Welcome back to Physics 215

1. Welcome back to Physics 215 Today’s agenda: Applying Newton’s Laws with motion Weight, elevators, and normal forces Static and kinetic friction

2. Current homework assignment • HW5: • Knight textbook Ch.6: 38, 58, 60, 70 • Ch.7: 46, 52 • due Friday, Oct. 8th in recitation

3. A force P is applied by a hand to two blocks which are in contact on a frictionless, horizontal table as shown in the figure. The blocks accelerate to the right. Block A has a smaller mass than block B. (a) Draw free body diagrams for each block. (b) Which block experiences the larger net force? (c) Suppose that initially the mass of block A were half that of block B. If in a subsequent experiment the mass of block A were doubled, by what factor would the acceleration change assuming the pushing force remained constant? B A

4. A block is held in place on a frictionless incline by a massless string, as shown. The force on the blockby the incline is 1. a normal force given by vector A. 2. a normal force given by vector B. 3. the force given by vector A, but it’s not a normal force. 4. the force given by vector B, but it’s not a normal force.

5. Free-body diagram: Block on frictionless incline • Show all forces exerted on the block. • Do not show forces exerted by the block on anything else.

6. Geometry… vertical component of NBP is NBPcos(q) horizontal is NBPsin(q) y N q x T W q

7. Force components: Block on frictionless incline F = 0 implies all components of F are zero! Horizontal and vertical

8. When the block is held in place as shown, is the magnitude of the normal force exerted on the block by the incline 1. greater than the magnitude of the weight force (on the block by the Earth), 2. equal to the magnitude of the weight, or 3. less than the magnitude of the weight. 4. Can’t tell.

9. Solving system • Vertical equilibrium: NBPcos(q) + WBE = 0 • Horizontal equilibrium: NBPsin(q) + TBR = 0 2 equations  solve for NBP and TBE in terms of WBE and q

10. What have we learned? • If at rest – net force = 0, since a = 0 ! • Write down FBD – identify all forces present • Take force components in 2 (in 2D) directions to find unknowns • Don’t plug in numbers until end

11. The string that holds the block breaks, so there is no more tension force exerted on the block. Will the magnitudes of either of the other two forces on the block (i.e., the weight and the normal force) change? 1. Both weight and normal force will change. 2. Only the weight force will change. 3. Only the normal force will change. 4. Neither one of the two forces will change.

12. Discussion • Say normal does not change. What will be the new net force? Same as old, but “minus” the tension, i.e., straight to the left (and not zero). • So what would the block do? Accelerate. OK -- Straight to the left? NO, can’t be correct answer. • Acceleration is down the incline, so normal gets smaller.

13. Geometry (2) *component of WBE at 900 to plane is WBEcos(q) *component along plane is WBEsin(q) N q q W q

14. Force components: Block on frictionless incline • The string that holds the block breaks. • What does happen? • Take components now along and perpendicular to incline

15. Solving for second case • Since acceleration is down incline, component of net force at 90 degrees to slope is zero. NBP + WBEcos(q) = 0 • Using second law applied to components along incline: a = WBEsin(q)/mblock

16. Conclusion • Normal forces can change when small changes are made to the situation. • Can choose any 2 directions to find force components, but it pays to pick ones that simplify equations

17. Suppose we chose to resolve forces vertically and horizontally? • Vertically: • Ncos(q) + W = -masin(q) • Horizontally: • Nsin(q) = ma cos(q) • Multiply (1) by cos(q), (2) by sin(q) and add:  N + Wcos(q) = 0 as before!!

18. If the block lies 0.5 m up such a plane inclined at an angle of 30how long will it take for the block to reach the ground? (take g = 10 m/s2)

19. Summary • To solve problems in mechanics, identify all forces and draw free body diagrams for all objects • If more than one object, use Newton’s Third law to reduce number of independent forces • Use Newton’s Second law for all components of net force on each object • Choose component directions to simplify equations

20. Weight, mass, and acceleration • What does a weighing scale ``weigh’’? • Does it depend on your frame of reference? • Consider elevators….

21. A person is standing on a bathroom scale while riding an express elevator in a tall office building. When the elevator is at rest, the scale reads about 160 lbs. While the elevator is moving, the reading is frequently changing, with values ranging anywhere from about 120 lbs to about 200 lbs. At a moment when the scale shows the maximum reading (i.e. 200 lbs) the elevator 1. must be going up 2. must be going down 3. could be going up or going down 4. I’m not sure.

