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4-4 everyday forces and
application of Newton’s Laws
Objectives
1.Explain the difference between mass
and weight.
2.Find the direction and magnitude of the
normal force.
3.Describe air resistance as a form of
friction.
4.Use coefficients of friction to calculate
frictional force.
Type of forces
• We are concerned with four types of forces:
•
•
•
•
•
Gravity (weight)
Normal force
Friction force
Tension force
Applied force
Gravity Force (Weight) Fgrav
• The force of gravity is the force with which the earth,
moon, or other massively large object attracts another
object towards itself. By definition, this is the weight
of the object. All objects upon earth experience a force
of gravity that is directed "downward" towards the
center of the earth. The force of gravity on earth is
always equal to the weight of the object as found by
the equation:
• Fgrav = m • g
• where g = 9.81 N/kg (on Earth) and m = mass (in kg)
• Note: g is different at different locations
Comparing Mass and Weight
Weight
• The force of gravity.
• Vector, its direction is
downward.
• W = mg
• The weight of an object
(measured in Newton)
will vary according to
where in the universe
the object is.
Mass
• The mass of an object
refers to the amount of
matter that is contained
by the object;
• Scalar, has no direction
• The mass of an object
(measured in kg) will be
the same no matter where
in the universe that object
is located.
The direction of gravity force is always downward
Fg
Fg
Fg
Fg
Fg
Fg
The slope of weight vs. mass
weight
Weight = mg
Slope is Gravitational
acceleration g
mass
practice
• Complete the following table showing the relationship between mass
and weight.
1.
Object
Mass (kg)
melon
1
apple
Joe Samall
Fred
Weight (N)
0.98
25
980
2. Different masses are hung on a spring scale calibrated in Newtons.
a) The force exerted by gravity on 5 kg = ______ N.
b) The force exerted by gravity on _______ kg = 98 N.
c) The force exerted by gravity on 70 kg = ________ N.
3. When a person diets, is their goal to lose mass or to lose weight?
Explain.
Normal Force (FN )
• The normal force is the support force exerted upon an
object that is in contact with another stable object (usually a
surface).
• The direction of the normal force is perpendicular to the
surface, from the surface toward the object and on the
object.
Practice- indicate FN on each box with an arrow
FN
FN
FN
Fg
Fg
Fg
FN
FN
FN
Fg
Fg
Fg
What is the direction of normal
force?
The magnitude of normal force
• The magnitude of normal force can be
determined using Newton’s Laws.
• Example: a car is moving along a
horizontal road.
Because the car is
F
N
not moving up or
down,
FN  mg
mg
Alert!
FN  Fnet
Normal
Force
Net
Force
Friction Force (Ff)
• The friction force is the force exerted by a surface as an
object moves across it or makes an effort to move
across it.
• Friction results from the two surfaces being pressed
together closely, causing intermolecular attractive forces
between molecules of different surfaces.
• Friction force can be reduced by lubricant such as oil.
The direction and magnitude
of friction force:
– The friction force often opposes the motion
of an object.
– Friction depends upon the nature of the two
surfaces (μ, dimensionless) and upon the
degree to which they are pressed
together(FN) .
Ff = μFN
Kinetic versus Static Friction
• kinetic friction results
when an object moves
across a surface.
Ffrict = μk • Fnorm
• The symbol μ represents
the coefficient of kinetic
friction between the two
surfaces. The coefficient
value is dependent
primarily upon the nature
of the surfaces that are in
contact with each other. It
does not depends on area
of contact, the angle of the
area, or the temperature,
etc.
• Static friction results when
the surfaces of two objects
are at rest relative to one
another and a force exists on
one of the objects to set it
into motion relative to the
other object.
• The static friction force
balances the force that you
exert on the box such that the
stationary box remains at
rest.
• Static friction can change.
Ffrict-static ≤ μs• Fnorm
Alert: which box has more friction?
A and B have the friction.
