1. 10/1 Friction  Text: Chapter 4 section 9  HW 9/30 “Block on a Ramp” due Friday 10/4  Suggested Problems: Ch 4: 56, 58, 60, 74, 75, 76, 79, 102  Talk about friction  Lab Questions?  Chapter 6 (energy) next then 5 (circular motion) for next exam (10/17)

2. Static Friction The maximum static friction force depends on the type of surface and the contact force between the surfaces. The actual static friction force depends on Newton’s second law! f T,B MAX =  s N T,B note subscripts!  s, called the “coefficient of static friction”, is a number that that is small for slippery surfaces. Note this is not a vector equation as N is 90° to f and only relates their magnitudes.

3. Kinetic Friction:  Always acts parallel to the surface of contact.  Relative motion of surfaces (sliding)  f a,b =  k N a,b always (Remember, static by second law only)   k is the coefficient of kinetic friction.

4. Kinetic Friction Jack pulls the box at constant speed Draw a FBD of the box. W E,B N F,B a = ? T S,B f F,B Box Since a = 0, F net = 0 and T S,B = f F,B =  k N F,B v = Jack pulls harder, what happens? 0

5. Kinetic Friction Jack pulls harder Draw a FBD of the box. W E,B a = ? T S,B f F,B Box v = The tension force increases but the kinetic friction force stays the same! Kinetic friction does not depend on velocity or acceleration. f F,B =  k N F,B F net = T S,B - f F,B = ma x y N F,B

6. Kinetic Friction Jack pulls with the same force but in a different direction. Draw a FBD of the box. W E,B N F,B a = ? f F,B Box v = How will the string angle change the FBD? Tension force now has a vertical component which means N F,B is smaller. f F,B is proportional to the normal force so it is smaller also. T S,B N F,B f F,B

7. Static vs. Kinetic Friction f F,B =  k N F,B f F,B max =  s N F,B Static Kinetic Equation only gives the maximum possible, not what is is. Equation gives what is is. Find the force with the 2nd and 3rd laws always! Still may need to use the 2nd and 3rd laws, but will always satisfy the above equation.

8. 36.8 ° A 70 kg skier skis down a slope that is angled 36.8 ° with respect to the horizontal. The coefficient of kinetic friction between the skis and the snow is 0.10. What is the skiers acceleration? Homework Problem If the instantaneous velocity is 16m/s, how far up the slope did he start from rest? Start with FBD, find W x and W y, apply 2nd law to x and y separately (hint: must start with y first) (hint2: what is a y ? x y