1. Forces
In One Dimension

2. Chapter 4 assignments 4.1: 2,3,8,9,10 4.2: 16,17,19 4.3: 28 and
4.1: 2,3,8,9,10
4.2: 16,17,19
4.3: and
42,45,48,52,53,82,84,85

3. In this chapter you will
Use Newton’s laws to solve problems
Determine the magnitude and direction of a net force that causes a change in the motion of an object
Classify forces according to their cause

4. Section 4.1 Force & Motion
Force - a push or pull acting on an object that can cause the object to speed up, slow down, or change direction
Forces have both magnitude and direction - they are ___.
Forces are divided into contact and field forces
Fields: electric, magnetic

5. Free-body diagrams Draw vectors away from objects
Table pushing up on books
Books pushing down on table

6. Determine net force Add forces acting in the same direction
Subtract forces acting in opposite directions.
3n E + 2n W n E
F1 = 3.0 n E F2 = 2.0 n E F net = n E

7. Vectors at Right Angles
Sin q = opp hyp
Cos q = adj hyp
Tan q = opp adj

8. Newton’s 2nd Law a = Fnet / m
The acceleration on an object is equal to the sum of the forces acting on an object divided by the mass of the object.

9. Newton’s 1st Law (Inertia)
An object has a tendency to resist a change in its motion unless there is an outside net force acting on it.
No net force can result in
No motion or constant motion
Known as equilibrium
A net force can result in
Speeding up or slowing down

10. Problem
1. A rock falls freely from a cliff. Draw vectors and label each. +y
V a Fnet

11. A skydiver falling towards earth at a constant rate… +y
F air resistance on diver
F net = 0
F Earth’s mass on diver v
a = 0 v v

12. A rope pulls a box at a constant speed across a horizontal surface
A rope pulls a box at a constant speed across a horizontal surface. The surface provides a force that resists the box’s motion. +x v v v v
F friction on box
F pull on box
F net = 0

13. Problem
Two horizontal forces, 255 n and 184 n, are exerted on a boat. If these forces are applied in the same direction, find the net horizontal force on the boat.
F net = 255n + 184n = 4.39 x 102 n in the direction of the forces.

15. 4.2 Objectives
Describe how the weight and mass of an object are related.
Differentiate between actual weight and apparent weight
Use Newton’s 2nd Law in two forms:
F = ma
F = mg

16. A ball in mid-air in free fall has only the force of gravity acting on it. Air resistance can be neglected.
System V a
Known:
a = g m
Unknown: Fg Fg
Fnet = ma
Fnet = Fg a = g
Therefore:
Fg = mg

17. The only force acting on the falling ball is Fg.
Fg is the weight force.
Fg is acting down as are the velocity and the acceleration
Newton’s 2nd law has become:
Fg = mg

18. How a bathroom scale works.
When you stand on the scale the spring exerts an upward force on you while are in contact with the scale.
You are not accelerating, so the net force acting on you must be zero.
The spring force, Fsp upwards must be opposite and equal to your weight Fg that is acting downward.

19. Newton’s 2nd Law Problem
Two girls are fighting over a stuffed toy (mass = 0.30 kg). Sally (on left) pulls with a force of 10.0 n and Susie pulls right with a force of 11.0 n. What is the horizontal acceleration of the toy?

20. Sally n
M toy = 0.30 kg
Susie 11.0 n

21. Solution to Susie’s & Sally’s dilemma.
In this chapter you will:
Find the net Force:
11.0n R + (-10.0n L)
Fnet = 1.0 n R
Fnet = ma a = Fnet / mw
a = 1.0 n / 0.30 kg = 3.33 m/s2 Right

22. Apparent Weight
The force an object experiences as a result of the contact forces acting on it, giving the object an acceleration

23. Real and Apparent Weight
same when a body is traveling either up or down at a constant rate, in an elevator, for example.
Apparent weight < real weight when the elevator is slowing while rising or speeding up while descending.
Apparent weight > real weight when speeding up while rising or slowing while going down.

