1. Friction
Ffriction = μFNormal

2. Ff = μFN
Ff = friction force (direction is always the opposite of velocity causing it to slow objects down)

3. Ff = μFN
Ff = friction force (direction is always the opposite of velocity causing it to slow objects down)
μ = coefficient of friction (experimentally determined value based on the two surfaces in contact with one another)

4. Ff = μFN
Ff = friction force (direction is always the opposite of velocity causing it to slow objects down)
μ = coefficient of friction (experimentally determined value based on the two surfaces in contact with one another)
FN = normal force (force perpendicular to the surface boundary)

5. Types of Friction
static friction – friction that prevents an object from moving.

6. Types of Friction
static friction – friction that prevents an object from moving.
kinetic (sliding) friction – friction that slows down an object in motion.

7. Types of Friction
static friction – friction that prevents an object from moving.
kinetic (sliding) friction – friction that slows down an object in motion.
static friction is always greater than kinetic friction for the same object.

8. Types of Friction
static friction – friction that prevents an object from moving.
kinetic (sliding) friction – friction that slows down an object in motion.
static friction is always greater than kinetic friction for the same object.
This means that it takes more force to get an object moving than it takes to keep it moving.

9. Friction on Horizontal Surfaces
On a horizontal surface, the normal force will always be equal to the weight of the object.

10. Friction on Horizontal Surfaces
On a horizontal surface, the normal force will always be equal to the weight of the object.
Example: You push a 25.0 kg wooden box across a wooden floor at a constant speed. How much force do you exert on the box?

11. Example: You push a 25.0 kg wooden box across a wooden floor at a constant speed. How much force do you exert on the box?
Start by drawing a free-body diagram.

12. Example: You push a 25.0 kg wooden box across a wooden floor at a constant speed. How much force do you exert on the box?
Start by drawing a free-body diagram.

13. Example: You push a 25.0 kg wooden box across a wooden floor at a constant speed. How much force do you exert on the box?
Start by drawing a free-body diagram.
All of the forces must cancel out because the box is moving at a constant speed.

14. This means that the pushing force must be the same magnitude as the friction force.

15. This means that the pushing force must be the same magnitude as the friction force.
The coefficient of friction can be obtained from the table on p We want the value for kinetic friction since the box is moving.

16. This means that the pushing force must be the same magnitude as the friction force.
The coefficient of friction can be obtained from the table on p We want the value for kinetic friction since the box is moving.
The normal force will be the weight of the box since the surface boundary is horizontal.

17. This means that the pushing force must be the same magnitude as the friction force.
The coefficient of friction can be obtained from the table on p We want the value for kinetic friction since the box is moving.
The normal force will be the weight of the box since the surface boundary is horizontal.
Ff = (0.20)(25.0 kg x 9.80 m/s2) = 49.0 N

18. Calculate the minimum force required to get the box moving from rest in the previous example.

19. Calculate the minimum force required to get the box moving from rest in the previous example.
This time we must use the coefficient of static friction from the table. The normal force remains the same.

20. Calculate the minimum force required to get the box moving from rest in the previous example.
This time we must use the coefficient of static friction from the table. The normal force remains the same.
Ff = (0.50)(25.0 kg x 9.80 m/s2) = 123 N

21. Calculate the minimum force required to get the box moving from rest in the previous example.
This time we must use the coefficient of static friction from the table. The normal force remains the same.
Ff = (0.50)(25.0 kg x 9.80 m/s2) = 123 N
It takes 123 N to get the box moving, but just 49 N to keep it moving.

22. Why do you suppose it is bad to lock the brakes on a moving vehicle?

23. Why do you suppose it is bad to lock the brakes on a moving vehicle?
When the wheel is not spinning, the friction produced is kinetic friction.

24. Why do you suppose it is bad to lock the brakes on a moving vehicle?
When the wheel is not spinning, the friction produced is kinetic friction.
When the wheel is spinning, static friction is taking place.

25. Why do you suppose it is bad to lock the brakes on a moving vehicle?
When the wheel is not spinning, the friction produced is kinetic friction.
When the wheel is spinning, static friction is taking place.
Because static friction is always greater, there is more friction between your tires and the road when the wheels are spinning. This gives you greater control over the vehicle.

26. What pushing force is needed to accelerate the 25
What pushing force is needed to accelerate the 25.0 kg box at a rate of 1.0 m/s2?

27. What pushing force is needed to accelerate the 25
What pushing force is needed to accelerate the 25.0 kg box at a rate of 1.0 m/s2?
Draw another free body diagram.

28. What pushing force is needed to accelerate the 25
What pushing force is needed to accelerate the 25.0 kg box at a rate of 1.0 m/s2?
Draw another free body diagram.

29. What pushing force is needed to accelerate the 25
What pushing force is needed to accelerate the 25.0 kg box at a rate of 1.0 m/s2?
Draw another free body diagram.
To get the box to accelerate, the pushing force has to be greater than the friction force.

