Friction

1. Friction

2. Consider An Object Coming to Rest • Aristotle’s Idea: At rest is the “natural state” of terrestrial objects

3. Consider An Object Coming to Rest • Aristotle’s Idea: At rest is the “natural state” of terrestrial objects • Newton’s View: (Galileo’s too!) A moving object comes to rest because a force acts on it.

4. Consider An Object Coming to Rest • Aristotle’s Idea: At rest is the “natural state” of terrestrial objects • Newton’s View: (Galileo’s too!) A moving object comes to rest because a force acts on it. • Most often, this stopping force is Due to a phenomenon called friction.

5. Friction • It must be accounted forwhen doing realistic calculations! • It exists between any 2 • sliding surfaces. • There are 2 types friction: • Static(no motion) friction • Kinetic(motion) friction • Frictionis alwayspresent when 2 solid surfaces slide along each other. See figure.

6. Two types of friction: • Static(no motion) friction • Kinetic(motion) friction • The size of the friction force depends on the microscopic details of the 2 sliding surfaces. • These details aren’t fully understood & depend on the materials they are made of • Are the surfaces smooth or rough? Are they wet or dry? Etc., etc., etc.

7. Kinetic Frictionis the same as Sliding Friction. • The kinetic friction force Ffropposes the motion of a mass. Experiments find the relation used to calculate Ffr. • Ffr is proportional to the magnitude of the normal forceN = FNbetween 2 sliding surfaces. The DIRECTIONSof Ffr & N areeach other!! FfrN • We write their relation as Ffr kFN(magnitudes) k  Coefficient of Kinetic Friction

8. Properties of k • Depends on the surfaces & their conditions. • Is different for each pair of sliding surfaces. • Values for μkfor various materials can be looked up in a table (shown soon).Further, • kisdimensionless • Usually, k < 1 The Kinetic Coefficient of Friction k

9. Set up the problem as usual, including the force of friction. For example, the hockey puck in the figure: Problems Involving Friction Newton’s 2nd Law for the Puck:

10. Set up the problem as usual, including the force of friction. For example, the hockey puck in the figure: Problems Involving Friction Newton’s 2nd Law for the Puck: (In the horizontal (x) direction): ΣF = Ffr = -μkN = ma (1) (In the vertical (y) direction): ΣF = N – mg = 0 (2)

11. Set up the problem as usual, including the force of friction. For example, the hockey puck in the figure: Problems Involving Friction Newton’s 2nd Law for the Puck: (In the horizontal (x) direction): ΣF = Ffr = -μkN = ma (1) (In the vertical (y) direction): ΣF = N – mg = 0 (2) • Combining (1) & (2) gives -μkmg = ma so a = -μkg

12. Set up the problem as usual, including the force of friction. For example, the hockey puck in the figure: Problems Involving Friction Newton’s 2nd Law for the Puck: (In the horizontal (x) direction): ΣF = Ffr = -μkN = ma (1) (In the vertical (y) direction): ΣF = N – mg = 0 (2) • Combining (1) & (2) gives -μkmg = ma so a = -μkg • Once a is known, we can do kinematics, etc. • Values for coefficients of friction μkfor various materials can be looked up in a table (shown later). These values depend on the smoothness of the surfaces

13. Static Friction • In many situations, the 2 surfaces are not slipping (moving) with respect to each other. This situation involves Static Friction • The amount of the pushing force Fpush can vary without the object moving. • The static friction force Ffr is as big as it needs to be to prevent slipping, up to a maximum value. Usually it is easier to keep an object sliding than it is to get it started.

14. Static Friction • The static friction force Ffr is as big as it needs to be to prevent slipping, up to a maximum value. Usually it is easier to keep an object sliding than it is to get it started. • Consider Fpush in the figure. Newton’s 2nd Law: (In the horizontal (x) direction): ∑F = Fpush - Ffr = ma = 0

15. Static Friction • The static friction force Ffr is as big as it needs to be to prevent slipping, up to a maximum value. Usually it is easier to keep an object sliding than it is to get it started. • Consider Fpush in the figure. Newton’s 2nd Law: (In the horizontal (x) direction): ∑F = Fpush - Ffr = ma = 0 so Ffr = Fpush

16. Static Friction • The static friction force Ffr is as big as it needs to be to prevent slipping, up to a maximum value. Usually it is easier to keep an object sliding than it is to get it started. • Consider Fpush in the figure. Newton’s 2nd Law: (In the horizontal (x) direction): ∑F = Fpush - Ffr = ma = 0 so Ffr = Fpush • This remains true until a large enough pushing force is applied that the object starts moving. That is, there is a maximum static friction force Ffr.

17. Experiments find that the maximum static friction forceFfr (max)is proportional to the magnitude(size) of the normal forceN between the 2 surfaces. • The DIRECTIONS of Ffr & N areeach other!! FfrN • Write the relation as Ffr(max) = sN(magnitudes) s  Coefficient of Static Friction • Always find s > k  Static friction force:Ffr  sN

18. Static Coefficient of Friction s • Depends on the surfaces & their conditions. • Is different for each pair of sliding surfaces. • Values for μs for various materials can be looked up in a table (next slide).Further, sisdimensionless Usually, s < 1 Always, k < s

19. Coefficients of Friction μs > μk Ffr (max, static) > Ffr(kinetic)

20. Conceptual Example • A hockey puck is moving at a constant • speed v, with NO friction. Which free • body diagram is correct?

21. Static & Kinetic Friction

22. Kinetic Friction Compared to Static Friction • Consider both the kinetic and static friction cases • Use the different coefficients of friction • The force of Kinetic Frictionis Ffriction = μk N • The force of Static Frictionvaries: Ffriction ≤ μs N • For a given combination of surfaces, generally μs > μk • It is more difficult to start something moving than it is to keep it moving once started

23. Friction & Walking • The person “pushes” off during each step. • The bottoms of his shoes exert a force on the ground. This is Fon ground . • If the shoes do not slip, the force is due to static friction • The shoes do not move relative to the ground

24. Newton’s Third Law • This tells us that there is a reaction force Fon shoe • This force propels the person as he moves • If the surface was so slippery that there was no frictional force, the person would slip

25. Friction & Rolling • The car’s tire does not slip. So, there is a friction forceFon ground between the tire & road. • There is also a Newton’s 3rd Law reaction force Fon tire on the tire. This is the force that propels the car forward

26. A box, mass m =10.0-kg rests on a horizontal floor. The coefficient of static friction is s = 0.4; the coefficient of kinetic friction is k = 0.3. Calculate the friction force on the box for a horizontal external applied force of magnitude: (a) 0, (b) 10 N, (c) 20 N, (d) 38 N, (e) 40 N. Example: Friction; Static & Kinetic

27. Conceptual Example • You can hold a box • against a rough wall & • prevent it from slipping • down by pressing hard • horizontally. • How does the application • of a horizontal force keep • an object from moving • vertically?