Physics

1. Physics Chapter 7 Forces and Motion in Two Dimensions

2. Forces and Motion in Two Dimensions • 7.1 Forces in Two Dimensions • 7.2 Projectile Motion • 7.3 Circular Motion • And…Simple Harmonic Motion from Chapter 6!

3. Forces in Two Dimensions • We have already studied a few forces in two dimensions • When dealing with friction acting on an object (parallel), the normal force on the object (perpendicular) also plays a role in friction • In this chapter we will be looking at forces on a object that are at angles other than 90°

4. Equilibrium and the Equilibrant • Equilibrium—when the net forces on an object is equal to zero • This means an object can be stationary or moving but it can’t be _________. • When graphically adding vectors together, if a closed geometric shape is formed, there is no resultant (?) so net force is zero

5. Equilibrium and the Equilibrant If an object is not in equilibrium this means that: • There is a net force on it • When the vectors are added together, there is a resultant • It is accelerating

6. Equilibrium and the Equilibrant • If add a vector to the resultant that is equal and opposite to it, there will be no net force and the object will be in equilibrium • This vector is called the equilibrant • The equilibrant is the vector that puts an object in equilibrium • It is equal and opposite to the resultant of the existing vectors

7. Creating Equilibrium Hanging a Sign • A 168 N sign is equally supported by two ropes that make an angle between them of 135°. What is the tension in the rope? 168 N

8. Creating Equilibrium Hanging a Sign • Is 168 N the sign’s weight, resultant force, and/or equilibrant? 168 N

9. Creating Equilibrium Hanging a Sign • So both ropes together must pull up with a force of 168 N to put the sign into equilibrium 168 N

10. Creating Equilibrium Hanging a Sign • So each ropes together must pull up with a force of 84 N (1682) to put the sign into equilibrium 168 N 84 N

11. Creating Equilibrium Hanging a Sign • So to find the tension in the rope (hypotenuse of a right triangle) use trig! • cos  = a/h so h = a/cos  • H = 84/cos 67.5 • H = 219.5 N • Why was  = 67.5° and not 135°? 84 N

12. Creating Equilibrium Try this….. Hanging a Sign • A 150 N sign is equally supported by two ropes that make an angle between them of 96°. What is the tension in the rope? • Answer: • 112 N

13. Motion Along an Inclined Plane • When an object sits on a flat surface, there are four forces that determine if there is a net force acting on the object. • What are these forces? (five force equation)

14. Motion Along an Inclined Plane • If the object is in equilibrium, what can you say about the relationship between Fg, Fn, Fa and Ff?

15. Motion Along an Inclined Plane • If the object is in equilibrium on a sloping surface, which of the following forces doesn’t change; Fg, Fn, Fa and Ff?

16. Motion Along an Inclined Plane • If the object is in equilibrium on a sloping surface, which of the following forces doesn’t depend on the angle of the slope; Fg, Fn, Fa and Ff?

17. Motion Along an Inclined Plane • The object weight (Fg) is always the same (regardless of the slope of the surface) and it always directed straight down (toward center of the earth).

18. Motion Along an Inclined Plane • Fn, Fa, Ff all change with the angle of the slope

19. Motion Along an Inclined Plane • So lets look how Fg, Fa, Fn, and Ff are related

20. Motion Along an Inclined Plane • When an object sits on an inclined plane it is being pulled into the plane and down the plane by the force of gravity Fg

21. Motion Along an Inclined Plane • Fg is always the hypotenuse of the right force triangle formed Fg

22. Motion Along an Inclined Plane • The force with which the object is pulled into the plane is called the perpendicular force (F) F Fg

23. Motion Along an Inclined Plane • The force with which the object is pulleddown the plane is called the parallel force (F) F Fg F

24. Motion Along an Inclined Plane • These forces are perpendicular and parallel to what? F Fg F

25. Motion Along an Inclined Plane • We could solve for F and F|| using Fg and angle  if we knew what  was. • Angle  is always equal to the slope of the plane F  Fg F

26. Motion Along an Inclined Plane • So for F; cos  = a/h • Since a = F and h = Fg • F = Fg cos  F = Fg cos   Fg F

27. Motion Along an Inclined Plane • So for F; sin  = o/h • Since o = F and h = Fg • F = Fg sin  F = Fg cos   Fg F = Fg sin 

28. Motion Along an Inclined Plane • So if gravity pulls the object down into the plane with a force F , which force counteracts it? F = Fg cos   Fg F = Fg sin 

29. Motion Along an Inclined Plane • So F , is equal and opposite Fn Fn F = Fg cos   Fg F = Fg sin 

30. Motion Along an Inclined Plane • So if gravity pulls the object down the plane with a force F ,which force counteracts it? F = Fg cos   Fg F = Fg sin 

31. Motion Along an Inclined Plane • So F is equal and opposite to Ff • If there is no motion (or acceleration)! Ff F = Fg cos   Fg F = Fg sin 

32. Motion Along an Inclined Plane • So F , is equal and opposite Fn • and F is equal and opposite to Ff • Only if there is no ________or_______.

