Status: Unit 3

1. Status: Unit 3 • Force, Mass, & Newton’s 1st, 2nd, 3rd Laws of Motion (4-1 through 4-5) • Weight – the Force of Gravity: and the Normal Force, Problem solving (4-6, 4-8, 4-7) • Solving Problems w/ Newton’s Laws, Problems w/ Friction (4-7) • Problems w/ Friction and Terminal Velocity revisited (4-7, 5-1) Physics 253

2. But First a Bit about Test One • Total Number of Exams Taken: 116 • Low : 5 / Average: 30 / High: 49 • Frequency • 0-10: 3 • 11-20: 19 • 21-30: 36 • 31-40: 41 • 41-50: 17 • Pick them up from the stage -- after class. Physics 253

3. Friction • To get a handle on the principles we’ve ignored friction. However this is a big oversimplification. Now we have the tools to properly deal with the force. • Friction exist between any two solid surfaces. • When one or both of the surfaces are very coarse and bumpy, like a cobble stone street impeding on the runners of a sled, it’s pretty clearly a mechanical effect. • But even for very smooth surface, friction exists due to the microscopic imperfections or roughness. Just like the cobblestones these bumps impede motion. Most likely this is an electromagnetic effect. • There are two kind so friction: kinetic and static. Physics 253

4. Kinetic Friction • When two objects are sliding past one another they are said to have kinetic fiction. • The force of friction acts opposite the direction of an object’s velocity. You’ve felt this before, perhaps when shoving a couch across a carpet. • The magnitude of the force depends on the characteristics of the two surface. • It turns out experimentally that the force of friction is • Proportional to the normal force between the two surfaces • Independent of surface area • Independent of the velocity Physics 253

5. We deal with the different surfaces by imposing a constant of proportionality: The constant or coefficient of kinetic friction depends only on the surfaces. Caveats: NOT a vector equation! NOT a theory - but a model that works well Rubber on solid surfaces: 1 Steel on steel: 0.6 Rubber on wet concrete: 0.5 Wood on wood: 0.2 Ice on ice: 0.03 Human joints: 0.01 Oiled Ball Bearings: <0.01 Physics 253

6. Remember shoving the couch across the carpet? You’ll also remember it’s actually harder and takes more force to get it moving. That’s because the static friction, the force parallel to the surfaces of two motionless objects is larger than kinetic friction. What’s more it depends on how hard you push! The couch doesn’t accelerate right a way when a small force is applied, it’s not until later when you exceed some limiting force. Since the entire time a=0, the force of friction must be balancing and changing to match your force! It turns out that the maximum force of static friction is also proportional to the normal force: Once again the constant, in this case the coefficient of static friction, depends on the surfaces. Since in this case the frictional force varies it’s more complete to say Static Friction Physics 253

7. Rubber on solid surfaces: 1 Steel on steel: 0.6 Rubber on wet concrete: 0.5 Wood on wood: 0.2 Ice on ice: 0.03 Human joints: 0.01 Oiled Ball Bearings: <0.01 Rubber on solid surfaces: 1-4 Steel on steel: 0.7 Rubber on wet concrete: 0.7 Wood on wood: 0.4 Ice on ice: 0.1 Human Joints: 0.01 Oiled Ball Bearings: <0.01 Kinetic Coefficients: mk Static Coefficients: ms Physics 253

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9. Consider the freebody diagram of the 10.0kg box with ms= 0.40 and mk= 0.30 What is the magnitude of the force of friction, Ffr, if FA = 0, 20, 38, 40 N? In the vertical direction there will never be acceleration For FA = 0, no force is applied, the box doesn’t move and Ffr =0. So (FA , Ffr)=(0,0) Physics 253

10. Now consider FA = 20N, the force of static friction will oppose an applied force up to a max of: The applied force is 20N so the box will not move. From Newton’s 2nd Law: So (FA , Ffr)=(20N,20N) The same argument holds for 38N so (FA , Ffr)=(38N,38N) Last consider FA = 40N. This will start the box in motion since it exceeds the maximum force from static friction. Now we need to switch to kinetic friction. In this case: (FA , Ffr)=(40N,29N) Physics 253

11. We return to Newton’s law to calculate the acceleration of the moving box: Let’s collect our points: (FA , Ffr)=(0,0) (FA , Ffr)=(20N,20N) (FA , Ffr)=(38N,38N) (FA , Ffr)=(40N,29N) Physics 253

12. Consider a truck carrying a crate up a 10-degree hill. The coefficient of static friction between the truck bed and the crate is ms = 0.35. Find the maximum acceleration that the truck can reach before the crate begins to slip backward. How do we measure ms? Physics 253

13. The crate will not slip if it’s acceleration is equal to that of the truck. For it to be accelerated some force must act upon it. One of these is certainly the static friction force. This force must be directed upward to keep the crate from sliding backwards. As the acceleration of the truck increases so must the frictional force. Once Ffr(max) = msFN is reached the crate will start sliding. This will be the point at which the maximum acceleration of the crate and truck is obtained. Physics 253

14. Note that the free-body diagram has been drawn for the moment at which slippage occurs. • Concentrating on the x comp.: • But, we can’t go any further without FM. Concentrating on the y-direction. • Substituting this 2nd result into the 1st: Physics 253

15. Note how the mass cancels (this is a hallmark of friction). • Substituting for g, ms, and the trig functions we see: • If we just consider any angle q then, • No acceleration will be tolerated when the maximum acceleration =0. To find the coefficient of static friction just put an object on a plane and tilt it! Physics 253

16. How do we measure ms? Consider the skier at rest accelerating down a hill. If mk=0.10 what is her acceleration and speed after 4.0 s? Physics 253

17. First the Free-body Diagram • Note: • Three forces • Axes require only • one decomposition Physics 253

18. Next Decomposition • FN=+FNj • Ffr=-Ffri= -mkFNi • FG =+mgsinqi - mgcosqj Physics 253

19. Now Apply Newton’s 2nd Law Note that the acceleration doesn’t depend on mass! It’s the same for anyone. Physics 253

20. Well, just consider what happens if the skier is moving at constant velocity: That is, the slope at which the skier moves at constant velocity will give you the coefficient of kinetic friction. This will be true for any system. Note the similarity between ms and mk! What about determining ms? Physics 253

21. A Problem with It All! • Consider a box of mass m1=10.0kg resting on a plane inclined at 37o. The 1st box is connected via a pulley by a perfect cord to a hanging box of mass m2. • If ms = 0.40 what range of masses for m2 will keep the system at rest? • If ms = 0.40 and m1=m2=10.0kg, what is the acceleration of the system? Physics 253

22. M1<<M2 M1>>M2 Physics 253

23. Now Apply Newton’s 2nd Law Physics 253

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27. Summary • We’ve explored Newton’s Laws in a number of situations and with conventional forces. • There is a clear path from force to acceleration and to the equations of motions given by SF=ma • Next we’ll start exploring circular motion and gravity. Physics 253