22. Motion of elevator (if a < 0 ) Moving upward and slowing down, OR Moving downward and speeding up. Motion of elevator (if a > 0 ) Moving upward and speeding up, OR Moving downward and slowing down.

23. A person is standing on a bathroom scale while riding an express elevator in a tall office building. When the elevator is at rest, the scale reads about 160 lbs. While the elevator is accelerating, a different reading is observed, with values ranging anywhere from about 120 lbs to about 200 lbs. At a moment when the scale shows the maximum reading (i.e. 200 lbs) the acceleration of the elevator is approximately 1. 1 m/s2 2. 2.5 m/s2 3. 5 m/s2 4. 12.5 m/s2

24. Conclusions • Scale reads magnitude of normal force |NPS| • Reading on scale does not depend on velocity (principle of relativity again!) • Depends on acceleration only * a > 0  normal force bigger * a < 0  normal force smaller

25. Reminder of free-fall experiment • Objects fall even when there is no atmosphere (i.e., weight force is not due to air pressure). • When there is no “air drag” things fall “equally fast.” i.e. same acceleration • From Newton’s 2nd law, a = W/m is independent of m -- means W = mg • The weight (i.e., the force that makes objects in free fall accelerate downward) is proportional to mass.

26. Inertial and gravitational mass • Newton’s second law: F = mIa • For an object in “free fall” W = mG g • If a independent of mI, must have mI = mG Principle of equivalence

27. Forces of friction • There are two types of situations in which frictional forces occur: • Two objects “stick to each other” while at rest relative to one another (static friction). • Two objects “rub against each other” while moving relative to each other (kinetic friction). • We will use a macroscopic description of friction which was obtained by experiment.

28. Friction demo • Static friction: depends on surface and normal force for pulled block • Kinetic friction: generally less than maximal static friction

29. The maximum magnitude of the forceofstatic friction between two objects • depends on the type of surfaces of the objects • depends on the normal force that the objects exert on each other • does not depend on the surface area where the two objects are touching The actual magnitude of the force of static friction is generally less than the maximum value.

30. A 2.4-kg block of wood is at rest on a concrete floor. (Using g = 10 m/s2, its weight force is about 24 N.) No other object is in contact with the block. If the coefficient of static friction is ms = 0.5, the frictional force on the block is: 1. 0 N 3. 12 N 2. 8 N 4. 24 N

31. A 2.4-kg block of wood is at rest on a concrete floor. (Using g = 10 m/s2, its weight force is about 24 N.) Somebody is pulling on a rope that is attached to the block, such that the rope is exerting a horizontal force of 8 N on the block. If the coefficient of static friction is ms = 0.5, the frictional force on the block is: 1. 0 N 3. 12 N 2. 8 N 4. 24 N

32. Having no choice, you have parked your old heavy car on an icy hill, but you are worried that it will start to slide down the hill. Would a lighter car be less likely to slide when you park it on that icy hill? 1. No, the lighter car would start sliding at a less steep incline. 2. It doesn’t matter. The lighter car would start sliding at an incline of the same angle. 3. Yes, you could park a lighter car on a steeper hill without sliding.

33. Block on incline revisited F N W q

34. Initially at rest • What is the largest angle before the block slips? • Resolve perpendicular to plane  N = Wcosq • Resolve parallel F = Wsinq • Since F ≤ msN, we have Wsinq ≤ msWcosq i.e. tanq ≤ ms

35. What if  > tan-1ms ? The magnitude of the forceofkinetic friction between two objects • depends on the type of surfaces of the objects • depends on the normal force that the objects exert on each other • does not depend on the surface area where the two objects are touching • does not depend on the speed with which one object is moving relative to the other

36. What if  > tan-1ms ? • Block begins to slide • Resolve along plane: Wsinq- mKWcosq= ma • Or: a = g(sinq- mKcosq)

37. Summary of friction • 2 laws of friction: static and kinetic • Static friction tends to oppose motion and is governed by inequality Fs ≤ msN • Kinetic friction is given by equality FK = mKN

38. Reading assignment • More friction, tension, pulleys • Chapters 6, 7 in textbook • Start circular motion • Ch. 8 in textbook