Fluid Mechanics Lecture

01:05:01
https://www.youtube.com/watch?v=NaCjTp2-GPw

Resumen

TLDRThis video provides an overview of fluid mechanics, a fundamental subject for civil and mechanical engineering students. It begins with defining fluid mechanics and distinguishes between fluid statics and dynamics—statics being the study of fluids at rest, and dynamics the study of fluids in motion. Key properties such as density and pressure are explained, demonstrating how they affect fluid behavior. The instructor then covers Pascal's Principle, which describes how pressure applied to a fluid is transmitted through it, and Bernoulli's equation, which relates pressure, velocity, and elevation in moving fluids. Practical problems are solved to demonstrate these principles, including calculations involving density, pressure, and buoyancy, with emphasis on the practical application of these concepts such as in hydraulic presses. Key takeaways from the video include understanding how fluid characteristics and principles like Pascal's and Bernoulli’s are applied in real-world scenarios, such as calculating the forces in hydraulic systems or analyzing buoyant forces.

Para llevar

  • 💧 Fluid mechanics studies fluid (liquid and gas) behavior.
  • ⚖️ Fluid statics concerns fluids at rest; dynamics is about fluids in motion.
  • 🌡️ Density is crucial for determining if objects sink or float.
  • 📏 Pressure: force per unit area, measured in pascals (Pa).
  • 🔄 Pascal's Principle: pressure applied to a fluid transmits throughout.
  • ✈️ Bernoulli's Equation ties together pressure, velocity, and elevation in fluids.
  • 🚢 Buoyancy determines if an object will float based on density.
  • ⚙️ Hydraulic systems use fluid principles in mechanisms and tools.
  • 🧪 SI unit of pressure is the pascal (Pa).
  • 🌊 Archimedes' Principle: buoyant force equals the weight of the fluid displaced.

Cronología

  • 00:00:00 - 00:05:00

    Today's discussion introduces fluid mechanics, an introductory topic for civil and mechanical engineering. Fluid mechanics studies the behavior of fluids, both gases and liquids, contrary to the common misconception that fluids are only in liquid form. The key branches include fluid statics (fluids at rest) where forces are in equilibrium, and fluid dynamics (moving fluids), which is more complex. Key concepts include density, defined as mass per unit volume, crucial for understanding whether an object will sink or float. Common densities include water and air, important for solving problems in fluid mechanics.

  • 00:05:00 - 00:10:00

    The problem illustrated involves calculating the mass and weight of air in a specified room using density, derived from the equation density = mass/volume. The density of air is known, and the room's volume is calculated from its dimensions. By cross-multiplying in the equation, mass is found and subsequently weight, using the gravity constant (9.81 m/s²). The weight of air is then provided in newtons. The discussion then shifts to pressure, defined as a force per unit area and measured in pascals, and its direct relationship with force but inverse with area. Real-life examples illustrate these principles, like wearing high heels and feeling pressure due to the small area of contact.

  • 00:10:00 - 00:15:00

    Pressure's inverse relationship with area means that as area decreases, pressure increases, affecting real-life scenarios. The presentation discusses multiple units for pressure and scenarios such as the increased pressure experienced in high heels due to the reduced contact area. The calculation demonstrates how pressure changes with force or area. Atmospheric pressure, the pressure exerted by the weight of air, is explained, and an illustration of changes in pressure with elevation or depth is given with scenarios like swimming underwater or climbing in an elevator.

  • 00:15:00 - 00:20:00

    Atmospheric pressure is significant, especially in open containers. The formula for total pressure in such contexts includes atmospheric pressure and pressure due to the weight of the fluid column (rho g h). Examples show differential pressures when comparing freshwater and seawater due to their respective densities. For calculation, atmospheric pressure is typically 101.3 kPa, equivalent in other units like mmHg or atm, crucial for problems involving pressure at various depths or altitudes.

  • 00:20:00 - 00:25:00

    Conversion between pressure units and Pascal's principle, key elements of fluid mechanics, are discussed. According to Pascal's principle, pressure applied to a confined fluid is transmitted undiminished throughout. This principle enables hydraulic systems, where a small force applied at one point can result in a larger force elsewhere, as illustrated with hydraulic presses. Calculations show how changing cross-sectional areas influences the forces and pressures involved, useful in automotive lifts and other mechanical applications.

  • 00:25:00 - 00:30:00

    Illustrative problems apply Pascal's principle to determine required forces or pressures in hydraulic systems. Given dimensions and forces, calculations reveal how small inputs can generate substantial outputs, key to understanding hydraulic mechanisms. The role of cross-sectional area and pressure transfer without loss is vital in designing such systems. The specific calculations involve cross-sectional areas, force application, and pressure determination, accentuating how pressures equalize in hydraulics, allowing for efficient force distribution.

  • 00:30:00 - 00:35:00

    Hydrostatic pressure is introduced, derived from considerations of forces acting on fluid elements at rest. Key derivations conclude with the relationship between pressure and depth, expressed as ΔP = ρgΔh, indicating that deeper fluids experience greater pressure. The analysis of a fluid element within a container illustrates how static fluid pressures are calculated, pivotal for understanding buoyancy. The principle relates directly to determining buoyant forces acting on submerged objects.

  • 00:35:00 - 00:40:00

    Buoyant forces, essential for determining whether objects float or sink, depend on fluid density. Archimedes' principle states that buoyant force is equal to the weight of the displaced fluid. Comparisons illustrate why some objects float while others sink based on relative densities. An object's ability to float or sink is independent of its size but depends on its density relative to the fluid. Examples suggest why a dense object may sink while a less dense one floats, grounded in classic fluid mechanics principles.

  • 00:40:00 - 00:45:00

    Archimedes' principle details the conditions for sinking or floating based on object and fluid densities. Illustrative examples clarify these conditions, showing how density dictates buoyancy irrespective of object dimensions. The distinction between static states and buoyant dynamics highlights how objects interact with fluid environments, crucial for naval architecture and similar fields. Explanations clarify the common misconception—size or mass alone doesn't define buoyancy.

  • 00:45:00 - 00:50:00

    Bernoulli’s equation connects pressure, velocity, and height in moving fluids, contrasting with static principles. The continuity equation (A1V1 = A2V2) ensures mass conservation in fluid flow through varied cross-sections, explaining changes in velocity and pressure. In pipes, as fluid moves from wider to narrower sections, velocity increases, impacting pressure and energy distribution as explained by Bernoulli's theorem. This understanding is important for systems including pipelines and airflows.

  • 00:50:00 - 00:55:00

    Bernoulli's and continuity equations provide frameworks for solving fluid flow problems. Combined, they calculate outcomes like fluid velocity and pressure differentials. Worked examples offer insights into pressure and speed calculations as fluids transit through pipes. Mass conservation and energy principles govern these flow dynamics, foundational for engineering applications. Thorough analysis of pipe flow scenarios clarifies how cross-sectional areas and gravitational effects articulate in engineering solutions.

  • 00:55:00 - 01:05:01

    Example problems, including pressure unit conversions and applications of Bernoulli’s principle in dynamic systems, guide understanding. These include calculating fluid speed and pressure changes, integrating principles of mass and energy conservation. Such problem-solving skills are crucial for engineering students and professionals dealing with fluid dynamics. By exploring various scenarios and deriving insights, learners equip themselves to manage practical challenges in fields like civil, mechanical, and aerospace engineering.

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Mapa mental

Mind Map

Preguntas frecuentes

  • What is fluid mechanics?

    Fluid mechanics is a branch of science that deals with the behavior and properties of fluids (liquids and gases).

  • What is the difference between fluid statics and fluid dynamics?

    Fluid statics studies fluids at rest, while fluid dynamics studies fluids in motion.

  • What is density?

    Density is the mass per unit volume of a substance, usually expressed in kg/m3.

  • How is pressure defined in fluid mechanics?

    Pressure is defined as the force exerted per unit area on a surface, measured in pascals (Pa).

  • What is Pascal's Principle?

    Pascal's Principle states that a change in pressure applied to an enclosed fluid is transmitted undiminished throughout the fluid.

  • What is Bernoulli's equation?

    Bernoulli's equation relates the pressure, velocity, and elevation in a moving fluid, assuming no energy loss.

  • How does density affect buoyancy?

    An object will float if it is less dense than the fluid it is submerged in, and sink if it is denser.

  • What is the unit of pressure in the SI system?

    The SI unit of pressure is the pascal (Pa), equivalent to one newton per square meter.

  • What is the relationship between force, area, and pressure?

    Pressure is equal to force divided by area.

  • What is Archimedes' Principle?

    Archimedes' Principle states that the buoyant force on an object submerged in fluid is equal to the weight of the fluid displaced by the object.

