ENG 201 Lecture, Spring 2012

10-0 CHAPTER 10 Force and Force-Related Parameters 10-1 Spring’s forces and Hooke’s law 10-2 Measuring a spring’s constant, k, the setup • Hooke’s law • F=K X 10-3 Measuring a spring’s constant, k. The results 10-4 Friction 10-5 Friction Friction force is not constant F=μ N • μ is the coefficient of static friction (unitless) • N is the normal force 10-6 Example 10.2 A book on a table: weight 20N μ = 0.6 • A horizontal force of 10N is applied, will it move the book? • What is the minimum horizontal force that will move the book? 10-7 Newton’s first law • ∑ F = 0 • Object at rest , remains at rest • Object in motion, remains moving at same speed, same direction → → 10-8 Newton’s second law • ∑ F = m a • Where a is the acceleration • The resultant of the forces applied on an object creates an acceleration proportional to this force → → 10-9 Newton’s third law • Forces of action and reaction have the same magnitude and act on the same line, but in opposite directions 10-10 Newton’s law of gravitation G m1m2 r2 F = ------------ 10-11 • Weight on earth G Mearthmobject R2earth W = ------------------- 10-12 Example 10.3 • Determine the weight of an exploration vehicle, mass = 250Kg • On earth • On moon (g=1.6 m/s2) • On Mars (g=3.7 m/s2) 10-13 • On earth : 2450 N • On moon : 400 N • On Mars : 925 N Example 10.3 - results 10-14 Example 10.4 • Space shuttle orbits Earth at altitudes from 250 Km to 965Km. Determine g • If mass of astronaut = 70Kg, what is his weight in orbit? (Mass of Earth = 5.97 x 1024 Kg) (Radius of Earth = 6378 km) (G=6.673 x 10-11 m3/kg.s2) • Answer: (g=9.07 m/s2 to 7.38 m/s2) 10-15 Example 10.4 - solution • Space shuttle orbits Earth at altitudes from 250 Km to 965Km. Determine g • If mass of astronaut = 70Kg, what is his weight in orbit? (Mass of Earth = 5.97 x 1024 ) (Radius of Earth = 6378 km) • Answer: (g=9.07 m/s2 to 7.38 m/s2) 10-16 Units of force • 1 newton = 1 kg x 1 m/s2 • 1 lb = 1 slug x 1 ft/s2 (pound force) • 1 lb = 4.448 N 10-17 Pressure and stress, forces acting over an area Pa = 1271 Pa Pb = 4289 Pa 10-18 Common units of pressure • 1 Pa = 1 N/m2 • 1 psi (lb/in2) = 6894.7 Pa • 1 lb/ft2 = 144 psi 10-19 Pascal’s Law • P = ρ g h 10-20 Pascal’s Law • P = ρ g h • Pascal’s law is in close relationship with the buoyant force: • FB = ρ V g 10-21 Example 10.5 – water tower • Create a table showing the relationship between height and pressure (in psi) in a water tower. • What should be the height of water to create a pressure of 50 psi at the bottom of the tower? • Given: ρwater = 1.94 slugs/ft3 • g = 32.2 ft/s2 10-22 Example 10.5 • P(lb/ft2) = ρ g h = 1.94 (slug/ft3) x 32.2 (ft/s2) x h (ft) (1lbf=1slug x 1ft/s2) • Convert to lb/in2: • P(lb/in2) = P(lb/ft2) /144 10-23 Example 10.5 10-24 Atmospheric pressure 10-25 Atmospheric pressure • 1 atm = 101.3KPa • 1 atm = 14.69 lb/in2 (psi) • 1 atm = 760 mmHg = 29.92 inHg 10-26 Example 10.6 • Derive 1 atm in : • mmHg (ρHg = 13550Kg/m3) • Inches of Hg (1mm = 0.03937in) • Meters of water • Feet of water (1m = 3.28ft) • Answers : 762mmHg = 30inHg = 10.328mWater = 33.87ftWater 10-27 Absolute pressure and gauge pressure • Note: pressure gauges are the instruments that measure pressure • Pabsolute = Pgauge + Patmospheric • If Pgauge is negative ==> vacuum • If Pabsolute = 0 ==> absolute vacuum (this is theoretical) 10-28 Example 10.7 • Pressure inside a car tire = 35.0 psi • 1. Find absolute pressure if – a) car at sea level – b) car at elevation 1500m (use atmospheric pressure table; Patm,1500=84.55 KPa) – (1psi=6895Pa) • 2. If you measure the pressure in your car tires at a station in antelias (sea level) and you find that you have 35.0 psi. what will this pressure be when you go to Faraya at 1500m height? • Ans: 1.a-342 325 Pa, 1.b-325 875 Pa, 2- 37.4 psi 10-29 Example 10.8 • A vacuum pressure gauge reads 200mmHg vacuum in a container. What are gauge and absolute pressures (in Pa)? (1mmHg = 133.3Pa) • Answers: Pgauge = -26 660Pa; Pabs = 74 665Pa 10-30 • Vapor pressure of a liquid – Liquids with higher vapor pressure require a higher pressure at their surface to stop them from evaporating. Therefore they evaporate faster than liquids with low vapor pressure. • Blood pressure – Systolic/diastolic blood pressures (pressure in arteries when heart contracts/relaxes) – Example: 115/75 mmHg = 15/10 KPa 10-31 Hydraulic systems • If the liquid inside the cylinder is incompressible, pressure is transmitted: • P1= F1/A1 = F2/A2 10-32 Hydraulic systems • Fluid volume displaced: A1 L1 = A2 L2 • Speed of piston displacement: V1 A1 =V2 A2 10-33 Example 10.9 (load =?) 10-34 Manually activated hydraulic system 10-35 Pump-driven hydraulic system 10-36 Stress in material (intensity of force over an area) 10-37 Modulus of elasticity • Young’s modulus, E (N/m2; lb/in2) = computed from the slope of the stress-strain curve when the material is in elastic deformation. • The larger E is, the larger the force required to stretch or shorten the material by a certain amount. 10-38 10-39 Modulus of elasticity 10-40 Hooke’s law in a material • σ = E ε • Where σ is normal stress (N/m2; lb/in2) • E is the modulus of elasticity, or Young’s modulus (N/m2; lb/in2) • ε is the normal strain OR: • F/A= E (x/L) (x is elongation) • E = FL/(Ax) (F in N, L in m, A in m2) 10-41 Elastic strength of a material • The elastic strength of a material is the maximum load a material can carry without permanent deformation (calculated on the elastic point on the graph) 10-42 Elastic point 10-43 Modulus of rigidity (shear modulus) • The value of the modulus of rigidity shows the resistance of a given material to shear or twisting deformation. 10-44 10-45 10-46 Ductile and brittle material • When subject to tensile force: – A ductile material deforms significantly before fracture – A brittle material deforms little before fracture NOTE: If the forces pull the body producing an elongation, it is termed tensile force, if on the other hand the forces push the body reducing its length, it is termed compressive stresses. 10-47 Tensile strength • The tensile strength (TS) = maximum tensile load a material specimen can carry without failure • TS = tensile force max/original C.S. Area (C.S. Area= cross-sectional area) 10-48 • Note: lower yield strength 10-49 10-50 Example 10.10 • Load = 4000N • What type of material carries the load safely? 10-51 Example 10.10-solution σ = 4000N/(0.05m x 0.005m) σ = 16 MPa • Aluminium alloy of yield strength 50 MPa can carry the load safely 10-52 Factor of safety • FS = Pmax/Pallowable • Where Pallowable is the allowable or the installed load and Pmax is the maximum load or the load that can cause failure. 10-53 Example 10.11 - Shear modulus calculation 10-54 Example 10.11 • Shear modulus : G = 32TL/πD4φ • T = 3450 N • L = 20 cm = 0.2m • D = 5 cm = 0.05m • φ = 0.015 rad Answer: G = 75 GPa 10-55 Bulk Modulus of Compressibility dP dP -dV/V dρ/ρ Where Ev is the modulus of compressibility (N/m2) dP is the increase in pressure, V is the original volume, -dV is the decrease in volume, dρ and ρ are the increase in density and the density. Ev = ------- = ------- 10-56 10-57 Example 10.13 • Given: 1m3 water, Ev,water = 2.24 x 109 N/m2. • What is the pressure that would decrease the volume by 1% • dP = Ev,water x 0.01 = 2.24 x 107 N/m2 10-58 Moment, torque M = distance x Force (cw) 10-59 • MA = d1 F (ccw) • MB = d2 F (cw) • MC = 0