Newton’s Laws

1. Newton’s Laws Applications

2. Friday, October 21, 2011 Introduction to Friction

3. Friction • Friction is the force that opposes a sliding motion. • Friction is due to microscopic irregularities in even the smoothest of surfaces. • Friction is highly useful. It enables us to walk and drive a car, among other things. • Friction is also dissipative. That means it causes mechanical energy to be converted to heat. We’ll learn more about that later.

4. N (friction) Fpush f W Microscopic View Big view: Surfaces look perfectly smooth. Small view: Microscopic irregularities resist movement. Friction may or may not exist between two surfaces. If it exists, it opposes the direction object “wants” to slide. It is parallel to the surface.

5. Friction depends on the normal force. • The friction that exists between two surfaces is directly proportional to the normal force. • Increasing the normal force increases friction; decreasing the normal force decreases friction. • This has several implications, such as… • Friction on a sloping surface is less than friction on a flat surface (since the normal force is less on a slope). • Increasing weight of an object increases the friction between the object and the surface it is resting on. • Weighting down a car over the drive wheels increases the friction between the drive wheels and the road (which increases the car’s ability to accelerate).

6. Static Friction • This type of friction occurs between two surfaces that are not slipping relative to each other. • fssN • fs : static frictional force (N) • s: coefficient of static friction • N: normal force (N)

7. fs<msN is an inequality! • The fact that the static friction equation is an inequality has important implications. • Static friction between two surfaces is zero unless there is a force trying to make the surfaces slide on one another. • Static friction canincrease as the force trying to push an object increases until it reaches its maximum allowed value as defined by ms. • Once the maximum value of static friction has been exceeded by an applied force, the surfaces begin to slide and the friction is no longer static friction.

8. Force Diagram N Physics W Static friction and applied horizontal force surface There is no static friction since there is no applied horizontal force trying to slide the book on the surface.

9. Force Diagram N fs F Physics W Static friction and applied horizontal force surface Static friction is equal to the applied horizontal force, and there is no movement of the book since SF = 0.

10. Force Diagram N fs F Physics W Static friction and applied horizontal force surface Static friction is at its maximum value! It is still equal to F, but if F increases any more, the book will slide.

11. Force Diagram N Physics W Static friction and applied horizontal force fk F surface Static friction cannot increase any more!The book accelerates to the right. Friction becomes kinetic friction, which is usually a smaller force.

12. fs Static friction on a ramp surface N Physics Without friction, the book will slide down the ramp. If it stays in place, there is sufficient static friction holding it there. W = mg q At maximum angle before the book slides, let’s prove that ms = tanq.

13. Kinetic Friction • This type of friction occurs between surfaces that are slipping past each other. • fk=kN • fk : kinetic frictional force (N) • k: coefficient of kinetic friction • N: normal force (N) • Kinetic friction (sliding friction) is generally less than static friction (motionless friction) for most surfaces.

14. Sample Problem A 10-kg box rests on a ramp that is laying flat. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is 0.30. • What is the maximum horizontal force that can be applied to the box before it begins to slide? • What force is necessary to keep the box sliding at constant velocity?

15. Sample Problem A 10-kg wooden box rests on a ramp that is lying flat. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is 0.30. What is the friction force between the box and ramp if • no force horizontal force is applied to the box? • a 20 N horizontal force is applied to the box? • a 60 N horizontal force is applied to the box?

16. October 2011 Strings and Tension Springs

17. Tension • Tension is a pulling force that arises when a rope, string, or other long thin materialresists being pulled apart without stretching significantly. • Tension always pulls away from a body attached to a rope or string and toward the center of the rope or string.

18. A physical picture of tension Imagine tension to be the internal force preventing a rope or string from being pulled apart. Tension as such arises from the center of the rope or string. It creates an equal and opposite force on objects attached to opposite ends of the rope or string.

19. Tension examples Note that the pulleys shown are magic! They affect the tension in any way, and serve only to bend the line of action of the force.

20. Sample problem • A 1,500 kg crate hangs motionless from a crane cable. What is the tension in the cable? Ignore the mass of the cable. • Suppose the crane accelerates the crate upward at 1.2 m/s2. What is the tension in the cable now?

21. Problem A 10-kg wooden box rests on a wooden ramp. The coefficient of static friction is 0.50, and the coefficient of kinetic friction is 0.30. What is the friction force between the box and ramp if • the ramp is at a 25o angle? • the ramp is at a 45o angle? • what is the acceleration of the box when the ramp is at 45o?

22. Springs (Hooke’s Law) • The magnitude of the force exerted by a spring is proportional to the amount it is stretched. • F = -kx • F: force exerted by the spring (N) • k: force constant of the spring (N/m or N/cm) • x: displacement from equilibrium (unstretched and uncompressed) position (m or cm) • The direction of the force is back toward the equilibrium (or unstretched) position.

23. Sample problem • A 1.50 kg object hangs motionless from a spring with a force constant of k = 250 N/m. How far is the spring stretched from its equilibrium length?

