1. FRICTIONAL FORCE MU OR MOO?

2. FRICTION OPPOSES THE APPLIED FORCE OPPOSES THE APPLIED FORCE KINETIC FRICTION KINETIC FRICTION –FORCE THAT RETARDS MOTION STATIC FRICTION STATIC FRICTION –FORCE THAT KEEPS THE OBJECT FROM MOVING

3. OTHER TYPES OF FRICTION ROLLING FRICTION – THE RESISTANCE FORCE BETWEEN A SURFACE AND A ROLLING OBJECT. ROLLING FRICTION – THE RESISTANCE FORCE BETWEEN A SURFACE AND A ROLLING OBJECT. FLUID FRICTION – THE RESISTANCE FROCE OF A GAS OR A LIQUID AS AN OBJECT PASSES THROUGH. FLUID FRICTION – THE RESISTANCE FROCE OF A GAS OR A LIQUID AS AN OBJECT PASSES THROUGH. –EXAMPLE: AIR RESISTANCE

4. STATIC FRICTION AS LONG AS AN OBJECT DOES NOT MOVE STATIC FRICTION IS EQUAL AND OPPOSITE TO THE APPLIED FORCE. AS LONG AS AN OBJECT DOES NOT MOVE STATIC FRICTION IS EQUAL AND OPPOSITE TO THE APPLIED FORCE. - Fs = Fapplied - Fs = Fapplied WHEN THE APPLIED FORCE IS AS GREAT AS IT CAN BE WITHOUT CAUSING THE OBJECT TO MOVE THE FORCE OF STATIC FRICTION REACHES ITS MAXIMUM VALUE WHEN THE APPLIED FORCE IS AS GREAT AS IT CAN BE WITHOUT CAUSING THE OBJECT TO MOVE THE FORCE OF STATIC FRICTION REACHES ITS MAXIMUM VALUE

5. Fk V.S. Fs WHICH ONE IS GREATER? WHICH ONE IS GREATER? THE NET EXTERNAL FORCE ON AN OBJECT IS EQUAL TO THE DIFFERENCE BETWEEN THE APPLIED FORCE AND THE FORCE OF KINETIC FRICITON. THE NET EXTERNAL FORCE ON AN OBJECT IS EQUAL TO THE DIFFERENCE BETWEEN THE APPLIED FORCE AND THE FORCE OF KINETIC FRICITON.

6. WHERE DOES FRICTION COME FROM? FRICTION COMES FROM THE INTERACTION OF SURFACES ON THE MICROSCOPIC LEVEL. FRICTION COMES FROM THE INTERACTION OF SURFACES ON THE MICROSCOPIC LEVEL.

7. IT IS PROPORTIONAL THE NORMAL FORCE IS PROPORTIONAL TO FORCE OF FRICTION THE NORMAL FORCE IS PROPORTIONAL TO FORCE OF FRICTION

8. DEPENDS ON CONTACT FRICTION DEPENDS UPON WHAT TYPE OF SURFACES ARE IN CONTACT FRICTION DEPENDS UPON WHAT TYPE OF SURFACES ARE IN CONTACT AND AND HOW MUCH OF THE SURFACES ARE TOUCHING HOW MUCH OF THE SURFACES ARE TOUCHING

9. MU OR MOO MU (μ) THE COEFFICIENT OF FRICTION MU (μ) THE COEFFICIENT OF FRICTION THE HIGHER THE COEFFICIENT OF FRICTION THE LESS LIKELY THE OBJECT WILL MOVE. THE HIGHER THE COEFFICIENT OF FRICTION THE LESS LIKELY THE OBJECT WILL MOVE. THE RATIO OF NORMAL FORCE TO THE SLIDING FRICTIONAL FORCE THE RATIO OF NORMAL FORCE TO THE SLIDING FRICTIONAL FORCE Ff = μFn OR μ=Ff/Fn Ff = μFn OR μ=Ff/Fn DOES IT HAVE UNITS? DOES IT HAVE UNITS?

