1. Chapter 7-1 Forces and Motion in Two-Dimensions

2. Equilibrium An object is in equilibrium when all the forces on it add up to zero

3. Equilibrant The Equilibrant is the single additional force that, when exerted on an object, will produce equilibrium same magnitude as the resultant but opposite in direction

4. “Statics” Force Problems With “Statics” By definition all forces are balanced Net Force = zero

5. Some Classic Problems:

7. Ah, the classic sign problem! Classic! Heed it well….

9. Practice Problems, continued

10. Motion along an Inclined Plane Please Please Please make sure you understand everything up to this point Now we will be applying what we know about two dimensional vectors You must be very clear on the use of vector components The coordinate system is about to TILT

11. Weight Components This is a very important skill: You will use this in almost all Chap 7 problems—make sure you are clear on how to do this!

12. Let’s go through P1 of The Chapter 7 Worksheet: Analyzing forces in 2-dimensions

13. Let’s analyze how the weight vector can be broken down into two components: a component parallel to the plane (which tends to pull the object down the plane) and a component perpendicular to the plane (the magnitude of this component equals the Normal Force) (#32) A 10 kg object is sitting on an inclined plane which makes a 25 degree angle to the horizontal. a.What is the weight of the object (down/up)? b.What angle does the weight vector make with a line which is perpendicular to the plane? c.What is the component of the weight force which is perpendicular to the plane (down/up)? d.What is the Normal force that the plane exerts on the object (down/up)? e.What is the component of the weight force which is parallel to the plane (left/right)? f. How much force is acting on the object, tending to pull it down the plane (left/right)?

14. Important to note that the thetas shown are equal because of similar triangles—we will use this informa- tion a lot.

15. Multiple forces on an angle It’s starting to get interesting now!

16. These are the variables we will be using (look on the back of your yellow packet)

17. Type 1: Object not moving or sliding down the incline

18. Type 1 In this class of problems, The object is Not moving or Moving at constant velocity The net force = zero The acceleration = 0 Friction force = Parallel Fg Normal force = Perpendicular Fg (#37) (MP #3, #7) If an object weighs 10 N and is placed on an inclined plane tilted at 30 degrees to the horizontal (the object is not moving): The angle between the weight force vector and a line perpendicular to the plane is _____ a. What is the component of the weight force parallel to the plane(left, right)? b. What is the component of the weight force perpendicular to the plane (up/down)? c. What is the Normal force (down/up)? d. What is the Friction Force (right/left)? e.μ = f. If the object is now moving at constant velocity down the plane, what of the above forces has changed and why?

19. Type 2 In this class of problem: There is acceleration DOWN the inclined plane. Net Force = ma Parallel Fg is greater than Friction Force Normal force = Perpendicular Fg (#38) (P153 Ex Prob) An object weighing 45 N is accelerating under its own weight at 2 m/s 2 down an inclined plane set at 15 degrees to the horizontal. a. What is the component of the weight force parallel to the plane (right/left)? b. What is the component of the weight force perpendicular to the plane (up/down)? c. The mass of the object is d. What is the net Force (right/left)? e. What is the Normal Force (up/down)? f. What is the total Force down the plane? g. What is the Friction Force (right/left)? h. μ = i. What is the total force up the plane?

20. Type 3 In this class of problem, there is a force up the plane (F h in the drawing) which just balances the parallel component of the weight force and the friction force. (#36) (MP #4, #8) An object weighing 5 N is being pulled up an inclined plane set at 10◦ to the horizontal. The object is not moving. The coefficient of friction is 0.1. a. What is the Normal Force (up/down)? b. What is the Friction force (right/left)? c. What is the parallel component of weight force (left/right)? d.What is the force pulling or pushing up the plane on the object? e What is the net force? f…What is the acceleration?