24. Friday, October 23, 2009 Connected Objects Lot’s of Tension Today!

25. Sample problem • A 1.80 kg object is connected to a spring of force constant 120 N/m. How far is the spring stretched if it is used to drag the object across a floor at constant velocity? Assume the coefficient of kinetic friction is 0.60. <NUM 13>

26. Sample problem A 5.0 kg object (m1) is connected to a 10.0 kg object (m2) by a string. If a pulling force F of 20 N is applied to the 5.0 kg object as shown, A) what is the acceleration of the system? <NUM 14> B) what is the tension in the string connecting the objects? (Assume a frictionless surface.) <NUM 15>

27. Gravity A very common accelerating force is gravity. Here is gravity in action. The acceleration is g.

28. Slowing gravity down The pulley lets us use gravity as our accelerating force… but a lot slower than free fall. Acceleration here is a lot lower than g.

29. N T -x m1g T x m2g Magic pulleys on a flat table • Magic pulleys bend the line of action of the force without affecting tension. SF = ma m2g + T – T = (m1 + m2)a a = m2g/(m1+m2) m1 Frictionless table m2

30. Sample problem Mass 1 (10 kg) rests on a frictionless table connected by a string to Mass 2 (5 kg). Find • the acceleration of each block <Num 16>. • the tension in the connecting string <Num 17>. m1 m2

31. Sample problem Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg). Find the minimum coefficient of static friction for which the blocks remain stationary <NUM 18>. m1 m2

32. Monday, October 8, 2007 More with Pulleys

33. Tuesday, October 9, 2007 Pulleys and Ramps -- Together at Last!

34. Sample problem Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg). If ms = 0.30 and mk = 0.20, what is • the acceleration of each block <NUM 23>? • the tension in the connecting string<NUM 24>? m1 m2

35. N T T m1gsinq m1g m1gcosq m2g Magic pulleys on a ramp • It’s a little more complicated when a magic pulley is installed on a ramp. SF = ma m2g -T + T – m1gsinq = (m1+m2)a m2g – m1gsinq = (m1+m2)a a = (m2 – m1sinq)g/(m1+m2) m1 m2 q

36. 10 kg 5 kg 45o Sample problem • Two blocks are connected by a string as shown in the figure. What is the acceleration, assuming there is no friction?

37. N T T m1gsinq m1g m1gcosq 10 kg 5 kg m2g 45o Sample problem - solution SF = ma m2g -T + T – m1gsinq = (m1+m2)a m2g – m1gsinq = (m1+m2)a a = (m2 – m1sinq)g/(m1+m2) a = [(5 – 10sin45o)(9.8)]/15 a = - 1.35 m/s2

38. N T T m1gsinq m1g m1gcosq 10 kg 5 kg m2g 45o Sample problem - solution SF = ma m2g -T + T – m1gsinq = (m1+m2)a m2g – m1gsinq = (m1+m2)a a = (m2 – m1sinq)g/(m1+m2) a = [(5 – 10sin45o)(9.8)]/15 a = - 1.35 m/s2 How would this change if there is friction on the ramp?

39. Thursday, November 3, 2011 Uniform Circular Motion

40. Uniform Circular Motion • An object that moves at uniform speed in a circle of constant radius is said to be in uniform circular motion. • Question: Why is uniform circular motion accelerated motion? • Answer: Although the speed is constant, the velocity is not constant since an object in uniform circular motion is continually changing direction.

41. Centrifugal Force • Question: What is centrifugal force? • Answer: That’s easy. Centrifugal force is the force that flings an object in circular motion outward. Right? • Wrong!Centrifugal force is a myth! • There is no outward directed force in circular motion. To explain why this is the case, let’s review Newton’s 1st Law.

42. Newton’s 1st Law and cars • When a car accelerates forward suddenly, you as a passenger feel as if you are flung backward. • You are in fact NOT flung backward. Your body’s inertia resists acceleration and wants to remain at rest as the car accelerates forward. • When a car brakes suddenly, you as a passenger feel as if you are flung forward. • You are NOT flung forward. Your body’s inertia resists acceleration and wants to remain at constant velocity as the car decelerates.

43. When a car turns • You feel as if you are flung to the outside. You call this apparent, but nonexistent, force “centrifugal force”. • You are NOT flung to the outside. Your inertia resists the inward acceleration andyour body simply wants to keep moving in straight line motion! • As with all other types of acceleration, your body feels as if it is being flung in the opposite direction of the actual acceleration. The force on your body, and the resulting acceleration, actually point inward.

44. Centripetal Acceleration • Centripetal (or center-seeking) acceleration points toward the center of the circle and keeps an object moving in circular motion. • This type of acceleration is at right angles to the velocity. • This type of acceleration doesn’t speed up an object, or slow it down, it just turns the object.

45. ac v Centripetal Acceleration • ac = v2/r • ac: centripetal acceleration in m/s2 • v: tangential speed in m/s • r: radius in meters Centripetal acceleration always points toward center of circle!

46. Centripetal Force • A force responsible for centripetal acceleration is referred to as a centripetal force. • Centripetal force is simply mass times centripetal acceleration. • Fc = m ac • Fc = m v2 / r • Fc: centripetal force in N • v: tangential speed in m/s • r: radius in meters Fc Always toward center of circle!

47. Any force can be centripetal • The name “centripetal” can be applied to any force in situations when that force is causing an object to move in a circle. • You can identify the real force or combination of forces which are causing the centripetal acceleration. • Any kind of force can act as a centripetal force.

48. Static friction As a car makes a turn on a flat road, what is the real identity of the centripetal force?

49. Tension As a weight is tied to a string and spun in a circle, what is the real identity of the centripetal force?

50. Gravity As the moon orbits the Earth, what is the real identity of the centripetal force?