Download Forces and the Laws of Motion

Forces and the Laws of
Motion
Ch 4
FORCES
• The cause of an acceleration or the
change in an object’s velocity
Forces are measured in Newtons
• SI system
• 1N = 1 kg m/s2
1N =(approx)
¼ lb or 1 stick of butter
Types of Forces
1. Contact
– Force that arises from physical contact
of two objects
2. Field
Force that can exist between objects,
even in the absence of physical contact
between objects
Contact Forces
EX: pull on a spring- it stretches
pull on a wagon- the wagon moves
Field Force
Field- region of influence
EX: gravity, attraction and repulsion
between electrical charges, particle
physics
Force Diagrams
The arrows represent forces, which act in
pairs on an object.
Force Diagram
• Vectors- use to identify forces
• Free body diagrams
– Help analyze a situation
used to analyze only the forces affecting the
motion of a single object
– Isolate an object and the forces acting on it
NewtoN’s First Law (1687)
An object at rest remains at rest;
An object in motion continues to remain in
motion;
Unless the object is acted upon by an
unbalanced force
NewtoN’s First Law
• Also called the “ Law of Inertia”
Weight
Weight - mass X gravity W = m g
• Elevation: further away from the center of
the earth gravity decreases
• -9.81m/s2 is an average (round to 10
m/s2)
• Weight is downward vector in force diagrams.
Mass
Mass- the amount of matter in a substance
-same no matter where you go in the
universe
W = m g or m = W / g
HW: Weight vs mass WS
NewtoN’s secoNd Law
• Force = mass
X acceleration
F = ma
EX: F = measured in Newtons
m = measured in kg
a = measured in m/s2
Predicting
• If mass remains the same and the force needs
to be increased then …
F
= m
a
• If the force remains the same and the mass
increases then..
F
= m
a
Vertical Problems
• F Net = F App - W
F Net = ma
F App
W
Applied Force
Fapp is the applied force.
It is the force that you apply to the object
Net Force
• FNET is the net force
• F = ma
is
FNET = ma
• Net force is the total force.
• If the variable is just listed as F it is the net
force.
Vertical Problems
• Example 1: A 5 kg ball is struck upward with a
force of 70 N ( Use g = 10 m/s2). What is the
acceleration?
• W = m g = 5 kg (10 m/s2 ) = 50 N
70 N
20 N
FNET
50N
Vertical Problems – EX:1
FNet = Fapp - W
FNet = 70 N - 50 N
FNet = 20 N
FNet = ma
a = FNet / m
a = 20 N / 5 kg = 4 m/s2
Vertical Problems-EX: 2
EX 2: A 5 kg ball is struck downward with a force of
70 N ( Use g = 10 m/s2). What is the
acceleration?
• W = m g = 5 kg (10 m/s2 )= 50 N
-50 N
-70 N
-120 N
FNET
Vertical #2
FNet = ma
a = FNet / m
a = -120 N / 5 kg = -24 m/s2
-
Vertical Problems-EX: 3
EX 3: A 5 kg ball is struck upward with a force of
40 N ( Use g = 10 m/s2). What is the
acceleration?
W = m g = 5 kg (10 m/s2 )= 50 N
40N
-10 N
FNet
-50 N
Vertical Problems – EX: 3
FNet = Fapp - W
FNet = 40 N - 50 N
FNet = -10 N
FNet = ma
a = FNet / m
a = -10 N / 5 kg = -2 m/s2
Vertical WS
-
Elevator Problems
What would a bathroom scale read on an
elevator?????
Fapp = ma + W
up
a =+
down
a = -
Elevator Problems - EX 1
A 100 kg person is accelerating upward at 3 m / s2.
What would the scale read?
W = mg = 100kg (10 m / s2) = 1000 N
Fapp = ma + W
Fapp = (100kg)(+ 3 m / s2 ) + 1000 N
Fapp = 1300 N
Elevator Problems - EX 2
A 150 kg person is accelerating downward at
2m/s2. What would the scale read?
W = mg = 150kg (10 m / s2) = 1500 N
Fapp = ma + W
Fapp = (150kg)(- 2 m / s2 ) + 1500 N
Fapp = 1200 N
Elevator Problems - EX 3
The elevator cable breaks and a 120 kg person in the
elevator is free falling down the elevator shaft . What
would the scale read?
W = mg = 120kg (10 m / s2) = 1200 N
Fapp = ma + W
Fapp = (120kg)(- 10 m / s2 ) + 1200 N
Fapp = 0 N
“weightLess”
Elevator WS
Forces of Friction
When an object is moved how much resistance
you feel depends on
1. how heavy the object is
2. The surfaces
Forces of Friction
FN
FF
Fapp
W
Forces of friction
Surfaces- look it up in table 2 pg 138
coefficient of static friction ( ms) - ratio of
friction when an object is just starting to
move
Coefficient of kinetic friction ( mk) - ratio
of friction when an object is already
moving and you want to keep it moving
Forces are all measured in Newtons.
Normal is a mathematical term for a line that
is perpendicular to a surface
FN – Normal Force
FF – the Force of Friction or resistance
Forces of Friction Equations
W = mg
FF = FN m
FNet = Fapp - FF
FNet = ma
(a = FNet / m )
Force of Friction- Example
A 10 kg wooden box is pushed from rest
with a force of 120 N on a wooden
surface. What is the acceleration of the
wooden box?
1.Determine the mass and weight.
Mass = 10 kg
Weight = mg = 10 kg ( 10 m/s2) = 100 N
Force of Friction- Example
2.
If the surface is perpendicular to the the object then
the normal force is the same as the weight. The
weight and the normal force need to be the same so
that the box does not move vertically.
100 N
100 N
Force of Friction- Example
3. Find the force of friction by using
FF = m FN = 0.4 (100 N) = 40 N
FN = 100 N
FF = 40 N
Fapp
W = 100 N
Force of Friction- Example
4. Find the net force.
FNet = Fapp - FF = 120 N - 40 N = 80 N
100 N
40 N
120 N
100 N
Force of Friction- Example
5. Calculate the acceleration.
FNet = ma
(a = FNet / m )
a = 80 N / 10 kg = 8 m /s2
8 m /s2
Horizontal WS
Horizontal with an Angle
An object is being moved but the force is
applied at an angle.
FN
Fapp
y
FF
x
W
Horizontal with an Angle
A 10 kg wooden box is pushed from rest
with a force of 120 N at an angle of 30o on
a wooden surface. What is the
acceleration of the wooden box?
1.Determine the mass and weight.
Mass = 10 kg
Weight = mg = 10 kg ( 10 m/s2) = 100 N
Horizontal with an Angle
2. Find the x and y.
120 N (sin 30o ) = y = 60 N
120 N (cos 30o ) = x = 104 N
3. Determine FN
W = FN + y
FN = W – y = 100N – 60 N = 40 N
Horizontal with an Angle
4. Find the force of friction by using
FF = m FN = 0.4 (40 N) = 16 N
5. Find the FNET.
FNET = x - FF = 104 N - 16 N = 88 N
6. Find the acceleration. FNET = ma
a = FNET / m = 88 N / 10 kg = 8.8 m/s2
Horizontal with an angle WS
Incline
Fll
W
Fl
Incline
Sin q = Fll / W
Cos q = F / W
q
q
Incline
A 30o incline has a 5 kg box on it. Find the parallel
and perpendicular forces.
W = mg
W = 5 kg (10 m/s2) = 50 N
50 N (sin 30o) = Fll = 25 N
50 N (cos 30o) = F = 43 N
Incline WS
Newton’s Third Law
For every action there is an equal and opposite
reaction
Forces always act in pairs
Action-Reaction pair
a pair of simultaneous equal but opposite forces
resulting from the interaction of two objects
Action- Reaction Pairs
Bullet being fired is action and the recoil of
the gun is the reaction
Pulling out from a stop sign you go forward
and your head goes backward