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Desplazamiento automático:
  • 00:00:00
    hi so for today we're going to talk
  • 00:00:01
    about fluid mechanics and the basics of
  • 00:00:04
    fluid mechanics hey this is an
  • 00:00:07
    introductory subject for civil
  • 00:00:09
    engineering or mechanical engineering
  • 00:00:10
    their major course fluid mechanics okay
  • 00:00:15
    so we'll go to discuss in this video I'm
  • 00:00:17
    going to discuss the basics on the other
  • 00:00:19
    with mechanics so let us start with some
  • 00:00:22
    definition of terms what is fluid
  • 00:00:25
    mechanics fluid mechanics is basically
  • 00:00:27
    either branch on science that which
  • 00:00:30
    deals with the study or end behavior of
  • 00:00:33
    fluids such as gases or liquids
  • 00:00:36
    misconception all about fluids that when
  • 00:00:39
    we talk about fluids the first thing
  • 00:00:41
    that comes to our mind is that fluid is
  • 00:00:45
    limited to a liquid form but in the case
  • 00:00:48
    of the study of fluid mechanics fluid is
  • 00:00:51
    not only in liquid form but basically
  • 00:00:54
    fluid is any substance that can flow
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    either a liquid or a gas meaning to say
  • 00:01:00
    even a gas can be considered as fluid
  • 00:01:03
    why because it can flow okay so in this
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    kind of fluid that is the basic
  • 00:01:08
    definition of fluid mechanics so we have
  • 00:01:11
    two branches of fluid mechanics and we
  • 00:01:14
    have the fluid statics with this which
  • 00:01:15
    is the study of fluids at rest meaning
  • 00:01:18
    the fluid is not moving Ray's just
  • 00:01:22
    resting all right so the magnitude of
  • 00:01:25
    forces acting along that fluid is zero
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    because it is at rest rest in
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    equilibrium situations and we have the
  • 00:01:33
    fluid dynamics once this fluids starts
  • 00:01:36
    to move then we are talking about fluid
  • 00:01:38
    dynamics here and this fluid dynamics is
  • 00:01:41
    much more complicated or complex than
  • 00:01:45
    the fluid statics because now the fluid
  • 00:01:47
    is now moving right so here is a sample
  • 00:01:52
    important quantities when talking about
  • 00:01:54
    flowing fluid mechanics so we have
  • 00:01:57
    worked on the density of the material so
  • 00:01:59
    what's this density basically density is
  • 00:02:02
    defined as the mass per unit volume okay
  • 00:02:06
    so it is basically represented by a
  • 00:02:09
    Greek alphabet Rho and that is
  • 00:02:11
    equivalent to mass of this
  • 00:02:14
    the volume of the substance they were in
  • 00:02:17
    our unit for density is SI unit is
  • 00:02:20
    kilogram per cubic meter because all it
  • 00:02:23
    is cubic per meter okay a homogeneous
  • 00:02:26
    material that is a material that has the
  • 00:02:30
    same material composition of material
  • 00:02:36
    has the same density in other words for
  • 00:02:39
    example we have the same steel range
  • 00:02:44
    here and steel nail they are both still
  • 00:02:46
    though they have different shape and
  • 00:02:50
    different mass okay their density is the
  • 00:02:53
    same because they are a homogeneous
  • 00:02:56
    material they are composed of the same
  • 00:02:58
    material so different mass same density
  • 00:03:01
    world steel and still they regardless of
  • 00:03:04
    their size and regardless of their mass
  • 00:03:07
    so because the wrench in made are both
  • 00:03:09
    made of steel they have the same density
  • 00:03:11
    must per unit input and again density is
  • 00:03:15
    a important parameter or whether we the
  • 00:03:19
    object would actually sink or float away
  • 00:03:22
    when submerged into a liquid okay so
  • 00:03:25
    density that is basically density so
  • 00:03:28
    here are some densities of some common
  • 00:03:30
    substances we have the most common is
  • 00:03:34
    water wherein the density of water is 1
  • 00:03:37
    times 10 raised to 3 kilogram per cubic
  • 00:03:40
    meter so that is the density of water
  • 00:03:43
    for every element we have the density in
  • 00:03:47
    the constant density all right so do we
  • 00:03:51
    need to memorize all of this not
  • 00:03:52
    necessarily but we have to take look
  • 00:03:54
    what is the density of water because
  • 00:03:56
    sometimes in the problem it's not given
  • 00:03:58
    so that is basically 1 times 10 raised
  • 00:04:01
    to 3 kilogram per cubic meter and we
  • 00:04:05
    have the air 1.2 kilograms per cubic
  • 00:04:07
    meter
  • 00:04:07
    you have the ethanol the benzene the
  • 00:04:09
    eyes and different elements here okay so
  • 00:04:13
    we have the densities of some common
  • 00:04:17
    materials okay so let's try to solve a
  • 00:04:21
    problem okay
  • 00:04:22
    I've written the problem here so for our
  • 00:04:26
    problem number one
  • 00:04:30
    find the mass and weight of the air at
  • 00:04:33
    20 degrees Celsius that is in Celsius in
  • 00:04:36
    a living room with the four meters by 5
  • 00:04:38
    meters in a ceiling of 3 meters high
  • 00:04:41
    so basically we're asked to find the
  • 00:04:44
    mass of the air and the weight of the
  • 00:04:48
    air inside the room so what is the mass
  • 00:04:51
    of the air what is the weight of that
  • 00:04:54
    air all right
  • 00:04:56
    using the formula density is equal to
  • 00:04:59
    mass over volume we can calculate the
  • 00:05:02
    air the mass of the air inside the room
  • 00:05:06
    so of course inside the air inside the
  • 00:05:10
    room we have the density of air is
  • 00:05:12
    equals to the mass of air divided by the
  • 00:05:16
    volume of the room ok so what are we
  • 00:05:21
    going to do is to get the mass of here
  • 00:05:23
    see we cross multiply that a mass of the
  • 00:05:27
    air is simply density of the air and the
  • 00:05:31
    volume of the room okay so the density
  • 00:05:34
    of air is shown I wanna go 1.2 kilograms
  • 00:05:38
    per cubic metre and the volume of the
  • 00:05:43
    room is that that is a cube so because
  • 00:05:46
    we have not a tube necessarily so we
  • 00:05:48
    have a horror meter by 5 meter by 3
  • 00:05:51
    meter so length times width times height
  • 00:05:53
    of the boom so we have 4 meters
  • 00:05:56
    multiplied by 5 meters multiplied by 3
  • 00:06:00
    meters all right and this should give us
  • 00:06:03
    this meter cubic meter will cancel
  • 00:06:07
    nobody and that should give us 72
  • 00:06:12
    kilograms if I'm not mistaken so let's
  • 00:06:16
    try to verify so 72 kilograms that is
  • 00:06:20
    the mass of the air inside that room how
  • 00:06:23
    about the weight of the air well
  • 00:06:24
    actually we know that the weight of the
  • 00:06:26
    air is simply the mass times the
  • 00:06:30
    acceleration due to gravity that is 72
  • 00:06:33
    kilograms multiplied by 9.81 meters per
  • 00:06:39
    second square
  • 00:06:40
    and weight should be neutral so that is
  • 00:06:43
    75 times point eighty one that should be
  • 00:06:46
    706 706 point 32 that should be our and
  • 00:06:55
    so for the weight and the mass of e here
  • 00:07:00
    hey
  • 00:07:01
    so this is our computation thing we have
  • 00:07:06
    the same answer okay so how about
  • 00:07:09
    pressure pressure is also an important
  • 00:07:11
    quantity when we are talking about fluid
  • 00:07:14
    mechanics and pressures basically
  • 00:07:16
    defined as the amount of force exerted
  • 00:07:19
    on a given area this force is absolutely
  • 00:07:24
    perpendicular to the normal or the
  • 00:07:27
    surface area of the object right so and
  • 00:07:34
    it's SI unit is given in Pascal and
  • 00:07:39
    Pascal is denoted by E a and of course
  • 00:07:43
    the unit of pressure is denoted by one
  • 00:07:48
    Newton meter per meter squared is
  • 00:07:50
    absolutely equal to one Pascal so in
  • 00:07:53
    other words if you're going to write
  • 00:07:55
    oops for going to write the pressure
  • 00:08:01
    equation so pressure is defined as the
  • 00:08:03
    force per unit area and the unit of
  • 00:08:07
    force is actually one Newton divided by
  • 00:08:10
    this SI unit of area is meters fear so
  • 00:08:13
    we say that the unit of pressure is that
  • 00:08:16
    say one Pascal denoted by 1 p.m. and
  • 00:08:20
    that is basically equivalent to 1 Newton
  • 00:08:24
    per meter squared okay and we say that
  • 00:08:28
    this force is actually 1 it's actually
  • 00:08:31
    always perpendicular to the surface to
  • 00:08:35
    which it acts for example we have this
  • 00:08:38
    plate of a object with an area a so the
  • 00:08:42
    force acting should be always
  • 00:08:45
    perpendicular to the normal or the
  • 00:08:48
    surface of the object so that is the
  • 00:08:51
    pressure so a
  • 00:08:53
    in our equation if you are going to know
  • 00:08:55
    what this equation if we increase the
  • 00:08:58
