Friction

Newton’s Second Law W = mg W = mg Gives us F = ma Gives us F = ma

Friction A force which resists sliding or rolling motion by opposing the movement of one body over or through another A force which resists sliding or rolling motion by opposing the movement of one body over or through another Can be useful: Can be harmful: - Needed for walking, biking, driving - Slows down machinery - Without it, couldn’t stop a car; would look funny walking - Makes us apply more effort

Types of Friction A) Kinetic Friction – A) Kinetic Friction – Exists as an object slides across a surface Exists as an object slides across a surface Aka – “sliding friction” (kinetic = moving) Aka – “sliding friction” (kinetic = moving) F k F k B) Static Friction – B) Static Friction – A force present b/w two surfaces even when objects are not sliding A force present b/w two surfaces even when objects are not sliding F s F s

Coefficient of Friction Amount of “grip” of a surface Amount of “grip” of a surface A value ranging from 0-1, rating 2 surfaces A value ranging from 0-1, rating 2 surfaces The higher the value (closer to 1) the greater the grip The higher the value (closer to 1) the greater the grip Symbol:  Symbol:   k = coefficient of kinetic friction  k = coefficient of kinetic friction  s = coefficient of static friction  s = coefficient of static friction

Realtionship to FBD’s F f =  F N F f =  F N Friction Force Coefficient of Friction (This is either static or kinetic) Normal Force (Force acting _|_ to the surface)

Example 1 - “A 10-kg box rests on a horizontal floor. The  k = 0.30 and  s = 0.40 Determine how much force is needed to move the box and comment on whether or not there is motion when the applied force, F A, is exerted on it at these magnitudes.” “A 10-kg box rests on a horizontal floor. The  k = 0.30 and  s = 0.40 Determine how much force is needed to move the box and comment on whether or not there is motion when the applied force, F A, is exerted on it at these magnitudes.” a)0 N: b)20 N: c)38 N d)40 N F f =  F N

EXAMPLE 2 A 10 kg box is on a horizontal surface, but an applied force of 40 N acts at an angle 30 o to the horizontal. A) What will be the acceleration of the box if friction is ignored? A 10 kg box is on a horizontal surface, but an applied force of 40 N acts at an angle 30 o to the horizontal. A) What will be the acceleration of the box if friction is ignored? B) What is the normal force? B) What is the normal force? Solving Process: First make a FBD. Solving Process: First make a FBD.

Free Body Diagram FOR PART A: FOR PART A: Solve for vector F x which stands for F in the x direction; then plug into F = ma Solve for vector F x which stands for F in the x direction; then plug into F = ma F x = cos 30 = adj / 40; 35N F x = cos 30 = adj / 40; 35N F = ma F = ma 35N = 10 kg(a) 35N = 10 kg(a) a = 3.5 m/s 2 a = 3.5 m/s 2

Free Body Diagram FOR PART B: FOR PART B: Solve for vector F y which simply stands for the F in the y direction; then plug into F N = mg - F y Solve for vector F y which simply stands for the F in the y direction; then plug into F N = mg - F y Sin 30 = opp / 40; 20N Sin 30 = opp / 40; 20N F N = 98N – 20N F N = 98N – 20N F N = 78N F N = 78N