30. The net force can be calculated using Newton’s 2nd Law (Fnet = ma)

31. The net force can be calculated using Newton’s 2nd Law (Fnet = ma)
Fnet = (25.0 kg)(1.0 m/s2) = 25 N

32. The net force can be calculated using Newton’s 2nd Law (Fnet = ma)
Fnet = (25.0 kg)(1.0 m/s2) = 25 N
This means that the pushing force is 25 N greater than the friction force.

33. The net force can be calculated using Newton’s 2nd Law (Fnet = ma)
Fnet = (25.0 kg)(1.0 m/s2) = 25 N
This means that the pushing force is 25 N greater than the friction force.
The friction force is always the same for this box on this surface
Ff = (0.20)(25.0 kg x 9.80 m/s2) = 49.0 N

34. The net force can be calculated using Newton’s 2nd Law (Fnet = ma)
Fnet = (25.0 kg)(1.0 m/s2) = 25 N
This means that the pushing force is 25 N greater than the friction force.
The friction force is always the same for this box on this surface
Ff = (0.20)(25.0 kg x 9.80 m/s2) = 49.0 N
The pushing force must be 74.0 N ( ).

35. Homework Read section 5.2 (p.126-130)
Problems beginning on p.128

36. Friction on an incline

37. Friction on an incline
The normal force is no longer the weight of the object.

38. Friction on an incline
The normal force is no longer the weight of the object.
The weight vector can be broken down into 2 components, parallel to the incline and perpendicular to the incline.

39. Friction on an incline
The normal force is no longer the weight of the object.
The weight vector can be broken down into 2 components, parallel to the incline and perpendicular to the incline.

40. Friction on an incline
The normal force is no longer the weight of the object.
The weight vector can be broken down into 2 components, parallel to the incline and perpendicular to the incline.
The perpendicular component is the normal force.

41. Friction on an incline
The normal force is no longer the weight of the object.
The weight vector can be broken down into 2 components, parallel to the incline and perpendicular to the incline.
The perpendicular component is the normal force.
The parallel component represents a pulling force.

42. Calculating normal force
SOH CAH TOA can be used to determine both the normal force and the pulling force.

43. Calculating normal force
SOH CAH TOA can be used to determine both the normal force and the pulling force.
cos21o = FN/weight (w = mg)

44. Calculating normal force
SOH CAH TOA can be used to determine both the normal force and the pulling force.
cos21o = FN/weight (w = mg)
sin21o = Fpull/weight (w = mg)

45. Determine the coefficient of static friction for a metal box on a wooden ramp if the box has a mass of 13.5 kg and begins to slide down the ramp when the incline reaches 38o.

46. Determine the coefficient of static friction for a metal box on a wooden ramp if the box has a mass of 13.5 kg and begins to slide down the ramp when the incline reaches 38o.
Start by drawing a free-body diagram. It is helpful to draw the weight force as 2 component forces (parallel and perpendicular)

47. Determine the coefficient of static friction for a metal box on a wooden ramp if the box has a mass of 13.5 kg and begins to slide down the ramp when the incline reaches 38o.
Start by drawing a free-body diagram. It is helpful to draw the weight force as 2 component forces (parallel and perpendicular)
The box will begin to slide when the friction force and the parallel force are equal.

48. Determine the coefficient of static friction for a metal box on a wooden ramp if the box has a mass of 13.5 kg and begins to slide down the ramp when the incline reaches 38o.
Start by drawing a free-body diagram. It is helpful to draw the weight force as 2 component forces (parallel and perpendicular)
The box will begin to slide when the friction force and the parallel force are equal.
sin38o = (F||)/(132 N)

49. Determine the coefficient of static friction for a metal box on a wooden ramp if the box has a mass of 13.5 kg and begins to slide down the ramp when the incline reaches 38o.
Start by drawing a free-body diagram. It is helpful to draw the weight force as 2 component forces (parallel and perpendicular)
The box will begin to slide when the friction force and the parallel force are equal.
sin38o = (F||)/(132 N)
F|| = 81.3 N

50. F|| = 81.3 N
Next, we need to determine the normal force (perpendicular component) to use in the friction equation.

51. F|| = 81.3 N
Next, we need to determine the normal force (perpendicular component) to use in the friction equation.
cos38o = (FN)/(132 N)

52. F|| = 81.3 N
Next, we need to determine the normal force (perpendicular component) to use in the friction equation.
cos38o = (FN)/(132 N)
FN = 104 N

53. F|| = 81.3 N
Next, we need to determine the normal force (perpendicular component) to use in the friction equation.
cos38o = (FN)/(132 N)
FN = 104 N
Now, we can solve for μ.
81.3 N = μ(104 N)

54. F|| = 81.3 N
Next, we need to determine the normal force (perpendicular component) to use in the friction equation.
cos38o = (FN)/(132 N)
FN = 104 N
Now, we can solve for μ.
81.3 N = μ(104 N)
μ = 0.782