    force acting per unit area well of
  • 00:09:01
    course if we if the pressure here is
  • 00:09:04
    also the force over area what if we
  • 00:09:06
    don't want the force like what will
  • 00:09:10
    happen is that since this is the scene
  • 00:09:12
    sports is directly proportional to the
  • 00:09:13
    pressure so we will be having pores our
  • 00:09:17
    pressure is equal to twice the F over a
  • 00:09:20
    so our pressure will increase our
  • 00:09:22
    actually by two also because of the
  • 00:09:25
    increase in the force so what will
  • 00:09:28
    happen if if we try to apply a force to
  • 00:09:34
    a very small area of a much smaller than
  • 00:09:38
    this so what will happen let's say we
  • 00:09:40
    have here an object okay it is acted by
  • 00:09:43
    a force this is an area a prime which is
  • 00:09:47
    basically smaller than this so we have
  • 00:09:49
    the same force of life so the pressure
  • 00:09:52
    here is absolutely much greater than the
  • 00:09:55
    pressure here because the area is now
  • 00:09:59
    smaller and we if we decrease the area
  • 00:10:02
    to which the force is acting what will
  • 00:10:04
    happen to the pressure it would increase
  • 00:10:06
    alright so because this has the area and
  • 00:10:10
    the pressure has a indirectly
  • 00:10:12
    proportional or indirectly proportional
  • 00:10:13
    to each other or inversely proportional
  • 00:10:16
    to each other so that if we're going to
  • 00:10:18
    decrease the area the pressure will
  • 00:10:20
    increase and of course increasing the
  • 00:10:22
    area to which the force is acting upon
  • 00:10:24
    well will also decrease the pressure
  • 00:10:27
    right so that's the basic concept of the
  • 00:10:30
    pressure so in other words if the
  • 00:10:32
    applied force is acting on a small area
  • 00:10:35
    then the pressure will be large and vice
  • 00:10:38
    versa all right so take note that
  • 00:10:42
    pressure has many lives its SI units
  • 00:10:44
    only given by Pascal but we can convert
  • 00:10:49
    pressure in different types of units
  • 00:10:52
    such a store mmHg
  • 00:10:54
    bar over a atm we have many units for
  • 00:10:58
    pressure but for now we're going to talk
  • 00:11:00
    about the facts cos okay again the area
  • 00:11:04
    here is the prof sectional
  • 00:11:06
    the object to which the voice is up so
  • 00:11:10
    something to think about when we are
  • 00:11:12
    dealing with pressure for ladies do you
  • 00:11:15
    feel pain or pressure whenever you wear
  • 00:11:18
    your hiatus for a long time
  • 00:11:20
    well of course some of our friends was
  • 00:11:23
    going to party though those ladies are
  • 00:11:25
    wearing high heels maybe five inches so
  • 00:11:30
    what happens the tendency is that
  • 00:11:33
    whenever they wear that for a very long
  • 00:11:37
    time they will feel pain in their feet
  • 00:11:41
    high or even in relaxed why because
  • 00:11:44
    there is the presence of pressure
  • 00:11:45
    if you're going to recall from our
  • 00:11:47
    equation a while ago that the pressure
  • 00:11:50
    is inversely proportional to the area
  • 00:11:53
    cross sectional area so for example your
  • 00:11:56
    friend is wearing a high heels so the
  • 00:11:58
    tendency is that the weight now become
  • 00:12:01
    becomes the force because weight is also
  • 00:12:04
    a force so it is now acting on the very
  • 00:12:07
    small cross sectional area of the hills
  • 00:12:11
    imagine force if I'm going to draw that
  • 00:12:15
    so let's just make this as a as an
  • 00:12:18
    exaggeration okay suppose this is the
  • 00:12:21
    heels of your friend and this is the
  • 00:12:23
    force acting away on this hill of the
  • 00:12:28
    shoes of the five-inch shoes of your
  • 00:12:31
    heels of your friend so this force is
  • 00:12:35
    after the per week okay and it let's say
  • 00:12:39
    that if the weight is of the user of
  • 00:12:42
    your friend is really larger than in
  • 00:12:44
    other words it's a little chubby
  • 00:12:46
    what will happen is that the the
  • 00:12:49
    pressure that your feet will feel is
  • 00:12:54
    very large because the cross section
  • 00:12:56
    here is an S is actually a very small
  • 00:13:01
    squared away area of a sphere so very
  • 00:13:06
    small area so that's why we cannot
  • 00:13:09
    actually be R or ladies you're not
  • 00:13:12
    absolutely bear wearing that highest for
  • 00:13:16
    a very long time because the
  • 00:13:18
    area which daily acting is acting is on
  • 00:13:22
    a very very small area my compared to
  • 00:13:26
    when we wear shoes the normal shoes the
  • 00:13:29
    area of our shoes is really really large
  • 00:13:36
    compared to the area when wearing high
  • 00:13:38
    heels okay so the point is that there is
  • 00:13:42
    a great pressure in the high heels of
  • 00:13:45
    your friend who is going to wear this
  • 00:13:48
    idea sorry so number two do you feel
  • 00:13:50
    friendship when you are at the drugs
  • 00:13:52
    average sometimes and then use the
  • 00:13:53
    elevator to go to a much higher level in
  • 00:13:56
    a building all right so when of course
  • 00:13:59
    sometimes we are tough at the ground
  • 00:14:01
    level and we try to ride the elevator
  • 00:14:03
    and from ground to xxx let's say that is
  • 00:14:07
    constantly going up okay so what happens
  • 00:14:11
    is that the tendency is we feel pressure
  • 00:14:13
    in our ears
  • 00:14:14
    I mean sometimes there is a thickening
  • 00:14:17
    sound in our ears because of the
  • 00:14:21
    pressure okay so not only because the
  • 00:14:25
    pressure is limited by this formula but
  • 00:14:29
    also the pressure is absolutely has
  • 00:14:32
    another formula of which the pressure to
  • 00:14:36
    which our body feels is absolutely equal
  • 00:14:39
    also into what we call the Rho gh and it
  • 00:14:42
    is a function of height so Rho gh is
  • 00:14:45
    actually the Rho is the density G is the
  • 00:14:49
    gravitational force
  • 00:14:52
    alright levitational force observation
  • 00:14:57
    due to gravity and this is now the
  • 00:14:59
    height so whenever we go to a higher
  • 00:15:03
    elevation we experience pressure for
  • 00:15:06
    example we write the flight from Manila
  • 00:15:10
    kadavul were actually riding a an
  • 00:15:13
    airplane so at the very high height know
  • 00:15:19
    we feel pressure in our ears because
  • 00:15:21
    pressure is also a function of height so
  • 00:15:25
    we have this following formula pressure
  • 00:15:26
    is also equal to Rho G each so number
  • 00:15:29
    three do you feel pressure when you have
  • 00:15:31
    to swim at the very bottom of the
  • 00:15:33
    swimming pool well actually we always
  • 00:15:35
    feared that whenever we try to swim at
  • 00:15:38
    the very bottom of the swimming hole if
  • 00:15:40
    you don't know how to swim good for you
  • 00:15:43
    you don't feel any pressure but but most
  • 00:15:46
    of us maybe have experienced this that
  • 00:15:49
    whenever we go deeper okay in the into
  • 00:15:53
    the deep of the swimming pool what
  • 00:15:54
    happens is that the tendencies we feel
  • 00:15:56
    pressure in our nose okay maybe in our
  • 00:16:00
    ears and even in our head because of the
  • 00:16:03
    pressure and that is all because of the
  • 00:16:07
    formula Rosie eats so suppose we have a
  • 00:16:10
    pan filled with a fluid density ro that
  • 00:16:13
    just like what I would say the wine
  • 00:16:14
    about that French rule is equals to Rho
  • 00:16:17
    gh okay so at the very bottom of this
  • 00:16:21
    pan right there is a pressure and that
  • 00:16:24
    pressure is actually Rho G H they not
  • 00:16:27
    including the what we call t atmospheric
  • 00:16:30
    pressure but what later on we'll be
  • 00:16:32
    discussing that
  • 00:16:33
    so this equation helps us understand why
  • 00:16:36
    the deeper we go underwater
  • 00:16:37
    the greater the pressure our body fills
  • 00:16:40
    think so atmospheric pressure there's
  • 00:16:44
    also what we call the atmospheric
  • 00:16:46
    pressure and that is basically defined
  • 00:16:47
    as the P sub atm
  • 00:16:50
    alright and when do we use that constant
  • 00:16:55
    alright for atmospheric pressure well of
  • 00:16:57
    course if we're going to use that for an
  • 00:17:00
    open tank we should include the pressure
  • 00:17:02
    outside the plan
  • 00:17:03
    pressing on the top of the surface of
  • 00:17:05
    the fluid or in other words whenever we
  • 00:17:08
    have an open of a an open Pam and it is
  • 00:17:13
    actually adapted by the atmospheric
  • 00:17:16
    pressure and if we're going to make sure
  • 00:17:18
    the very pressure of the very bottom of
  • 00:17:21
    that pan we should improve the
  • 00:17:23
    atmospheric pressure for you to analyze
  • 00:17:25
    what I'm trying to say is that whenever
  • 00:17:27
    we have an open path and this time for
  • 00:17:31
    example is actually filled with water
  • 00:17:34
    right up to this level to a height H if
