Download In this chapter you will

Chapter 4:
Forces In One Dimension
Chapter 4 assignments
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

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4.1: 2,3,8,9,10
4.2: 16,17,19
4.3: 28 and
42,45,48,52,53,82,84,85
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
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
Free-body diagrams
 Draw vectors away from objects
Table pushing up on
books
Books pushing down
on table
Determine net force
 Add forces acting in the same
direction
F1 = 3.0 n E
F2 = 2.0 n E
F net = 5.0 n E
 Subtract forces acting in opposite
directions.
 3n E + 2n W
1n E
Vectors at Right Angles
Sin q = opp
hyp
Cos q = adj
hyp
Tan q = opp
adj
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.
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
Problem
 1. A rock falls freely from a cliff. Draw
vectors and label each.
 +y
V
a
Fnet
 A skydiver falling towards earth at a
constant rate…
+y
F air resistance on diver
v
v
v
a=0
F net = 0
F Earth’s mass on diver
 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
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.
http://www.physicsclassroom.
com/Class/newtlaws/u2l2d3.g
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4.2 Using Newton’s Laws
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
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:
Fg
a=g
m
Unknown:
Fg
Fnet = ma
Fnet = Fg
Therefore:
Fg = mg
a=g
 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
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.
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?
Sally
10.0 n
M toy = 0.30 kg
Susie
11.0 n
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
Apparent Weight
 The force an object experiences as a
result of the contact forces acting on
it, giving the object an acceleration
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.
Appar
ent
weight
is
Fscale
less
when
…
Fg
Slowly rising or
speeding up while
descending.
Fscale
Fg
Appare
nt
weight
is
greater
when
Speeding up
while rising or
slowing while
going down
www.ux1.eiu.edu/.../Images/elevator.gif
Going
Up?
v=0
a=0
v>0
a>0
v>0
a=0
v>0
a<0
Heavy feeling
Normal feeling
Normal feeling
Light feeling
Wapp
Wapp
Wapp
Wapp
W
W
W
W
Ground
floor
Just
starting up
Between
floors
Arriving at
top floor
Going
Down?
v<0
a<0
v=0
a=0
v<0
a>0
v<0
a=0
Heavy feeling
Normal feeling
Normal feeling
Light feeling
Wapp
Wapp
Wapp
Wapp
W
W
W
W
Beginning
descent
Between Arriving at
floors Ground floor
Top
floor
Turn to page 100
 Let’s look at Example Problem 2
 Refer to 19 &20
Grain is stored in grain
elevators like these
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)
A. What is your mass?
 m = Fg / g
 m = 585 n /9.8 m/s2
 m = 59.7 kg
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
Back on Earth…
m = 59.7 kg
m = 131 lb
x
2.2 lb
1 kg
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
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
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.
A
F B on A
B
F A on B
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
 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.
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
Find Earth’s Acceleration




F ball on Earth
a Earth on ball
a Earth on ball
a Earth on ball
the ball
=
=
=
=
- F Earth on ball = - 1.8 n
Fnet/ m Earth
1.8 n / 6.0 x 1024kg
2.9 x 10-25 m/s2 toward
Tension
 Tension, the specific name for the
force exerted by a string or rope is an
interaction force.
Normal Force
 The perpendicular contact force
exerted by a surface on another
object.