1. Friction! Recall, Friction is a force that opposes all motion. Friction is the result of contact between 2 surfaces. The “Normal force” is the force that acts perpendicular to the contact surfaces. The “Applied force” is the push or pull acting on the object. The “Gravitational” force (weight) = mg The “Frictional” force acts opposite the direction of motion.

2. “ Normal” Forces: The Normal force equals the weight of the object (mg) times the cosine of the angle the contact surface makes to the ground. F n = mg(cos  ) The surfaces in contact will determine the amount of friction the system will have. The ratio of Frictional force : Normal force is known as the Coefficient of Friction and is given the variable “  ” (mu, mew)  F f /F n *Similarly to density, there are constants between surfaces and can be found in reference tables.

3. Static vs. Kinetic Friction Static Friction : When a force is applied to an object at rest the object will not move until the applied force exceeds the maximum static frictional force between those surfaces. F Static equals F applied (the force parallel to the surfaces) until the object starts to move. F static =  s (F n ) F static =  s (mg)(cos  ) Kinetic Friction : Also known as sliding friction, Kinetic friction is the frictional force that an object experiences while it is in motion. F Kinetic = F applied when an object is in motion and is moving at a constant velocity (speed and direction). F kinetic =  k (F n ) F kinetic =  k (mg)(cos  ) Static Friction > Kinetic Friction!

4. Problems: 1. A 65 N force is required to place a 6400 N truck in motion from rest. What is the coefficient of friction between the surfaces? Is this static friction or kinetic? F A = 65N F f =  (F n ) F g = 6400N  = F f /F n  = 65 N/6400N  =.01 Since the truck is being put INTO motion from rest, Static. 2. A waxed wooden sled has a mass of 16 Kg. It is sitting on a level mound of snow. The 3 yr-old & 12 yr-old passengers have masses of 18 Kg & 57 Kg respectively. What force must be applied to cause an acceleration in the sled? How much more or less force must be applied to keep the sled in motion over the level surface than to place it in motion?  s =.085  k =.062

5. F g = F N = mg(Cos  ) = 891.8 N F Fs = (.085)(891.8N) = 75.8 N F Fk = (.062)(891.8N) = 55.29 N  F f = 20.51 N Demo

6. Friction Problem… A 65 Kg wood block is pulled across a flat wood floor by a force of 312N. What is the Frictional Force? What is the acceleration of the block? coeff p131 PS7/PS3 p. 146 #30, 33, 34, 35, 37 ans. Sheet Potomac.Forces.2008 5 Frictionprob

7. Fluid Friction Fluid Friction/Air Resistance occurs when objects pass through any fluid (substance that can flow ….liquid or gas!) Free fall: An object will accelerate towards the ground at a rate of 9.8 m/s 2 until the fluid frictional force becomes so great that it balances with the force due to gravity (weight). When forces are balanced, what happens? We call this Terminal Velocity (F g = F f ).

8. Terminal Velocity: Weight (force due to gravity) = Fluid friction F g = F f F f =  Av 2 mg =  Av 2 Ex. A meteor is falling from the planet Krypton. It has a surface area of 25m 2 and a mass of 2000Kg. The density of air (assuming it remains constant during descent) is 1.183 Kg/m 3. What is its maximum velocity? A = 25m 2 F g = F f m = 2000 Kgmg =  Av 2  = 1.183 Kg/m 3 (2000Kg)(9.8 m/s 2 ) = (1.183 Kg/m 3 )(25 m 2 )(v 2 ) 19600 Kgm/s 2 = (29.575Kg/m)(v 2 ) v 2 = 662.72 m 2 /s 2 v t = 25.74 m/s

9. Terminal Velocity Problem #2 A drug dealer is thrown out of an airplane without a parachute. His mass is 75 Kg and his surface area is.4m 2. The density of air is 1.183 Kg/m 3. Assuming he can minimally survive a fall hitting the ground at 9.41 m/s and assuming he can flap his arms 5 times per second to slow his descent by 3 m/s (per set of 5); How fast must he flap in order to survive? Assuming he can flap 48 times/second, will he deal again? A =.4m 2 F g = F f m = 75 Kgmg =  Av 2  = 1.183 Kg/m 3 (75Kg)(9.8 m/s 2 ) = (1.183 Kg/m 3 )(.4 m 2 )(v 2 ) 735 Kgm/s 2 = (.4732 Kg/m)(v 2 ) v 2 = 1553.25 m 2 /s 2 v t = 39.41 m/s v f -v s = 39.41m/s – 9.41m/s =  v = 30.00 m/s (30m/s) x (5 flaps/sec) = 50 flaps per second 3 m/s