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Physics 172H Lecture 6 – Ball-Spring Model of Solids, Friction Read 4.1-4.8 d radial force (N) Model of solid: chemical bonds G F ≈ linear 0 If atoms don’t move too far away from equilibrium, force looks like a spring force! A ball-spring model of a solid Ball-spring model of a solid To model need to know: - spring length s - spring stiffness - mass of an atom Initial conditions for circular motion Length of a bond: “diameter” of copper atom density ρ = 8.94 g/cm3: molecular weight = 63.55 g/mole NA molecules 1. Number of atoms in one cm3 8.94 g/cm3 atoms atoms N= ⋅ 6.022 × 1023 = 8.47 × 1022 63.55 g/mole mole cm3 2. Volume per one atom: 3 1 −23 cm VCu = = 1.18 ×10 22 3 8.47 ×10 atoms/cm atom 3. Interatomic spacing: dCu = 3 1.18 × 10−23 cm 3 = 2.27 × 10−8 cm=2.27 × 10−10 m=2.27 Ball-Spring Model of a Wire How is the stiffness of the wire related to the stiffness of one of the short springs (bonds)? Two Springs in Series Spring constant k Mass M Each spring must supply an upward force equal to Mg, thus, each stretches by s giving a total stretch of 2s, or an effective spring constant of k/2. Two Springs in Parallel Mass M Each spring provides an upward force of Mg/2, so each stretches s/2, giving an effective spring constant of 2k. Stiffness of a Copper Wire 2-meter long Cu wire 8.8 x 109 bonds in series Each side = 1 mm 1.94 x 1013 chains in parallel The stiffness of the wire is much greater than the effective spring stiffness between atoms due to the large number of chains in parallel compared to the number of bonds in series. Estimating interatomic “spring” stiffness ΔL strain = L stress = FT A tension stress = Y ⋅ strain FT ΔL =Y A L Y - Young’s modulus depends only on material Compare: Larger Y means stiffer wire G Fspring = k s s G Fspring s A = ks L L GA Fspring L s = ks A A L ks = A Y L Effective interatomic spring stiffness ks = A Y L d2 ks = Y d Interatomic spring stiffness k s = Yd Limits of applicability of Young’s modulus Plastic deformation stress = Y ⋅ strain FT ΔL =Y A L When the stretch of the material exceeds the “elastic limit”, the linear relation between stress and strain fails, and the material deforms plastically. This is an irreversible process. Elastic region Aluminum alloy Brick on a table: compression G FN G Mg If the weight of the brick exceeds a limit, the table will either break (if it’s brittle) or bend, (if it’s ductile). Friction Exert a force so that the brick moves to the right at a constant speed. What is the net force on the brick? Friction Doesn’t Always Oppose Motion—Just RELATIVE motion of the “frictioning” surfaces! QUIZ Box dropped onto moving conveyor belt. What happens? What is the state of motion of the box when the belt force has a horizontal component? A Decelerating B Accelerating C Steady motion If the green v vector is the belt’s velocity, what can you say about the velocity of the box at the instant of the diagram? How is it that a sprinter can accelerate? Friction Doesn’t Always Oppose Motion—Just RELATIVE motion of the “frictioning” surfaces! Box dropped onto moving conveyor belt. What happens? There are no other objects or agents shown. We are shown ALL the forces involved. BUT Fbelt includes two components, FN and Fk , each acting on the box. The force Fk is unbalanced, and so at the moment diagrammed, the box is necessarily accelerating to the right. It will eventually match velocity with the belt, at which point Fk will VANISH You can further argue that the existence of the Fk component means that there MUST be relative motion between the belt and the box at the instant shown. So the box is moving slower than the belt at this instant. The box’s “inertia” is NOT sufficient to keep it moving slower than the belt (or standing still) – that would take an extra agent pulling to the left. Sliding Friction • When one object slides on another, the component of force exerted by one object on the other has a component parallel (or antiparallel) to the motion: • NOTE – the friction always acts against the relative motion. It acts on both the object and the surface. • ffriction ~ μkFN μk is the coefficient of kinetic friction FN is the “normal force” – the perpendicular component of the force that is squeezing the two objects into each other Static Friction • If the applied force Fapplied < μkFN the object will SLOW DOWN (relative to the surface it’s frictioning with.) What is the kinetic friction at this point? • The block will eventually come to rest, and THEN, any (sideways) force applied < μsFN will not cause the block to move (relative to the surface it’s touching).. • Generally the static friction is a bit greater than the kinetic friction: μk < μs • Once the static friction breaks, the block will speed up (if the applied force does not change) since the applied force is then somewhat greater than the retarding force of the kinetic friction.