#set document(title: "12.3 Stress, Strain, and Elastic Modulus", 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")) == 12.3#h(0.6em)Stress, Strain, and Elastic Modulus A model of a rigid body is an idealized example of an object that does not deform under the actions of external forces. It is very useful when analyzing mechanical systems—and many physical objects are indeed rigid to a great extent. The extent to which an object can be #emph[perceived] as rigid depends on the physical properties of the material from which it is made. For example, a ping-pong ball made of plastic is brittle, and a tennis ball made of rubber is elastic when acted upon by squashing forces. However, under other circumstances, both a ping-pong ball and a tennis ball may bounce well as rigid bodies. Similarly, someone who designs prosthetic limbs may be able to approximate the mechanics of human limbs by modeling them as rigid bodies; however, the actual combination of bones and tissues is an elastic medium. For the remainder of this chapter, we move from consideration of forces that affect the motion of an object to those that affect an object’s shape. A change in shape due to the application of a force is known as a #strong[deformation]. Even very small forces are known to cause some deformation. Deformation is experienced by objects or physical media under the action of external forces—for example, this may be squashing, squeezing, ripping, twisting, shearing, or pulling the objects apart. In the language of physics, two terms describe the forces on objects undergoing deformation: #emph[stress] and #emph[strain]. Stress is a quantity that describes the magnitude of forces that cause deformation. Stress is generally defined as #emph[force per unit area]. When forces pull on an object and cause its elongation, like the stretching of an elastic band, we call such stress a #strong[tensile stress]. When forces cause a compression of an object, we call it a #strong[compressive stress]. When an object is being squeezed from all sides, like a submarine in the depths of an ocean, we call this kind of stress a #strong[bulk stress] (or #strong[volume stress]). In other situations, the acting forces may be neither tensile nor compressive, and still produce a noticeable deformation. For example, suppose you hold a book tightly between the palms of your hands, then with one hand you press-and-pull on the front cover away from you, while with the other hand you press-and-pull on the back cover toward you. In such a case, when deforming forces act tangentially to the object’s surface, we call them ‘shear’ forces and the stress they cause is called #strong[shear stress]. The SI unit of stress is the pascal (Pa). When one newton of force presses on a unit surface area of one meter squared, the resulting stress is one pascal: #math.equation(block: true, alt: "one pascal equals 1.0 Pa equals the fraction 1.0 N over 1.0 m squared .")[$"one pascal" = 1.0 #h(0.2em) "Pa" = frac(1.0 #h(0.2em) "N", 1.0 #h(0.2em) "m"^(2)) .$] In the Imperial system of units, the unit of stress is ‘psi,’ which stands for ‘pound per square inch’ #math.equation(block: false, alt: "open parenthesis lb/in squared close parenthesis .")[$( "lb/in"^(2) ) .$] Another unit that is often used for bulk stress is the atm (atmosphere). Conversion factors are #math.equation(block: true, alt: "1 psi equals 6895 Pa and 1 Pa equals 1.450 times 10 to the power −4 psi; 1 atm equals 1.013 times 10 to the power 5 Pa equals 14.7 psi.")[$1 #h(0.2em) "psi" = 6895 #h(0.2em) "Pa" #h(0.5em) "and" #h(0.5em) 1 #h(0.2em) "Pa" = 1.450 #h(0.2em) × #h(0.2em) 10^(−4) "psi" \ 1 #h(0.2em) "atm" = 1.013 #h(0.2em) × #h(0.2em) 10^(5) "Pa" = 14.7 #h(0.2em) "psi."$] An object or medium under stress becomes deformed. The quantity that describes this deformation is called #strong[strain]. Strain is given as a fractional change in either length (under tensile stress) or volume (under bulk stress) or geometry (under shear stress). Therefore, strain is a dimensionless number. Strain under a tensile stress is called #strong[tensile strain], strain under bulk stress is called #strong[bulk strain] (or #strong[volume strain]), and that caused by shear stress is called #strong[shear] #strong[strain]. The greater the stress, the greater the strain; however, the relation between strain and stress does not need to be linear. Only when stress is sufficiently low is the deformation it causes in direct proportion to the stress value. The proportionality constant in this relation is called the #strong[elastic] #strong[modulus]. In the linear limit of low stress values, the general relation between stress and strain is #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "stress equals (elastic modulus) times strain.")[$"stress" = "(elastic modulus)" #h(0.2em) × #h(0.2em) "strain."$] ] As we can see from dimensional analysis of this relation, the elastic modulus has the same physical unit as stress because strain is dimensionless. We can also see from that when an object is characterized by a large value of elastic modulus, the effect of stress is small. On the other hand, a small elastic modulus means that stress produces large strain and noticeable deformation. For example, a stress on a rubber band produces larger strain (deformation) than the same stress on a steel band of the same dimensions because the elastic modulus for rubber is two orders of magnitude smaller than the elastic modulus for steel. The elastic modulus for tensile stress is called #strong[Young’s modulus]; that for the bulk stress is called the #strong[bulk modulus]; and that for shear stress is called the #strong[shear modulus]. Note that the relation between stress and strain is an #emph[observed] relation, measured in the laboratory. Elastic moduli for various materials are measured under various physical conditions, such as varying temperature, and collected in engineering data tables for reference . These tables are valuable references for industry and for anyone involved in engineering or construction. In the next section, we discuss strain-stress relations beyond the linear limit represented by , in the full range of stress values up to a fracture point. In the remainder of this section, we study the linear limit expressed by . #figure(table( columns: 4, align: left, inset: 6pt, table.header([Material], [Young’s modulus #linebreak() #math.equation(block: false, alt: "times 10 to the power 10 Pa")[$× #h(0.2em) 10^(10) "Pa"$]], [Bulk modulus #linebreak() #math.equation(block: false, alt: "times 10 to the power 10 Pa")[$× #h(0.2em) 10^(10) "Pa"$]], [Shear modulus #linebreak() #math.equation(block: false, alt: "times 10 to the power 10 Pa")[$× #h(0.2em) 10^(10) "Pa"$]]), [Aluminum], [7.0], [7.5], [2.5], [Bone (tension)], [1.6], [0.8], [8.0], [Bone (compression)], [0.9], [], [], [Brass], [9.0], [6.0], [3.5], [Brick], [1.5], [], [], [Concrete], [2.0], [], [], [Copper], [11.0], [14.0], [4.4], [Crown glass], [6.0], [5.0], [2.5], [Granite], [4.5], [4.5], [2.0], [Hair (human)], [1.0], [], [], [Hardwood], [1.5], [], [1.0], [Iron], [21.0], [16.0], [7.7], [Lead], [1.6], [4.1], [0.6], [Marble], [6.0], [7.0], [2.0], [Nickel], [21.0], [17.0], [7.8], [Polystyrene], [3.0], [], [], [Silk], [6.0], [], [], [Spider thread], [3.0], [], [], [Steel], [20.0], [16.0], [7.5], [Acetone], [], [0.07], [], [Ethanol], [], [0.09], [], [Glycerin], [], [0.45], [], [Mercury], [], [2.5], [], [Water], [], [0.22], [], )) === Tensile or Compressive Stress, Strain, and Young’s Modulus Tension or compression occurs when two antiparallel forces of equal magnitude act on an object along only one of its dimensions, in such a way that the object does not move. One way to envision such a situation is illustrated in . A rod segment is either stretched or squeezed by a pair of forces acting along its length and perpendicular to its cross-section. The net effect of such forces is that the rod changes its length from the original length #math.equation(block: false, alt: "L sub 0")[$L_(0)$] that it had before the forces appeared, to a new length #emph[L] that it has under the action of the forces. This change in length #math.equation(block: false, alt: "Δ L equals L minus L sub 0")[$"Δ" L = L − L_(0)$] may be either elongation (when #emph[L] is larger than the original length #math.equation(block: false, alt: "L sub 0 close parenthesis")[$L_(0) \)$] or contraction (when #emph[L] is smaller than the original length #math.equation(block: false, alt: "L sub 0 close parenthesis .")[$L_(0) \) .$] Tensile stress and strain occur when the forces are stretching an object, causing its elongation, and the length change #math.equation(block: false, alt: "Δ L")[$"Δ" L$] is positive. Compressive stress and strain occur when the forces are contracting an object, causing its shortening, and the length change #math.equation(block: false, alt: "Δ L")[$"Δ" L$] is negative. In either of these situations, we define stress as the ratio of the deforming force #math.equation(block: false, alt: "F sub ⊥")[$F_(⊥)$] to the cross-sectional area #emph[A] of the object being deformed. The symbol #math.equation(block: false, alt: "F sub ⊥")[$F_(⊥)$] that we reserve for the deforming force means that this force acts perpendicularly to the cross-section of the object. Forces that act parallel to the cross-section do not change the length of an object. The definition of the tensile stress is #math.equation(block: true, alt: "tensile stress equals the fraction F sub ⊥ over A .")[$"tensile stress" = frac(F_(⊥), A) .$] Tensile strain is the measure of the deformation of an object under tensile stress and is defined as the fractional change of the object’s length when the object experiences tensile stress #math.equation(block: true, alt: "tensile strain equals the fraction Δ L over L sub 0 .")[$"tensile strain" = frac("Δ" L, L_(0)) .$] Compressive stress and strain are defined by the same formulas, and , respectively. The only difference from the tensile situation is that for compressive stress and strain, we take absolute values of the right-hand sides in and . #figure(figph[Figure A is a schematic drawing of a cylinder with a length L0 that is under the tensile strain. Two forces at the different sides of cylinder increase its length by Delta L. Figure B is a schematic drawing of a cylinder with a length L0 that is under the compressive strain. Two forces at the different sides of cylinder reduce its length by Delta L.], alt: "Figure A is a schematic drawing of a cylinder with a length L0 that is under the tensile strain. Two forces at the different sides of cylinder increase its length by Delta L. Figure B is a schematic drawing of a cylinder with a length L0 that is under the compressive strain. Two forces at the different sides of cylinder reduce its length by Delta L.", caption: [When an object is in either tension or compression, the net force on it is zero, but the object deforms by changing its original length #math.equation(block: false, alt: "L sub 0 .")[$L_(0) .$] (a) Tension: The rod is elongated by #math.equation(block: false, alt: "Δ L .")[$"Δ" L .$] (b) Compression: The rod is contracted by #math.equation(block: false, alt: "Δ L .")[$"Δ" L .