Topic 2.2 Extended A – Friction

1. tension T T T T friction f f f f Force f T Time Topic 2.2 ExtendedA – Friction Recall that friction acts opposite to the intended direction of motion, and parallel to the contact surface. Suppose we begin to pull a crate to the right, with gradually increasing force. We plot the applied force, and the friction force, as functions of time: kinetic friction static friction static kinetic

2. fk tension friction Force Time static kinetic Topic 2.2 ExtendedA – Friction fs,max Observe the graph of the friction force. During the static friction phase, the static friction forcefs matches exactly the applied force. fs increases until it reaches a maximum value fs,max. The friction force then almost instantaneously decreases to a constant value fk, called the kinetic frictionforce. Make note of the following general properties of the friction force: fk= a constant fk< fs,max 0≤ fs≤ fs,max

3. Topic 2.2 ExtendedA – Friction So what causes friction? People in the manufacturing sector who work with metals know that the more you smoothen and polish two metal surfaces, the more strongly they stick together if brought in contact. In fact, if suitably polished in a vacuum, they will stick so hard that they cannot be separated. We say that the two pieces of metal have been cold-welded.

4. surface 1 surface 1 surface 1 surface 2 surface 2 surface 2 cold welds Topic 2.2 ExtendedA – Friction At the atomic level, when two surfaces come into contact, small peaks on one surface bind with small peaks on the other surface, in a process similar to cold welding. Applying the initial sideways force, all of the cold welds oppose the motion. Then suddenly, the cold welds break, and new peaks contact each other and cold weld. But if the surfaces remain in relative sliding motion, fewer welds have a chance to form. Of course, the friction force depends on what materials the two surfaces are made of. We define the unitless constant called the coefficient of frictionμ. μdepends on the two surface materials.

5. fs ≤ μsN static friction fk = μkN kinetic friction Topic 2.2 ExtendedA – Friction Since there are two types of friction, static and kinetic, every pair of materials will have two coefficients of friction, μs and μk. In addition to the "roughness" or "smoothness" of the materials, the friction force depends on the normal force N. The harder the two surfaces are squished together (this is what the normal force measures) the more cold welds can form. Here are the relationships between the friction force f, the coefficient of friction μ, and the normal force N:

6. y N x fs mg mgsin15° mgcos15° μs= FBD, coin Topic 2.2 ExtendedA – Friction One might ask how you can find the coefficient of friction between two materials: Here is one way: A piece of wood with a coin on it is raised on one end until the coin just begins to slip. The angle the wood makes with the horizontal is θ = 15°. 15° What is the coefficient of static friction? θ = 15° ∑Fx= 0 ∑Fy= 0 N – mgcos15° = 0 fs – mgsin15° = 0 fs = mgsin15° N = mgcos15° fs = μsN = tan15° = 0.268 mgsin15° = μsmgcos15° Thus the coefficient of static friction between the metal of the coin and the wood of the plank is 0.268.

7. y N x fk mg FBD, coin Topic 2.2 ExtendedA – Friction Now suppose the plank of wood is long enough so that you can lower it to the point that the coin keeps slipping, but no longer accelerates (v = 0). 12° If this new angle is 12°,what is the coefficient of kinetic friction? θ = 12° ∑Fx= 0 ∑Fy= 0 N – mgcos12° = 0 fk – mgsin12° = 0 fk = mgsin12° N = mgcos12° fk = μkN μk= tan12° = 0.213 mgsin12° = μkmgcos12° Thus the coefficient of kinetic friction between the metal of the coin and the wood of the plank is 0.213.

8. a y N x fk mg FBD, coin Topic 2.2 ExtendedA – Friction If the plank of wood is now raised to 16°, what is the coin’s acceleration? ∑Fy= 0 N – mgcos16° = 0 N = mgcos16° ∑Fx= -ma 16° fk – mgsin16° = -ma fk = mgsin16° - ma fk = μkN mgsin16° - ma = μkmgcos16° ma = mgsin16° - μkmgcos16° a = (sin16° - μkcos16°)g a = [ 0.276 – 0.213(0.961)](10) a = 0.7 m/s2

9. y 30° x F F FBD, crate N N 30° a a f f mg mg Topic 2.2 ExtendedA – Friction Since friction is proportional to the normal force, be aware of problems where an applied force increases or diminishes the normal force. A 100-n crate is to be dragged across the floor by an applied force F of 60 n, as shown. The coefficients of static and kinetic friction are 0.75 and 0.60, respectively. What is the acceleration of the crate? Static friction will oppose the applied force until it is overcome.

10. y 30° x F FBD, crate N a f mg Topic 2.2 ExtendedA – Friction Determine if the crate even begins to move. Thus, find the maximum value of the static friction, and compare it to the horizontal applied force: The horizontal applied force is just = 60cos30° Fcos30° = 51.96 n. The maximum static friction force is fs,max=μsN =0.75N The normal force is found from... N + Fsin30°- mg =0 N + 60sin30°- 100=0 N =70 fs,max=0.75(70) fs,max=52.5 n Our analysis shows that the crate will not even begin to move!

