Welcome back to Physics 211

1. Welcome back to Physics 211 Today’s agenda: Weight, elevators, and normal forces Static and kinetic friction Internal forces

2. Current homework assignments • WHW5: • In blue Tutorials in Physics homework book • HW-31 #1, #2, HW-32 #3, HW-34 #5 • due Wednesday, Oct. 4th in recitation • FHW3: • From end of chapters 5 & 6 in University Physics • 5.55, 5.60, 5.108, 6.6 • due Friday, Oct. 13th in recitation

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

4. Cinema Classics Person on scale in elevator

5. 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.

6. Free body diagram for person on scale in elevator

7. Motion of elevator (if a ) Moving upward and slowing down, OR Moving downward and speeding up. Motion of elevator (if a ) Moving upward and speeding up, OR Moving downward and slowing down.

8. 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

9. 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

10. 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.

11. 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

12. 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.

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

14. 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.

15. 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

16. 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

17. 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.

18. Block on incline revisited F N W q

19. 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

20. 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

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

22. 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

23. Internal Forces • So far replaced macroscopic bodies by points – why is this OK? • Specifically, such body composed of (very many) parts – neglected all internal forces of these parts on each other • Also neglected rotational motion -- later

24. A B Simple example A: NAG Constant v NAB PAH FAG WAE B: NBG NBA FBG WBE

25. Composite system • Newton’s Third Law for A and B imply that we can consider combined system C=A+B in which NAB,etc.do not appear – internal forces NCG=NAG+NBG PCH FC=FAG+FBG WCE=WAE+WBE

26. Internal forces summary • Can apply Newton’s laws to a composite body • Can ignore internal forces of one part of body on another, since these cancel (Third law) • Justifies treating macroscopic bodies as point-like …

27. Two crates, one with mass 4.00 kg and the other with mass 6.00 kg, sit on a frictionless surface. A horizontal force F is applied to the 6.00 kg crate, giving it an acceleration of 2.50 m/s2. (a) What is the acceleration of the 4.00 kg crate? (b) Draw a free body diagram for the 4.00 kg crate and find the tension in the rope between the crates. (c) Draw a free body diagram for the 6.00 kg crate. (d) Calculate the force F.

28. Now consider the same two crates, but moving along at constant velocity with a kinetic friction coefficient with the floor K = 0.5. Calculate the tension T in the rope between the blocks and the force F on the rope attached to block A necessary to produce this motion.

29. Reading assignment • Tension, circular motion • 5.4 in textbook