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FORCES
 Inertia,
Mass, Weight
 Newton’s Law
 Types of Forces
 Free-body diagram
 Friction
FORCE
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Push/pull
Causes an object to accelerate or change its
velocity.
Net force: sum of all forces
Two types of forces:
a) Contact forces
b) Field forces
•
•
•
•
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Gravitational …objects
Electromagnetic…Charges
Nuclear…subatomic particle
Weak…radioactive decay
Vector
Inertia, Mass, and Weight

Inertia
 Resistance to change in its state of
motion
 Depends on mass

Example
• Pushing heavy things
• When moving…difficult to stop
• Seatbelts

Galileo and inertia
• Incline plane: initial height = final height
Inertia, Mass, and Weight
Mass
 Amount of matter it has
 Measures inertia
 Units: kg
 Remains constant
 Weight
 Changes with gravity
 Measures force
 Units: Newton
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Inertia, Mass, and Weight
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Examples
Mac and Tosh are arguing in the cafeteria. Mac says that if he
throws his jello with a greater speed it will have a greater
inertia. Tosh argues that inertia does not depend upon speed,
but rather upon mass. With whom do you agree? Why?
If you were in a weightless environment in space, would it
require a force to set an object in motion?
Mr. Wegley spends most Sunday afternoons at rest on the
sofa, watching pro football games and consuming large
quantities of food. What effect (if any) does this practice have
upon his inertia? Explain.
Ben Tooclose is being chased through the woods by a bull
moose which he was attempting to photograph. The enormous
mass of the bull moose is extremely intimidating. Yet, if Ben
makes a zigzag pattern through the woods, he will be able to
use the large mass of the moose to his own advantage.
Explain this in terms of inertia and Newton's first law of
motion.
Newton’s First Law
Law of inertia
 Two parts
 Object at rest, remains at rest
 Object in motion, remains in
motion
 Unless an unbalance force acts
upon it
 No force needed to keep an object in
motion
 Resists acceleration

Newton’s First Law
 Contradicted:
 Aristotle:
all objects have
natural tendency come to rest
 Newton: all objects come to
rest due to friction; w/o
friction object continues to
move
 Balanced and unbalanced forces
Newton’s First Law
http://www.physicsclassroom.com/Class/newtlaws/U2L1a.html
Newton’s First Law
Newton’s Second Law
• A net force or unbalanced forces
causes an object to accelerate in the
direction of the force.
• Acceleration is directly proportional to
the force
Newton’s Second Law
• Acceleration is inversely proportional
to the mass
• Fnet = ma
• Units: Newton (N) = kg·m / s2
• Vector sum of all forces
• FORCES CAUSE ACCELERATION!!
Newton’s Second Law
Equilibrium means “Zero” acceleration
Newton’s Second Law
Free Fall
 All objects will fall with the same
rate of acceleration, regardless of
their mass
 Air Resistance

Decreases the acceleration
 Depends on cross-sectional area and
speed


Terminal Velocity

Weight equals force due to air
resistance; a = 0 m/s/s
Newton’s Second Law
NEWTON’S SECOND LAW
NEWTON’S SECOND LAW
Problems: Pg 92
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1) When a shot-putter exerts a net force of
140 N on a shot, the shot has an
acceleration of 19 m/s/s. What is the mass
of the shot?
7.4 kg
2) Together a motorbike and rider have a
mass of 275 kg. The motorbike is slowed
down with an acceleration of –4.50 m/s/s.
What is the net force on the motorbike?
Describe the direction of this force and the
meaning of the negative sign.
- 1.24 x 103 N; direction of force is in
opposite direction of the velocity.
Problems: Pg 92
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3) A car, mass 1225 kg, traveling at 105
km/h, slows to a stop in 53 m. What is the
size and direction of the force that acted on
the car? What provided the force?
- 9.8 x 103 N; road surface pushing against
the car tires
4) Imagine a spider with mass 7.0 x 10-5 kg
moving downward on its thread. The
thread exerts a force that results in a net
upward force on the spider of 1.2 x 10-4 N.
a) What is the acceleration of the spider?
b) Explain the sign of the velocity and
describe in words how the thread changes
the velocity of the spider.
1.7 m/s/s
Newton’s Third Law
Two different objects
 Action-reaction pair
 For every action force there is a
equal and opposite reaction force
 F object A on object B = - F object B on object A

Newton’s Third Law
Newton’s Third Law
Newton’s Third Law
Types of Forces

Applied Force (FA)
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Force of Gravity (Fg or W)
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Force exerted by strings, ropes, and wires
Spring Force (Fs)
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Exerted by contact with a surface
Perpendicular to the surface
Tension (T)
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W = mg
Perpendicular to the Earth
Normal or Support Force (Fn or N)
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Applied to an object by another object
Force exerted by compressed or stretch spring
Friction Force (Ff )