A
B
Practice- indicate Ff on each box with an arrow
v
FN
Ff
FN
FN
Ff
v
Ff
v
Fg
Fg
Fg
Ff
FN
Ff
Ff
FN
FN
v
v
Fg
v
Fg
Fg
Example
• A 24 kg crate initially at rest on a
horizontal floor requires a 75 N horizontal
force to set it in motion. Find the
coefficient of static friction between the
crate and the floor.
s  0.32
Class work
• Text book page 145 Practice 4C #1-3
• Answers
1. 0.23
2. a. 1.5;
b. 1.3
3. a. 870 N, 670 N;
b. 110 N, 84 N;
c. 1000 N, 500 N;
d. 5N, 2 N
Air Resistance Force (Fair )
• The air resistance is a special type of frictional force that
acts upon objects as they travel through the air. The
force of air resistance is often observed to oppose the
motion of an object. This force will frequently be
neglected due to its negligible magnitude.
Tension Force (FT )
• The tension force is the force that is transmitted through
a string, rope, cable or wire when it is pulled tight by
forces acting from opposite ends. The tension force is
directed along the length of the wire and pulls equally on
the objects on the opposite ends of the wire.
Spring Force (Fspring )
• The spring force is the force exerted by a compressed or
stretched spring upon any object that is attached to it. An
object that compresses or stretches a spring is always
acted upon by a force that restores the object to its rest
or equilibrium position – directed toward equilibrium
position.
Fs  kx
Fs is the force
exerted by the spring
on the object, k is
spring constant and x
is displacement from
equilibrium position
Newton’s Laws Application
1. Drawing Free-Body Diagrams
2. Determining the Net Force
3. Apply Newton’s Laws
Drawing Free-Body Diagrams
• Free-body diagrams are used to show the
relative magnitude and direction of all forces
acting upon an object in a given situation.
• The size of the arrow in a free-body diagram
reflects the magnitude of the force. The arrow
shows the direction that the force is acting.
• Each force arrow in the diagram is labeled to
indicate the exact type of force.
• It is generally customary to draw the force arrow
from the center of the box outward in the
direction that the force is acting.
A block of wood is sitting motionless on a table.
What forces are acting on it?
Normal
FN
Fg  FN
Weight
Fg
Normal Force is a
REACTION
force that any
object exerts
when pushed on
Weight is the force of
gravity
pulling an object
toward the
CENTER OF THE
EARTH
Class work
• Practice on free body diagrams
Finding the External Net Force
and acceleration
• Fnet is the vector sum of all the individual
forces. The three major equations that will
be useful are
– Fnet = m•a,
– Fg = m•g,
– Ff = μ•FN
Example #1
• A man pushes a 50 kilogram crate across a
frictionless surface with a constant force of
100 Newtons.
WhatDraw
isWhat
the
What
What
anormal
free-body
isisisthe
the
thenet
force
crate’s
weight
force
diagram
that
acceleration?
of
on
pushes
the
the
of crate?
the
crate?
on
crate.
the crate?
FN
Fg = mg
Fg = (50 kg)(9.81 m/s2)
Fg = 490.5 N
FA
FN = Fg
FN = 490.5 N
Fg
Fnet will only be
the 100N horizontal
force
a = Fnet / m
a = (100 N) / (50 kg)
a = 2 m/s2
Example #2
• A horse pulls a 500 kilogram sled with a
constant force of 3,000 Newtons. The force of
friction between the sled and the ground is
500 Newtons.
WhatDraw
isWhat
What
the
What
anormal
free-body
isisisthe
the
thenet
sled’s
force
weight
force
diagram
that
acceleration?
of
onpushes
the
the
of sled?
the
sled?
on
sled.
the sled?
