1. Frictional Forces  Two types: - static – applies to stationary objects - kinetic – applies to sliding (moving) objects  Like F N, the Frictional Force is a contact force, but acts parallel to the interface of two objects Chapter 6. Applications of Newton’s Laws

2. y x FNFN mgmg FAFA f Apply Newton’s 2 nd Law  If applied force is small, book does not move (static), a x =0, then f=f s  Increase applied force, book still does not move  Increase F A more, now book moves, a x  0

3.  There is some maximum static frictional force, f s max. Once the applied force exceeds it, the book moves   s is the coefficient of static friction, it is a dimensionless number, different for each surface-object pair (wood-wood, wood-metal); also depends on surface preparation   s does not depend on the mass or surface area of the object  Has value: 0 <  s < 1.5  If no applied vertical force Magnitudes not vectors

4.  Push down on book  Apply Newton’s 2 nd Law y x FNFN mgmg FAFA f FPFP  What is F A needed just to start book moving?

5. When an Object is Moving?  f s max is exceeded so the object can move, but friction force is still being applied.  However, less force is needed to keep an object moving (against friction) than to get it started  We define kinetic friction   k is the coefficient of kinetic friction, similar to  S but always less than  S  Now, let’s consider incline plane problem, but with friction

6. y y FBD  mgmg FNFN x FNFN mgmg   x fsfs fsfs  Book is at rest  m = 1.00 kg,  = 10.0°,  s = 0.200

7.  Book can move (slide) if f s >f s max  What is f s max ? Book does not move.  What angle is needed to cause book to slide?  As  is increases, F N decreases, therefore f s max decreases

8.  Once book is moving, we need to use the kinetic coefficient of friction  Lets take,  = 15.0° and  k = 0.150 <  s

9. The Tension Force Crate, m rope T frictionless  Assume rope is massless and taut T -T FBD of rope FBD of crate T mgmg T m Frictionless pulley Crate at rest T mgmg

10.  Like the normal force, the friction and tension forces are all manifestations of the electromagnetic force  They all are the result of attractive (and repulsive) forces of atoms and molecules within an object (normal and tension) or at the interface of two objects Applications of Newton’s 2 nd Law  Equilibrium – an object which has zero acceleration, can be at rest or moving with constant velocity

11.  Example: book at rest on an incline with friction  Non-equilibrium – the acceleration of the object(s) is non-zero Example Problem Three objects are connected by strings that pass over massless and frictionless pulleys. The objects move and the coefficient of kinetic friction between the middle object and the surface of the table is 0.100 (the other two being suspended by strings). (a) What is the acceleration of the three objects? (b) What is the tension in each of the two strings?

12. m2m2 m1m1 m3m3  Given: m 1 = 10.0 kg, m 2 =80.0 kg, m 3 =25.0 kg,  k =0.100  Find: a 1, a 2, a 3, T 1, and T 2 Solution: 1) Draw free-body diagrams m1m1 T1T1 m1gm1g T2T2 m3m3 m3gm3g m2gm2g T1T1 FNFN T2T2 fkfk y x 2) Apply Newton’s 2 nd Law to each object

13. Also, a x1 =0, a x3 =0, a y2 =0

14.  Three equations, but five unknowns: a y1,a y3,a x2,T 1,and T 2  But, a y1 = a x2 = -a y3 = a  Substitute 2 nd and 3 rd equations into the 1 st equation

16. Example Problem A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 25.0° with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier’s mass is 55.0 kg, and the coefficient of kinetic friction between the skis and the snow is 0.120. Find the magnitude of the force that the tow bar exerts on the skier.

17.  Given: m = 55.0 kg,  k = 0.120,  = 25.0°  Infer: since velocity is constant, a x =0; also a y =0 since skier remains on slope  equilibrium Draw FBD, apply Newton’s 2 nd Law  mgmg FNFN x FNFN mgmg   x FpFp FpFp fkfk fkfk