Transcript Ch 4 Forces in 1D

Forces
In One Dimension
Chapter 4 assignments
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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.gif
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
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 x 2.2 lb
1 kg
m = 131 lb
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.
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
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
• 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
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F ball on Earth = - F Earth on ball = - 1.8 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 10-25 m/s2 toward the
ball
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.