10. NO μ=Ff/Fn μ=Ff/Fn μ=10 N/50 N μ=10 N/50 N μ=.2 μ=.2

11. PROBLEM 1 1. BRIAN IS WALKING THROUGH THE SCHOOL CAFETERIA BUT DOES NOT REALIZE THAT THE PERSON IN FRONT OF HIM HAS JUST SPILLED HIS GLASS OF CHOCOLATE MILK. AS BRIAN, WHO WEIGHS 420 N, STEPS IN THE MILK, THE COEFFICIENT OF SLIDING FRICITON BETWEEN BRIAN AND THE FLOOR IS SUDDENLY REDUCED TO 0.040. WHAT IS THE FORCE OF SLIDING FRICTION BETWEEN BRIAN AND THE SLIPPERY FLOOR? 1. BRIAN IS WALKING THROUGH THE SCHOOL CAFETERIA BUT DOES NOT REALIZE THAT THE PERSON IN FRONT OF HIM HAS JUST SPILLED HIS GLASS OF CHOCOLATE MILK. AS BRIAN, WHO WEIGHS 420 N, STEPS IN THE MILK, THE COEFFICIENT OF SLIDING FRICITON BETWEEN BRIAN AND THE FLOOR IS SUDDENLY REDUCED TO 0.040. WHAT IS THE FORCE OF SLIDING FRICTION BETWEEN BRIAN AND THE SLIPPERY FLOOR?

12. SOLUTION Ff = Ff = μ=.04 μ=.04 Fn=420 N Fn=420 N Ff = μFn Ff = μFn Ff =.04 X 420 N Ff =.04 X 420 N Ff = 17 N Ff = 17 N

13. PROBLEM 2 2. WHILE REDECORATING HER APARTMENT, KITTY SLOWLY PUSHES AN 82 KG CHINA CABINET ACROSS THE WOODEN DINING ROOM FLOOR, WHICH RESISTS THE MOTION WITH A FORCE OF FRICTION OF 320 N. WHAT IS THE COEFFICIENT OF SLIDING FRICITON BETWEEN THE CHINA CABINET AND THE FLOOR? 2. WHILE REDECORATING HER APARTMENT, KITTY SLOWLY PUSHES AN 82 KG CHINA CABINET ACROSS THE WOODEN DINING ROOM FLOOR, WHICH RESISTS THE MOTION WITH A FORCE OF FRICTION OF 320 N. WHAT IS THE COEFFICIENT OF SLIDING FRICITON BETWEEN THE CHINA CABINET AND THE FLOOR?

14. SOLUTION W = 820 NM = 82 KG W = 820 NM = 82 KG W = Fn G = 10.0M/S 2 W = Fn G = 10.0M/S 2 Ff = 320 N W=MG Ff = 320 N W=MG Fn = 820 N W = 82 KG. X 10M/S 2 Fn = 820 N W = 82 KG. X 10M/S 2 μ = Ff/ Fn μ = Ff/ Fn μ = 320 N/ 820 N μ = 320 N/ 820 N μ = 0.39 μ = 0.39

15. PROBLEM 3 3. AT SEA WORLD, A 900 KG POLAR BEAR SLIDES DOWN A WET SLIDE INCLINED AT AN ANGLE OF 25 DEGREES TO THE HORIZONTAL. THE COEFFICIENT OF FRICITON BETWEEN THE BEAR AND THE SLIDE IS.0500. WHAT FRICITONAL FORCE IMPEDES THE BEAR’S MOTION DOWN THE SLIDE? 3. AT SEA WORLD, A 900 KG POLAR BEAR SLIDES DOWN A WET SLIDE INCLINED AT AN ANGLE OF 25 DEGREES TO THE HORIZONTAL. THE COEFFICIENT OF FRICITON BETWEEN THE BEAR AND THE SLIDE IS.0500. WHAT FRICITONAL FORCE IMPEDES THE BEAR’S MOTION DOWN THE SLIDE?

16. SOLUTION M = 900 kg G= 10 m/s2 M = 900 kg G= 10 m/s2 Fg = W =mg=9000 N Fg = W =mg=9000 N μ = 0.0500 μ = 0.0500 θ = 25° θ = 25° Fn=COS θ= Fn/ Fg Ff = μFn Fn=COS θ= Fn/ Fg Ff = μFn Fn=COS 25(9000N) Ff =.05(8160N) Fn=COS 25(9000N) Ff =.05(8160N) Fn=8160 N Ff = 408 N Fn=8160 N Ff = 408 N Fn FnFg θ θ