Fg = mg
Fg = (500 kg)(9.81 m/s2)
Fg = 4905 N
Ff
FN = Fg
FN = 4905 N
Fg
FN
Fnet = ΣFx
Fnet = 3000 N – 500 N
Fnet = 2500 N
FA
a = Fnet / m
a = (2500 N) / (500 kg)
a = 5 m/s2
Example #3
the object is moving horizontally. Use the diagram to
determine the normal force, the net force, the mass,
and the acceleration of the object. (g =10 m/s2)
80 N
8 kg
5 m/s2 right
40 N right
Example #4
• Edwardo applies a 4.25-N rightward force to a 0.765-kg
book to accelerate it across a tabletop. The coefficient of
friction between the book and the tabletop is 0.410.
Determine the acceleration of the book.
Example #5
• Lee Mealone is sledding with his friends when he
becomes disgruntled by one of his friend's comments.
He exerts a rightward force of 9.13 N on his 4.68-kg sled
to accelerate it across the snow. If the acceleration of the
sled is 0.815 m/s/s, then what is the coefficient of friction
between the sled and the snow?
Class work
• Application of Newton’s Laws – pages 1&2
Free Fall and Air Resistance
Free Fall
Falling with air resistance
• Objects that are said to be
undergoing free fall, are
• not encountering air
resistance;
• falling under the sole
influence of gravity. All
objects will fall with the
same rate of acceleration,
regardless of their mass.
This is due to that the
acceleration is The ratio of
force to mass (Fnet/m)
• As an object falls through air, it
usually encounters some degree of
air resistance - the result of collisions
of the object's leading surface with air
molecules.
• The two most common factors that
have a direct affect upon the amount
of air resistance are
– the speed of the object: Increased
speeds result in an increased
amount of air resistance.
– the cross-sectional area of the
object: Increased cross-sectional
areas result in an increased
amount of air resistance.
Falling with air resistance – terminal velocity
• As an object falls, it picks up speed. The increase in
speed leads to an increase in the amount of air
resistance. Eventually, the force of air resistance
becomes large enough to balances the force of gravity.
At this instant in time, the net force is 0 Newton; the
object will stop accelerating. The object is said to have
reached a terminal velocity.
Net Force Problems in 2D
• When forces acting at angles to
the horizontal, Newton’s 2nd law
still applies:
∑F = ma
• Force is a vector quantity.
Adding forces in 2 dimensions
follows the rules for adding
vectors.
Determine the Fnet mathematically
1. Resolve the vectors at an angle into x
and y components.
2. Add all the x components together
3. Add all the y components together –
usually the
4. Use Pythagorean Theorem to find the
resultant (hypotenuse)
Resultant2 = x2 + y2
5. Use trigonometric function to determine
the direction: tanθ = opp / adj
Example 1 - Pulling on an Angle
A block of 10 kg mass is pushed
along a frictionless, horizontal
surface with a force of 100 N at an
angle of 30° above horizontal.
FN
FAY
FA
30˚
This applied force (FA)
FAxcan
= 100cos(30
) = 87 N
be broken ointo
COMPONENTS
FAy = 100sin(30o) = 50 N
X verticalYforce must
The total
be 0, so
FAY
RyFAX
= FN + FAY
–Fg = 0
FN = Fg F–g FAY
FN
FAX
Fg
Total =R
FAX
= Rx Total
= Fax= 0
Acceleration depends only on
FAX
Example 2
• A man pulls a 40 kilogram crate across a
smooth, frictionless floor with a force of 20 N
that is 45˚ above horizontal.
What is the net force on the sled?
How could the
Fnetacceleration
= FA cos θ be increased?
Fnet = (20
N)(cos
45°) F greater and
Pushing at a smaller
angle
will make
net
Fnetincrease
= 14.14acceleration.
N
therefore
What is the crate’s acceleration?
a = Fnet / m
a = (14.14 N) / (40 kg)
a = 0.35 m/s2
Pushing on an Angle
A block is pushed along a
frictionless, horizontal surface with a
force of 100 newtons at an angle of
30° below horizontal.
