1. Warmup

2. Affects on Velocity and Acceleration
Force
Affects on Velocity and Acceleration

3. Force Causes Change in Velocity
Oh! Skeeter!
Ha, ha, good one
This one time at band camp…

4. Force and Motion Newton’s First Law
At rest – Stays at rest (until force is applied)
In motion – Stays in motion (until force is applied)
Force causes change in velocity
Force causes acceleration
Force causes a change in direction

6. Warm-up
A blue one with a nubbin was moving at 5m/s straight down when the problem started. The difference between the bottom and the top is 3 times the height of a trans-atlantic 10m building. This nubbin sporting thing is earthbound. (include units)

7. Types of Fundamental Force
Gravitational Force
Force we use in this section
Electromagnetic Force
Includes the contact forces we work with in this section
Nuclear Force
Weak Force
(summarize)

8. Electromagnetic Force
Contact Forces
Normal
Friction
Static: friction when the object is not in motion
Sliding: friction when the object is in motion
Tension
Spring
(Use article to organize with topic oval)

9. Force Newton’s Third Law
Every action has an opposite and equal reaction.

10. Force
Experiment with force sensors.
equal and opposite

11. Key Vocab Net force (Fnet) Vector addition/subtraction 5 N + 7N
The resultant is the net force
5N N =
12N

13. Newton’s Second Law
Two men pull a 50-kg box with forces 19.7 N and 15.6 N in the directions shown below. Find the net force of the box.
19.7 N N or
-19.7N

14. Force
The pound-force or simply pound (abbreviations: lb, lbf, or lbf) is a unit of force
1N=0.225lb; 1lb=4.45N
Normal plus lift (Fnet)
Weight from force gauge.
Upward force must be greater than gravity to have upward acceleration.
HW:
16,17, (pg 97),
Example problem 2 pg 99 with elevator 3 times with a = 2.50, 3.00, 15.0m/s^2 and t=1.50, 3.25, 3.00s
19, 20, 22, 25 (pg 100 & 101)
Demo

15. Force Newton’s Second Law………...........
Weight is a Force………………
Defined with the universal constant ‘G’.
1 lbf ≈ N

16. Force and Gravity

17. Key Vocab
Normal Force
Force due to gravity and mass is referred to as a normal force or ‘the normal vector’.
Normal vector also refers to a vector that intersects at 90 degrees.
(also stated as: A vector that is
perpendicular to the tangent
line at the interface)

18. Newton’s Second Law
Two men pull a 50-kg box with forces 19.7 N and 15.6 N in the directions shown below. Find the resultant acceleration of the box and the direction in which the box moves.
19.7 N N or
-19.7N

19. Practice -
A large helicopter is used to lift a heat pump to the roof of a new building. The mass of the helicopter is 7.0x10^3 kg and the mass of the heat pump is 1700 kg.
a. How much force must the air exert on the helicopter to lift the heat pump with an acceleration of 1.2 m/s^2?
b. Two chains connected to the load each can withstand 95,000 N. Can the load be safely lifted at 1.2 m/s^2?

20. Air
Drag Force: exerted by a fluid on an object moving through the fluid.
Terminal Velocity: drag force is equal to force of gravity

21. Friction Static friction Ff,s
Friction Coefficients:
Table 5.1 pg 129
Static friction Ff,s
Friction when there is no motion between the objects.
Ff,s <= usFN
Sliding friction (or kinetic friction) Ff,k
Friction when surfaces are rubbing against each other (in motion).
Ff,k = ukFN

22. Friction A wood block on a wood plank. m= kg us = 0.5 uk = 0.2 FN = N
Normal
A wood block on a wood plank.
m= kg
us = 0.5
uk = 0.2
FN = N
Friction

23. Review Elevator Example
Fnet = FE + (-Fg)
Fnet = ma (represents the force on the system as a whole)
FE = Fnet + Fg

24. Tension Force exerted by pulling (usually a string or rope).
For now, consider strings, ropes and pulleys (also called sheaves or blocks) to be massless and frictionless.
Provide a change in direction of force.