Opposes motion of object
FRICTIONAL FORCE
Fluids or surface drag
 Opposes motion
 Two types:
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Static friction…friction at rest
• Ff  sFn
• Maximum value

Kinetic/sliding friction…friction in motion
• Ff = kFn
FRICTIONAL FORCE

Coefficient of static friction ()
Measures how difficult it is to slide one
material over another.
 Dimensionless
 Independent

• surface area
• speed (except when v =0)
• temperature
• roughness/texture

Dependent
• nature of the surfaces (material)
Proportional to normal force
 k < s

FRICTIONAL FORCE
FREE-BODY DIAGRAM
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Diagrams showing all forces that are
found on the object with direction
and magnitude
Arrows start from the center of the
object.
Each arrow is labeled.
FREE-BODY DIAGRAM EXAMPLES
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The force of gravity on a wagon is 230 N. The force
of friction between the tires and the ground is 10 N.
Sheila is pulling the wagon with a force 15 N. Draw
the free body diagram showing the forces on the
wagon.
A book is at rest on a table top.
A girl is suspended motionless from a bar which
hangs from the ceiling by two ropes.
An egg is free-falling from a nest in a tree. Neglect
air resistance.
A flying squirrel is gliding (no wing flaps) from a
tree to the ground at constant velocity. Consider air
resistance.
FREE-BODY DIAGRAM EXAMPLES
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A rightward force is applied to a book in order to
move it across a desk with a rightward acceleration.
Consider frictional forces. Neglect air resistance.
A rightward force is applied to a book in order to
move it across a desk at constant velocity. Consider
frictional forces. Neglect air resistance.
A college student rests a backpack upon his
shoulder. The pack is suspended motionless by one
strap from one shoulder.
A skydiver is descending with a constant velocity.
Consider air resistance.
A force is applied to the right to drag a sled across
loosely-packed snow with a rightward acceleration.
FREE-BODY DIAGRAM EXAMPLES
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A football is moving upwards towards its peak after
having been booted by the punter. Neglect air
resistance.
A car is coasting to the right and slowing down.
FREE-BODY DIAGRAM

Write a Fnet equation

Fx and Fy equations.
Fnet = ma
 Balanced Forces

At rest
 Constant velocity…a = 0 m/s/s
 Fnet = ma = 0
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Unbalanced Forces
a0
 Fnet = ma

Free Body Diagrams Calculations
Fx = Fa – Ff = ma
Fy = Fn – Fg = ma
Free Body Diagrams Calculations
Fx = Ff = ma
Fy = Fn – Fg = ma
Free Body Diagrams Calculations
Fx = Fa - Ff = ma
Fy = Fn – Fg = ma
Free Body Diagrams Calculations
Fy = A – Fg = ma = 0
Free Body Diagrams Calculations
Fx = Fa - A = ma
Fy = Fn – B = ma
Problems: Pg 94
5) What is the weight of each of the
following objects?
a) 0.113 kg hockey puck
b) 108 kg football player
c) 870kg automobile
 1.11 N; 1.06 x 103 N; 8.50 x 103N
 6) Find the mass of each of these
weights.
a) 98 N
b) 80 N
c) 0.98 N
 10 kg; 8.2 kg; 0.10 kg

