1. Objectives Investigate the relationship between mass, force, and acceleration. State Newton’s second law and give examples to illustrate the law. Draw an accurate free body diagram locating each of the forces acting on an object or a system of objects. Use free body diagrams and Newton's laws of motion to solve word problems.

2. Warm-up Find the weight in Newtons for: 10 kg = _____ N 100 kg = _____ N 1 kg = _____ N

3. Engagement Activity Toliet Paper roll

4. Everything in the world does one of two things:

5. Or……

6. Demo http://www.youtube.com/watch?v=iwP4he WDhvwhttp://www.youtube.com/watch?v=iwP4he WDhvw Demo with carts (varying force, varying mass)

7. Objective To design two experiments to show the effect mass has on acceleration keeping force constant and the effect force has on acceleration keeping mass constant.

8. Exploring Newton’s Second Law Exploring Forces (discuss how mass is constant in a system) Design two separate data collection strategies to determine how two factors affect the acceleration of the system: the net force on the system and the total mass of the system. (design for homework Period 3)

9. Newton's 2nd law lab due November 10. Lab Report includes: 2 separate experiments in which one experiment changes force keeping mass constant and finding acceleration and the other one changes mass keeping force constant and finding acceleration. -Problem or question -Hypothesis for each -Methods for each - Data Table for each done in Excel if possible - Graph for each done in Excel - Conclusion

10. Activity Computer Activity on Exploring Forces

11. Objectives Investigate the relationship between mass, force, and acceleration. State Newton’s second law and give examples to illustrate the law. Draw an accurate free body diagram locating each of the forces acting on an object or a system of objects. Use free body diagrams and Newton's laws of motion to solve word problems.

12. Newton’s Second Law of Motion The relation between acceleration and force. Acceleration is proportional to force and inversely proportional to mass. Force is vector Acceleration is a vector “sigma” symbol means summation If there is only one force present, then you can leave it out.

13. Newton’s Second law F net = ma (N) = (kg) (m/s 2 ) Force problems always must use these units! If you prefer triangles “net” means “the total sum” or you can use the symbol Σ a F net m

14. Newton’s Second Law Force = Mass x Acceleration Force is measured in Newtons ACCELERATION of GRAVITY(Earth) = 9.8 m/s 2 Weight (force) = mass x gravity (Earth) Moon’s gravity is 1/6 of the Earth’s If you weigh 420 Newtons on earth, what will you weigh on the Moon? 70 Newtons If your mass is 41.5Kg on Earth what is your mass on the Moon?

15. Newton’s Second Law WEIGHT is a measure of the force of ________ on the mass of an object measured in __________ gravity Newtons

16. Newton’s Second Law One rock weighs 5 Newtons. The other rock weighs 0.5 Newtons. How much more force will be required to accelerate the first rock at the same rate as the second rock? Ten times as much

17. Net Force & the 2 nd Law For a while, we’ll only deal with forces that are horizontal or vertical. When forces act in the same line, we can just add or subtract their magnitudes to find the net force. 2 kg 15 N 32 N F net = 27 N to the right a = 13.5 m/s 2 10 N

18. Units F net = m a 1 N = 1 kg m/s 2 The SI unit of force is the Newton. A Newton is about a quarter pound. 1 lb = 4.45 N

19. Graph of F vs. a In the lab various known forces are applied—one at a time, to the same mass— and the corresponding accelerations are measured. The data are plotted. Since F and a are directly proportional, the relationship is linear. F a

20. Slope F a Since slope = rise / run =  F /  a, the slope is equal to the mass. Or, think of y = m x + b, like in algebra class. y corresponds to force, m to mass, x to acceleration, and b (the y-intercept) is zero. FF aa

21. W = mg Weight = mass  acceleration due to gravity. This follows directly from F = m a. Weight is the force of gravity on a body. Near the surface of the Earth, g = 9.8 m/s 2.

22. Video http://science360.gov/obj/video/58e62534- e38d-430b-bfb1-c505e628a2d4/science- nfl-football-newtons-second-law-motionhttp://science360.gov/obj/video/58e62534- e38d-430b-bfb1-c505e628a2d4/science- nfl-football-newtons-second-law-motion

23. A scalar is simply a number, a magnitude alone. A force is usually shown as a vector, which includes both magnitude and a direction. Force (or free-body) diagrams show the relative magnitude and direction of all forces acting upon an object. The object must be isolated and “free” of its surroundings.