25. Tension Problem
The blocks shown are placed on a smooth horizontal surface and connected by a piece of string. If a 8.8-N force is applied to the 8.8-kg block, what is the tension in the string?
6.1N

26. Tension
Three blocks A, B, and C are connected by two massless strings passing over smooth pulleys as shown below, with the 3.4-kg block on a smooth horizontal surface. Calculate the tension in the strings connecting A and B, and B and C.
54N
48N

27. Tension

28. Tension Practice

29. Sketch Problem
68N
40N A B C
68N
40N
ABC

30. Sketch Problem
68N
40N A B C
40N
68N C AB
68N
40N A BC

31. Fnet Fnet = sum of all forces acting on a system = ma
Fnet = Ff,k + FT + Fpush + Fgravity = ma

32. Force Newton’s Third Law
Every action has an opposite and equal reaction.

33. Force
Newton’s Second Law………

34. Force Newton’s first law
When the net forces are zero, an object at rest remains at rest and an object in motion remains in motion in the same direction at the same speed.
For Fnet = 0
velocity is constant: v1 = v2
acceleration is zero

35. Friction Static friction Ff,s
Friction Coefficients:
Table 5.1 pg 129
Static friction Ff,s
Friction when there is no motion between the objects.
Ff,s <= usFN
Sliding friction (or kinetic friction) Ff,k
Friction when surfaces are rubbing against each other (in motion).
Ff,k = ukFN

36. Friction Static friction Ff,s
Friction when there is no motion between the objects.
Ff,s <= usFN
Ff,s <= usmg
10N
5kg

37. Review Elevator Example
Fnet = FE + (-Fg)
Fnet = ma (represents the force on the system as a whole)
FE = Fnet + Fg

38. Setting up the Problem
Connected by a massless string, pulled along a surface with a coefficient of friction
100N

39. Steps What forces are present? What is the value of each force?
Draw the free body diagram
100N

40. Steps
List known variables, solve for unknown

41. Break apart Find the tension in each string
First, identify the forces of each part
100N

42. Consider each section as a system
Draw a free body diagram for this system

43. Consider each section as a system
Draw a free body diagram for this system

44. Break apart
Draw a sketch (free body diagram) for each string

45. Trig Identities SOHCAHTOA Adjacent Leg = Hypotenuse * cos
Opposite Leg = Hypotenuse * sin
Opposite Leg = Adjacent Leg * tan

46. Force components
The magnitude of the force 1 is 87 N, of force 2 is 87 N, and of 3 is 87 N. The angles θ 1 and θ 2 are 60° each. Use the Pythagorean theorem and trig identities to find the resultant of the forces 1, 2 , and 3 .

47. Drag Force
Terminal velocity is when an object in free fall has reached equilibrium. That is, the drag force is equal to the force of gravity.
The Fnet for a system in equilibrium is always zero.

48. Friction with Force components
The system shown below is in equilibrium. Calculate the force of friction acting on the block A. The mass of block A is 7.10 kg and that of block B is 7.30 kg. The angle θ is 48.0°.

49. Break Apart
The system shown below is in equilibrium: This means Fnet=0, since Fnet=ma, then ma=0 and a=0
In order for the net force to be zero on block A (no acceleration) the tension in the rope pulling block A has to be matched by the friction between block A and the surface.
The tension here must be equal to
the friction resisting motion
The friction here must be equal to
the tension pulling A

50. Break Apart There is a force from gravity acting on block B.
Convert this force to its x and y components.
Using the concept of equilibrium, realize that the sum of x components is 0
θ is 48.0°.
7.30kg

51. Two wire tension

52. Familiar Problem
Three blocks A, B, and C are connected by two massless strings passing over smooth pulleys as shown below, with the B block on a smooth horizontal surface.
A=2kg, B=3kg, C=5kg

53. Redraw the problem
Force of gravity on A
Force of gravity on C A B C
The problem may be redrawn to help visualize which direction the forces are acting and which direction is positive.

54. Breaking the problem down
Find the total mass of the system ABC.
Find the net force of the system ABC.
Find the acceleration of the system ABC.
Find the acceleration of the system C.
Find the total mass of the system C.
Calculate the force due to the acceleration of system C.
Calculate the force due to gravity on system C.
Calculate the tension in the strings connecting B and C.

55. Spring (restoring force)
F=-kx
k=constant
x=displacement

56. Force Experiment with force sensors. equal and opposite
normal plus lift (Fnet)
friction (static and sliding)