21. Type 4 In this class of problem, there is a force up the plane which is greater than the parallel component of the weight force plus the friction force. There is a net force up the plane and an acceleration up the plane An object weighing 5 N is being pulled up an inclined plane set at 10 degrees to the horizontal with a force of 10N. The coefficient of friction is 0.1. a. What is the Normal Force (up/down)? b. What is the Friction force (right/left)? c. What is the parallel component of weight force (right/left)? d. What is the force pulling up the plane on the object? e What is the net force (right/left)? f..What is the mass? g..What is the acceleration (right/left)?

22. Type 5 Another class of 2-dimensional problem: The object is on the horizontal but it is being pulled by a rope on an angle. Normal Force is less than Fg Because the y-component of the pulling Force, in effect, lightens the object If the object is not moving or is moving at constant velocity, then Friction Force = x-component of pulling Force (#35) (MP #5, #6) A 4 kg object is being pulled at constant velocity across a horizontal surface with a force of 10 N by a rope which is at a 37 degree angle to the horizontal. a. What is the object’s weight (up/down)? b. What is the vertical (y-) component of the pulling force (up/down)? c. What is the normal force (up/down)? d. What is the Friction Force (right/left)? e.What is the coefficient of friction? f…What is the Net Force? g. What is the acceleration?

23. Type 6 A variation on the previous scenario: If the object is accelerating, then Friction Force is less than the x- component of the pulling force and the object accelerates in the direction of motion, in the direction of the Net Force. A 4 kg object is being pulled across a horizontal surface with a force of 10 N by a rope which is at a 37 degree angle to the horizontal. The coefficient of friction is 0.1 a. What is the object’s weight (up/down)? b. What is the vertical (y-) component of the pulling force (up/down)? c. What is the normal force (up/down)? d. What is the Friction Force (right/left)? e. What is the Net Force (right/left)? f. What is the acceleration (right/left)?

24. Let’s consider some Classic Problems:

25. The skier—a classic problem—anything (a box, a skier) accelerating down an incline with friction (Type 3) Here’s the vector representation Here’s the Picture: The skier is accelerating down The slope

26. Strategy FREAK OUT!!!! NO, NO, NO---- Resolve all forces Parallel to the plane (all X forces) Perpendicular to the plane (all Y forces) List of variables: m = 62 kg θ = 37◦ μ = 0.15 v 0 = 0.0 m/s t = 5.0 s

27. Since you need the Normal Force, F N before you Can calc the Friction Force, F f Resolve the weight vector, F g first [The Normal Force, F N is the Y-component of Fg Another way of saying that is that F N is the component of the weight which is perpendicular to the plane] Now calc the Friction Force, F f, using F N that you just calculated List of variables: m = 62 kg θ = 37◦ μ = 0.15 v 0 = 0.0 m/s t = 5.0 s

28. We are no longer interested in the Y- components of anything. All of our interest is in resolving forces on the X-axis. We just calculated the only force that points up the X-axis: The Friction Force. Two things point down the X-axis: The X-component of the weight, Fg in the X-direction The ma which is the Net Force accelerating the skiier downhill Find the Net Force (=ma) by taking the difference between F f and F gx : Find the acceleration by dividing the Net Force by the mass of the skier List of variables: m = 62 kg θ = 37◦ μ = 0.15 v 0 = 0.0 m/s t = 5.0 s

29. Calculate the velocity using v = v 0 +`at List of variables: m = 62 kg θ = 37◦ μ = 0.15 v 0 = 0.0 m/s t = 5.0 s

30. Other Classic Problems: Tension (Force) applied horizontally to an object whose weight points straight down:

31. Other Classic Problems: A rope attached on an angle to an object that is being pulled horizontally (Type 1): Ans: 51 ◦,140N,0.19

32. Other Classic Problems: (Type 4) Object pulled up a plane: MP,p84: A crate weighing 1800 N is pulled along a floor by a force of 990 N. The floor is inclined 15◦ to the horizontal and th crate moves with constant velocity. What is the coefficient of sliding friction between the crate and the floor? Ans. 0.31

34. End Chapter 7-1