Friction. Newton’s Second Law W = mg W = mg Gives us F = ma Gives us F = ma.
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Transcript of Friction. Newton’s Second Law W = mg W = mg Gives us F = ma Gives us F = ma.
FrictionFriction
Newton’s Second LawNewton’s Second Law
W = mgW = mgGives us F = maGives us F = ma
FrictionFriction A force which resists sliding or A force which resists sliding or
rolling motion by opposing the rolling motion by opposing the movement of one body over or movement of one body over or through anotherthrough another
Can be useful:Can be useful: Can be harmful:Can be harmful:
- Needed for - Needed for walking, biking, walking, biking, drivingdriving
- Slows down - Slows down machinerymachinery
- Without it, couldn’t - Without it, couldn’t stop a car; would stop a car; would look funny walkinglook funny walking
- Makes us apply - Makes us apply more effortmore effort
Types of FrictionTypes of Friction A) Kinetic Friction –A) Kinetic Friction – Exists as an object slides across a Exists as an object slides across a
surface surface Aka – “sliding friction” (kinetic = Aka – “sliding friction” (kinetic =
moving)moving) FFkk
B) Static Friction – B) Static Friction – A force present b/w two surfaces even A force present b/w two surfaces even
when objects are not slidingwhen objects are not sliding FFss
Coefficient of FrictionCoefficient of Friction
Amount of “grip” of a surfaceAmount of “grip” of a surface A value ranging from 0-1, rating 2 A value ranging from 0-1, rating 2
surfacessurfaces The higher the value (closer to 1) the The higher the value (closer to 1) the
greater the gripgreater the grip Symbol: Symbol:
kk = coefficient of kinetic friction = coefficient of kinetic friction ss = coefficient of static friction = coefficient of static friction
Table of Coefficients of Table of Coefficients of FrictionFriction
Static Kinetic
Relationship to FBD’sRelationship to FBD’s
FFff = = FFNNFriction Force
Coefficient of Friction (This is either static or kinetic)
Normal Force (Force acting _|_ to the surface)
Static Friction increases steadily as force is applied until the objects begins moving.
This represents the kinetic friction once the object is moving at a constant rate.
Static vs. KineticStatic vs. Kinetic
Example 1 Example 1 - - ““A 10-kg box rests on a horizontal floor. The A 10-kg box rests on a horizontal floor. The kk = 0.30 and = 0.30 and
ss = 0.40 Determine how much force is needed to move = 0.40 Determine how much force is needed to move the box and comment on whether or not there is motion the box and comment on whether or not there is motion when the applied force, Fwhen the applied force, FAA, is exerted on it at these , is exerted on it at these magnitudes.”magnitudes.”
a)0 N:
b)20 N:
c)38 N
d)40 N
Ff = FN
EXAMPLE 2EXAMPLE 2 A 10 kg box is on a horizontal A 10 kg box is on a horizontal
surface, but an applied force of 40 surface, but an applied force of 40 N acts at an angle 30N acts at an angle 30oo to the to the horizontal. A) What will be the horizontal. A) What will be the acceleration of the box if friction acceleration of the box if friction is ignored? is ignored?
B) What is the normal force?B) What is the normal force?
Solving Process: First make a Solving Process: First make a FBD.FBD.
Free Body DiagramFree Body Diagram FOR PART A:FOR PART A: Solve for vector FSolve for vector Fxx which which
stands for F in the x stands for F in the x direction; then plug into F direction; then plug into F = ma= ma
FFxx = cos 30 = adj / 40; 35N = cos 30 = adj / 40; 35N
F = maF = ma
35N = 10 kg(a)35N = 10 kg(a)
a = 3.5 m/sa = 3.5 m/s22
Free Body DiagramFree Body Diagram FOR PART B:FOR PART B: Solve for vector FSolve for vector Fyy which which
simply stands for the F simply stands for the F in the y direction; then in the y direction; then plug into Fplug into FNN = mg - F = mg - Fyy
Sin 30 = opp / 40; 20NSin 30 = opp / 40; 20N
FFNN = 98N – 20N = 98N – 20N
FFNN = 78N = 78N