  • 00:17:40
    we want to measure the pressure at this
  • 00:17:43
    point at the very
  • 00:17:44
    bottom of the of the time what is the
  • 00:17:47
    pressure there so there is what we call
  • 00:17:50
    the atmospheric pressure and for every
  • 00:17:53
    open and we must include that in our
  • 00:17:56
    [Music]
  • 00:17:58
    competition so the pressure at the very
  • 00:18:01
    bottom of this time is the atmospheric
  • 00:18:04
    pressure of a multiplied by the pressure
  • 00:18:08
    due to regimes so the total pressure at
  • 00:18:11
    the very bottom for an open and acted by
  • 00:18:14
    the pressure or the air is actually the
  • 00:18:17
    atmospheric pressure plus the what we
  • 00:18:21
    call the Rho G H okay so atmospheric
  • 00:18:25
    pressure is actually equal to 101.3 kilo
  • 00:18:29
    Pascal's and that is equivalent to 760
  • 00:18:33
    Torr equal to 760 mmHg and that is
  • 00:18:37
    basically equal to 1 atm all of these
  • 00:18:40
    are units of pressure raised 11.3
  • 00:18:45
    kilopascal or in other word one one
  • 00:18:47
    other words one one point three to five
  • 00:18:50
    kilo Pascal's for accurate combinations
  • 00:18:54
    right so for this problem the swimming
  • 00:18:59
    pool fresh or saltwater so compare the
  • 00:19:01
    total pressure at the bottom of a
  • 00:19:03
    swimming pool of a depth three meters
  • 00:19:05
    for about 99 feet fi or eight feet if it
  • 00:19:09
    is filled with fresh water and sea water
  • 00:19:11
    so first we're going to compute the
  • 00:19:15
    pressure when we are when we go we are
  • 00:19:19
    going to use the fresh water all right
  • 00:19:22
    so let's calculate the pressure so for
  • 00:19:25
    example here is our pool okay here is
  • 00:19:30
    our pool this pool is filled with what
  • 00:19:35
    we call the fresh water a fresh water
  • 00:19:37
    first and it's high array is actually
  • 00:19:44
    three meters so that is for a three
  • 00:19:49
    meters height so we're going to compute
  • 00:19:51
    now the total pressure at the very
  • 00:19:54
    bottom of the spoon so the total
  • 00:19:58
    pressure
  • 00:19:58
    is also the people that is P atm because
  • 00:20:03
    this is an open swimming pool so we have
  • 00:20:05
    to include the pressure due to
  • 00:20:09
    atmospheric pressure oops
  • 00:20:11
    so that is added to the pressure Rho G H
  • 00:20:19
    so we're going to include pressure
  • 00:20:21
    atmospheric pressure this for number
  • 00:20:23
    three
  • 00:20:24
    so 101.3 oh five or 11.3 let's be
  • 00:20:30
    consistent
  • 00:20:31
    then the pressure atmospheric pressure
  • 00:20:34
    is 101 point three kilo Pascal's Plus D
  • 00:20:39
    Rho what is this the density of water is
  • 00:20:42
    1 times 10 raised to 3 kilogram per
  • 00:20:45
    cubic meter multiplied by 9.81 m/s^2 and
  • 00:20:51
    that should be multiplied by 3 meters so
  • 00:20:56
    what are we going to get here is
  • 00:20:58
    actually 11.3 Giga Pascal's 1 times 10
  • 00:21:02
    this would be okay let me just conclude
  • 00:21:06
    9.81 they're three meters so one 30
  • 00:21:11
    point seventy three okay so let's check
  • 00:21:17
    by or one point 31 times 10 raised to 5
  • 00:21:21
    Pascal's one point 31 times 10 raised to
  • 00:21:25
    5 Pascal's or engineering in engineering
  • 00:21:29
    notation 130 point 73 kilo Pascal's so
  • 00:21:40
    what will happen if we're going to
  • 00:21:41
    replace this fresh water this fresh
  • 00:21:45
    water by soft water no for soft water
  • 00:21:50
    what will happen is that the same
  • 00:21:53
    competition we have to include the
  • 00:21:56
    atmospheric pressure plus the role of
  • 00:21:59
    now the sea or the soft water times the
  • 00:22:03
    G times the height well of course the
  • 00:22:06
    only thing that changed here is the
  • 00:22:09
    density of
  • 00:22:11
    the object so from fresh water you see
  • 00:22:15
    water so we have total pressure is
  • 00:22:19
    absolutely equal to 101.3 Ito Pascal's
  • 00:22:23
    plus the density of water or sea water
  • 00:22:26
    is one point zero three times ten days
  • 00:22:30
    to three kilogram per cubic meter
  • 00:22:37
    multiplied by G 9.81 times the height
  • 00:22:41
    the same that is three so the total
  • 00:22:43
    pressure we're going to compute for da
  • 00:22:46
    is 101.3 a dope-ass cos plus one times
  • 00:22:51
    one point zero three times 10 raised to
  • 00:22:54
    3 9.81 times three and that is one point
  • 00:23:00
    what 1.30 to one point 32 times 10
  • 00:23:07
    raised to 5 fast palace or engine in
  • 00:23:10
    engineering notation we have what 30 1.1
  • 00:23:17
    1.6 1 to 9 Pascal's or kilo Pascal's and
  • 00:23:24
    slowly kilo Pascal's this is in terms of
  • 00:23:30
    email passwords all right so there is a
  • 00:23:35
    big difference in the pressure at the
  • 00:23:38
    very bottom of the spirit for the VTC
  • 00:23:40
    fresh water and so far as you can see
  • 00:23:42
    here the depressor the very bottom of
  • 00:23:44
    the swimming pool if it is fresh water
  • 00:23:46
    is given by 1 point 31 times 10 raised
  • 00:23:49
    to 5 Pascal we're asked for salt water
  • 00:23:51
    that is 1 point 32 times 10 raised to 5
  • 00:23:53
    Pascal there is a great difference so
  • 00:23:56
    that the determining difference here is
  • 00:23:58
    actually because of the tens in the
  • 00:24:00
    difference of their days so that's why
  • 00:24:03
    we have to include that whenever we
  • 00:24:05
    screen we try to stream at the very
  • 00:24:07
    bottom of a fresh water the pressure is
  • 00:24:10
    not really that great but when you try
  • 00:24:13
    to swim the very bottom of in salt water
  • 00:24:16
    forcing water what will happen is that
  • 00:24:19
    the tendency is that you will feel a
  • 00:24:22
    greater pressure why because
  • 00:24:23
    the density again of the seawater is
  • 00:24:25
    much greater than the density or in
  • 00:24:28
    freshwater okay let's go back to our
  • 00:24:32
    discussion so conversion of conversion
  • 00:24:36
    of pressure units we have different
  • 00:24:38
    conversion of pressure units because we
  • 00:24:40
    have many units of fractures or Pascal
  • 00:24:45
    denoted by P a1 e is equals to 100 per
  • 00:24:48
    square meter and we have what people do
  • 00:24:50
    bar bar is a unit of pressure also in
  • 00:24:54
    one part is equal to 0.1 mega Pascal's
  • 00:24:58
    that's a huge amount of pressure so 0.1
  • 00:25:02
    mega Pascal is absolutely equal to 0.1
  • 00:25:05
    times 10 raised to six Oscars we have
  • 00:25:09
    the water column meter way in that is
  • 00:25:11
    the equivalent conversion units honestly
  • 00:25:14
    I don't know that way for the irony
  • 00:25:18
    haven't encountered that during my time
  • 00:25:21
    in my college but I hope someday we'll
  • 00:25:25
    know that that in your major subjects so
  • 00:25:28
    we have the atmospheric pressure what 1
  • 00:25:30
    atm ATM 118 is basically equal to this
  • 00:25:34
    the pressure atmospheric pressure is 101
  • 00:25:38
    3 to 5 Pascal's or what we have record a
  • 00:25:42
    while ago that is one one point three
  • 00:25:44
    times ten days to one one point three
  • 00:25:49
    kilo Pascal's so the most are more
  • 00:25:52
    accolade is 101.325 kilo Pascal's or 1 1
  • 00:25:57
    3 to 5 Pascal's so we have the mercury
  • 00:26:00
    kilometer given by this conversion and
  • 00:26:03
    we have the third 37 the pressure in
  • 00:26:06
    that one 4 is equal to 1 millimeter HD
  • 00:26:09
    of pressure okay
  • 00:26:11
    that's basic the basic conversion of
  • 00:26:14
    pressure units so let's go to Pascal's
  • 00:26:17
    principle so Pascal principle states
  • 00:26:19
    that a change in the pressure applied to
  • 00:26:23
    improve in I start smitten undiminished
  • 00:26:26
    to every point in the fluid and to the
  • 00:26:29
    walls of the container in other words
  • 00:26:30
    Pascal's principle whenever we have a
  • 00:26:33
    fluid inside and we apply force the
  • 00:26:37
    Oh pressure on the left side the
  • 00:26:39
    pressure on the left side must be always
  • 00:26:41
    equal to the pressure on the right side
  • 00:26:44
    of the container okay
  • 00:26:46
    the pressure is transmitted undiminished
  • 00:26:49
    to every point in the fluid so in the
  • 00:26:53
    absence of the gravity the pressure is
  • 00:26:54
    the same everywhere in the vessel so
  • 00:26:57
    remember the pressure is a scalar
  • 00:26:59
    quantity but the force is actually a
  • 00:27:02
    vector so but the pressure is scalar
  • 00:27:06
    even though the force is vector and the
  • 00:27:09
    fall is always perpendicular to the
  • 00:27:10
    cross sectional area so what will happen
  • 00:27:14
    here is that if we have for example one
  • 00:27:19
    application of Pascal's principle is
  • 00:27:22
    actually what we call the hydraulic jaw
  • 00:27:25
    this hydraulic or hydraulic press or
  • 00:27:28
    hydraulic jump this hydraulic presses
  • 00:27:30
    actually I used to give first in a hard
  • 00:27:32
