1. Friction Outline Newton’s 2 nd Law with Friction Air drag Examples

2. QQ38: Inequality Example: A car is moving at a constant speed of 24m/s. The driver hits the brakes which causes the car to slide. Given that μ k =0.8 for rubber on concrete: a) what is the acceleration of the car ? b) how far will it move before stopping ?

3. Demo Example: Two blocks are pulled with a force of 28N over a rough surface. Block 1 has a mass of 1kg and a coefficient of friction of 0.3 with the surface while Block 2 a mass of 2kg and a coefficient of friction of 0.2 with the surface. What is the acceleration of the system ?

4. QQ38: Inequality Quick Quiz: A horizontal force of 150 N is applied to a 50 kg box on a level floor. The coefficient of static friction between the box and the floor is 0.5. The magnitude of the frictional force on the box is: (A) 150 N (B) 250 N (C) 500 N (D) 3750 N (E) 15000 N

5. QQ38: F prop N Quick Quiz: The pilot of the helicopter that you are in has fallen asleep, and the helicopter is now in freefall. Can you run to the cockpit and save the day? a) Yes b) No c) Depends of μ k of the floor d) Depends of μ s of the floor There is no normal force needed to keep you from falling, since you and the floor of the helicopter are both in freefall. If there is no normal force, there is no friction. No friction means that you can’t run.

6. QQ38: Inequality Quick Quiz: A force of 150 N is applied at an angle of 30 0 to the horizontal to a 20 kg box on a level floor. How does this force affect the normal force: (A) No change (B)Decrease the normal force (C)Increase the normal force

7. QQ38: Inequality Terminal Speed A falling object in the presence of air resistance will reach some maximum velocity beyond which it cannot accelerate. This is due to the fact that air friction (drag) increases as your velocity increases. Once this drag force is equal to the force of gravity, there is not net force and acceleration is zero: FgFg D FgFg D F g > D a > 0 F g = D a = 0

8. QQ38: Inequality Terminal Speed The air drag is equal to D=¼ Av 2, where A is the surface area of the object. Hence, the maximum speed will be when D=F g: ¼ Av 2 = mg v term = Note: more massive objects have a larger v term, as do objects with a smaller surface Area

9. QQ38: Inequality Example a) Estimate the terminal velocity of a 60kg person falling down in a horizontal position. b) Estimate the terminal velocity of a 60kg person falling down in a vertical position.

10. Do for next class: Read: Sections 5.6 Suggested problems: 5.50, 5.51, 5.53 Do as many more of the problems from the end of the chapter as it takes you to realise that they are all the same. If this takes you more than two full problems, make sure that you are following the procedure that I followed for the examples in class.