$] In both cases, the deforming force acts along the length of the rod and perpendicular to its cross-section. In the linear range of low stress, the cross-sectional area of the rod does not change.]) Young’s modulus #emph[Y] is the elastic modulus when deformation is caused by either tensile or compressive stress, and is defined by . Dividing this equation by tensile strain, we obtain the expression for Young’s modulus: #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "Y equals the fraction tensile stress over tensile strain equals the fraction F sub ⊥ / A over Δ L / L sub 0 equals the fraction F sub ⊥ over A the fraction L sub 0 over Δ L .")[$Y = frac("tensile stress", "tensile strain") = frac(F_(⊥) "/" #h(0.1em) A, "Δ" L #h(0.1em) "/" #h(0.1em) L_(0)) = frac(F_(⊥), A) #h(0.2em) frac(L_(0), "Δ" L) .$] ] #examplebox("Example 1")[Compressive Stress in a Pillar][ A sculpture weighing 10,000 N rests on a horizontal surface at the top of a 6.0-m-tall vertical pillar . The pillar’s cross-sectional area is #math.equation(block: false, alt: "0 .20 m squared")[$0 attach(".20 m", t: 2)$] and it is made of granite with a mass density of #math.equation(block: false, alt: "2700 kg/m cubed .")[$attach(2700 #h(0.2em) "kg/m", t: 3) .$] Find the compressive stress at the cross-section located 3.0 m below the top of the pillar and the value of the compressive strain of the top 3.0-m segment of the pillar. #figure(figph[Picture shows a photograph of Nelson’s Column in Trafalgar Square.], alt: "Picture shows a photograph of Nelson’s Column in Trafalgar Square.", caption: [Nelson’s Column in Trafalgar Square, London, England.]) Strategy First we find the weight of the 3.0-m-long top section of the pillar. The normal force that acts on the cross-section located 3.0 m down from the top is the sum of the pillar’s weight and the sculpture’s weight. Once we have the normal force, we use to find the stress. To find the compressive strain, we find the value of Young’s modulus for granite in and invert . Solution The volume of the pillar segment with height #math.equation(block: false, alt: "h equals 3.0 m")[$h = 3.0 #h(0.2em) "m"$] and cross-sectional area #math.equation(block: false, alt: "A equals 0.20 m squared")[$A = 0.20 #h(0.2em) "m"^(2)$] is #math.equation(block: true, alt: "V equals A h equals open parenthesis 0.20 m squared close parenthesis open parenthesis 3.0 m close parenthesis equals 0.60 m cubed .")[$V = A h = ( 0.20 #h(0.2em) "m"^(2) ) ( 3.0 #h(0.2em) "m" ) = 0.60 #h(0.2em) "m"^(3) .$] With the density of granite #math.equation(block: false, alt: "ρ equals 2.7 times 10 cubed kg/m cubed ,")[$ρ = 2.7 #h(0.2em) × #h(0.2em) 10^(3) #h(0.2em) "kg/m"^(3) ,$] the mass of the pillar segment is #math.equation(block: true, alt: "m equals ρ V equals open parenthesis 2.7 times 10 cubed kg/m cubed close parenthesis open parenthesis 0.60 m cubed close parenthesis equals 1.60 times 10 cubed kg .")[$m = ρ V = ( 2.7 #h(0.2em) × #h(0.2em) 10^(3) #h(0.2em) "kg/m"^(3) ) ( 0.60 #h(0.2em) "m"^(3) ) = 1.60 #h(0.2em) × #h(0.2em) 10^(3) "kg" .$] The weight of the pillar segment is #math.equation(block: true, alt: "w sub p equals m g equals open parenthesis 1.60 times 10 cubed kg close parenthesis open parenthesis 9.80 m/s squared close parenthesis equals 1.568 times 10 to the power 4 N.")[$w_(p) = m g = ( 1.60 #h(0.2em) × #h(0.2em) 10^(3) "kg" ) ( 9.80 #h(0.2em) "m/s"^(2) ) = 1.568 #h(0.2em) × #h(0.2em) 10^(4) "N."$] The weight of the sculpture is #math.equation(block: false, alt: "w sub s equals 1.0 times 10 to the power 4 N ,")[$w_(s) = 1.0 #h(0.2em) × #h(0.2em) 10^(4) "N" ,$] so the normal force on the cross-sectional surface located 3.0 m below the sculpture is #math.equation(block: true, alt: "F sub ⊥ equals w sub p plus w sub s equals open parenthesis 1.568 plus 1.0 close parenthesis times 10 to the power 4 N equals 2.568 times 10 to the power 4 N.")[$F_(⊥) = w_(p) + w_(s) = ( 1.568 + 1.0 ) #h(0.2em) × #h(0.2em) 10^(4) "N" = 2.568 #h(0.2em) × #h(0.2em) 10^(4) "N."$] Therefore, the stress is #math.equation(block: true, alt: "stress equals the fraction F sub ⊥ over A equals the fraction 2.568 times 10 to the power 4 N over 0.20 m squared equals 1.284 times 10 to the power 5 Pa equals 128.4 kPa.")[$"stress" = frac(F_(⊥), A) = frac(2.568 #h(0.2em) × #h(0.2em) 10^(4) "N", 0.20 #h(0.2em) "m"^(2)) = 1.284 #h(0.2em) × #h(0.2em) 10^(5) "Pa" = "128.4 kPa."$] Young’s modulus for granite is #math.equation(block: false, alt: "Y equals 4.5 times 10 to the power 10 Pa equals 4.5 times 10 to the power 7 kPa .")[$Y = 4.5 #h(0.2em) × #h(0.2em) 10^(10) "Pa" = 4.5 #h(0.2em) × #h(0.2em) 10^(7) "kPa" .$] Therefore, the compressive strain at this position is #math.equation(block: true, alt: "strain equals the fraction stress over Y equals the fraction 128.4 kPa over 4.5 times 10 to the power 7 kPa equals 2.85 times 10 to the power −6 .")[$"strain" = frac("stress", Y) = frac(128.4 #h(0.2em) "kPa", 4.5 #h(0.2em) × #h(0.2em) 10^(7) "kPa") = 2.85 #h(0.2em) × #h(0.2em) 10^(−6) .$] Significance Notice that the normal force acting on the cross-sectional area of the pillar is not constant along its length, but varies from its smallest value at the top to its largest value at the bottom of the pillar. Thus, if the pillar has a uniform cross-sectional area along its length, the stress is largest at its base. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Find the compressive stress and strain at the base of Nelson’s column. #solutionbox[ #math.equation(block: true, alt: "206.8 kPa ; 4.6 times 10 to the power −5")[$206.8 #h(0.2em) "kPa" ; #h(0.2em) 4.6 #h(0.2em) × #h(0.2em) 10^(−5)$] ] ] #examplebox("Example 2")[Stretching a Rod][ A 2.0-m-long steel rod has a cross-sectional area of #math.equation(block: false, alt: "0.30 cm squared .")[$0.30 #h(0.2em) "cm"^(2) .$] The rod is a part of a vertical support that holds a heavy 550-kg platform that hangs attached to the rod’s lower end. Ignoring the weight of the rod, what is the tensile stress in the rod and the elongation of the rod under the stress? Strategy First we compute the tensile stress in the rod under the weight of the platform in accordance with . Then we invert to find the rod’s elongation, using #math.equation(block: false, alt: "L sub 0 equals 2.0 m .")[$L_(0) = 2.0 #h(0.2em) "m" .$] From , Young’s modulus for steel is #math.equation(block: false, alt: "Y equals 2.0 times 10 to the power 11 Pa .")[$Y = #h(0.2em) 2.0 #h(0.2em) × #h(0.2em) 10^(11) "Pa" .$] Solution Substituting numerical values into the equations gives us #math.equation(block: true, alt: "the fraction F sub ⊥ over A, equals, the fraction open parenthesis 550 kg close parenthesis open parenthesis 9.8 m/s squared close parenthesis over 3.0 times 10 to the power −5 m squared equals 1.8 times 10 to the power 8 Pa; Δ L, equals, the fraction F sub ⊥ over A the fraction L sub 0 over Y equals open parenthesis 1.8 times 10 to the power 8 Pa close parenthesis the fraction 2.0 m over 2.0 times 10 to the power 11 Pa equals 1.8 times 10 to the power −3 m equals 1.8 mm.")[$frac(F_(⊥), A) & = & frac(( 550 #h(0.2em) "kg" ) ( 9.8 #h(0.2em) "m/s"^(2) ), 3.0 #h(0.2em) × #h(0.2em) 10^(−5) #h(0.2em) "m"^(2)) = 1.8 #h(0.2em) × #h(0.2em) 10^(8) "Pa" \ "Δ" L & = & frac(F_(⊥), A) #h(0.2em) frac(L_(0), Y) = ( 1.8 #h(0.2em) × #h(0.2em) 10^(8) "Pa" ) #h(0.2em) frac(2.0 #h(0.2em) "m", 2.0 #h(0.2em) × #h(0.2em) 10^(11) "Pa") = 1.8 #h(0.2em) × #h(0.2em) 10^(−3) "m" = #h(0.2em) 1.8 #h(0.2em) "mm."$] Significance Similarly as in the example with the column, the tensile stress in this example is not uniform along the length of the rod. Unlike in the previous example, however, if the weight of the rod is taken into consideration, the stress in the rod is largest at the top and smallest at the bottom of the rod where the equipment is attached. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ A 2.0-m-long wire stretches 1.0 mm when subjected to a load. What is the tensile strain in the wire? #solutionbox[ #math.equation(block: true, alt: "5.0 times 10 to the power −4")[$5.0 #h(0.2em) × #h(0.2em) 10^(−4)$] ] ] Objects can often experience both compressive stress and tensile stress simultaneously . One example is a long shelf loaded with heavy books that sags between the end supports under the weight of the books. The top surface of the shelf is in compressive stress and the bottom surface of the shelf is in tensile stress. Similarly, long and heavy beams sag under their own weight. In modern building construction, such bending strains can be almost eliminated with the use of #strong[I-beams] . #figure(figph[Figure A is a schematic drawing of forces experienced by the object during bending downward. It experiences tensile stress (stretching) in the upper section and compressive stress (compressing) in the lower section. Figure B shows a photograph of weightlifter during the lifting. The iron bar that he is holding is bent.], alt: "Figure A is a schematic drawing of forces experienced by the object during bending downward. It experiences tensile stress (stretching) in the upper section and compressive stress (compressing) in the lower section. Figure B shows a photograph of weightlifter during the lifting. The iron bar that he is holding is bent.", caption: [(a) An object bending downward experiences tensile stress (stretching) in the upper section and compressive stress (compressing) in the lower section. (b) Elite weightlifters often bend iron bars temporarily during lifting, as in the 2012 Olympics competition.]) #figure(figph[Figure is a photograph of steel I-beams are used in construction.], alt: "Figure is a photograph of steel I-beams are used in construction.", caption: [Steel I-beams are used in construction to reduce bending strains.]) #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ A heavy box rests on a table supported by three columns. View this #link("https://openstax.org/l/21movebox")[demonstration] to move the box to see how the compression (or tension) in the columns is affected when the box changes its position. ] === Bulk Stress, Strain, and Modulus When you dive into water, you feel a force pressing on every part of your body from all directions. What you are experiencing then is bulk stress, or in other words, #strong[pressure]. Bulk stress always tends to decrease the volume enclosed by the surface of a submerged object. The forces of this “squeezing” are always perpendicular to the submerged surface . The effect of these forces is to decrease the volume of the submerged object by an amount #math.equation(block: false, alt: "Δ V")[$"Δ" V$] compared with the volume #math.equation(block: false, alt: "V sub 0")[$V_(0)$] of the object in the absence of bulk stress. This kind of deformation is called bulk strain and is described by a change in volume relative to the original volume: #math.equation(block: true, alt: "bulk strain equals the fraction Δ V over V sub 0 .")[$"bulk strain" = frac("Δ" V, V_(0)) .$] #figure(figph[Figure is a schematic drawing of forces experienced by an object under the bulk stress. Equal forces perpendicular to the surface act from all directions and reduce the volume by the amount delta V compared to the original volume, V0.], alt: "Figure is a schematic drawing of forces experienced by an object under the bulk stress. Equal forces perpendicular to the surface act from all directions and reduce the volume by the amount delta V compared to the original volume, V0.", caption: [An object under increasing bulk stress always undergoes a decrease in its volume. Equal forces perpendicular to the surface act from all directions. The effect of these forces is to decrease the volume by the amount #math.equation(block: false, alt: "Δ V")[$"Δ" V$] compared to the original volume, #math.equation(block: false, alt: "V sub 0 .")