#set document(title: "11.5 Pascal’s Principle", author: "OpenStax / XYZ Homework") #set page(width: 8.5in, height: auto, margin: 1in) #import "@preview/cetz:0.5.2" #set text(font: ("STIX Two Text", "Libertinus Serif", "New Computer Modern"), size: 10.5pt, lang: "en") #show math.equation: set text(font: ("STIX Two Math", "New Computer Modern Math")) #set par(justify: true, leading: 0.62em, spacing: 0.9em) #set enum(spacing: 1.1em) // room between list items so tall inline fractions don't collide #set list(spacing: 1.1em) #set table(stroke: 0.5pt + rgb("#c7ccd3")) #let BLUE = rgb("#183B6F") // brand navy — section bars + example/solution labels (white on navy 11.09:1) #let ORANGE = rgb("#A94509") // brand primary-700 — AA-safe deep orange for TEXT (5.93:1 on white; raw brand #F37021 is 2.94:1 and must never carry text) #let RED = rgb("#DC2626") // brand error-600 #let GREEN = rgb("#059669") // brand success-600 (decoration only; small green text uses green-text #007942) #show heading.where(level: 1): it => block(width: 100%, above: 0pt, below: 16pt, fill: gradient.linear(BLUE, rgb("#2C5AA0")), inset: (x: 14pt, y: 12pt), radius: 3pt, text(fill: white, weight: "bold", size: 19pt, it.body)) #show heading.where(level: 2): it => block(width: 100%, above: 18pt, below: 10pt, fill: BLUE, inset: (x: 10pt, y: 6pt), radius: 2pt, text(fill: white, weight: "bold", size: 12pt, it.body)) #show heading.where(level: 3): it => text(fill: ORANGE, weight: "bold", size: 12.5pt, it.body) #show heading.where(level: 4): it => text(fill: BLUE, weight: "bold", size: 10.5pt, it.body) #let examplebox(label, title, body) = block(width: 100%, breakable: true, fill: rgb("#EFF1F5"), stroke: 0.5pt + rgb("#CFDDF0"), radius: 4pt, inset: 10pt, above: 12pt, below: 12pt)[ #block(below: 6pt)[#box(fill: BLUE, inset: (x: 6pt, y: 2pt), radius: 2pt, text(fill: white, weight: "bold", size: 8.5pt, label)) #h(0.4em) #strong[#title]] #body] // rail = decorative left rule (raw brand token); labelcolor = AA-safe label text shade #let notebox(label, rail, labelcolor, tint, body) = block(width: 100%, breakable: true, fill: tint, stroke: (left: 3pt + rail), inset: (left: 10pt, rest: 8pt), radius: (right: 4pt), above: 11pt, below: 11pt)[ #text(fill: labelcolor, weight: "bold", size: 7.5pt, tracking: 0.5pt)[#upper(label)] #linebreak() #body] #let solutionbox(body) = block(above: 4pt, below: 8pt)[ #text(fill: BLUE, weight: "bold", size: 8.5pt)[Solution] #linebreak() #body] #let figph(msg) = block(width: 100%, height: 60pt, fill: rgb("#f6f7f9"), stroke: (paint: rgb("#c7ccd3"), dash: "dashed"), radius: 4pt, inset: 10pt)[ #align(center + horizon, text(fill: rgb("#889"), style: "italic", size: 9pt, msg))] // Standardize inlined figure sizes: measure the natural CeTZ canvas, then scale to a // consistent envelope (aspect-aware; see build_typst.py FIG_* constants). Unlike the // print preamble, dimensions are FLOORED: in an editor a user can trim a figure to a // degenerate 1-D shape (a bare line), and w/h or tw/w would then divide by zero. #let _STD_W = 3.5 #let _WIDE_W = 5.6 #let _MAX_H = 3.4 #let _ASPECT_WIDE = 2.2 #let _UPSCALE_MAX = 1.15 #let stdfig(body) = context { let m = measure(body) let w = calc.max(m.width / 1in, 0.01) let h = calc.max(m.height / 1in, 0.01) let tw = if w / h > _ASPECT_WIDE { _WIDE_W } else { _STD_W } let s = calc.min(tw / w, _MAX_H / h, _UPSCALE_MAX) align(center, box(scale(x: s * 100%, y: s * 100%, reflow: true, body))) } #show figure: set block(breakable: false) #set figure(gap: 8pt) #show figure.caption: set text(size: 8.5pt, fill: rgb("#555")) == 11.5#h(0.6em)Pascal’s Principle === Learning Objectives By the end of this section, you will be able to: - Define pressure. - State Pascal’s principle. - Understand applications of Pascal’s principle. - Derive relationships between forces in a hydraulic system. #strong[Pressure] is defined as force per unit area. Can pressure be increased in a fluid by pushing directly on the fluid? Yes, but it is much easier if the fluid is enclosed. The heart, for example, increases blood pressure by pushing directly on the blood in an enclosed system (valves closed in a chamber). If you try to push on a fluid in an open system, such as a river, the fluid flows away. An enclosed fluid cannot flow away, and so pressure is more easily increased by an applied force. What happens to a pressure in an enclosed fluid? Since atoms in a fluid are free to move about, they transmit the pressure to all parts of the fluid and to the walls of the container. Remarkably, the pressure is transmitted #emph[undiminished]. This phenomenon is called #strong[Pascal’s principle], because it was first clearly stated by the French philosopher and scientist Blaise Pascal (1623–1662): 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. #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Pascal’s Principle] 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. ] Pascal’s principle, an experimentally verified fact, is what makes pressure so important in fluids. Since a change in pressure is transmitted undiminished in an enclosed fluid, we often know more about pressure than other physical