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Physics 231 Lecture 23 • • Main points of today’s lecture: Gravitation potential energy PE grav = − • GM1M 2 r12 Tensile stress and strain ΔF ΔL =Y A L0 • Bulk stress and strain: ΔF ΔV = ΔP = B A V • Pressure in fluids: P= • F ; Pbot = Ptop + ρgh A Buoyancy FB = Wdisplaced = ρ f Vdisplacedg Gravitational Potential Energy • • ΔPE = mgh is valid only near the earth’s surface F objects For bj t high hi h above b th the earth’s th’ surface, f an alternate expression is needed (You would need calculus to derive it.) M Em PE2 = −G r • – Zero potential energy is defined to be the value infinitelyy far from the earth If the total mechanical energy of an object exceeds zero, the object can escape the gravitation field of the earth. The escape velocity is when the Emech=0. At the Earth's surface: GM Earth m 1 E mech = mv 2 − 2 R Earth If E mech ≥ 0 corresponding to v ≥ vescape , the object can escape the Earth's gravitational field 0= GM Earth m 1 2 mvescape − vescape = 2 R Earth 2GM Earth R Earth you will have loncapa problems on this. 9 Conceptual question • A rock, k iinitially iti ll att restt with ith respectt to t Earth E th and d located l t d an infinite distance away is released and accelerates toward Earth. An observation tower is built 3 Earth-radii high to observe the rock as it plummets to Earth. Earth Neglecting friction, friction the rock’s rock s speed when it hits the ground is KE f + PE f = KE 0 + PE 0 = PE 0 – a) twice GM E m PE = − PE 0 = 0 – b) three th times ti r – c) four times also: KE 0 = 0 KE f + PE f = 0 – d) six times GM E m KE f = − PE f = – e) eight times its speed at the top of the tower. KE groundd 4= KE ground KEtower 1 2 mvground v 2 ground =2 = 2 1 2 vtower mvtower 2 KE tower rf GM E m 1 rgroundd r ground rtower = 4rE = 4 = = GM E m 1 rground rE rtower rtower Stress and strain in solids • All materials deform when under stress, stress ie ie. when a force is applied to them. The atoms in a solid are held in place by forces that act like springs. p g • If you pull on a bar of cross-sectional area A with a force F, the bar will stretch a distance ΔL.. In this t s case, the t e force o ce iss perpendicular pe pe d cu a to the surface, and the force F and distance ΔL are related by Young’s modulus Y: ΔL F ΔL F =Y ; = Stress; = Strain A L0 A L0 • The amount of force you need to apply to achieve the same strain increases with the cross sectional area of the bar. The force will be linearly proportional to the strain up to the elastic limit. Beyond this point, the bar will be permanently deformed. Stress units Pa=N/m2. It is the force divided by the area, and has the units of pressure. Example • Bone h B has a Y Young’s ’ modulus d l off about b t 18 x 109 Pa. P Under U d compression, it can withstand a stress of about 160 x 106 Pa before breaking. Assume that a femur (thigh bone) is 0.50 m long and calculate the length by which this bone can be compressed before breaking. Y 18 x 109 Pa – a) 4.4 mm F/A 160 x 106 Pa – b) 6.4 6 4 mm L0 0.50 m – c) 1.6 cm ΔL ? – d) 2.0 cm ΔL F = 160x106 N / m 2 = Y L0 A 0.5m F L0 = 160x10 6 2 160 10 N / m = 44.4mm 4 ΔL = 9 2 18x10 N / m A Y If this force is uniformly applied, then F/A F/A=P, P, where P is the pressure or force per unit area applied to the end of the femur. Units for pressure are N/m2 = Pa. (Pa = Pascals) Bulk modulus • If you drop d a metal t l block bl k in i the th ocean, it will ill sink i k to t bottom b tt where h the th pressure is much greater than that of the atmosphere. This will compress the block; the amount of compression is governed by the bulk modulus B. B ΔV ΔV ΔP = B ; ΔP = pressure change ; = Volume strain V V Reading Quiz 1. The SI unit of pressure is A. B. C. D D. N kg/m2 Pa kg/m3 Slide 13-6 Density Nuclear matter 2.66x1017 Slide 13-12 Pressure in fluids, Pascal’s principle • Consider a vessel filled with liquid and a smaller rectangular volume l V off the th same liquid li id contained t i d within ithi th the vessel. l If the liquid is at rest, there can be no shear force on V, because it would cause the liquid to move, but there is a pressure force F=P⋅A F=P A perpendicular to all 6 surfaces. surfaces y x In equilibrium: F = 0 z x : Fleft − Fright = 0; Fleft = Pleft A x ; Fright = Pright A x Pleft = Pright z: Ffront − Fback = 0; Ffront = Pfront A z ; Fback = Pback A z Pfront = Pback W yy: Fbot − Ftop − W = 0;; Fbot = Pbot A y ; Ftopp = Ptopp A y Pbot = Ptopp + Ay • • More detailed considerations show that pressure always exerts a force perpendicular to a surface and that pressure is the same everywhere at the same height i.e. height, i e Pfront =Pback, and Pleft =Pright and Pfront= Pright, etc. etc Defining density ρ of the liquid by ρ=M/V , and using V=Ayh we can simplify the dependence of P on height as follows: W = Mg = ρVg = ρA hg Pbot = Ptop + ρA y hg Ay y Pbot = Ptop + ρgh Pressure and Depth equation Pbot = Ptop + ρgh • • • • If the top is open, Pbot is normal atmospheric pressure – 1.013 x 105 Pa = 14.7 lb/in2 The p pressure does not depend p upon the shape of the container Gauge pressure is the difference between the pressure in the fluid and atmospheric pressure. If you connected a gauge which measures the pressure relative to atmospheric pressure to the fluid at depth h, then for this specific case: Pguage = Pbottom,h _ below _ surface − Patmosphere = ρgh Pascal’s principal and hydraulic machines • • • Because F=PA, B F PA one can make k hhydraulic d li machines that can generate large forces. Units: One atmosphere (1 atm) = – 76.0 cm of mercury – 1.013 x 105 Pa = 1.013 x 105 N/m2 – 14.7 lb/in Quiz: A simple hydraulic lift is employed to lift a truck that weighs 30000N. If A1=1 cm2 and A2= 1000 cm2, what force F1 is needed to lift the truck? (Hint: Are the pressures equal under each piston?) – a) 3N P1 = P2 ; this is pascal pascal'ss principal – b) 300N – c) 30N F2 F1 F2 2 30000N F1 = A1 = 1cm = = 30N 2 – d) .0.33N 0 33N A2 A A 1000cm 1 2