[$V_(0) .$]]) The bulk strain results from the bulk stress, which is a force #math.equation(block: false, alt: "F sub ⊥")[$F_(⊥)$] normal to a surface that presses on the unit surface area #emph[A] of a submerged object. This kind of physical quantity, or pressure #emph[p], is defined as #math.equation(block: true, alt: "pressure equals p ≡ the fraction F sub ⊥ over A .")[$"pressure" = p ≡ frac(F_(⊥), A) .$] We will study pressure in fluids in greater detail in Fluid Mechanics. An important characteristic of pressure is that it is a scalar quantity and does not have any particular direction; that is, pressure acts equally in all possible directions. When you submerge your hand in water, you sense the same amount of pressure acting on the top surface of your hand as on the bottom surface, or on the side surface, or on the surface of the skin between your fingers. What you are perceiving in this case is an increase in pressure #math.equation(block: false, alt: "Δ p")[$"Δ" p$] over what you are used to feeling when your hand is not submerged in water. What you feel when your hand is not submerged in the water is the #strong[normal pressure] #math.equation(block: false, alt: "p sub 0")[$p_(0)$] of one atmosphere, which serves as a reference point. The bulk stress is this increase in pressure, or #math.equation(block: false, alt: "Δ p ,")[$"Δ" p ,$] over the normal level, #math.equation(block: false, alt: "p sub 0 .")[$p_(0) .$] When the bulk stress increases, the bulk strain increases in response, in accordance with . The proportionality constant in this relation is called the bulk modulus, #emph[B], or #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "B equals the fraction bulk stress over bulk strain equals minus the fraction Δ p over Δ V / V sub 0 equals − Δ p the fraction V sub 0 over Δ V .")[$B = frac("bulk stress", "bulk strain") = − frac("Δ" p, "Δ" V #h(0.1em) "/" #h(0.1em) V_(0)) = "−" "Δ" p #h(0.2em) frac(V_(0), "Δ" V) .$] ] The minus sign that appears in is for consistency, to ensure that #emph[B] is a positive quantity. Note that the minus sign #math.equation(block: false, alt: "open parenthesis minus close parenthesis")[$( − )$] is necessary because an increase #math.equation(block: false, alt: "Δ p")[$"Δ" p$] in pressure (a positive quantity) always causes a decrease #math.equation(block: false, alt: "Δ V")[$"Δ" V$] in volume, and decrease in volume is a negative quantity. The reciprocal of the bulk modulus is called #strong[compressibility] #math.equation(block: false, alt: "k ,")[$k ,$] or #math.equation(block: true, alt: "k equals the fraction 1 over B equals minus the fraction Δ V / V sub 0 over Δ p .")[$k = frac(1, B) = − frac("Δ" V #h(0.1em) "/" #h(0.1em) V_(0), "Δ" p) .$] The term ‘compressibility’ is used in relation to fluids (gases and liquids). Compressibility describes the change in the volume of a fluid per unit increase in pressure. Fluids characterized by a large compressibility are relatively easy to compress. For example, the compressibility of water is #math.equation(block: false, alt: "4.64 times 10 to the power −5 /atm")[$4.64 #h(0.2em) × #h(0.2em) 10^(−5) "/atm"$] and the compressibility of acetone is #math.equation(block: false, alt: "1.45 times 10 to the power −4 /atm .")[$1.45 #h(0.2em) × #h(0.2em) 10^(−4) "/atm" .$] This means that under a 1.0-atm increase in pressure, the relative decrease in volume is approximately three times as large for acetone as it is for water. #examplebox("Example 3")[Hydraulic Press][ In a hydraulic press , a 250-liter volume of oil is subjected to a 2300-psi pressure increase. If the compressibility of oil is #math.equation(block: false, alt: "2.0 times 10 to the power −5 / atm ,")[$2.0 #h(0.2em) × #h(0.2em) 10^(−5) #h(0.1em) "/" #h(0.1em) #h(0.1em) "atm" ,$] find the bulk strain and the absolute decrease in the volume of oil when the press is operating. #figure(figph[Figure is a schematic drawing of a hydraulic press. A small piston is displaced downward and causes the large piston holding object to move upward.], alt: "Figure is a schematic drawing of a hydraulic press. A small piston is displaced downward and causes the large piston holding object to move upward.", caption: [In a hydraulic press, when a small piston is displaced downward, the pressure in the oil is transmitted throughout the oil to the large piston, causing the large piston to move upward. A small force applied to a small piston causes a large pressing force, which the large piston exerts on an object that is either lifted or squeezed. The device acts as a mechanical lever.]) Strategy We must invert to find the bulk strain. First, we convert the pressure increase from psi to atm, #math.equation(block: false, alt: "Δ p equals 2300 psi equals 2300 / 14.7 atm approximately equals 160 atm ,")[$"Δ" p = 2300 #h(0.2em) "psi" = 2300 #h(0.1em) "/" #h(0.1em) 14.7 #h(0.2em) "atm" ≈ #h(0.2em) 160 #h(0.2em) "atm" ,$] and identify #math.equation(block: false, alt: "V sub 0 equals 250 L .")[$V_(0) = #h(0.2em) 250 #h(0.2em) "L" .$] Solution Substituting values into the equation, we have #math.equation(block: true, alt: "bulk strain equals the fraction Δ V over V sub 0 equals the fraction Δ p over B equals k Δ p equals open parenthesis 2.0 times 10 to the power −5 /atm close parenthesis open parenthesis 160 atm close parenthesis equals 0.0032; answer: Δ V equals 0.0032 V sub 0 equals 0.0032 open parenthesis 250 L close parenthesis equals 0.78 L.")[$"bulk strain" = frac("Δ" V, V_(0)) = frac("Δ" p, B) = k "Δ" p = ( 2.0 #h(0.2em) × #h(0.2em) 10^(−5) "/atm" ) ( 160 #h(0.2em) "atm" ) = 0.0032 \ "answer:" #h(0.5em) "Δ" V = 0.0032 #h(0.2em) V_(0) = 0.0032 ( 250 #h(0.2em) "L" ) = 0.78 #h(0.2em) "L."