Physics 172H. Lecture 6 Ball-Spring Model of Solids, Friction. Read

Transcription

1 Physics 172H Lecture 6 Ball-Spring Model of Solids, Friction Read

2 Model of solid: chemical bonds d radial force (N) 0 F linear If atoms don t move too far away from equilibrium, force looks like a spring force!

3 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

4 Initial conditions for circular motion

5 Length of a bond: diameter of copper atom 1. Number of atoms in one cm 3 density ρ = 8.94 g/cm 3 : molecular weight = g/mole N A molecules g/cm 23 atoms 22 atoms N = = g/mole mole cm 2. Volume per one atom: V Cu 1 cm = = atoms/cm atom 3. Interatomic spacing: d Cu = cm 3 = cm= m=2.27

6 Ball-Spring Model of a Wire How is the stiffness of the wire related to the stiffness of one of the short springs (bonds)?

7 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.

8 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.

9 Stiffness of a Copper Wire 2-meter long Cu wire 8.8 x 10 9 bonds in series 1.94 x chains in parallel Each side = 1 mm 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.

10 Estimating interatomic spring stiffness strain stress = = ΔL L FT A tension stress = Y strain FT A = Y ΔL L Y - Young s modulus depends only on material Larger Y means stiffer wire Compare: Fspring = ks s Fspring s A = ks L A L Fspring L s = k s A A L k s = A Y L

11 Effective interatomic spring stiffness k s = A Y L k s = d d 2 Y Interatomic spring stiffness ks = Yd

12 Limits of applicability of Young s modulus stress = Y strain Plastic deformation FT A = Y ΔL L Elastic region 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. Aluminum alloy

13 Brick on a table: compression F N 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).

14 Friction Exert a force so that the brick moves to the right at a constant speed. What is the net force on the brick?

15 Friction Doesn t Always Oppose Motion Just RELATIVE motion of the frictioning surfaces! Box dropped onto moving conveyor belt. What happens? QUIZ What is the state of motion of the box when the belt force has a horizontal component? 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? A Decelerating B Accelerating C Steady motion

16 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 F belt includes two components, F N and F k, each acting on the box. The force F k 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 F k will VANISH You can further argue that the existence of the F k 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.

17 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. f friction ~ μ k F N μ k is the coefficient of kinetic friction F N is the normal force the perpendicular component of the force that is squeezing the two objects into each other

18 Static Friction If the applied force F applied < μ k F N 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 < μ s F N 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: μ k < μ s 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.