1. Forces A box is being pulled across a surface by a horizontal rope. The force: applied by the rope is called _________. resisting the motion is called _________. of gravitational attraction is called _________. applied by the surface perpendicular to the object is called _________.

2. Forces A box is being pulled across a surface by a horizontal rope. The force: applied by the rope is called tension. resisting the motion is called friction. of gravitational attraction is called weight. applied by the surface perpendicular to the object is called the normal force.

3. An Introduction to Forces: Student Learning Goal The student will use free-body diagrams to determine the net force on an object and be able to apply Newton’s 1 st Law to explain the effect of forces acting on objects. (C2.2 and C3.3)

4. FBDs and Newton’s 1 st Law SPH3U

5. Free-Body Diagrams Usually an object will have more than one force acting upon it.

6. Free-Body Diagrams Usually an object will have more than one force acting upon it. A free-body diagram (FBD) shows all the forces acting on an object

7. Free-Body Diagrams Usually an object will have more than one force acting upon it. A free-body diagram (FBD) shows all the forces acting on an object – and only the forces acting on the object.

8. Free-Body Diagrams Usually an object will have more than one force acting upon it. A free-body diagram (FBD) shows all the forces acting on an object – and only the forces acting on the object. A representation of the object is drawn in the centre of the diagram

9. Free-Body Diagrams Usually an object will have more than one force acting upon it. A free-body diagram (FBD) shows all the forces acting on an object – and only the forces acting on the object. A representation of the object is drawn in the centre of the diagram and the forces acting on it are drawn as arrows pointing outwards.

10. Free-Body Diagrams Usually an object will have more than one force acting upon it. A free-body diagram (FBD) shows all the forces acting on an object – and only the forces acting on the object. A representation of the object is drawn in the centre of the diagram and the forces acting on it are drawn as arrows pointing outwards. The arrows must be labelled!

11. FBD: Example 1 A ball is falling downward through the air. Draw a FBD for the ball.

12. FBD: Example 1 A ball is falling downward through the air. Draw a FBD for the ball.

13. FBD: Example 1 A ball is falling downward through the air. Draw a FBD for the ball. FgFg

14. FBD: Example 1 A ball is falling downward through the air. Draw a FBD for the ball. FgFg F air

15. FBD: Example 2 A book is being pushed rightward across a table. Draw a FBD for the book.

16. FBD: Example 2 A book is being pushed rightward across a table. Draw a FBD for the book.

17. FBD: Example 2 A book is being pushed rightward across a table. Draw a FBD for the book. FAFA

18. FBD: Example 2 A book is being pushed rightward across a table. Draw a FBD for the book. FAFA FfFf

19. FBD: Example 2 A book is being pushed rightward across a table. Draw a FBD for the book. FgFg FAFA FfFf

20. FBD: Example 2 A book is being pushed rightward across a table. Draw a FBD for the book. FgFg FNFN FAFA FfFf

21. Net Force FBDs are drawn to help determine the net force (the sum of all forces) acting on an object. Resolve all vertical and horizontal vectors and determine the resultant.

22. FBD: Example 2 with Numbers A book is being pushed across a table with a force of 5 N [right]. The force of friction is 2 N [left], the gravitational force is 10 N [down], and the normal force is 10 N [up]. Find the net force on the book.

23. FBD: Example 2 with Numbers A book is being pushed across a table with a force of 5 N [right]. The force of friction is 2 N [left], the gravitational force is 10 N [down], and the normal force is 10 N [up]. Find the net force on the book. F g = 10 N F A = 5 NF f = 2 N F N = 10 N

24. FBD: Example 2 with Numbers A book is being pushed across a table with a force of 5 N [right]. The force of friction is 2 N [left], the gravitational force is 10 N [down], and the normal force is 10 N [up]. Find the net force on the book. Horizontal forces: F A + F f = 5 N + (- 2 N) = 3 N Vertical forces: F g + F N = (- 10 N) + (10 N) = 0 The net force F net = 3 N [right]

25. Zero Net Force: Example A weightlifter holds a weight above his head by exerting a force of 1.6 kN [up]. The force of gravity acting on the weight is 1.6 kN [down]. Draw a FBD of the weight. What is the net force on the weight?

26. Zero Net Force: Example F A =1.6 kN F g =1.6 kN

27. Zero Net Force: Example The net force is zero. F A =1.6 kN F g =1.6 kN

28. Newton’s First Law of Motion If there is no net force acting on an object, the object will remain at rest

29. Newton’s First Law of Motion If there is no net force acting on an object, the object will remain at rest

30. Newton’s First Law of Motion If there is no net force acting on an object, the object will remain at rest or will keep moving at the same constant velocity.

31. Newton’s First Law of Motion If there is no net force acting on an object, the object will remain at rest or will keep moving at the same constant velocity. (Conversely, if an object is at rest or is moving at constant velocity, there is no net force acting upon it.)

32. Inertia This is the principle of inertia, first articulated by Galileo:

34. Newton’s First Law of Motion Question: An object is pushed along at constant velocity by a force of 5 N [left]. What is the force of friction acting on the object?

35. Newton’s First Law of Motion Question: An object is pushed along at constant velocity by a force of 5 N [left]. What is the force of friction acting on the object? If the velocity is constant, there is no net force, so the force of friction must be equal in magnitude and opposite in direction to the applied force:

36. Newton’s First Law of Motion : An object is pushed along at constant velocity by a force of 5 N [left]. What is the force of friction acting on the object? Question: An object is pushed along at constant velocity by a force of 5 N [left]. What is the force of friction acting on the object? If the velocity is constant, there is no net force, so the force of friction must be equal in magnitude and opposite in direction to the applied force: F f = 5 N [right]

37. More Practice Accelerometer Lab Activity Accelerometer Lab Activity Homework: FBDs and Newton’s 1 st Law Homework: FBDs and Newton’s 1 st Law