$] Significance Notice that since the compressibility of water is 2.32 times larger than that of oil, if the working substance in the hydraulic press of this problem were changed to water, the bulk strain as well as the volume change would be 2.32 times larger. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ If the normal force acting on each face of a cubical #math.equation(block: false, alt: "1 .0-m cubed")[$1 ".0-m"^(3)$] piece of steel is changed by #math.equation(block: false, alt: "1.0 times 10 to the power 7 N ,")[$1.0 #h(0.2em) × #h(0.2em) 10^(7) "N" ,$] find the resulting change in the volume of the piece of steel. #solutionbox[ 63 mL ] ] === Shear Stress, Strain, and Modulus The concepts of shear stress and strain concern only solid objects or materials. Buildings and tectonic plates are examples of objects that may be subjected to shear stresses. In general, these concepts do not apply to fluids. Shear deformation occurs when two antiparallel forces of equal magnitude are applied tangentially to opposite surfaces of a solid object, causing no deformation in the transverse direction to the line of force, as in the typical example of shear stress illustrated in . Shear deformation is characterized by a gradual shift #math.equation(block: false, alt: "Δ x")[$"Δ" x$] of layers in the direction tangent to the acting forces. This gradation in #math.equation(block: false, alt: "Δ x")[$"Δ" x$] occurs in the transverse direction along some distance #math.equation(block: false, alt: "L sub 0 .")[$L_(0) .$] Shear strain is defined by the ratio of the largest displacement #math.equation(block: false, alt: "Δ x")[$"Δ" x$] to the transverse distance #math.equation(block: false, alt: "L sub 0")[$L_(0)$] #math.equation(block: true, alt: "shear strain equals the fraction Δ x over L sub 0 .")[$"shear strain" = frac("Δ" x, L_(0)) .$] Shear strain is caused by shear stress. Shear stress is due to forces that act #emph[parallel] to the surface. We use the symbol #math.equation(block: false, alt: "F sub ∥")[$F_(∥)$] for such forces. The magnitude #math.equation(block: false, alt: "F sub ∥")[$F_(∥)$] per surface area #emph[A] where shearing force is applied is the measure of shear stress #math.equation(block: true, alt: "shear stress equals the fraction F sub ∥ over A .")[$"shear stress" = frac(F_(∥), A) .$] The shear modulus is the proportionality constant in and is defined by the ratio of stress to strain. Shear modulus is commonly denoted by #emph[S]: #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ #math.equation(block: true, alt: "S equals the fraction shear stress over shear strain equals the fraction F sub ∥ / A over Δ x / L sub 0 equals the fraction F sub ∥ over A the fraction L sub 0 over Δ x .")[$S = frac("shear stress", "shear strain") = frac(F_(∥) "/" #h(0.1em) A, "Δ" x #h(0.1em) "/" #h(0.1em) L_(0)) = frac(F_(∥), A) #h(0.2em) frac(L_(0), "Δ" x) .$] ] #figure(figph[Figure is a schematic drawing of an object under shear stress: Two antiparallel forces of equal magnitude are applied tangentially to opposite parallel surfaces of the object. As the result, the object is transformed from the rectangle to the parallelogram, shape. While the height of the object remains the same, top corners move to the right by the Delta X.], alt: "Figure is a schematic drawing of an object under shear stress: Two antiparallel forces of equal magnitude are applied tangentially to opposite parallel surfaces of the object. As the result, the object is transformed from the rectangle to the parallelogram, shape. While the height of the object remains the same, top corners move to the right by the Delta X.", caption: [An object under shear stress: Two antiparallel forces of equal magnitude are applied tangentially to opposite parallel surfaces of the object. The dashed-line contour depicts the resulting deformation. There is no change in the direction transverse to the acting forces and the transverse length #math.equation(block: false, alt: "L sub 0")[$L_(0)$] is unaffected. Shear deformation is characterized by a gradual shift #math.equation(block: false, alt: "Δ x")[$"Δ" x$] of layers in the direction tangent to the forces.]) #examplebox("Example 4")[An Old Bookshelf][ A cleaning person tries to move a heavy, old bookcase on a carpeted floor by pushing tangentially on the surface of the very top shelf. However, the only noticeable effect of this effort is similar to that seen in , and it disappears when the person stops pushing. The bookcase is 180.0 cm tall and 90.0 cm wide with four 30.0-cm-deep shelves, all partially loaded with books. The total weight of the bookcase and books is 600.0 N. If the person gives the top shelf a 50.0-N push that displaces the top shelf horizontally by 15.0 cm relative to the motionless bottom shelf, find the shear modulus of the bookcase. Strategy The only pieces of relevant information are the physical dimensions of the bookcase, the value of the tangential force, and the displacement this force causes. We identify #math.equation(block: false, alt: "F sub ∥ equals 50.0 N , Δ x equals 15.0 cm ,")[$F_(∥) = 50.0 #h(0.2em) "N" , #h(0.2em) "Δ" x = 15.0 #h(0.2em) "cm" ,$] #math.equation(block: false, alt: "L sub 0 equals 180.0 cm ,")[$L_(0) = 180.0 #h(0.2em) "cm" ,$] and #math.equation(block: false, alt: "A equals (30.0 cm) (90.0 cm) equals 2700.0 cm squared ,")[$A = "(30.0 cm)" #h(0.1em) "(90.0 cm)" = 2700.0 #h(0.2em) "cm"^(2) ,$] and we use to compute the shear modulus. Solution Substituting numbers into the equations, we obtain for the shear modulus #math.equation(block: true, alt: "S equals the fraction F sub ∥ over A the fraction L sub 0 over Δ x equals the fraction 50.0 N over 2700.0 cm squared the fraction 180.0 cm . over 15.0 cm . equals the fraction 2 over 9 the fraction N over cm squared equals the fraction 2 over 9 times 10 to the power 4 the fraction N over m squared equals the fraction 20 over 9 times 10 cubed Pa equals 2.222 kPa.")