24. This is a free-body diagram of the Statue of Liberty. She is represented by a simple box. The forces acting on her are labeled with a magnitude and the arrow shows direction. Notice the surrounding objects are stripped away and the forces acting on the object are shown. 495,704 lb

25. W W here represents the force of the weight of the statue. N N is the normal force, which represents the force Liberty Island is pushing back up on the statue. The island has a great resistance to compression. The ground is exerting a force upward on the statue perpendicular, or normal, to the surface. 495,704 lb N = W =

26. Think of the diagram on an XY plane. If “up” is assumed to be the positive direction, then N is positive and W is negative. lb 495,704 lb N = N = W = W = (Positive y-direction) +y +x (Positive x-direction)

27. The first line of this calculation reads, The sum of the Forcesin the positive y direction WN “The sum of the Forces in the positive y direction is W + N” (  is the Greek symbol for “sum” ) +   F y WN +   F y = W + N - lb+ lb  F y = (-495704 lb) + (+495704 lb )  F y = 0 lb 495,704 lb N = N = W = W = (Positive y-direction) +y +x (Positive x-direction) The sum of the forces in the y is zero. The forces acting on the object cancel each other out.

28. aWe know F = m * a, where “a” is acceleration. a000If a = 0, then F = m * 0 = 0. When  F = 0, the object is not accelerating. We we can then say that the forces acting on the object cancel each other out and it is in a state of static equilibrium.

29. Sitting Gorilla Free Body Diagram of the Sitting Gorilla (The box represents the gorilla, W = weight of the gorilla (Earth’s mass on gorilla), N = Normal force (ground on gorilla) W N Create a free body diagram (FBD) for each of the following situations. Draw a FBD of the gorilla:

30. This is also an acceptable diagram. N W Sitting Gorilla Create a free body diagram (FBD) for each of the following situations. Draw a FBD of the gorilla:

31. Parrot on wooden swing hung by ropes Draw a FBD of the wooden swing: Free Body Diagram of the wooden swing (The box represents the wooden swing, W = weight of the swing (Earth’s mass on swing) and the parrot, T represents the ropes that are in tension supporting the weight) W T2T1

32. Bungee jumping from crane Draw a FBD of bucket the bungee jumper leaped from:

33. Bungee jumping from crane Draw a FBD of bucket the bungee jumper leaped from: Free Body Diagram of the bucket (T represents the tensile force of the cable the bucket is suspended from, and W is the weight of the diver and the bucket or Earth’s mass on diver and bucket) WT

34. Traffic Light supported by cables Draw a FBD of the ring at point C: A B C D

35. Traffic Light supported by cables Draw a FBD of the ring at point C: A B C D Free Body Diagram of the ring at point C (T represents the force of the cables that are in tension acting on the ring) T CA T CD T CB

36. Draw a FBD of the traffic light: Traffic Light supported by cables A B C D

37. Draw a FBD of the traffic light: Free Body Diagram of the traffic light (T CD represents the force of the cables acting on the light and W is the weight acting on the light or Earth’s mass on traffic light) WT CD Traffic Light supported by cables A B C D

38. Pin-Connected Pratt Through Truss Bridge Draw a FBD of the pin at point A: A B E D C

39. Pin-Connected Pratt Through Truss Bridge Draw a FBD of the pin at point A: A B E D C Free Body Diagram of pin A (If you consider the third dimension, then there is an additional force acting on point A into the paper: The force of the beam that connects the front of the bridge to the back of the bridge.) T AE T AC T AB T AD

40. Class work Chapter 4 Questions on page 89 #s 1-5 P.93 6-8 Free body diagram worksheet

41. Homework Chapter 4 Finish classwork. Section Review page 95. 9-14

42. 42 Closure If a loaded truck that can accelerate at 1 m/s 2 loses its load and has three-fourths of the original mass, what acceleration can it attain from the same driving force. Kahoot