Topic 2.2 Extended A – 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: T f T f T f T f T f Force Time tension friction static friction kinetic friction statickinetic

Topic 2.2 Extended A – Friction  Observe the graph of the friction force.  During the static friction phase, the static friction force f s matches exactly the applied force. Force Time tension friction statickinetic  f s increases until it reaches a maximum value f s,max. f s,max  The friction force then almost instantaneously decreases to a constant value f k, called the kinetic friction force.  Make note of the following general properties of the friction force: 0 ≤ f s ≤ f s,max f k < f s,max f k = a constant fkfk

Topic 2.2 Extended A – 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.

Topic 2.2 Extended A – 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. surface 1 surface 2 surface 1 surface 2 cold welds  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. surface 1 surface 2  Of course, the friction force depends on what materials the two surfaces are made of.  We define the unitless constant called the coefficient of friction μ.  μ depends on the two surface materials.

Topic 2.2 Extended A – Friction  Since there are two types of friction, static and kinetic, every pair of materials will have two coefficients of friction, μ s and μ k.  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: f s ≤ μ s N static friction f k = μ k N kinetic friction

Topic 2.2 Extended A – 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°. θ = 15° x y FBD, coin mgmg fsfs N 15°  What is the coefficient of static friction? N – mg cos 15° = 0 N = mg cos 15° f s – mg sin 15° = 0 f s = mg sin 15° f s = μ s N ∑F y = 0 ∑F x = 0 mg sin 15° = μ s mg cos 15° = tan 15° = Thus the coefficient of static friction between the metal of the coin and the wood of the plank is mgsin15° mgcos15° μs =μs =

Topic 2.2 Extended A – 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). θ = 12° x y FBD, coin mgmg fkfk N 12°  If this new angle is 12°, what is the coefficient of kinetic friction? N – mg cos 12° = 0 N = mg cos 12° f k – mg sin 12° = 0 f k = mg sin 12° f k = μ k N ∑F y = 0 ∑F x = 0 mg sin 12° = μ k mg cos 12°  μ k = tan 12° = Thus the coefficient of kinetic friction between the metal of the coin and the wood of the plank is

Topic 2.2 Extended A – Friction  If the plank of wood is now raised to 16°, what is the coin’s acceleration? x y FBD, coin mgmg fkfk N 16° N – mg cos 16° = 0 N = mg cos 16° f k – mg sin 16° = -ma f k = mg sin 16° - ma f k = μ k N ∑F y = 0 ∑F x = -ma mg sin 16° - ma = μ k mg cos 16° a ma = mg sin 16° - μ k mg cos 16° a = (sin 16° - μ k cos 16°)g a = [ – 0.213(0.961) ](10) a = 0.7 m/s 2

Topic 2.2 Extended A – 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? F 30° mgmg N f a F mgmg N f a FBD, crate x y 30°  Static friction will oppose the applied force until it is overcome.

Topic 2.2 Extended A – Friction F mgmg N f a FBD, crate x y 30°  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 F cos 30° = 60 cos 30° = n.  The maximum static friction force is f s,max = μ s N= 0.75N  The normal force is found from... N + F sin 30° - mg = 0 N + 60 sin 30° = 0 N = 70 f s,max = 0.75(70) f s,max = 52.5 n Our analysis shows that the crate will not even begin to move!

Topic 2.2 Extended A – Friction F mgmg N f a FBD, crate x y 30°  If someone gives the crate a small push (of how much) it will “break” loose.  What will its acceleration be then? F cos 30° = 60 cos 30° = n.  The kinetic friction force is fk = μkNfk = μkN= 0.60N  The normal force is still N = 70.  The horizontal applied force is still  Thus f k = 0.60(70) = 42 n.  The crate will accelerate. F cos 30° - f = ma = (100/10)a a = m/s 2

Topic 2.2 Extended A – Friction T HE D RAG F ORCE AND T ERMINAL S PEED  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. D D  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.

Topic 2.2 Extended A – Friction T HE D RAG F ORCE AND T ERMINAL S PEED  The drag force depends on many things.  D is proportional to the cross-sectional area A. This is how much of the body actually cuts into the fluid.  D is proportional to the fluid density ρ. This is how much mass the fluid has per unit volume. D is proportional to the cross-sectional area A D D D is proportional to the fluid density ρ D D ρ water = 1000 kg/m 3 ρ air = 1.2 kg/m 3 Some fluid densities

Topic 2.2 Extended A – Friction T HE D RAG F ORCE AND T ERMINAL S PEED  The drag force depends on many things.  D is proportional to the drag coefficient C. This is a unitless, experimentally derived quantity that represents how aerodynamic a body is.  The more aerodynamic the body, the smaller the drag coefficient C. D is proportional to the drag coefficient C D D  D is proportional to the square of the relative velocity v, of the body and the fluid. D D D is proportional to the SQUARE of the relative velocity v 0 < C < 1

Topic 2.2 Extended A – Friction T HE D RAG F ORCE AND T ERMINAL S PEED  Putting it all together we have… D = CρAv The drag force  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.

Force and Motion 4-6 Friction

Topic 2.2 Extended A – Friction T HE D RAG F ORCE AND T ERMINAL S PEED  Suppose a minivan has a cross-sectional area of 4 m 2 and an experimentally-established drag coefficient of 0.8. Compare the drag forces on the van at 30 mph and 70 mph. v 30 = 30 mi h 5280 ft mi 1 m 3.28 ft 1 h 3600 s = 13.4 m/s v 70 = 70 mph m/s 30 mph = 31.3 m/s D 30 = CρAv = (0.8)(1.2)(4)( ) 1212 = 345 n D 70 = CρAv = (0.8)(1.2)(4)( ) 1212 = 1881 n D 70 D 30 = 5.45 =  D 70 = 5.45 D 30 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.

Topic 2.2 Extended A – Friction T HE D RAG F ORCE AND T ERMINAL S PEED  Suppose a blue whale suddenly materializes high above the ground.  Obviously, it will begin to fall.  1 metric ton is 1000 kg. y mgmg At first, v = 0.  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 v terminal. "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." Guinness World RecordsGuinness World Records. Falkland Islands Philatelic Bureau. 2 March y mgmg D v reaches a maximum value, called terminal speed. D = mg. v terminal y mgmg D Then, as v increases, so does D. v

Topic 2.2 Extended A – Friction T HE D RAG F ORCE AND T ERMINAL S PEED  At terminal speed v t, D = mg. Thus D = mg CρAv t 2 = mg 1212 v t = 2mg CρA √ terminal speed

Topic 2.2 Extended A – Friction T HE D RAG F ORCE AND T ERMINAL S PEED v t = 2mg CρA √  So how do we estimate the terminal speed of the whale? 27 m  We need the mass: m = 190(1000 kg) = kg.  We need the drag coefficient: C = 0.60 (an estimate).  We need the cross-sectional area A: - Since the whale is 27 m long, a simple estimate makes out the central diameter to be d = 27/5 = 5.4 m. - We will assume that the whale is approximately ellipsoid in shape: ellipse A = πab a b - Thus A = π(5.4/2)(27/2) = 115 m 2. = 2(190000)(10) (0.60)(1.2)(115) √ = 214 m/s