2. Forces due to Friction Friction: A force between the contacted surfaces of two objects that resists motion. If an object is not moving, that does not mean that there are no friction forces at work: mg N In this case, there is no motion, nor any additional external force that may produce motion, so there is not friction present!

3. No Motion  F = f static F fsfs  Accelerated Motion F fkfk a and then: F > f kinetic  Uniform Motion F fkfk F = f k If F max > f s-max then acc.

4. Friction (N) time (s) f s max f s fkfk Static friction builds to a maximum value at which point the object “breaks free” -- it accelerates and kinetic friction is present! No Acceleration  Static Friction

5. The ratio of the maximum force of static friction to the magnitude of the normal force is called the coefficient of static friction:  s = f s N The ratio of the magnitude of the force of kinetic friction to the magnitude of the normal force is called the coefficient of kinetic friction:  k = f k N

7. F N.5mg fsfs ∑F x = 0 = F – N ∑F y = 0 = f s –.5mg µ s = f s N N = F = f s µ s =.5(75)(9.80).41 =.5mg µ s = 900 N Examine the right side:

8. m1m1 m2m2 Ø Ø = 34˚ m 1 = 9.5 kg m 2 = 2.6 kg Will the blocks accelerate, and if so, what will that acceleration be? µ s =.24 µ k =.15

9. m1gm1g m 1 gcosøø m 1 gsinø T1T1 N f s m2gm2g T2T2 T 1 = T 2 = T If the block will stay, then: ∑F x1 = 0 = T - (f s + m 1 gsinø) ∑F y1 = 0 = N - m 1 gcosø N = m 1 gcosø ∑F y2 = 0 = m 2 g - T 2

10. T = f s + m 1 gsinø m 2 g = µ s (m 1 gcosø) + m 1 gsinø m 2 = m 1 (µ s cosø + sinø) 2.6 < 8.5 The tension in the string is not strong enough! OR: ∑F x = 0 = T + f s - m 1 gsinø T = m 1 gsinø - f s m 2 g = m 1 gsinø - µ s m 1 gcosø 2.6 < 7.2 again, T is not enough to hold therefore: T < f s + m 1 gsinø

11. To find the acceleration: ∑F x = m 1 a 1 = T + f k - m 1 gsinø ∑F y = m 2 a 2 = T - m 2 g ∑F y = 0 = N - m 1 gcosø in terms of the magnitude of a: a 1 = a 2 = a T = m 1 a + m 1 gsinø - f k T = m 2 a + m 2 g m 1 a + m 1 gsinø - f k = m 2 a + m 2 g m 1 a + m 1 gsinø - µ k m 1 gcosø = m 2 a + m 2 g

12. a(m 1 + m 2 ) = m 2 g + m 1 g(µ k cosø - sinø) a = [m 2 g + m 1 g(µ k cosø - sinø)] (m 1 + m 2 ) a = 25 + 93[(.15)(.83) - (.56)] 12.1 = - 1.3 m/s 2 Note well what the negative sign means: for block 1, acceleration is in the negative x direction- opposite of up (+) the incline. for block 2, opposite of g (which we used as +)

13. The dynamics of Uniform Circular Motion Remember that for any object traveling in uniform circular motion, the magnitude of the net (centripetal) force acting upon the object is: ∑F = ma = mv 2 /r For the purposes of discussing motion of an object in uniform circular motion, we will not use x and y axes! Instead, vector will be described in the z direction (perpendicular to the plane of motion) and r direction (along the radius of the path)!

14. We will consider 3 specific examples of forces that act centripetally: The Conical Pendulum L m ø z r mg T ø R

15. ∑F r = ma = mv 2 /R = Tsinø ∑F z = 0 = Tcosø - mg Setting the components in ratio form: Tsinø = mv 2 /R Tcosø mg Solving for v: v = √ Rgtanø This equation gives the constant speed needed for the pendulum to maintain its circle.

16. The constant speed of a body would be: v = 2πR T Substituting and solving for time: T = 2π (Lcosø)/g This would give the period of the motion, which is independent of the mass of the bob!

17. The Rotor mg fsfs N

18. The Rotor An object is kept in uniform circular motion by frictional forces! ∑F z = 0 = f s - mg ∑F r = ma r = N f s = mg N = mv 2 /R Solving these equations together we can derive an equation for speed needed to prevent slipping: v = √ gR/µ s

19. The Banked Curve

20. ø mg N ∑F z = 0 = Ncosø - mg ∑F r = ma r = Nsinø manipulating these equations will give: tanø = v 2 /Rg this is the angle-speed combination where friction is not present!

21. tanø = v 2 /Rg NOTE: the angle of the bank is independent of the mass of the vehicle or even friction! when a banked curve is constructed, the road is banked according to an average calculated speed. as with the conical pendulum, the constant speed needed to maintain a given banked curve (neglecting any frictional forces): v = Rgtanø

22. Drag Forces and Projectiles Due to air resistance, objects will only accelerate until reaching a final /maximum speed. That max. speed is called terminal velocity. Terminal velocity depends on the properties of the falling object (size and shape and density) as well as the properties of the fluid (esp. density). One particular characteristic of drag forces is that they depend on velocity.

23. Air Resistance (Aerodynamic Drag) Unlike kinetic friction, air resistance (Drag) is not constant but increases as speed increases. The amount of Drag will depend on the shape of the object (drag coefficient [C]), the density of the air (ρ), the cross sectional area (A) and the square of the velocity of the object: D =.5CρAv 2 Terminal Velocity occurs when the object stops accelerating and the Drag equals weight!

24. ∑F y = 0 = mg - D mg =.5CρAv 2

25. D = bv For a relatively tiny particle traveling through a thick fluid, Drag is proportional to only velocity: b is a constant value dependent upon the characteristics of the object and the fluid bv T = mg Terminal velocity in the fluid would still occur when the Drag equals the weight force: