplished by using a tiny compressor to pump oil into this little cylinder

Ques: In order to lift an automobile of 5000 kg, a hydraulic lift with a piston of 800 cm 2 in area has been used. Determine the force applied to a small piston of area 5 cm 2 to achieve the same. (5 marks)

Ans: Force applied on the large piston (F 1 ) = mg

Thus, F 1 = 5000 × 9.8 N = 4.9 × 10 4 N

The area of large piston, as given, A 1 = 800 cm 2

Pressure of the large piston (P) = $\frac{F_1}{A_1} = \frac{4.9 \times 10^4}{800 \times 10^{-4}}$ = 6.125 x 10 5 Pa

Now, because the liquid transfers pressure equally, the pressure on the piston is going to be equivalent to the pressure on the larger piston.

It can be said, $\frac{F_1}{A_1} = \frac{F_2}{A_2} = P$

Thus, the given area of the smaller piston A 2 = 5 cm 2

Force on smaller piston is (F 2 ) = P x A 2 = 6.125 × 10 5 × (5 × 10 −4 ) = 306.25 N

Previous Year Questions

A soap bubble, having radius of 1mm, is blown from a detergent solution… [NEET 2019]
A liquid is filled upto a height of 20cm20cm in a cylindrical vessel… [KEAM]
The force acting on a window of area 50cm×50cm of a submarine… [KEAM]
A small spherical ball falling through a viscous medium of negligible density… [KEAM]
A spherical ball of diameter 1cm and density… [KEAM]
Ice pieces are floating in a beaker A containing water and also… [KEAM]
A boy's catapult is made of rubber cord which is… [JEE Main 2019]
A man grows into a giant such that his linear dimensions… [JEE Main 2017]
If the ratio of diameters, lengths and Young's modulus of steel… [NEET 2013]
Two wires are made of the same material and have the same volume…. [NEET 2018]
When a block of mass M is suspended by a long wire of length… [NEET 2019]
Two wires A and B are stretched by the same load… [KCET 2018]
Young's modulus is defined as the ratio of… [KCET 2017]
Which of the following substances has the highest elasticity… [KCET 2010]
The ratio of hydraulic stress to the corresponding strain is known as… [KEAM]
Hydraulic Lift works on the basis of… [UPSEE 2017]

CBSE CLASS XII Related Questions

1. figure shows tracks of three charged particles in a uniform electrostatic field. give the signs of the three charges. which particle has the highest charge to mass ratio, 2. a boy of mass 50 kg is standing at one end of a, boat of length 9 m and mass 400 kg. he runs to the other, end. the distance through which the centre of mass of the boat boy system moves is, 3. (a) a circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 a is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 t. the field lines make an angle of 60° with the normal of the coil. calculate the magnitude of the counter torque that must be applied to prevent the coil from turning. (b) would your answer change, if the circular coil in (a) were replaced by a planar coil of some irregular shape that encloses the same area (all other particulars are also unaltered.), 4. a series lcr circuit with r = 20 w, l = 1.5 h and c = 35 μf is connected to a variable-frequency 200 v ac supply. when the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle, 5. two charges 5 × 10 –8 c and –3 × 10 –8 c are located 16 cm apart. at what point(s) on the line joining the to charges is the electric potential zero take the potential at infinity to be zero., 6. a convex lens of glass is immersed in water compared to its power in air, its power in water will.

decrease for red light increase for violet light

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6 experiments for fluid dynamics.

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Introduction

AP Physics 1 has added Unit 8: Fluids. This video and article will give you some timely tips on how to deliver this content.

The thing about teaching fluids in introductory physics is that there are three main principles: Pascal, Archimedes, and Bernoulli. Below we will outline six Fluid Physics Experiments – two for each – that will help you instruct on these important ideas.

Figure 1. Pascal’s Vases show us that pressure depends on depth and not area.

#1 The Definition of Pressure

Pascal gives us the definition of pressure, that P = F/A it is a force over an Area. Pressure is something that exists in a fluid, it is everywhere equal, at a given height. And, pressure, being a force causes acceleration. Fluids will flow from a place of high pressure to a place of low pressure, by Newton’s 2nd Law, it’s a net force. And because of area difference, it is possible to establish a mechanical advantage. Either using hydraulic pressure (liquids) or pneumatic pressure (gases).