quantities in fluids. Moreover, Pascal’s principle implies that #emph[the total pressure in a fluid is the sum of the pressures from different sources]. We shall find this fact—that pressures add—very useful. Blaise Pascal had an interesting life in that he was home-schooled by his father who removed all of the mathematics textbooks from his house and forbade him to study mathematics until the age of 15. This, of course, raised the boy’s curiosity, and by the age of 12, he started to teach himself geometry. Despite this early deprivation, Pascal went on to make major contributions in the mathematical fields of probability theory, number theory, and geometry. He is also well known for being the inventor of the first mechanical digital calculator, in addition to his contributions in the field of fluid statics. === Application of Pascal’s Principle One of the most important technological applications of Pascal’s principle is found in a #emph[hydraulic system], which is an enclosed fluid system used to exert forces. The most common hydraulic systems are those that operate car brakes. Let us first consider the simple hydraulic system shown. #figure(figph[A small force can be converted into a larger force when pressure is transmitted through liquids in different containers with pistons that are connected.], alt: "A small force can be converted into a larger force when pressure is transmitted through liquids in different containers with pistons that are connected.", caption: [A typical hydraulic system with two fluid-filled cylinders, capped with pistons and connected by a tube called a hydraulic line. A downward force #math.equation(block: false, alt: "F sub 1")[$"F"_(1)$] on the left piston creates a pressure that is transmitted undiminished to all parts of the enclosed fluid. This results in an upward force #math.equation(block: false, alt: "F sub 2")[$"F"_(2)$] on the right piston that is larger than #math.equation(block: false, alt: "F sub 1")[$"F"_(1)$] because the right piston has a larger area.]) === Relationship Between Forces in a Hydraulic System We can derive a relationship between the forces in the simple hydraulic system shown in by applying Pascal’s principle. Note first that the two pistons in the system are at the same height, and so there will be no difference in pressure due to a difference in depth. Now the pressure due to #math.equation(block: false, alt: "F sub 1")[$F_(1)$] acting on area #math.equation(block: false, alt: "A sub 1")[$A_(1)$] is simply #math.equation(block: false, alt: "P sub 1 equals the fraction F sub 1 over A sub 1")[$P_(1) = frac(F_(1), A_(1))$], as defined by #math.equation(block: false, alt: "P equals the fraction F over A")[$P = frac(F, A)$]. According to Pascal’s principle, this pressure is transmitted undiminished throughout the fluid and to all walls of the container. Thus, a pressure #math.equation(block: false, alt: "P sub 2")[$P_(2)$] is felt at the other piston that is equal to #math.equation(block: false, alt: "P sub 1")[$P_(1)$]. That is #math.equation(block: false, alt: "P sub 1 equals P sub 2")[$P_(1) = P_(2)$]. But since #math.equation(block: false, alt: "P sub 2 equals the fraction F sub 2 over A sub 2")[$P_(2) = frac(F_(2), A_(2))$], we see that #math.equation(block: false, alt: "the fraction F sub 1 over A sub 1 equals the fraction F sub 2 over A sub 2")[$frac(F_(1), A_(1)) = frac(F_(2), A_(2))$]. This equation relates the ratios of force to area in any hydraulic system, providing 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 and the right one has an area five times greater, then the force out 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. #examplebox("Example 1")[Calculating Force of Wheel Cylinders: Pascal Puts on the Brakes][ Consider the automobile hydraulic system shown. #figure(figph[When the driver applies force on the brake pedal the pedal cylinder transmits the same pressure to the wheel cylinders but results in a larger force due to the larger area of the wheel cylinders.], alt: "When the driver applies force on the brake pedal the pedal cylinder transmits the same pressure to the wheel cylinders but results in a larger force due to the larger area of the wheel cylinders.", caption: [Hydraulic brakes use Pascal’s principle. The driver exerts a force of 100 N on the brake pedal. This force 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 #math.equation(block: false, alt: "F sub 2")[$F_(2)$]. The circular cross-sectional areas of the pedal and wheel cylinders are represented by #math.equation(block: false, alt: "A sub 1")[$A_(1)$] and #math.equation(block: false, alt: "A sub 2")[$A_(2)$], respectively]) A force of 100 N is applied to the brake pedal, which acts on the pedal cylinder through a lever. A force of 500 N is exerted on the pedal cylinder. (The reader can verify that the force is 500 N using techniques of statics from Applications of Statics, Including Problem-Solving Strategies.) Pressure created in the pedal cylinder is transmitted to 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 force #math.equation(block: false, alt: "F sub 2")[$F_(2)$] created at each of the