    lifting service station or in a company
  • 00:27:37
    where cars are being manufactured
  • 00:27:41
    okay so consider a hydraulic press or
  • 00:27:44
    hydraulic job this hydraulic job is
  • 00:27:46
    filled with fluid inside so Pascal
  • 00:27:49
    principle states that the pressure on
  • 00:27:51
    the left side this left side must be
  • 00:27:53
    equal to the pressure on the right side
  • 00:27:56
    so when of course for example if you
  • 00:28:00
    have this hydraulic press okay if we
  • 00:28:03
    apply force here on the left side okay
  • 00:28:06
    for this cross sectional area in sub 1
  • 00:28:09
    we are actually producing the pressure
  • 00:28:12
    because this force is acting
  • 00:28:14
    perpendicular to the cross sectional
  • 00:28:16
    area in someone and according to Pascal
  • 00:28:18
    if we create pressure here the pressure
  • 00:28:20
    we created it must be also equal to the
  • 00:28:23
    pressure that is produced here so often
  • 00:28:27
    if we're going to apply force here fluid
  • 00:28:30
    will be the space so that there will be
  • 00:28:32
    a force upward F sub 2 that is acting
  • 00:28:35
    went of much bigger cross sectional area
  • 00:28:38
    of this hydraulic press a sub 2 so what
  • 00:28:41
    the pressure remains the same according
  • 00:28:44
    to Pascal's
  • 00:28:45
    alright so what are we I'm trying to say
  • 00:28:48
    is that because of the
  • 00:28:51
    pressure on the left side and on the
  • 00:28:53
    right side we have this following
  • 00:28:54
    formula okay
  • 00:28:56
    so f1 is equals to a 1 that is pressure
  • 00:28:59
    1 that is equal to f2 over a 2 because
  • 00:29:03
    the pressures are equal so because the
  • 00:29:05
    increase in pressure is the same on the
  • 00:29:07
    both sides is more force on f1 to do
  • 00:29:10
    this a much greater force f2 so let me
  • 00:29:14
    just go back to our image okay so let us
  • 00:29:18
    assume that the smaller area we call
  • 00:29:21
    that a 1 and the smaller force to go the
  • 00:29:23
    f1 and the bigger area we produce or the
  • 00:29:26
    bigger force we produce is f2 and the
  • 00:29:29
    bigger area side of the hydraulic press
  • 00:29:31
    is what we call the a - so let's just
  • 00:29:34
    try to ok so for us to really see the
  • 00:29:41
    beauty of Pascal is full so we have
  • 00:29:44
    let's say in the p1 is equal to p2
  • 00:29:47
    according to Pascal's principle and if
  • 00:29:49
    we're going to elaborate this F 1 over a
  • 00:29:53
    1 that is equal to F 2 over a tune
  • 00:29:58
    suppose we want to find the f1 this f1
  • 00:30:02
    is the force that we are going to apply
  • 00:30:04
    on the left side of a to carry or to
  • 00:30:07
    lift half a car with a higher area a sub
  • 00:30:11
    2 so let's go let's let's say we we need
  • 00:30:15
    to the isolate F 1 so we have a1 / a2 x
  • 00:30:21
    f2 by and let's say class that the area
  • 00:30:26
    of one or the smaller disc on or on the
  • 00:30:29
    left side of the hydraulic press is
  • 00:30:32
    equal let's say - what do we call this
  • 00:30:41
    let's say 100
  • 00:30:46
    meter squared and well of course a to
  • 00:30:51
    let's say one meters square me okay so
  • 00:31:01
    let me just make this clear so the what
  • 00:31:06
    are we trying to get here is f2 instead
  • 00:31:08
    of F 1 I'm so sorry
  • 00:31:10
    so we have to get F 2 because F 2 is
  • 00:31:13
    where we have to leave the card that is
  • 00:31:15
    the right side where we have to leave
  • 00:31:17
    the heart so when we get F 2 F 2 or the
  • 00:31:21
    first
  • 00:31:21
    Anto should be what a2 over a1 x the
  • 00:31:27
    first one let's say that a 2 or the area
  • 00:31:30
    to which the car is being to be lifted
  • 00:31:34
    up let's say that the area of that is
  • 00:31:36
    that say 100 square meter
  • 00:31:38
    alright let's just exaggerate this right
  • 00:31:41
    so and the area of 1/2 which we are
  • 00:31:44
    going to apply the force of a this area
  • 00:31:48
    is the smaller area on the left side of
  • 00:31:50
    our hydraulic jack let's say that is 1
  • 00:31:52
    square meter okay
  • 00:31:55
    and what what if what are we going to do
  • 00:31:58
    is to apply 1 Newton of force on the
  • 00:32:00
    left side of the hydraulic press so if
  • 00:32:04
    we're going to compute for the first two
  • 00:32:06
    we have a to the area is 100 meter
  • 00:32:09
    square over 1 meter squared then
  • 00:32:12
    multiplied that by 1 Newton
  • 00:32:14
    all right let's say 1 unit on what will
  • 00:32:21
    happen is that upon evaluating the f2
  • 00:32:24
    what will happen is that this would
  • 00:32:26
    cancel and we are left with 100 times 1
  • 00:32:30
    Newton so that we are able to produce
  • 00:32:33
    100 Newton only by applying 1 Newton so
  • 00:32:38
    no wonder why this hydraulic jump would
  • 00:32:41
    actually perform the lifting of the
  • 00:32:44
    horse because small pores that we are
  • 00:32:48
    going to apply on the left side of the
  • 00:32:50
    hydraulic press would create a greater
  • 00:32:54
    force on the right
  • 00:32:56
    because of the Equality impression okay
  • 00:33:00
    so what I'm trying to say is this is our
  • 00:33:02
    first one one new phone available opted
  • 00:33:06
    by a small areas of what that is the our
  • 00:33:08
    one square meter then the force that
  • 00:33:11
    will be produced on the other side or
  • 00:33:13
    the right side of the hydraulic press
  • 00:33:15
    would be 100 movin imagine from one year
  • 00:33:19
    told to 100 Newton with with perfect
  • 00:33:21
    design or be the good design of the
  • 00:33:23
    diameter or the area of this hydraulic
  • 00:33:27
    presses way we can actually live up cars
  • 00:33:31
    in a very easy way okay so of course the
  • 00:33:36
    same amount of fluid dibs on the left
  • 00:33:39
    and enters on the right side this is
  • 00:33:40
    actually before you a one factor
  • 00:33:42
    displacement one is equal to a 1 need no
  • 00:33:45
    volume of the fluid is left or is
  • 00:33:49
    actually the whole this looks a force of
  • 00:33:59
    magnitude 1 well of course we have seen
  • 00:34:01
    this a while ago
  • 00:34:02
    oops so well of course the work done on
  • 00:34:04
    the left side of the hydraulic press is
  • 00:34:07
    also equal to the work done on the right
  • 00:34:09
    side of the hydraulic press so let's try
  • 00:34:13
    to have some sample problems okay so we
  • 00:34:16
    have a part that weighs 10 kilo Newton
  • 00:34:18
    is space on the 1 meter ranges based on
  • 00:34:21
    a hydraulic press how much force should
  • 00:34:24
    be exerted in the 5 cm Vegas piston to
  • 00:34:27
    lift the bar let me just see if I have a
  • 00:34:30
    face here okay so I haven't updated here
  • 00:34:34
    so we have to answer this accordant
  • 00:34:40
    weighs 10 kilo Newton is placed on the 1
  • 00:34:43
    meter radius well actually in in solving
  • 00:34:47
    hydraulic presses problems so we must
  • 00:34:51
    consider the high the viscosity or even
  • 00:34:56
    is equals to people and we must always
  • 00:34:58
    do 4 is the smaller radius all right if
  • 00:35:03
    we find the smaller radius which will in
  • 00:35:06
    this case 1 meter
  • 00:35:08
    and five centimeters the smaller radius
  • 00:35:10
    is 5 centimeter so five centimeters
  • 00:35:14
    should be our r1
  • 00:35:16
    that should be the left side okay so
  • 00:35:20
    what I'm trying to say is that we have
  • 00:35:22
    our 1 is given by 5 centimeters the
  • 00:35:27
    problem is asking is how much force
  • 00:35:30
    should be exerted in the 5 centimeters
  • 00:35:32
    so if that is the R 1 this should be the
  • 00:35:35
    first one to which the force is being
  • 00:35:37
    applied
  • 00:35:38
    okay in order to lift the car on the
  • 00:35:41
    other side so that's the unknown DF 1
  • 00:35:43
    and we have here 10 kilonewtons
  • 00:35:46
    by the way that could you kilonewtons is
  • 00:35:48
    also force that is the F 2 10 kilo
  • 00:35:52
    Newtons and our R 2 is also D 1 meter
  • 00:35:57
    okay
  • 00:35:58
    so applying Pascal's principle we have P
  • 00:36:00
    1 is equals to P 2 that is F 1 over a 1
  • 00:36:04
    is equals to F 2 over a 2 and we are
  • 00:36:08
    asked to find for the F 1 so f 1 will
  • 00:36:12
    now be equal to a1 over a2 x d F 2 so we
  • 00:36:20
    have now the area we have to be really
  • 00:36:22
    careful in this we are given that I do
  • 00:36:25
    snap the area but to get the area we are
  • 00:36:30
    asked to find the cross section area so
  • 00:36:33
    the cross-sectional area is actually a
  • 00:36:36
    circle okay so we have pi we have to
  • 00:36:40
    convert this 5cm into meters so that we
  • 00:36:44
    have 0.05 meters squared over pi
  • 00:36:50
    multiplied by area of 2 that is 1 meter
  • 00:36:54
    squared we have nothing we have no
  • 00:36:57
    problem with that and that is basically
  • 00:36:59
    multiplied by the weight of the card
  • 00:37:02
    okay so what are we going to do here is
  • 00:37:05