Figure 2. A small syringe can have a mechanical advantage over a large one syringe because the pressure in the fluid, be it air or water, is constant throughout. P = F/A = F / A

But, pressure can also be measured by depth, Pascal explains this too. P = ρgh means that pressure doesn’t depend on area, only depth h and density ρ (usually constant), so we can use depth to predict pressure. Torricelli found the atmospheric pressure to be strong enough to lift Mercury to a height of 760 mm, and less so on the mountains. But if you use water, you can lift it to a height of 10 meters.

Figure 3. Torricelli’s experiment, the atmosphere can lift liquid mercury 760 mm, but no further. The space above is vacuum.

Figure 4. Depth increases the pressure in a fluid, this is explained by the formula P = ρgh. Where ρ is density. The manometer at right shows the pressure is higher than the atmosphere: the right end of the U-shaped tube is open for direct comparison.

Thus, if you were to dive 10 meters deep, you would experience an additional atmosphere of pressure (by ρgh) and you would definitely feel it. Because the ratio of 760mm to 10 meters is about 13 to 1, we conclude that mercury has a density that is 13 about times greater than water.

#2 The Concept of Fluid Density

If we are going to understand fluid physics - we need to use the MKS units of kilograms per meter cubed. I suggest a lab on these density cubes – which all have different masses - despite being the same size.

Figure 5. The density blocks can be a fun window into teaching the MKS units of kg/m3 that are used extensively in physics.

As a lab, try to measure their densities in the new units of kg/m3 which turns out to be 1000 times bigger than g/cm 3 . For example the density of water is 1000 kg / m 3 . In the lab, you can try to guess which block is which based on the density you measure and comparing to a chart.

Figure 6. The float test is one way to distinguish the blocks nylon (sinks) and polypropylene (floats) which otherwise look almost identical.

For a liquids example, comparing the density of salt water to regular water can be done easily with a density manometer. It helps to keep the waters cold so that they don’t mix quickly. The ratio is about 6 to 5 for fully saturated salt water (be sure to dye them different colors, and use salt without iodine). Thus, saturated salt water is 20% denser, or 1200 kg/ m 3 . The equilibrium statement of P = P becomes ρ 1 gh 1 = ρ 2 gh 2 .

Figure 7. Salt water (green) vs. Pure Acetone (pink) the density ratio is 3 to 2. The zero line is established at the height at which the substances meet. The green is just food dye but the pink is from a little nail polish. The salt must be pure and not iodized to get this high density of 1.2 g/mL.

My favorite demo with the density manometer is to compare saturated salt water to pure acetone… These two liquids will not mix and are -- easy to clean. I have added a little nail polish as a dye for the pure acetone. The ratio is a - quite large - 3 to 2, thus acetone has a density of only about .8 g/cm3 or 800 kg/ m 3 .

#3 Floating and Bouyancy

Archimedes Principle is that the weight of the displaced fluid is the same as the buoyant force. F = pVg But, this too is a consequence of pressure increasing with depth.

Figure 8. The pressure increase with depth is canceled out left and right, but not up and down. Thus, the net force from the water is Fnet = Fup – Fdown = P2*A – P1*A = ρgh2*A - ρgh1*A = ρgΔh*A = ρgV = Mg where M is the mass of the displaced fluid. which is the formula for buoyancy. (Recall that in fluids h means depth.) So, we see that buoyancy is really caused by depth pressure.

As we get deeper underwater the pressure increases, therefore the force is greater on the bottom than the top. These left and right side arrows cancel out. The result is that the pressure difference between the top and bottom, equal to ρgh, becomes a force equal to the volume of the object times the fluid density. Since density times volume is mass and with g weight, we easily see the weight of the fluid is equal to the net fluid force on the object, this is the buoyant force.

Figure 9. Pine floats about 50% below the surface of water and from that we immediately know its density is near .50 g/mL = 500 kg/m3 half that of water.

Floating objects of course are completely lifted by the fluid, so their buoyant force is the same as their weight. But not all objects float the same. Here, we see pine floats about halfway, while Styrofoam floats about 90% above the surface. Ice, famously is about 90% below, it depends if the water is fresh or salty.

Figure 10. Ice floats 92% below the surface, thus it has a density of .92 g/mL or 920 kg/ m 3 .