This applied force (FA)
canXbe broken
Y into
COMPONENTS
FAX
F
FN
FAX
FAY
The total verticalg force must
N
be 0, Fso
= FgTotal
+ FAY
Total F
=N
FAX
=0
-30˚
FA
Fg
FAY
Acceleration depends only on
FAX
Example 3
• A girl pushes a 30 kilogram lawnmower
with a force of 15 Newtons at an angle of
60˚ below horizontal.
Assuming there is no friction, what is the
acceleration of the lawnmower?
Fnet = FA cos θ
Fnet = (15 N)(cos 60°)
Fnet = 7.5 N
a = Fnet / m
a = (7.5 N) / (30 kg)
a = 0.25 m/s2
What could she do to reduce her acceleration?
Push at an greater angle
Example 4 – determine Fnet and a
•
•
•
•
The vertical forces are balanced (Fgrav, Fy, and Fnorm add up to 0 N),
The horizontal forces add up to 29.3 N, right
The net force is 29.3 N, right
a = Fnet / m = 29.3 N / 10 kg = 2.93 m/s2, right
Sample problem 4D
• A student moves a box of books by attaching a rope to
the box and pulling with a force of 90.0 N at an angle of
30.0 degrees. The box of books has a mass of 20.0 kg,
and the coefficient of kinetic friction between the bottom
of the box and the sidewalk is 0.50. find the acceleration
of the box.
Class work
• Newton’s 2nd law practice – page 3 & 4
Equilibrium and Static
• When all the forces that act upon an object are
balanced, then the object is said to be in a state of
equilibrium.
• An object at equilibrium is either ...
– at rest and staying at rest, or
– in motion and continuing in motion with the same
speed and direction.
• "static equilibrium." refers to an object at rest
Example
• A frame is shown with the given tension. Determine the
weight of the frame.
Ax = 50cos(150o) = -43 N
Ay = 50sin(150o) = 25 N
Bx = 50cos(30o) = 43 N
A
B
30o
C=?
By = 50sin(30o) = 25 N
Rx = Ax + Bx + Cx = 0
-43 N + 43 N – Cx =0
Cx = 0
C2 =
Cx2+
R = 50. N
Cy2
Ry = Ay + By + Cy = 0
25 N + 25 N + Cy = 0
Cy = -50 N
example
• A sign is shown with the given mass of 5 kg.
Determine the tension of each cable. (g = 10 m/s2)
Tsin140o
Tsin40o
A=T
Tcos140o
B=T
40o
40o
Ry = Ay + By + Cy = 0
Tsin40o + Tsin140o – Fg = 0
T = 38 N
C = Fg = 50 N
Tcos40o
An important principle
• As the angle with the horizontal increases, the amount of
tensional force required to hold the sign at equilibrium
decreases.
Fg = 10 N
• A tool used to move objects from one height to
another.
• Allows for the movement of an object without
lifting it directly against gravity.
Down the Slope
• The object accelerate
downward due to the
component gravity that is
parallel to the plane.
Fg on Inclined Plane
Forces on an Incline Calculations
• Consider forces:
– Perpendicular
• F┴ = Fg cos θ
• Cancel out Normal (FN )
– Parallel
• F// = Fg sin θ
• Could be in the same or
opposite of Friction (Ff )
θ
Tilt you head method
θ
Essential Knowledge
• What happens to the component of weight that is
perpendicular to the plane as the angle is increased?
Decreases – Fg perpendicular
• What happens to the component of weight that
points ALONG the plane as the angle is increased?
Increases – Fg parallel
• What happens to the normal force as the angle is
increased?
Decreases – depends on Fg perpendicular
• What happens to the friction force as the angle is
increased?
Decreases – depends on normal force
• The net force is the vector sum of all the
forces.
– All the perpendicular components (including
the normal force) add to 0 N.
– All the parallel components (including the
friction force) add together to yield the net
force. Which should directed along the
incline.
