Last Time: Applications of Newton’s Laws of Motion  Free-Body Diagrams, Tension, Inclined Planes Today:  Forces of Friction: Static and Kinetic HW #3 due tonight, 11:59 p.m. Solutions will appear on Bb after the deadline Recitation Quiz #4 tomorrow

Exam #1 Details : Thursday, September 23, regular class time (75 minutes) Closed-book, closed-note You CAN AND SHOULD bring a calculator Formula Sheet posted on Blackboard Will be provided (not allowed to bring your own copy) Multiple Choice: 5  5 points = 25 points total Short Answer [on concept(s)]: 10 points Problems: 4 problems, 65 points total Bonus Problem: 5 points Possible Total: 105/100

My advice to you on how to study for the exams : Make sure you understand how to do the homework problems and the recitation quizzes, beyond just plugging in numbers into equations. Make sure you understand the concepts AND example problems presented in lecture. If you have difficulties understanding the concepts, consult the textbook for more in-depth descriptions. Study the solutions to the examples worked out in lecture. Physics is best learned with practice! Try working through the various (solved) examples in the textbook, and/or other problems.

Friction Friction: The resistance that an object, moving on a surface, or through a viscous medium such as air or water, encounters as it interacts with its surroundings. Friction is actually quite useful in our everyday lives …

Force of Static Friction You have to tug with sufficient force in order to get the sled to first start moving. A “weak” tug won’t move the sled. Question : But F = ma, so why doesn’t even a very small force cause the sled to accelerate from rest ? Answer : There is a force of static friction which opposes the sled’s initial motion !

Force of Static Friction external applied horizontal force y sled x ground As long as the sled is not moving: As we keep increasing F , fs also increases. Right when the sled is on the verge of slipping (about to start moving) : Once F > fs,max , the sled accelerates in the +x-direction The magnitude of fs is at a maximum: fs,max

Force of Static Friction The magnitude of the force of static friction between any two surfaces can have values : μs : coefficient of static friction [dimensionless, has no units] n : magnitude of the normal force exerted by one surface on the other When an object is on the verge of slipping (about to move) : This condition is called “impending motion”. The inequality, fs < μsn , holds when the applied force F < μsn .

Force of Kinetic Friction Now when the sled is in motion, it still experiences a frictional force. The frictional force of an object in motion is called the force of kinetic friction, fk . motion force of kinetic friction external applied horizontal force y sled x ground When the object is in motion, the magnitude of the force of kinetic friction is, usually, less than the maximum force of static friction:

Force of Kinetic Friction The magnitude of the force of kinetic friction between any two surfaces has the value : μk : coefficient of kinetic friction [dimensionless, has no units] n : magnitude of the normal force exerted by one surface on the other Usually, μk < μs . For this class, the force of kinetic friction does not depend on the object’s velocity.

Putting It All Together applied force F friction force f slips fs,max “static region” “kinetic region” fk = μkn fs = F Direction of friction force opposite the direction of actual motion (kinetic friction) or impending motion (static friction).

Example: 4.11 (p. 103) A block with mass m = 2.50 kg is resting on a ramp. If the coefficient of static friction between the block and ramp is 0.350, what maximum angle θ can the ramp have before the block starts to slide down the ramp?

Example: 4.12 (p. 104) A hockey puck is given an initial speed of 20 m/s. The puck remains on the ice and slides 120 m, during which it slows with constant acceleration until coming to rest. Determine the coefficient of kinetic friction.

Example: 4.53 Find the acceleration of each of the two objects if the coefficient of kinetic friction between the 7-kg object and the ramp is 0.250.

Example: 4.61 A girl slides down a hill on a sled, reaching a level surface at the bottom with a speed of 7.0 m/s. If the coefficient of kinetic friction between the sled and the snow is 0.050, and the girl and sled together weigh 600 N, how far does the sled travel before it comes to rest?

Next Class Exam #1 Study Hard, and Good Luck !!