    simply as we can see PI's will cancel ok
  • 00:37:09
    and multiply that by 10 kilonewton so we
  • 00:37:14
    have the answered 25-year furnace so
  • 00:37:20
    it's very
  • 00:37:22
    amazing why because we can lift the car
  • 00:37:25
    on the right side of the hydraulic press
  • 00:37:28
    by just simply applying 25 Newtons on
  • 00:37:32
    the left side of the hydraulic press how
  • 00:37:35
    cool is that right so even if in even
  • 00:37:38
    you you can slowly lift her so the best
  • 00:37:43
    example is that is that the mechanics we
  • 00:37:45
    can see here we can see them trying to
  • 00:37:48
    lift the car by water for the hydraulic
  • 00:37:50
    press anybody continuously applying
  • 00:37:53
    force so that they are able to lift the
  • 00:37:56
    heart that's one basic application of
  • 00:37:58
    the hydraulic press number two for
  • 00:38:00
    problem number two in a fairly used in a
  • 00:38:03
    service station compressed air exerts a
  • 00:38:05
    force on a small teaspoon that has a
  • 00:38:07
    radius of 5 cm this pressure is
  • 00:38:10
    transmitted by a liquid to a piston that
  • 00:38:12
    has a radius of 15 cm so we have to
  • 00:38:14
    range assess again and 5 cm is the R 1
  • 00:38:18
    because it's smaller are twisting 15 cm
  • 00:38:21
    so what are we going to do is to write
  • 00:38:22
    the event R 1 is 5 cm R 2 is 15 cm okay
  • 00:38:29
    so what force must the compressed air
  • 00:38:32
    exert to lift a car weighing 13,300
  • 00:38:37
    Newtons about 13,000 300 Newton's where
  • 00:38:40
    the force is on the right side and that
  • 00:38:44
    should be the weight of the bar that
  • 00:38:46
    should be F 2 all right
  • 00:38:50
    30,000 300 Newton what air pressure
  • 00:38:52
    produces this 4 so we have two questions
  • 00:38:54
    here what force should be exerted on the
  • 00:38:58
    left side of the hydraulic press in
  • 00:39:00
    order for this 13,300 Newton off part of
  • 00:39:03
    it if that up that is again F 1 and we
  • 00:39:07
    have derive again the F 1 by the formula
  • 00:39:10
    F 1 is equal to a 1 over a 2 multiplied
  • 00:39:14
    by F 2 so we have now a 1
  • 00:39:18
    alright the area of one cross sectional
  • 00:39:20
    area we have to convert this again into
  • 00:39:22
    meters so I r1 0.05 meters squared over
  • 00:39:30
    area of 2 pi over 0 point
  • 00:39:34
    in meters squared and that should be
  • 00:39:37
    multiplied by the force to thirteen
  • 00:39:40
    thousand three hundred Newtons that is
  • 00:39:43
    the weight of the car 0.05 squared 0.15
  • 00:39:48
    square multiplied by 13 thousand three
  • 00:39:53
    hundred and that should be equal to one
  • 00:39:57
    four seven seven point seven seven eight
  • 00:40:04
    that's that's Arabic up to three
  • 00:40:07
    decimals right so that is the amount of
  • 00:40:10
    force that is needed in order to leave
  • 00:40:14
    the are in a bar lift in service station
  • 00:40:16
    so number two the question is actually
  • 00:40:19
    what air pressures produces this force
  • 00:40:21
    so simply the problem is asking for us
  • 00:40:24
    what is the pressure well since this is
  • 00:40:26
    hydraulic press the pressure in the left
  • 00:40:30
    side is equal to the pressure to the
  • 00:40:31
    right side so that we can calculate any
  • 00:40:34
    of the pressure let's say pressure what
  • 00:40:36
    is equals to F 1 over a 1 and our F 1 is
  • 00:40:40
    basically what we have computed here one
  • 00:40:43
    four seven seven point seven seven eight
  • 00:40:47
    Newton's divided by the area one area
  • 00:40:51
    one is by 0.05 meter squared and that
  • 00:40:58
    should result one answer or let's see
  • 00:41:04
    one eighty a so 180 a point sixteen
  • 00:41:11
    times ten bits of tree or that is low
  • 00:41:14
    phosphorus all right
  • 00:41:17
    that is 188 point 16 kilo Pascal's and
  • 00:41:21
    for pressure to we have f2 over a - all
  • 00:41:28
    right so what do you expect if we're
  • 00:41:30
    going to compute P - since this is a
  • 00:41:33
    hydraulic press the pressure on the left
  • 00:41:35
    side is equal to the pressure on the
  • 00:41:37
    right side so this two pressure must be
  • 00:41:39
    yes you are correct that must be equal
  • 00:41:42
    so we have thirteen thousand three
  • 00:41:44
    hundred Newton or f2 over eight
  • 00:41:47
    then glass is 0.15 meters width okay so
  • 00:41:55
    13300 is 5 times 0.15 squared that is
  • 00:42:00
    also equal to 1/8 8.16 kilo Pascal's
  • 00:42:08
    okay so that because that works on the
  • 00:42:12
    face of all of the Pascal's law or
  • 00:42:16
    principle so hydrostatic pressure so we
  • 00:42:22
    are going to derive the hydrostatic
  • 00:42:24
    pressure for this video and we're going
  • 00:42:29
    to actually verify if the change in
  • 00:42:33
    height as we go deeper into the fluid
  • 00:42:36
    actually affects deep pressure all right
  • 00:42:39
    so I would like you to consider and to
  • 00:42:43
    listen very carefully
  • 00:42:44
    all right in this discussion
  • 00:42:48
    for example class we have this figure
  • 00:42:52
    all right
  • 00:42:53
    so for example I have an open pan and
  • 00:42:56
    this open pump is absolutely one filled
  • 00:43:01
    with water and I get a small amount of
  • 00:43:05
    let's say a QB size of water here all
  • 00:43:09
    right that water is sitting still doing
  • 00:43:12
    nothing and if I'm going to magnify this
  • 00:43:15
    cube that I get a portion of small
  • 00:43:18
    portion of water if you portion of water
  • 00:43:21
    we're going to zoom it in this is the
  • 00:43:24
    figure hey I hope you understand what
  • 00:43:27
    I'm talking about and if we're going to
  • 00:43:29
    look at this figure class if you're
  • 00:43:31
    going to look at this figure okay what
  • 00:43:35
    will happen class is that according to
  • 00:43:38
    our discussion we have pressure is
  • 00:43:40
    equals to F over a and if we're going to
  • 00:43:43
    manipulate the equation we have the
  • 00:43:44
    force is supposed to do the pressure PA
  • 00:43:47
    so at the very bottom of the tube of the
  • 00:43:52
    small tube water when we have a force f1
  • 00:43:56
    acting on the very bottom faith that is
  • 00:44:00
    called the f1 here
  • 00:44:01
    and at the very top of the the cube a
  • 00:44:05
    small element water water element that
  • 00:44:08
    we got here is we have also for school
  • 00:44:11
    okay and of course this F is actually a
  • 00:44:15
    height Y let's label that as height Y
  • 00:44:18
    the very bottom of the view height Y and
  • 00:44:21
    the top where the F 2 is being
  • 00:44:25
    experienced let's call that y plus Delta
  • 00:44:27
    Y so that the height is become now what
  • 00:44:31
    Delta Y this height becomes the Delta Y
  • 00:44:35
    okay so what are we going to do class is
  • 00:44:38
    well of course since force is acting
  • 00:44:42
    perpendicular to that water or cube of
  • 00:44:47
    water smooth element of water well we
  • 00:44:50
    must have also a force acting on the
  • 00:44:52
    sides correct but what happens is that
  • 00:44:56
    if we have a force acting here on the
  • 00:44:58
    side there is also a force acting here
  • 00:45:00
    on the other side so that it is
  • 00:45:02
    perpendicular to the surface of the Pew
  • 00:45:04
    but these two forces cancel each other
  • 00:45:07
    so we need not to include that sir but
  • 00:45:10
    why why do we have to consider the force
  • 00:45:12
    is acting on the Morison and wearing the
  • 00:45:15
    y-axis those things those forces will
  • 00:45:18
    never cancel because this water element
  • 00:45:22
    this element that we got on this open am
  • 00:45:27
    is absolutely has been with that is what
  • 00:45:30
    we got the mg so that if we're going to
  • 00:45:33
    get the summation then we're going to
  • 00:45:36
    get the summation of forces along Y that
  • 00:45:38
    should be equal to 0 because this fluid
  • 00:45:40
    element is static not moving away so
  • 00:45:43
    that if we're going to get the summation
  • 00:45:45
    of forces along Y positive going upward
  • 00:45:49
    we have F 1 negative going downward they
  • 00:45:52
    have F 2 that is minus mg so that these
  • 00:45:55
    two F 1 F 2 will never concern because
  • 00:45:58
    of the presence of the way and that
  • 00:46:00
    should be equal to zero okay so if we're
  • 00:46:06
    going to this is a cube by the way this
  • 00:46:09
    is of view so the surface or the basis
  • 00:46:12
    of the few are all squares and they are
  • 00:46:14
    all in
  • 00:46:15
    okay so if we're going to manipulate
  • 00:46:17
    this force F sub 1 and F sub 2
  • 00:46:20
    toward birth this away so this F 1 is
  • 00:46:24
    also equal to PA and so we have to
  • 00:46:27
    include the a that is the area of the
  • 00:46:30
    faces of the view that we called up a
  • 00:46:32
    okay and this F 1 or the pressure is
  • 00:46:36
    absolutely use the F 1 F 1 is at the
  • 00:46:39
    bottom force at the bottom of the
  • 00:46:43
    element so we have what include thus add
  • 00:46:47
    the pressure up the height Y because it
  • 00:46:52
    is at the very bottom that in our height
  • 00:46:55
    there we designate as wife - again the
  • 00:46:59
    area is the same for all tools okay then
  • 00:47:02
    we have F to the prayer the force at the
  • 00:47:06
    very foot of the very fact that is what
  • 00:47:08
    we call the y plus delta y avait that is
  • 00:47:11
    the height so we have to have P the
  • 00:47:15
    pressure at P y plus Delta Y at the very
  • 00:47:19
    top of the liquid element minus mg so we
  • 00:47:25
    have to convert em as Delta Y times a
  • 00:47:29
    times D density because as we are going
  • 00:47:32
    to recall Rho is equals to mass over
  • 00:47:35
    volume and if you're going to manipulate
  • 00:47:37
    this or mass is equal to volume times
  • 00:47:40
    density and in this case to get the
  • 00:47:42
    volume of this cube simply we have to
  • 00:47:46
    multiply the area away the area any of
  • 00:47:50
    the base times height and the height we
  • 00:47:52
    have considered us Delta Y and of course
  • 00:47:56
    we have the included density of that
  • 00:47:59
    material right so P so this term a times
  • 00:48:03
    Delta Y is the volume and this Rho is
  • 00:48:06
    the density so that we are going to
  • 00:48:08
    replace that by our M with a delta Y ro
  • 00:48:14
    multiplied by force the G is equal to
  • 00:48:18
    zero okay now we can cancel what the
  • 00:48:22
    area's here a way to look this screen so
  • 00:48:25
    we have py minus py plus there
  • 00:48:29
    why minus Delta Y Rho G is equal to zero
  • 00:48:34
    and if you're going to rearrange this
  • 00:48:37
    equation a ramming sneeze equation I'm
  • 00:48:41
    going to rate reality negative Delta Y
  • 00:48:43
    Rho G okay that is equal to what
  • 00:48:47
    positive py plus Delta Y minus py and if
  • 00:48:52
    we're going to divide this both sides by
  • 00:48:55
    or multiply this by 1 over Delta Y what
  • 00:49:00
    will happen is we'll have negative Rho G
  • 00:49:02
    here on the left side so we have py plus
  • 00:49:06
    Delta Y minus py over the Hawaii alright
  • 00:49:11
    so and again if we are going to take the
  • 00:49:15
    limits of both sides for example let me
  • 00:49:18
    just rewrite this alright we're going to
  • 00:49:21
    derive the to get the limits of both
  • 00:49:24
    sides so the limit of Delta Y as Delta y
  • 00:49:29
    equal to 0 the limit of negative Rho G
  • 00:49:31
    and we have to apply him here the limit
  • 00:49:34
    of py plus Delta Y minus P Y divided by
  • 00:49:38
    Delta Y a step away approaches zero let
  • 00:49:41
    us recall that that this is the
  • 00:49:43
    definition of basically the derivative
  • 00:49:45
    okay so and if we offer a value in the
  • 00:49:49
    limit of this since that's how I know
  • 00:49:51
    that how are you then this should be
  • 00:49:53
    constant so what will happen is that we
  • 00:49:55
    have now negative Rho G is equal to what
  • 00:50:00
    the derivative of pressure with respect
  • 00:50:03
    to Delta Y right
  • 00:50:08
    so that what will happen here glass is
  • 00:50:11
    that let's continue let me just erase
  • 00:50:13
    this so we have now the Rho G negative
  • 00:50:17
    Rho G is equal to DP dy so let me just
  • 00:50:21
    write negative Rho G suppose we change
  • 00:50:25
    in pressure with respect to Y we're
  • 00:50:27
    going to cross multiply this negative
  • 00:50:29
    Rho G dy is equal to DP and if we're
  • 00:50:34
    going to integrate this to all right
  • 00:50:38
    integrating this from p1 to people
  • 00:50:42
    integrating this phone height 1 - hi -
  • 00:50:45
    what will happen we have a definite
  • 00:50:48
    integral so it is easy to evaluate the
  • 00:50:52
    definite integral integral we have the
  • 00:50:55
    integral of dy is y and that is
  • 00:50:58
    evaluating limits from y1 to y2 and the
  • 00:51:02
    integral of DS is simply P evaluating
  • 00:51:05
    limits from p1 to p2 so what will happen
  • 00:51:08
    is that the negative Rho G is now equal
  • 00:51:12
    to Y sub 2 minus y sub 1 and equal to P
  • 00:51:16
    sub 2 minus P sub 1 and if I'm going to
  • 00:51:20
    multiply this all by negative 1 what do
  • 00:51:24
    I have to get is let's say Rho G times
  • 00:51:28
    y2 minus y1 is equal to p1 minus p2 now
  • 00:51:35
    this is the hydrostatic or the change in
  • 00:51:38
    pressure whenever we change our height
  • 00:51:41
    all right so this Y plane y1 can be
  • 00:51:45
    replaced as I owe - height 1 because we
  • 00:51:49
    are dealing with the high okay so we
  • 00:51:53
    have proven that as we go on deeper the
  • 00:51:56
    there is a pressure change because of
  • 00:51:59
    the change in high alright so that is
  • 00:52:03
    our formula so I am I have derived it
  • 00:52:06
    here okay so now we have gotten this
  • 00:52:10
    formula right so there is a change in
  • 00:52:14
    pressure whenever there is a change in
  • 00:52:16
    high okay
  • 00:52:21
    sotto continue our discussion we're
  • 00:52:23
    going to talk about the who enforces and
  • 00:52:25
    neck turkey leaders principle on the
  • 00:52:27
    latter part of our argued to with
  • 00:52:29
    mechanics so have you ever tried to push
  • 00:52:31
    a boat down underwater but it's very
  • 00:52:33
    hard to do because there is absolutely
  • 00:52:36
    what people deeply reinforce it is
  • 00:52:38
    extremely difficult to do to hold down
  • 00:52:41
    push them in both underwater because of
  • 00:52:44
    the large upward force exerted by the
  • 00:52:47
    water on the board and that's what of
  • 00:52:49
    the buoyant force the buoyant force
  • 00:52:51
    actually the force that the water or the
  • 00:52:55
    liquid exerted on the object and
  • 00:52:58
    immersed object that is what totally
  • 00:53:00
    buoyant force and we of course once
  • 00:53:03
    again is the outward force exerted by a
  • 00:53:05
    fluid on any immersed object so for
  • 00:53:08
    example we have a cylinder here and we
  • 00:53:10
    have a witness in integral shape here
  • 00:53:13
    that is being immersed on this water
  • 00:53:16
    move it so we have two forces acting on
  • 00:53:19
    its top and on its bottom and the we are
  • 00:53:22
    forced by formula is actually the
  • 00:53:24
    magnitude or the resultant of the forces
  • 00:53:28
    exerted on the top and the bottom of the
  • 00:53:32
    pages of by the liquid so in other words
  • 00:53:34
    FB is equal to f1 or f2 minus f1 the
  • 00:53:39
    result plan of the forces acting on the
  • 00:53:42
    top and on the bottom of the object okay
  • 00:53:46
    and we have the Archimedes principle
  • 00:53:47
    Archimedes principle only states that
  • 00:53:49
    the magnitude of the buoyant force on an
  • 00:53:53
    object by the liquid is always equal to
  • 00:53:56
    the weight of the fluid displaced by the
  • 00:53:58
    object so we have here the formula
  • 00:54:00
    according to work a B this principle we
  • 00:54:02
    have the boolean force is equals to the
  • 00:54:04
    magnitude of the buoyant force equal to
  • 00:54:08
    the weight of the fluid displaced okay
  • 00:54:10
    so let's take note of this this is an
  • 00:54:13
    important concept or principle okay so
  • 00:54:18
    as I'm said available in the video what
  • 00:54:21
    if we have the determining factor
  • 00:54:24
    whether the object will float or sink
  • 00:54:27
    when it is submerged in point liquid is
  • 00:54:30
    that its density so here is the Vepr
  • 00:54:32
    meaning
  • 00:54:33
    if the density of the fluid is greater
  • 00:54:35
    than the density of the object then the
  • 00:54:37
    object will probe alright the Gardeners
  • 00:54:39
    of their size the gardens of the river
  • 00:54:41
    wait as long as the density of the fluid
  • 00:54:44
    is greater than the density of the
  • 00:54:45
    object the object will flow on the other
  • 00:54:47
    hand if the density of the object is
  • 00:54:50
    greater than the density of the fluid to
  • 00:54:53
    which it is immersed the object will see
  • 00:54:55
    so our grandfather would actually talk
  • 00:55:00
    something about us which is much massive
  • 00:55:04
    okay is it the one peso going or the
  • 00:55:07
    ship if you answer the ship why does the
  • 00:55:12
    one peso Polly sing and the ship nothing
  • 00:55:16
    it is much massive than the one pencil
  • 00:55:19
    body okay because the one peso point is
  • 00:55:22
    absolutely has a greater density than
  • 00:55:26
    the density of the fluid or the water so
  • 00:55:29
    that the one peso going on with sink and
  • 00:55:33
    the the ship will actually by convention
  • 00:55:37
    or without any large amount of waves
  • 00:55:41
    well technically that ship is made of
  • 00:55:44
    wood and the the density of wood is
  • 00:55:48
    greater than the density of the fluid
  • 00:55:50
    water or the sea water then we can
  • 00:55:52
    conclude that that ship would actually
  • 00:55:54
    one would actually prove okay so the
  • 00:55:57
    only determining factor whether the