Generally, we can tell the density based on how well the object is floating. The % amount an object will sink is the same as its density, compared to water or the fluid it is in. As an example, ice sinks 92% in water, so its density is 92%. A styrofoam boat will sink more and more as I add passengers (washers) until it has the same density as water at which point it completely sinks.

Figure 11. As a lab, we will fill a Styrofoam boat with passengers until it completely sinks. Predicting this number is a good challenge and it usually works perfectly. The sinking moment occurs when the density of the boat matches the density of water.

#4 Boyle's Law:

Now the compressibility of gases is NOT tested in AP Physics 1 – only AP 2, so why is Boyle’s Law still an appropriate lab for AP Physics students?

Well , students probably already know gases are compressible from 8th grade physical science, and they are also important examples of fluids. But the compression of a gas under pressure is one of the surest metrics by which to gauge the pressure of a fluid.

Figure 12. The Elasticity of Gases Demo is a great way to get to know the formula for pressure and to make a measurement of the ambient atmospheric pressure.

When we add books to an Elasticity of Gases Demo we are demonstrating a familiarity with the formula for pressure: P = F/A. Each book provides a unit force of about 20 N and the syringe plunger's the area is constant at .0004 m2, thus, the pressure is increasing in 50,000 pascals (n/m2) with each book. A pascal is the unit of pressure and the atmosphere itself is measured by this experiment to be about 100,000 N/m2 of pressure. And that value is on the AP Physics test. Therefore, we conclude that yes, this is a great experiment on the physics of fluids.

#5 The Spouting Cylinder:

The spouting cylinder is a classic physics demonstration and a staple problem in AP Physics. Holes at different heights are permitted to emit water when subjected to fluid pressure. The question is, which spout will shoot the furthest? The lower spout has more pressure due to depth and so goes faster, but the upper spout has the height advantage.

The answer is to apply Torricelli’s theorem: ρgh = ½ ρv 2 which is a simpler version of Bernoulli’s Theorem. In short, the conservation of energy can be applied to fluids. Now that’s not how Torricelli said it, but this, combined with the laws of projectile motion, results in the maximum distance occurring always when the depth is equal to the height of the spout, or exactly right in the middle.

Figure 13. The Spouting Cylinder is a classic demonstration in AP Physics and demonstrates that the principles of projectile motion and conservation of energy can be applied to fluids.

#5 Bernoulli's Principal:

A fast-moving fluid is at a lower pressure, that is Bernoulli’s Principle, and it is somewhere here in the Bernoulli Equation:

P + ρgy + ½ ρv 2 = P + ρgy + ½ ρv 2

This principle explains the airplane lift force, because if there is a faster moving fluid over the curved wing it will generate a pressure difference between the top and bottom. Now, while you might not have an airplane at the ready, you do have a frisbee.

Figure 14. Airplane lift is explained by Bernoulli’s formula. When there is a greater velocity v, there is lower pressure P. This pressure reduction occurs on the top because the larger curve over the top compared to the bottom.

The frisbee displays a similar curving and obeys the same principle. So, explain that frisbees, when thrown through the air, are seen to float easily because they receive lift from the Bernoulli’s principle of fast-moving fluids.

The purpose of the spin, on the other hand, is that the frisbee will maintain a flat orientation so that the air is always lifting it up, always a lower pressure on top.

Figure 15. The frisbee is actually a great example of the Bernoulli Principle in effect. With a fluid streamline almost identical to that of the airplane wing.

So, play frisbee with your students. But, let us now ask, what if I were to glue two frisbees together in a macaroon shape and give that a throw. How would the range compare to two frisbees nested together.

Figure 16. The washers are taped to the yellow frisbee to match the mass of the double frisbee on the right. But this is optional not essential. The double frisbee loses either way.

The result, we can prove by experiment, is that a double frisbee will not have a lift advantage over a single frisbee, even when accounting for the weight increase. The added washers, which are taped in place, are optional and the experiments works well in either case and is quite convincing: a double frisbee will never go as far because it lacks the lift force to keep it aloft.