24. Apparent weight is
less when…
Fscale
Apparent weight is greater when
Fscale Fg Fg
Slowly rising or speeding up while descending.
Speeding up while rising or slowing while going down

26. Going Up? v = 0 a = 0 v > 0 a > 0 v > 0 a = 0 v > 0W
Wapp
Ground
floor
Normal feeling
v = 0
a = 0 W
Wapp
Just
starting up
Heavy feeling
v > 0
a > 0 W
Wapp
Between
floors
Normal feeling
v > 0
a = 0 W
Wapp
Arriving at
top floor
Light feeling
v > 0
a < 0
Going Up?

27. Going Down? v = 0 a = 0 v < 0 a < 0 v < 0 a = 0 v < 0
Wapp
Top
floor
Normal feeling
v = 0
a = 0 W
Wapp
Beginning
descent
Light feeling
v < 0
a < 0 W
Wapp
Between
floors
Normal feeling
v < 0
a = 0 W
Wapp
Arriving at
Ground floor
Heavy feeling
v < 0
a > 0
Going Down?

28. Turn to page 100
Let’s look at Example Problem 2
Refer to 19 &20

29. Grain is stored in grain elevators like these

30. Problem On Earth, a scale shows that you weigh 585 n.
A. What is your mass?
B. What would the scale read on the Moon where g = 1.60 m/s2?
C. Back on Earth, what do you weigh in pounds? (1 kg = 2.2 kg)

31. A. What is your mass?
m = Fg / g
m = 585 n /9.8 m/s2
m = 59.7 kg

32. B. What would the scale read on the Moon where g = 1.60 m/s2?
Fg = mgmoon
Fg= (59.7 kg)(1.60m/s2)
Fg = 95.5 n

33. Back on Earth…
m = 59.7 kg x lb
1 kg
m = 131 lb

34. Drag Force and Terminal Velocity
When an object moves through a fluid (liquid or gas), the fluid exerts a drag force opposite to the direction of motion of the object.
The force is dependent upon the motion of the object and the properties of the fluid (temperature and viscosity - resistance to flow).
As the object’s velocity increases, so does the drag force. The terminal velocity is the maximum velocity reached by the object as it moves through the fluid.

36. 4.3 Interaction Forces In this section you will :
Define Newton’s Third law
Explain tension in strings and ropes in terms of Newton’s 3rd law
Define the normal force
Determine the value of the normal force by applying Newton’s 2nd law

37. Identifying Interactive forces
You are on skates and so is your friend. You push on their arm to move them forward and they exert an equal and opposite force on you which causes you to move backwards.
These forces are an interaction pair.
An interaction pair (or action and reaction) is two forces that have equal magnitude and act in opposite directions.

38. A B
F A on B
F B on A

39. The forces simply exist together or not at all.
They result from the contact between the two of you.
The two forces act on different objects and are equal and opposite
Numerically, F A on B = - F B on A

40. Practice Problem 32. Someone please read the problem.
Identify the bucket as the system and up as positive.
Fnet = Frope on bucket - FEarth’s mass on bucket = ma
a = (Frope on bucket - FEarth’s mass on bucket)/m
a = (Frope on bucket - mg)/m
a = [450n - (42kg)(9.80m/s 2 )] / 42 kg
a = 0.91 m/s2

41. When a softball of mass 0.18 kg is dropped, its acceleration toward Earth is g. What is the force on the Earth due to the ball and what is Earth’s resulting acceleration? Earth’s mass is 6.0 x 1024 kg.

42. Use Newton’s 2nd and 3rd laws to find a Earth
F Earth on Ball = m ball a
Substitute a = -g
F Earth on Ball = m ball (-g)
Substitute knowns
F Earth on Ball = (0.18kg)(9.8m/s2)
F Earth on Ball = 1.8 n

43. Find Earth’s Acceleration
F ball on Earth = - F Earth on ball = n
a Earth on ball = Fnet/ m Earth
a Earth on ball = 1.8 n / 6.0 x 1024kg
a Earth on ball = 2.9 x m/s2 toward the ball

44. Tension
Tension, the specific name for the force exerted by a string or rope is an interaction force.

45. Normal Force
The perpendicular contact force exerted by a surface on another object.