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Day 51 Friction Aim: What are the different types of Friction? LO: Relate friction to the normal force LO: Calculate friction for different surface combinations.

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Presentation on theme: "Day 51 Friction Aim: What are the different types of Friction? LO: Relate friction to the normal force LO: Calculate friction for different surface combinations."— Presentation transcript:

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2 Day 51 Friction Aim: What are the different types of Friction? LO: Relate friction to the normal force LO: Calculate friction for different surface combinations LO: AGENDA Do Now - Worksheet Notes Worksheet HW# Due

3 Friction

4  Friction is a special force that is caused by the surface roughness of an object.  It always acts in the opposite direction of the motion of the object.  There are two types of friction –Static, and kinetic

5 Coefficient of Friction  All surfaces exhibit friction, some more than others.  It depends on the roughness of the surface of the object.  It is represented by the symbol  –For static friction:  s –For kinetic friction:  k

6 Sliding Friction – Microscopic model  Depends on microscopic (electrostatic) bonding forces  Depends on roughness of the surface

7 Kinetic Friction  Kinetic friction is the force of friction on an object when it is moving  The formula is: F f =  k F N

8 Static Friction  Static Friction is the force of friciton on an object when it stands still.  We find that it is harder to start an object moving than it is to keep it moving.  The formula is: Fs  sFNFs  sFN

9 Graph of the behavior of sliding friction

10 A Table of coefficients of sliding friction

11 Example  A boy exerts a 36N horizontal force as he pulls a 52N sled across a cement sidewalk at a constant speed. What is the coefficient of friction between the sidewalk and the sled (ignoring air resistance)? 52N 36N

12 Solution  Known: F N = F g = 52 N F pull = F friction = 36N because the sled is moving at constant velocity F friction = F N  k Therefore  k = F f /F N  k = 36N/52N = ?

13 Example 2  Suppose the sled runs on packed snow. The coefficient of friction is now only 0.12. If a person weighing 650N sits on the sled what is the force needed to pull the sled across the snow at a constant speed?

14  = 0.12 Fw = mg= 650N What force to pull sled?

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22 Inclined Plane  A common free body diagram used is often the inclined plane.  Another name for an inclined plane is a ramp.  Look at the diagram to the right showing the usual forces on an inclined plane FNFN FfFf W

23 Vector Diagram  If we look at just the vector diagram we see some interesting things  We usually know the weight of the object, so we can find the normal force.  The normal force is perpendicular to the friction force and the force of the inclined plane FNFN FfFf W

24 Example 3  A skier (Ki) has just begun to descend a 30 o slope. Assuming the coefficient of kinetic friction is 0.10 calculate: (i) his acceleration and (ii) his speed after 4 s

25 Example 3  A skier (Ki) m = 7 kg has just begun to descend a 30 o slope. Assuming the coefficient of kinetic friction is 0.10 calculate: (i) his acceleration and (ii) his speed after 4 s Approach: (i) Resolve forces | | and to slope (ii) Calculate frictional force (iii) Find net force down the slope => acceleration (iv) Use v f = v i + at => v f

26 Solution  Force of gravity down the slope is: F gpara = F g Sin(  ) F gpara = 7kg*10 m/s/s*0.5 = 35 N  Calculate Normal = F gperp F gperp = F g Cos(  ) F gperp = 7kg * 10m/s/s *0.866 = 60.62 N

27 Solution Continued  Calculate Frictional Force: F f =  k F normal F f = 0.1 * 60.62 N = 0.6062 N  Caluculate Net force down the slope F net = F gperp – F f F net = 35 N – 0.6062 N = 34.4 N

28 Solution last page  Calculate Acceleration down the slope: F net = ma a = 34.4 N/7kg = 4.9 m/s/s  Calculate velocity at t = 4 seconds V f = V i + at V f = 0 m/s + 4.9 m/s/s * 4 s = 19.7 m/s

29 Inclined Plane  A common free body diagram used is often the inclined plane.  Another name for an inclined plane is a ramp.  Look at the diagram to the right showing the usual forces on an inclined plane FNFN FfFf FpFp W F p = the force caused by the ramp

30 Vector Diagram  If we look at just the vector diagram we see some interesting things  We usually know the weight of the object, so we can find the normal force.  The normal force is perpendicular to the friction force and the force of the inclined plane FNFN FfFf FpFp W


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