[$S = frac(F_(∥), A) #h(0.2em) frac(L_(0), "Δ" x) = frac(50.0 #h(0.2em) "N", 2700.0 #h(0.2em) "cm"^(2)) #h(0.2em) frac(180.0 #h(0.2em) "cm" ".", 15.0 #h(0.2em) "cm" ".") = frac(2, 9) #h(0.2em) frac("N", "cm"^(2)) = frac(2, 9) #h(0.2em) × #h(0.2em) 10^(4) frac("N", "m"^(2)) = frac(20, 9) #h(0.2em) × #h(0.2em) 10^(3) "Pa" = "2.222 kPa."$] We can also find shear stress and strain, respectively: #math.equation(block: true, alt: "the fraction F sub ∥ over A equals the fraction 50.0 N over 2700.0 cm squared equals the fraction 5 over 27 kPa equals 185.2 Pa; the fraction Δ x over L sub 0 equals the fraction 15.0 cm over 180.0 cm equals the fraction 1 over 12 equals 0.083 .")[$frac(F_(∥), A) = frac(50.0 #h(0.2em) "N", 2700.0 #h(0.2em) "cm"^(2)) = frac(5, 27) #h(0.2em) "kPa" = "185.2 Pa" \ frac("Δ" x, L_(0)) = frac(15.0 #h(0.2em) "cm", 180.0 #h(0.2em) "cm") = frac(1, 12) = 0.083 .$] Significance If the person in this example gave the shelf a healthy push, it might happen that the induced shear would collapse it to a pile of rubbish. Much the same shear mechanism is responsible for failures of earth-filled dams and levees; and, in general, for landslides. ] #notebox("Note", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[ Explain why the concepts of Young’s modulus and shear modulus do not apply to fluids. #solutionbox[ Fluids have different mechanical properties than those of solids; fluids flow. ] ] === Summary - External forces on an object (or medium) cause its deformation, which is a change in its size and shape. The strength of the forces that cause deformation is expressed by stress, which in SI units is measured in the unit of pressure (pascal). The extent of deformation under stress is expressed by strain, which is dimensionless. - For a small stress, the relation between stress and strain is linear. The elastic modulus is the proportionality constant in this linear relation. - Tensile (or compressive) strain is the response of an object or medium to tensile (or compressive) stress. Here, the elastic modulus is called Young’s modulus. Tensile (or compressive) stress causes elongation (or shortening) of the object or medium and is due to an external forces acting along only one direction perpendicular to the cross-section. - Bulk strain is the response of an object or medium to bulk stress. Here, the elastic modulus is called the bulk modulus. Bulk stress causes a change in the volume of the object or medium and is caused by forces acting on the body from all directions, perpendicular to its surface. Compressibility of an object or medium is the reciprocal of its bulk modulus. - Shear strain is the deformation of an object or medium under shear stress. The shear modulus is the elastic modulus in this case. Shear stress is caused by forces acting along the object’s two parallel surfaces. === Conceptual Questions #emph[Note:] Unless stated otherwise, the weights of the wires, rods, and other elements are assumed to be negligible. Elastic moduli of selected materials are given in . Why can a squirrel jump from a tree branch to the ground and run away undamaged, while a human could break a bone in such a fall? #solutionbox[ In contact with the ground, stress in squirrel’s limbs is smaller than stress in human’s limbs. ] When a glass bottle full of vinegar warms up, both the vinegar and the glass expand, but the vinegar expands significantly more with temperature than does the glass. The bottle will break if it is filled up to its very tight cap. Explain why and how a pocket of air above the vinegar prevents the bottle from breaking. A thin wire strung between two nails in the wall is used to support a large picture. Is the wire likely to snap if it is strung tightly or if it is strung so that it sags considerably? #solutionbox[ tightly ] Review the relationship between stress and strain. Can you find any similarities between the two quantities? What type of stress are you applying when you press on the ends of a wooden rod? When you pull on its ends? #solutionbox[ compressive; tensile ] Can compressive stress be applied to a rubber band? Can Young’s modulus have a negative value? What about the bulk modulus? #solutionbox[ no ] If a hypothetical material has a negative bulk modulus, what happens when you squeeze a piece of it? Discuss how you might measure the bulk modulus of a liquid. === Problems The “lead” in pencils is a graphite composition with a Young’s modulus of approximately #math.equation(block: false, alt: "1.0 times 10 to the power 9 N / m squared .")[$1.0 #h(0.2em) × #h(0.2em) 10^(9) "N" #h(0.1em) "/" #h(0.1em) "m"^(2) .$] Calculate the change in length of the lead in an automatic pencil if you tap it straight into the pencil with a force of 4.0 N. The lead is 0.50 mm in diameter and 60 mm long. #solutionbox[ 1.2 mm ] TV broadcast antennas are the tallest artificial structures on Earth. In 1987, a 72.0-kg physicist placed himself and 400 kg of equipment at the top of a 610-m-high antenna to perform gravity experiments. By how much was the antenna compressed, if we consider it to be equivalent to a steel cylinder 0.150 m in radius? By how much does a 65.0-kg mountain climber stretch her 0.800-cm diameter nylon rope when she hangs 35.0 m below a rock outcropping? (For nylon, #math.equation(block: false, alt: "Y equals 1.35 times 10 to the power 9 Pa .)")[$Y = 1.35 #h(0.2em) × #h(0.2em) 10^(9) "Pa" ".)"