James Lincoln

Physics Instructor

James Lincoln is an experienced physics teacher with graduate degrees in education and applied physics. He has become known nationally as a physics education expert specializing in original demonstrations, the history of physics, and innovative hands-on instruction. The American Association of Physics Teachers and the Brown Foundation have funded his prior physics film series and SCAAPT's New Physics Teacher Workshops. Lincoln currently serves as the Chair of AAPT's Committee on Apparatus and has served as President of the Southern California Chapter of the AAPT, as a member of the California State Advisory for the Next Generation Science Standards, and as an AP Physics Exam Reader. He has also produced Videos Series for UCLA's Physics Demos Project, Arbor Scientific, eHow.com, About.com, and edX.org.

September 17, 2024 James Lincoln

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COMMENTS

Lift a Load Using Hydraulics
All three model hydraulic lifts should have a primary and secondary syringe, and hydraulic fluid, or water, in the primary syringe and tubing. Make sure that the hydraulic fluid, or water, is in the tubing and primary syringe. The secondary piston must be depressed in its cylinder. Place the vise on a sturdy table.
Lift a Load using Liquids
On the left, a hydraulic lift where the secondary syringe is much wider than the primary syringe. A large toy truck is on the platform. The secondary piston does not stick out by much while the primary piston is fully pushed in. On the right, a similar hydraulic lift where the secondary syringe is much narrower. A toy car is on the platform.
Lift a Large Load Using Liquids
Place the end of the small syringe's tube back in the water, and suck in some more water. Very little air should be left in either of the tubes or helping syringes. The small syringe should be ...
fluid mechanics hydraulic lift experiment
fluid mechanics hydraulic lift experiment by me :)a project on fluid mechanics 1st semester s.y. 2013-2014
Pascal's Principle, Hydraulic Lift System, Pascal's Law of Pressure
This physics video tutorial provides a basic introduction into pascal's principle and the hydraulic lift system. It explains how to use pascal's law of pres...
How to Build a Hydraulic Lift
Using your other syringe, draw as much of the colored water into the syringe as possible by pulling the plunger up. Attach the other end of the plastic tubing to the full syringe, and your hydraulic lift is ready for use! Simply push the plunger of the full syringe down, forcing the liquid through the tubing and into the syringe glued into the ...
14.3 Pascal's Principle and Hydraulics
14 Fluid Mechanics. Introduction; 14.1 Fluids, Density, and Pressure; 14.2 Measuring Pressure; ... Watch a simulation of Pascal's 1646 experiment, ... A hydraulic jack is used to lift heavy loads, such as the ones used by auto mechanics to raise an automobile. It consists of an incompressible fluid in a U-tube fitted with a movable piston on ...
Hydraulic Lift Example Problem
This tutorial works through a hydraulic lift example..This video is part of a full free course on fluid mechanics. The course covers fluid statics, fluid dyn...
Simple Hydraulic Lifter
In car repair mechanics use a hydraulic lift to raise vehicles. Many industries use huge lifters to load and unload goods from vehicles or ships. They can be used to lift personnel to, for example, service street lights. Hydraulic systems are also used in lifts within buildings. Hydraulic lifts can be controlled remotely or manually.
14.3 Pascal's Principle and Hydraulics
14 Fluid Mechanics. Introduction. 14.1 Fluids, Density, and Pressure. ... Watch a simulation of Pascal's 1646 experiment, ... The hydraulic jack is such a hydraulic system. A hydraulic jack is used to lift heavy loads, such as the ones used by auto mechanics to raise an automobile. It consists of an incompressible fluid in a U-tube fitted ...
14.3 Pascal's Principle and Hydraulics
The hydraulic jack is such a hydraulic system. A hydraulic jack is used to lift heavy loads, such as the ones used by auto mechanics to raise an automobile. It consists of an incompressible fluid in a U-tube fitted with a movable piston on each side. One side of the U-tube is narrower than the other.
Fluid Mechanics
Youth will learn about hydraulics then build their own hydraulic lift! ... Fluid Mechanics, Hydraulic. Science Venture. August 13, 2020. ... Grade 6, Grade 7, Grade 8, Grade 9. Students will learn about laminar flow and conduct a cool experiment to explore an interesting property of laminar flow in viscous fluids. Tagged: Fluid Mechanics ...
Hands-on Activity Hydraulic Arm Challenge
Students design and build a mechanical arm that lifts and moves an empty 12-ounce soda can using hydraulics for power. Small design teams (1-2 students each) design and build a single axis for use in the completed mechanical arm. One team designs and builds the grasping hand, another team the lifting arm, and a third team the rotation base. The three groups must work to communicate effectively ...
Experiment #10: Pumps
Experiment #10: Pumps. 1. Introduction. In waterworks and wastewater systems, pumps are commonly installed at the source to raise the water level and at intermediate points to boost the water pressure. The components and design of a pumping station are vital to its effectiveness. Centrifugal pumps are most often used in water and wastewater ...
The Hydraulic Lift Theory [Physics of Fluid Mechanics #22]
We'll be taking a look at how Pascal's Principle can help us create really powerful systems in physics, such as the hydraulic lift! The hydraulic lift is abl...
Experiment #1: Hydrostatic Pressure
Experiment #1: Hydrostatic Pressure. 1. Introduction. Hydrostatic forces are the resultant force caused by the pressure loading of a liquid acting on submerged surfaces. Calculation of the hydrostatic force and the location of the center of pressure are fundamental subjects in fluid mechanics. The center of pressure is a point on the immersed ...
Hydraulic Lift: Working, Hydraulic System & Pascal's Law
Pascal's law is also known as Pascal's principle or the principle of fluid-pressure transfer. Hydraulic lift Pascal's Law states that hydraulic lift enables the usage of a relatively small input force to form a larger output force. It is because the pressure applied on one piston is equivalent to the pressure of the second piston.
Fluid Mechanics Experiment on Hydraulic Lift
A hydraulic lift is a powerful tool used for lifting heavy loads with the application of Pascal's law, which states that any change in pressure applied to an...
6 Fluids Experiments for AP Physics
Therefore, we conclude that yes, this is a great experiment on the physics of fluids. #5 The Spouting Cylinder: The spouting cylinder is a classic physics demonstration and a staple problem in AP Physics. Holes at different heights are permitted to emit water when subjected to fluid pressure.
Fluid Mechanics & Hydraulics Experiments & Lab Practicals
Fluid Mechanics & Hydraulics Experiments & Lab Practicals. Full Details of Methods, Procedures and Apparatus used for Laboratory Experiments. Calibrate Pressure Guage Apparatus. Study of the Parts of Hydraulic Bench. Thoery. Types of Pressures. Apparatus Used. Procedure.

