Welcome back to Physics 215

1. Welcome back to Physics 215 Today’s agenda: Weight Friction Internal forces Tension, pulleys

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

3. 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

4. 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.

5. 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

6. 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.

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

8. 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.

9. 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

10. 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

11. 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.

12. Block on incline revisited F N W q

13. 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

14. 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

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

16. 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

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. 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

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

20. 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

21. 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 …

22. For an ideal string or rope connecting two objects: • does not stretch  inextensible • has zero mass Tension • Let’s look at an example of a cart connected to a falling mass by an ideal string...

23. What if the mass of the rope/string is not zero? Two blocks are connected by a heavy rope. A hand pulls block A in such a way that the blocks move upward at increasing speed. The (downward) tension force on the upper block by the rope is 1. less than 2. equal to 3. greater than the (upward) tension force on the lower block by the rope. 4. Answer depends on which block is heavier.

24. Hand pulls block A so blocks move up at increasing speed.

25. Notice: for mR = 0, the tension forces exerted at either end are the same. The term “tension in the string” is therefore often used as a short-hand for the tension forces exerted on or bythe string at either end.

26. Blocks A and C are initially held in place as shown. After the blocks are released, block A will accelerate up and block C will accelerate down. The magnitudes of their accelerations are the same. Will the tension in the string be 1. equal to 1.0 N (i.e. the weight of A), 2. between 1.0 N and 1.5 N, 3. equal to 1.5 N (i.e. the weight of C), or 4. equal to 2.5 N (i.e.the sum of their weights)?

27. Demo: Pulleys *2 pulleys 2T = W F = T = W/2 T F *N pulleys F = W/N! W

28. Motion around circular track, constant speed (for now): Forces in circular motion arad = v2/r

29. Reading assignment • Circular motion • Chapter 8 in textbook