  • 00:56:00
    object will sink or will float is d
  • 00:56:03
    density okay so the ability to sink or
  • 00:56:07
    float is completely independent on the
  • 00:56:09
    dimension of the object or the shape of
  • 00:56:11
    the object the only determining factor
  • 00:56:14
    is the density so we have here for the
  • 00:56:16
    first events the sinking stone we can
  • 00:56:19
    conclude that the density of this stone
  • 00:56:21
    is greater than the density of the water
  • 00:56:23
    to which it is being immersed and we
  • 00:56:25
    have the flow thing we move regardless
  • 00:56:28
    of their size regard regardless of their
  • 00:56:31
    shape okay so it's only determining
  • 00:56:35
    factor whether it will float or not is d
  • 00:56:38
    that's it so we have now deeper newless
  • 00:56:41
    equation when fluids are moving the
  • 00:56:45
    we more complicated than when they are
  • 00:56:47
    starting so suppose we have a fluid
  • 00:56:50
    flowing and if I this is our pipe
  • 00:56:52
    alright if the fluid is starting then
  • 00:56:54
    forget about v1 and v2 then and forget
  • 00:56:58
    about this hydrostatic pressure all
  • 00:57:00
    right we will no longer use that because
  • 00:57:02
    this formula is only for fluids that are
  • 00:57:06
    studies so how about if they move for
  • 00:57:09
    example we have a pipe here okay and
  • 00:57:12
    this pipe is filled with fluid so I'm so
  • 00:57:16
    sorry so fluid so that when the fluid
  • 00:57:20
    flows here it is flowing to a
  • 00:57:22
    cross-sectional area one and it has a
  • 00:57:24
    velocity V sub one
  • 00:57:26
    so when this fluid flows here and
  • 00:57:29
    eventually flows you there will be a
  • 00:57:31
    change in pressure because actually the
  • 00:57:34
    cross-sectional area here is not the
  • 00:57:36
    same as the cross-sectional area here
  • 00:57:37
    and of course there will be a change in
  • 00:57:40
    velocity why because there is also a
  • 00:57:44
    change in the opening of the soap or the
  • 00:57:46
    area of the tube so the we have the
  • 00:57:48
    velocity one here and we have the
  • 00:57:50
    velocity v2 here so of course if we make
  • 00:57:53
    a height of which we use this as our
  • 00:57:56
    reference to y1 the white one and we
  • 00:57:59
    have y2 here so there is also the
  • 00:58:01
    presence of what what potential energy
  • 00:58:05
    because of the height the velocity
  • 00:58:07
    kinetic energy the presence of kinetic
  • 00:58:10
    energy because of the velocity and of
  • 00:58:12
    course the presence of pressure
  • 00:58:14
    so in Bernoulli's equation rugby players
  • 00:58:16
    we have the potential energy because
  • 00:58:18
    there is a height y 1 and y 2 and we
  • 00:58:21
    have velocity 1 and velocity - which
  • 00:58:23
    reminds us of kinetic energy and we have
  • 00:58:26
    force the pressure due to difference in
  • 00:58:29
    area and the force of the fluid acting
  • 00:58:31
    on that area okay so I will not be
  • 00:58:34
    really deriving you the Bernoulli's
  • 00:58:36
    equation because it is a hideous
  • 00:58:38
    there I felt wrong okay but rather I
  • 00:58:41
    would like to take note I would like you
  • 00:58:43
    to take note of this formula which is
  • 00:58:45
    the painless Bernoulli's equation they
  • 00:58:47
    are p1 plus 1/2 the density of that
  • 00:58:51
    fluid V sub 1 squared away the velocity
  • 00:58:53
    1 of that first velocity of the fluid
  • 00:58:56
    plus Rho which again
  • 00:58:58
    the density G is the gravitational force
  • 00:59:00
    and height one is the height of the tube
  • 00:59:03
    with respect to a certain reference that
  • 00:59:05
    is equal to p2 plus one half Rho V sub 2
  • 00:59:07
    squared plus Rho H so this is the famous
  • 00:59:11
    Bernoulli's equation F and of course
  • 00:59:13
    before we solve the the usually the
  • 00:59:16
    problem is solving for the velocity at
  • 00:59:18
    the other end of the tube or the pipe or
  • 00:59:21
    insulting the pressure so in other words
  • 00:59:23
    before we can solve the pressure we have
  • 00:59:28
    to take note of what republican-led
  • 00:59:30
    equation and the continuity equation is
  • 00:59:32
    is simply a 1 V 1 where a 1 is the
  • 00:59:36
    cross-sectional area for 1 and a 2 is
  • 00:59:38
    the cross-sectional area for 2 V 1 is
  • 00:59:41
    the velocity 1 and the velocity 2 during
  • 00:59:43
    the fluid is moving alright so let's try
  • 00:59:48
    to solve a problem about the gasoline
  • 00:59:50
    problem the gasoline entering pipe
  • 00:59:52
    entering the pipe below at 3 meters per
  • 00:59:55
    second here I add 140 kilo Pascal's
  • 00:59:59
    compute the speed and pressure at which
  • 01:00:01
    the liquid meets the Bible all right
  • 01:00:03
    compute the pressure and the velocity of
  • 01:00:06
    the fluid at this point the the density
  • 01:00:10
    of doesn't mean 680 kilogram per cubic
  • 01:00:12
    meter
  • 01:00:14
    why not mistaken I am happy paste it
  • 01:00:17
    will write again our formula is you have
  • 01:00:23
    taken note of that or continuty equation
  • 01:00:25
    we have a1 v1 is equal to a2 v2 we once
  • 01:00:30
    used first the continuity equation to
  • 01:00:33
    find the velocity of the fluid at that
  • 01:00:35
    point at the second point to which it
  • 01:00:37
    lives the pipe so in order for us to get
  • 01:00:40
    v2 v3 simply equal to a1 / a2 times V V
  • 01:00:48
    sub 1
  • 01:00:49
    okay so a1 is we can see here from the
  • 01:00:52
    event
  • 01:00:53
    the radius is 4 cm we have to convert
  • 01:00:55
    that into meters and the cross-sectional
  • 01:00:58
    area of the tube or the pipe is after
  • 01:01:01
    the game yes that's right encircled by
  • 01:01:05
    0.04 meters squared by 4 cm 0.04 me
  • 01:01:11
    / pi the area of the other one is 0.07
  • 01:01:17
    meters squared that's the Regis I'm so
  • 01:01:20
    sorry the area is pi times R squared PI
  • 01:01:24
    R squared so we have here the formula
  • 01:01:28
    and multiplied by the velocity 1
  • 01:01:31
    velocity 1 is also d13 meters per second
  • 01:01:37
    so if we're going to compute that PI's
  • 01:01:40
    will cancel so we have zero point zero
  • 01:01:44
    four squared divided by zero point zero
  • 01:01:46
    seven squared times 3 and that is 0.98
  • 01:01:50
    meters per second so this is now our
  • 01:01:55
    velocity pool all right how do we find
  • 01:01:58
    the pressure oh we're going to use the
  • 01:02:01
    Bernoulli's equation where in fresh or 1
  • 01:02:04
    plus 1/2 Rho P sub 1 squared plus Rho G
  • 01:02:10
    H 1 is equal to P sub 2 plus 1/2 Rho V
  • 01:02:15
    sub 2 squared plus Rho G H sub 2 ok and
  • 01:02:21
    we are up 30 we are after to find D in
  • 01:02:26
    sub 2 so we have to isolate y sub 2 okay
  • 01:02:30
    so we have P sub 1 all right plus 1/2
  • 01:02:35
    Rho if we're going to factor it out
  • 01:02:38
    transpose this on the left side of a and
  • 01:02:42
    factor out 1/2 Rho of a 1/2 Rho that
  • 01:02:45
    should be V 1 V 1 squared minus minus V
  • 01:02:50
    sub 2 squared plus Rho G away again
  • 01:02:55
    we're going to transpose this on the
  • 01:02:58
    left side all right so that we can
  • 01:03:00
    factor Rho G that is i1 minus i2 so the
  • 01:03:06
    pressure 1 is 140 kilo Pascal's plus 1/2
  • 01:03:11
    the ROE is 680 that's the density of the
  • 01:03:16
    gasoline V sub 1 is 3 squared 3 meter
  • 01:03:20
    squared
  • 01:03:21
    meter per second squared I would not
  • 01:03:24
    like the unit soon so that we could
  • 01:03:27
    consume detailed space so three squared
  • 01:03:30
    minus zero point nine a a squared plus
  • 01:03:34
    zero again 680 kilograms per cubic meter
  • 01:03:38
    times the G that is 9.81 m/s^2
  • 01:03:43
    multiplied by height one - hi - and
  • 01:03:45
    according to our figure height one is
  • 01:03:48
    six meters
  • 01:03:49
    all right the reference is the height is
  • 01:03:52
    zero so six minus zero or simply six so
  • 01:03:57
    we're going to compute the pressure - of
  • 01:03:59
    that okay let's try it on page 148 up
  • 01:04:03
    asked us plus one-half 680 times three
  • 01:04:07
    squared minus zero point nine eight
  • 01:04:09
    squared plus six eight T times 9.81
  • 01:04:13
    times 6 and that is 182 or 183 oops
  • 01:04:23
    so our answer should be one one hundred
  • 01:04:29
    eighty two point seventy six kilo
  • 01:04:32
    pascals one hundred eighty two point
  • 01:04:37
    seventy six kilo Pascal's so that's all
  • 01:04:42
    okay so thank you so much for listening
  • 01:04:45
    here are our references university
  • 01:04:47
    physics by survey and modern physics in
  • 01:04:51
    person physics by young in Friedman and
  • 01:04:53
    pieces look like great terrific Envato
  • 01:04:55
    that is a philippine order so thank you
  • 01:04:57
    so much for listening and god bless you
Etiquetas
  • Fluid Mechanics
  • Pressure
  • Density
  • Pascal's Principle
  • Bernoulli's Equation
  • Hydraulics
  • Buoyancy
  • Fluid Dynamics
  • SI Units
  • Archimedes' Principle