11. y 30° x F FBD, crate N a f mg Topic 2.2 ExtendedA – Friction If someone gives the crate a small push (of how much) it will “break” loose. What will its acceleration be then? The horizontal applied force is still = 60cos30° Fcos30° = 51.96 n. The kinetic friction force is fk=μkN =0.60N The normal force is still N =70. fk=0.60(70) = 42 n. Thus The crate will accelerate. Fcos30°- f =ma 51.96- 42=(100/10)a a = 0.996 m/s2

12. D D Topic 2.2 ExtendedA – Friction THE DRAG FORCE AND TERMINAL SPEED A fluid is anything that can flow. We generally think of a fluid as a liquid, like water. But air is also a fluid. And under certain conditions, solids can act like fluids. Whenever there is a relative velocity between a fluid and a body, a drag force D is experienced. That drag force always points in the direction of the relative velocity of the fluid. If a boat is moving through still water, it feels a drag force opposite to its motion. If a boat is moving against a current, it feels a drag force in the direction of the current. Think of the drag force as a fluid friction force.

13. D D Some fluid densities D D Topic 2.2 ExtendedA – Friction THE DRAG FORCE AND TERMINAL SPEED The drag force depends on many things. D is proportional to the cross-sectional areaA. This is how much of the body actually cuts into the fluid. D is proportional to the cross-sectional area A D is proportional to the fluiddensityρ. This is how much mass the fluid has per unit volume. D is proportional to the fluid density ρ ρwater = 1000 kg/m3 ρair = 1.2 kg/m3

14. D D D D Topic 2.2 ExtendedA – Friction THE DRAG FORCE AND TERMINAL SPEED The drag force depends on many things. D is proportional to the drag coefficientC. This is a unitless, experimentally derived quantity that represents how aerodynamic a body is. D is proportional to the drag coefficient C 0 < C < 1 The more aerodynamic the body, the smaller the drag coefficient C. D is proportional to the square of the relative velocityv, of the body and the fluid. D is proportional to the SQUARE of the relative velocity v

15. 1 2 The drag force D = CρAv2 Topic 2.2 ExtendedA – Friction THE DRAG FORCE AND TERMINAL SPEED Putting it all together we have… Note the factor of 1/2 in the formula. This is found from experiment and theory beyond the scope of this course. If v doubles, note that D quadruples.

16. Force and Motion4-6 Friction

17. 1 2 1 2 1 2 D30 = CρAv302 = (0.8)(1.2)(4)(13.42) = (0.8)(1.2)(4)(31.32) v70 = 70 mph 1 30 mi h 1 m 3.28 ft 5280 ft mi 1 h 3600 s 1 2 v30 = D70 = CρAv702 13.4 m/s 30 mph D70 D30 1881 345 = Topic 2.2 ExtendedA – Friction THE DRAG FORCE AND TERMINAL SPEED Suppose a minivan has a cross-sectional area of 4 m2 and an experimentally-established drag coefficient of 0.8. Compare the drag forces on the van at 30 mph and 70 mph. = 13.4 m/s = 31.3 m/s = 345 n = 1881 n D70 = 5.45 D30 = 5.45 This is why it is more economical to drive at lower speeds. This is also why you can only go so fast on a bicycle.

18. At first, v = 0. "A female Blue Whale weighing 190 metric tonnes (418,877lb) and measuring 27.6m (90ft 5in) in length was caught in the Southern Ocean on 20 March 1947." mg Then, as v increases, so does D. Guinness World Records. Falkland Islands Philatelic Bureau. 2 March 2002. D mg v y y y v reaches a maximum value, called terminal speed. D = mg. D vterminal mg Topic 2.2 ExtendedA – Friction THE DRAG FORCE AND TERMINAL SPEED Suppose a blue whale suddenly materializes high above the ground. 1 metric ton is 1000 kg. Obviously, it will begin to fall. At first v = 0, so the drag is also 0. Because of the freefall acceleration, v increases. Thus D increases. But as D increases, the acceleration decreases… …until D = mg, at which time a is 0 and v stops changing. v has reached its maximum value, called terminal speed vterminal.

19. 1 2 CρAvt2 = mg terminal speed √ 2mg CρA vt = Topic 2.2 ExtendedA – Friction THE DRAG FORCE AND TERMINAL SPEED At terminal speed vt, D = mg. Thus D = mg

20. √ 2mg CρA vt = ellipse A = πab a √ 2(190000)(10) (0.60)(1.2)(115) = b 27 m Topic 2.2 ExtendedA – Friction THE DRAG FORCE AND TERMINAL SPEED So how do we estimate the terminal speed of the whale? We need the mass: m = 190(1000 kg) = 190000 kg. We need the drag coefficient: C = 0.60 (an estimate). We need the cross-sectional area A: - We will assume that the whale is approximately ellipsoid in shape: - Since the whale is 27 m long, a simple estimate makes out the central diameter to be d = 27/5 = 5.4 m. - Thus A = π(5.4/2)(27/2) = 115 m2. = 214 m/s