Lift a Large Load Using Liquids

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Build a Hydraulic Lift

14.3 Pascal's Principle and Hydraulics

Pascal’s Principle

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Applications of Pascal’s Principle and Hydraulic Systems

Example 14.3

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Simple Hydraulic Lifter WIP

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14.3 Pascal’s Principle and Hydraulics

Pascal’s Principle

Applications of Pascal’s Principle and Hydraulic Systems

Calculating Force on Wheel Cylinders: Pascal Puts on the Brakes

Significance

Check Your Understanding

Conceptual Questions

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14 Fluid Mechanics

Pascal’s Principle

Applications of Pascal’s Principle and Hydraulic Systems

Calculating Force on Wheel Cylinders: Pascal Puts on the Brakes

Significance

Check Your Understanding

Conceptual Questions

Hands-on Activity Hydraulic Arm Challenge

TE Newsletter

International Technology and Engineering Educators Association - Technology

State Standards

Safety Issues

Contributors

Experiment #1: Hydrostatic Pressure

2. Practical Application

3. Objective

5. Equipment

6. Equipment Description

7.1 Hydrostatic Force

7.2 Theoretical Determination of Center of Pressure

7.3 Experimental Determination of Center of Pressure

8. Experimental Procedure

9. Results and Calculations

Raw Data Table

9.2 Calculations

Result Table

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Hydraulic Lift: Working, Hydraulic System & Pascal's Law

What is a Hydraulic Lift?

Examples of Hydraulic Lift

Pascal’s Law

Pascal Law Formula

Hydraulic Lift Parts

Hydraulic Circuit

Hydraulic Pump

Hydraulic Motor

Hydraulic Cylinder

Applications of Hydraulic Lift

Hydraulic Elevator

Hydraulic Brake

Things to Remember

Sample Questions

Previous Year Questions

CBSE CLASS XII Related Questions

Similar Physics Concepts

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Fluid Mechanics & Hydraulics Experiments & Lab Practicals