In the absence of friction
Fnet = F//
mgsinθ = ma
a = gsinθ
With friction - Object is at equilibrium – at
rest or moving with constant velocity
Ff
Horizontal:
Fnet = 0
F// = Ff
Vertical:
mgsinθ = μFN
F┴ = FN
mgsinθ = μ∙mgcosθ
mgcosθ = FN
tanθ = μ
Example
Fg = 50N
30°
• What is the magnitude of the normal force?
FN = Fg perpendicular = Fg cos θ = 43.3 N
• If the box is sliding with a constant velocity,
what is the magnitude of the friction force?
Ff = Fg parallel = Fg sin θ = 25 N
Example 1
• The free-body diagram shows the forces acting upon a 100kg crate that is sliding down an inclined plane. The plane is
inclined at an angle of 30 degrees. The coefficient of friction
between the crate and the incline is 0.3. Determine the net
force and acceleration of the crate.
F┴ = Fgrav∙cos30o = 850 N
F// = Fgrav∙sin30o = 500 N
In perpendicular direction:
Fnorm = F┴ = 850 N
In parallel direction:
Fnet = F// - Ff
Fnet = 500 N - µFnorm
Fnet = 235 N
a = Fnet / m = 2.35 m/s2
practice
Class work
Lab 10 – weight vs. mass
• Purpose: determine the relationship between
mass and weight
• Material: spring scale, known masses
• Data table: include heading in data table
• Data analysis: graph weight vs. mass
• Conclusion:
– What is the relationship between weight and mass?
– Determine the slope and indicate the meaning of the
slope in the graph.
Double Trouble (a.k.a., Two Body
Problems)
• Two body-problems can typically be approached using
one of two basic approaches.
– One approach is the system analysis, the two
objects are considered to be a single object moving
(or accelerating) together as a whole.
– Another approach is the individual object analysis,
either one of the two objects is isolated and
considered as a separate, independent object.
Example - system analysis
• A 5.0-kg and a 10.0-kg box are touching each other. A
45.0-N horizontal force is applied to the 5.0-kg box in
order to accelerate both boxes across the floor. Ignore
friction forces and determine the acceleration of the
boxes and the force acting between the boxes.
m = 15 kg
Fnet = 45 N
a = Fnet / m = 3 m/s2
Example - individual analysis
In vertical direction:
FN = Fg = (5 kg) (9.81 m/s2) = 49 N
In vertical direction:
FN = Fg = (10 kg) (9.81 m/s2) = 98 N
In horizontal direction:
Fnet = Fapp - Fcontact
(5 kg)a = 45 N - Fcontact
In horizontal direction:
Fnet = Fcontact
(10 kg)∙a = Fcontact
5a = 45 – 10a
a = 3 m/s2
Example: system analysis
• A 5.0-kg and a 10.0-kg box are touching each other. A 45.0N horizontal force is applied to the 5.0-kg box in order to
accelerate both boxes across the floor. The coefficient of
kinetic friction is 0.200. Determine the acceleration and the
contact force.
In vertical direction:
FN = Fg = (15 kg) (9.81 m/s2) = 147 N
In horizontal direction:
Fnet = Fapp - Ffrict = 45 N - μ•Fnorm
Fnet = 15.6 N
a = Fnet / m = (15.6 N/15.0 kg) = 1.04 m/s2
However, in order to find the contact force between the
objects, we must make individual analysis.
Example: individual analysis
In vertical direction:
FN = Fg = (10 kg) (9.81 m/s2) = 98 N
In horizontal direction:
Fnet = Fcontact - Ff
(10 kg)∙(1.04 m/s2) = Fcontact - μ•Fnorm
10.4 = Fcontact – (0.2)(9.8)
Fcontact = 8.44 N
Lab 10 – weight vs. mass
• Purpose: determine the relationship between
mass and weight
• Material: spring scale, known masses
• Data table: include heading in data table
• Data analysis: graph weight vs. mass
• Conclusion:
– What is the relationship between weight and mass?
– What does the slope in the graph mean?