wheel cylinders. Strategy We are given the force #math.equation(block: false, alt: "F sub 1")[$F_(1)$] that is applied to the pedal cylinder. The cross-sectional areas #math.equation(block: false, alt: "A sub 1")[$A_(1)$] and #math.equation(block: false, alt: "A sub 2")[$A_(2)$] can be calculated from their given diameters. Then #math.equation(block: false, alt: "the fraction F sub 1 over A sub 1 equals the fraction F sub 2 over A sub 2")[$frac(F_(1), A_(1)) = frac(F_(2), A_(2))$] can be used to find the force #math.equation(block: false, alt: "F sub 2")[$F_(2)$]. Manipulate this algebraically to get #math.equation(block: false, alt: "F sub 2")[$F_(2)$] on one side and substitute known values: Solution Pascal’s principle applied to hydraulic systems is given by #math.equation(block: false, alt: "the fraction F sub 1 over A sub 1 equals the fraction F sub 2 over A sub 2")[$frac(F_(1), A_(1)) = frac(F_(2), A_(2))$]: #math.equation(block: true, alt: "F sub 2 equals the fraction A sub 2 over A sub 1 F sub 1 equals the fraction πr 2 2 over πr 1 2 F sub 1 equals the fraction 1.25 cm squared over 0.250 cm squared times 500 N equals 1 . 25 times 10 to the power 4 N .")[$F_(2) = frac(A_(2), A_(1)) F_(1) = frac("πr"_(2)^(2), "πr"_(1)^(2)) F_(1) = frac(attach(1.25 "cm", t: 2), attach(0.250 "cm", t: 2)) × "500 N" = 1 "." "25" × "10"^(4) #h(0.25em) "N" .$] Discussion 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 #math.equation(block: false, alt: "1 . 25 times 10 to the power 4 N .")[$1 "." "25" × "10"^(4) #h(0.25em) "N" "."$] ] A simple hydraulic system, such as a simple machine, can increase force but cannot do more work than 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 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. The movement of the legs of a spider is achieved partly by hydraulics. Using hydraulics, a jumping spider can create a force that makes it capable of jumping 25 times its length! #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #emph[Making Connections: Conservation of Energy] Conservation of energy applied to a hydraulic system tells us that the system cannot do more work than is done on it. Work transfers energy, and so the work output cannot exceed the work input. Power brakes and other similar hydraulic systems use pumps to supply extra energy when needed. ] === Section Summary - 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 pedal cylinder in a hydraulic system is at a greater height than the wheel cylinder. Explain how this will affect the force produced at the wheel cylinder. === Problems & Exercises How much pressure is transmitted in the hydraulic system considered in ? Express your answer in pascals and in atmospheres. #solutionbox[ #math.equation(block: false, alt: "2.55 times 10 to the power 7 Pa")[$2.55 × 10^(7) #h(0.25em) "Pa"$]; or 251 atm ] What force must be exerted on the pedal cylinder of a hydraulic lift to support the weight of a 2000-kg car (a large car) resting on the wheel cylinder? The pedal cylinder has a 2.00-cm diameter and the wheel has a 24.0-cm diameter. A crass host pours the remnants of several bottles of wine into a jug after a party. He then inserts a cork with a 2.00-cm diameter into the bottle, placing it in direct contact with the wine. He is amazed when he 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. #solutionbox[ #math.equation(block: false, alt: "5 . 76 times 10 cubed N")[$5 "." "76" × "10"^(3) #h(0.25em) "N"$] extra force ] 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 wheel cylinder to the area of the pedal 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 to friction. (a) Verify that work input equals work output for a hydraulic system assuming no losses 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. (b) 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? #solutionbox[ (a) #math.equation(block: false, alt: "V equals d sub i A sub i equals d sub o A sub o ⇒ d sub o equals d sub i the fraction A sub i over A sub o .")[$V = d_("i") A_("i") = d_("o") A_("o") ⇒ d_("o") = d_("i") frac(A_("i"), A_("o")) "."$] Now, using equation: #math.equation(block: true, alt: "the fraction F sub 1 over A sub 1 equals the fraction F sub 2 over A sub 2 ⇒ F sub o equals F sub i the fraction A sub o over A sub i .")[$frac(F_(1), A_(1)) = frac(F_(2), A_(2)) ⇒ F_("o") = F_("i") frac(A_("o"), A_("i")) "."$] Finally, #math.equation(block: true, alt: "W sub o equals F sub o d sub o equals the fraction F sub i A sub o over A sub i the fraction d sub i A sub i over A sub o equals F sub i d sub i equals W sub i .")[$W_("o") = F_("o") d_("o") = frac(F_("i") A_("o"), A_("i")) frac(d_("i") A_("i"), A_("o")) = F_("i") d_("i") = W_("i") .$] In other words, the work output equals the work input. (b) If the system is not moving, friction would not play a role. With friction, we know there are losses, so that #math.equation(block: false, alt: "W sub out equals W sub in minus W sub f")[$W_("out") = W_("in") − W_("f")$]; therefore, the work output is less than the work input. In other words, with friction, you need to push harder on the input piston than was calculated for the nonfriction case. ]