$] #solutionbox[ 32.9 cm ] When water freezes, its volume increases by 9.05%. What force per unit area is water capable of exerting on a container when it freezes? A farmer making grape juice fills a glass bottle to the brim and caps it tightly. The juice expands more than the glass when it warms up, in such a way that the volume increases by 0.20%. Calculate the force exerted by the juice per square centimeter if its bulk modulus is #math.equation(block: false, alt: "1.8 times 1 0 to the power 9 N / m squared ,")[$1.8 #h(0.2em) × #h(0.2em) 1 0^(9) "N" #h(0.1em) "/" #h(0.1em) "m"^(2) ,$] assuming the bottle does not break. #solutionbox[ #math.equation(block: true, alt: "3.6 times 10 squared N/cm squared")[$3.6 #h(0.2em) × #h(0.2em) 10^(2) #h(0.2em) "N/cm"^(2)$] ] A disk between vertebrae in the spine is subjected to a shearing force of 600.0 N. Find its shear deformation, using the shear modulus of #math.equation(block: false, alt: "1.0 times 1 0 to the power 9 N/m squared .")[$1.0 #h(0.2em) × #h(0.2em) 1 0^(9) #h(0.2em) "N/m"^(2) .$] The disk is equivalent to a solid cylinder 0.700 cm high and 4.00 cm in diameter. A vertebra is subjected to a shearing force of 500.0 N. Find the shear deformation, taking the vertebra to be a cylinder 3.00 cm high and 4.00 cm in diameter. #solutionbox[ #math.equation(block: true, alt: "11.9 μm")[$11.9 #h(0.2em) "μm"$] ] Calculate the force a piano tuner applies to stretch a steel piano wire by 8.00 mm, if the wire is originally 1.35 m long and its diameter is 0.850 mm. A 20.0-m-tall hollow aluminum flagpole is equivalent in strength to a solid cylinder 4.00 cm in diameter. A strong wind bends the pole as much as a horizontal 900.0-N force on the top would do. How far to the side does the top of the pole flex? #solutionbox[ 0.57 mm ] A copper wire of diameter 1.0 cm stretches 1.0% when it is used to lift a load upward with an acceleration of #math.equation(block: false, alt: "2.0 m/s squared .")[$2.0 #h(0.2em) "m/s"^(2) .$] What is the weight of the load? As an oil well is drilled, each new section of drill pipe supports its own weight and the weight of the pipe and the drill bit beneath it. Calculate the stretch in a new 6.00-m-long steel pipe that supports a 100-kg drill bit and a 3.00-km length of pipe with a linear mass density of 20.0 kg/m. Treat the pipe as a solid cylinder with a 5.00-cm diameter. #solutionbox[ 8.59 mm ] A large uniform cylindrical steel rod of density #math.equation(block: false, alt: "ρ equals 7.8 g/cm cubed")[$ρ = 7.8 #h(0.2em) "g/cm"^(3)$] is 2.0 m long and has a diameter of 5.0 cm. The rod is fastened to a concrete floor with its long axis vertical. What is the normal stress in the rod at the cross-section located at (a) 1.0 m from its lower end? (b) 1.5 m from the lower end? A 90-kg mountain climber hangs from a nylon rope and stretches it by 25.0 cm. If the rope was originally 30.0 m long and its diameter is 1.0 cm, what is Young’s modulus for the nylon? #solutionbox[ #math.equation(block: true, alt: "1.35 times 10 to the power 9 Pa")[$1.35 #h(0.2em) × #h(0.2em) 10^(9) "Pa"$] ] A suspender rod of a suspension bridge is 25.0 m long. If the rod is made of steel, what must its diameter be so that it does not stretch more than 1.0 cm when a #math.equation(block: false, alt: "2.5 times 10 to the power 4 -kg")[$2.5 #h(0.2em) × #h(0.2em) 10^(4) "-kg"$] truck passes by it? Assume that the rod supports all of the weight of the truck. A copper wire is 1.0 m long and its diameter is 1.0 mm. If the wire hangs vertically, how much weight must be added to its free end in order to stretch it 3.0 mm? #solutionbox[ 259.0 N ] A 100-N weight is attached to a free end of a metallic wire that hangs from the ceiling. When a second 100-N weight is added to the wire, it stretches 3.0 mm. The diameter and the length of the wire are 1.0 mm and 2.0 m, respectively. What is Young’s modulus of the metal used to manufacture the wire? The bulk modulus of a material is #math.equation(block: false, alt: "1.0 times 10 to the power 11 N/m squared .")[$1.0 #h(0.2em) × #h(0.2em) 10^(11) #h(0.2em) "N/m"^(2) .$] What fractional change in volume does a piece of this material undergo when it is subjected to a bulk stress increase of #math.equation(block: false, alt: "10 to the power 7 N/m squared ?")[$10^(7) #h(0.2em) "N/m"^(2) "?"$] Assume that the force is applied uniformly over the surface. #solutionbox[ 0.01% ] Normal forces of magnitude #math.equation(block: false, alt: "1.0 times 10 to the power 6 N")[$1.0 #h(0.2em) × #h(0.2em) 10^(6) "N"$] are applied uniformly to a spherical surface enclosing a volume of a liquid. This causes the radius of the surface to decrease from 50.000 cm to 49.995 cm. What is the bulk modulus of the liquid? During a walk on a rope, a tightrope walker creates a tension of #math.equation(block: false, alt: "3.94 times 1 0 cubed N")[$3.94 #h(0.2em) × #h(0.2em) 1 0^(3) N$] in a wire that is stretched between two supporting poles that are 15.0 m apart. The wire has a diameter of 0.50 cm when it is not stretched. When the walker is on the wire in the middle between the poles the wire makes an angle of #math.equation(block: false, alt: "5.0 °")[$5.0 "°"$] below the horizontal. How much does this tension stretch the steel wire when the walker is this position? #solutionbox[ 1.44 cm ] When using a pencil eraser, you exert a vertical force of 6.00 N at a distance of 2.00 cm from the hardwood-eraser joint. The pencil is 6.00 mm in diameter and is held at an angle of #math.equation(block: false, alt: "20.0 °")[$20.0 "°"$] to the horizontal. (a) By how much does the wood flex perpendicular to its length? (b) How much is it compressed lengthwise? Normal forces are applied uniformly over the surface of a spherical volume of water whose radius is 20.0 cm. If the pressure on the surface is increased by 200 MPa, by how much does the radius of the sphere decrease? #solutionbox[ 0.63 cm ]