33. Motion Along an Inclined Plane Example • A trunk weighing 562 N is resting on a plane inclined 30°. Find the perpendicular and parallel components of its weight. • Answers F = - 487 N F = - 281 N • Why are they negative? • What other forces are they equal to? • Would the force change if the object was moving?

34. Motion Along an Inclined Plane • Under what set of circumstances could the object be accelerating down the plane? • F > Ff • Slope too slippery—not enough friction to hold it • We could be pushing or pulling the object • What can we say about the net force on the object?

35. Motion Along an Inclined Plane What is the Five Forces Equation? • Fnet = Fa + Ff + Fg + Fn • Lets see how it can be modified to deal with inclined planes

36. Motion Along an Inclined Plane Fnet = Fa + Ff + Fg + Fn • Does Fn determine if the object will accelerate down the plane? • So we get rid of it • Fnet = Fa + Ff + Fg

37. Motion Along an Inclined Plane Fnet = Fa + Ff + Fg • What are the two components of weight for an object on the plane? • We replace weight with these two • Fnet = Fa + Ff + F + F

38. Motion Along an Inclined Plane Fnet = Fa + Ff + F + F • Does F determine if the object accelerates down the plane? • Get rid of it • Fnet = Fa + Ff+ F

39. Motion Along an Inclined Plane Fnet = Fa + Ff+ F • Can we still push or pull the object? • Does friction still act on it? • So this is the new Four Forces Equation for inclined planes Fnet = Fa + Ff+ F

40. Motion Along an Inclined Plane Example • A trunk weighing 562 N is resting on a plane inclined 30°. Find the perpendicular and parallel components of its weight. If the force of friction is 200 N, find the trunk’s acceleration rate. • Answers F = - 487 N F = - 281 N a = -1.41 m/s2

41. Projectile Motion Projectile—a launched object that moves through air only under the force of gravity • Ignoring air resistance! Trajectory—the path of a projectile through space

42. Projectile Motion A projectile has both horizontal and vertical components of its velocity • These components are independent of each other • This is because the force of gravity acts on the vertical component (causing acceleration) but not on the horizontal component (constant velocity). • But one can equal zero!

43. Projectile Motion We will be learning how to solve three types of projectile motion problems: • Dropped Objects • Objects Thrown Horizontally • Objects Launched at an Angle

44. A Dropped Object • When an object is dropped from a height it will fall • It has two components of its velocity, horizontal and vertical and both are equal to zero

45. A Dropped Object • As it falls, it picks up speed (accelerates) in the vertical direction due to the force of gravity • Acceleration due to gravity =? • So when dropped any object will pick up negative vertical velocity at the rate of – 9.8 m/s for each second it falls

46. A Dropped Object • As it falls, its horizontal speed stays constant (why?) • So as an object falls its vertical speed changes but its horizontal speed doesn’t • These two perpendicular components of speed are independent of each other, they have no effect on each other (this is true of all perpendicular vectors)

47. A Dropped Object Example • A 2.5 kg stone is dropped from a cliff 44 m high. How long is it in the air? • Answer 3.0 sec • What is its velocity right before it hits the ground? • Answer - 29.4 m/s

48. An Object Thrown Horizontally • When an object is thrown horizontally, it has horizontal and vertical components to its velocity • The horizontal velocity is constant during the entire time its in the air (why?) • The vertical velocity starts as zero but increases as it falls just as if it were dropped! (why?) • These two components are independent of each other

49. An Object Thrown Horizontally Example • A 2.5 kg stone thrown horizontally at 15 m/s from a cliff 44 m high. • How long is it in the air? • What is its horizontal velocity right before it hits the ground? • What is its vertical velocity right before it hits the ground? • How far from the cliff does it land?

50. An Object Thrown Horizontally Example • A 2.5 kg stone thrown horizontally at 15 m/s from a cliff 44 m high. • How long is it in the air? • Answer 3.0 sec