Problems: Pg 94
7) A 20 N stone rests on a table. What is
the force the table exerts on the stone? In
what direction?
 20 N, upward
 8) An astronaut with mass 75 kg travels to
Mars. What is his weight
a) on Earth
b) What is the weight on Mars where g =
3.8 m/s/s?
c) What is the value of g on top of a
mountain if the astronaut weight 683 N?
 740 N; 290 N; 9.1 m/s/s
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Problems: Pg 99
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9) Suppose Joe, who weighs 625 N, stands on a
bathroom scale calibrated in Newton.
a) What force would the scale exert on Joe? In
what direction?
b) If Joe now holds a 50 N cat in his arms, what
force would the scale exert on him?
c) After Joe puts down the cat, his father comes up
behind him and lifts upward on his elbows with a
72 N force. What force does the scale now exert
on Joe?
625 N, upward; 675 N; 553 N upward
10) A 52 N sled is pulled across a cement sidewalk
at constant speed. A horizontal force of 36 N is
exerted. What is the coefficient of sliding friction
between the sidewalk and the metal runners of the
sled?
0.69
Problems: Pg 100
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10b) Suppose the sled now runs on packed
snow. The coefficient of friction is now only
0.12. If a person weighing 650 N sits on
the sled, what force is needed to slide the
sled across the snow at constant speed?
84 N
11) The coefficient of sliding friction
between rubber tires and wet pavement is
0.50. The brakes are applied to a 750 kg
car traveling 30 m/s, and the car skids to a
stop.
a) What is the size and direction of the
force of friction that the road exerts on the
car?
-3.7 x 103 N
Problems: Pg 100
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11b) What would be the size and direction of
the acceleration of the car? Why would it be
constant?
c) How far would the car travel before
stopping?
-4.9 m/s/s; frictional force is constant; 92 m
12) If the tires of the car in Practice Problem
11 did not skid, the coefficient of friction
would have been 0.70. Would the force of
friction have been larger, smaller, or the
same? Would the car have come to a stop
in a shorter, the same, or a longer distance?
Larger; shorter distance
Problems: Pg 102
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13) A rubber ball weighs 49 N.
a) What is the mass of the ball?
b) What is the acceleration of he ball if an
upward force of 69 N is applied?
5.0 kg; 4.0 m/s/s
14) A small weather rocket weighs 14.7 N.
a) What is its mass?
b) The rocket is carried up by a balloon.
The rocket is released from the balloon and
fired, but its engine exerts an upward force
of 10.2 N. What is the acceleration of the
rocket?
1.5 kg; -3.00 m/s/s
Problems: Pg 102
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15) The space shuttle has a mass of 2.0 x
106 kg. At lift-off the engines generate an
upward force of 30 x 106 N.
a) What is the weight of the shuttle?
b) What is the acceleration of the shuttle
when launched?
c) The average acceleration of the shuttle
during its 10 minute launch is 13 m/s/s.
What velocity does it attain?
d) As the space shuttle engines burn, the
mass of the fuel becomes the same, would
you expect the acceleration to increase,
decrease or remain the same? Why?
2 x 107 N; 5.0 m/s/s; 7.8 km/s; increase
Problems: Pg 102
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16) A certain sports car accelerates
from 0 to 60 mph in 9.0 s (average
acceleration = 3.0 m/s/s). The mass
of the car is 1354 kg. The average
backward force due to air drag during
acceleration is 280 N. Find the
forward force required to give the car
this acceleration.
4.4 x 103 N
Free Body Diagrams Calculations
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A rightward force is applied to a 6-kg
object to move it across a rough surface at
constant velocity. The object encounters 15
N of frictional force. Use the diagram to
determine the gravitational force, normal
force, net force, and applied force.
(Neglect air resistance.)
A rightward force is applied to a 10-kg
object to move it across a rough surface at
constant velocity. The coefficient of friction,
µ, between the object and the surface is
0.2. Use the diagram to determine the
gravitational force, normal force, applied
force, frictional force, and net force.
Free Body Diagrams Calculations
Free Body Diagrams Calculations
Free Body Diagrams Calculations
A force of 55 N[E] acts on a combination of two
boxes that are next to each other on a
frictionless
surface.
Find:
a. the acceleration of the boxes.
b. the force exerted by box A on box B.
c. the force exerted by box B on box A.
Free Body Diagrams Calculations
Three toboggans are connected by a rope. The first
toboggan has a mass of 75 kg, the second a mass
of 55 kg and the third a mass of 10 kg. If the force
exerted on the first toboggan is 310 N and surfaces
are considered to be
frictionless.
Find:
a. the acceleration of the toboggans
b. the force exerted by toboggan one on toboggan
two.
c. the force exerted by toboggan two on toboggan
three.
Free Body Diagrams Calculations
Free Body Diagrams Calculations
NEXT-TIME QUESTION

As she falls faster and faster through the
air, her acceleration
a) increases
b) decreases
c) remains the same
NEXT-TIME QUESTION
b) decreases
Acceleration decreases because the net force on her
decreases. Net force is equal to her weight minus her air
resistance, and since air resistance increases with increasing
speed, net force and hence acceleration decrease. By
Newton's 2nd law:
where mg is her weight, and R is the air resistance she
encounters. As R increases, a decreases. Note that if she falls
fast enough so that R = mg, a = 0, then with no acceleration
she falls at constant velocity.
NEXT-TIME QUESTION
If an elephant and a feather fall from a high
tree, which one will encounter the greater
force of air resistance while falling to the
ground below?
NEXT-TIME QUESTION
the elephant
A much greater force of air resistance acts on the
elephant simple because it has to "plow through"
more air to get to the ground. So air resistance is
greater on the elephant. The effect of air resistance
is more pronounced for the feather, however,
because air resistance doesn't have to build up very
much to counteract the weight of the feather. A tiny
fraction of one Newton of air resistance acting on a
feather that weighs a tiny fraction of one Newton,
more easily produces a zero net force and zero
acceleration, compared to several Newtons of air
resistance acting on thousands of Newtons of
elephant. Remember to distinguish between a force
itself and the effect it produces!
NEXT-TIME QUESTION
For every force there exists an
equal and opposite force.
Consider action and reaction
forces in the case of a rock
falling under the influence of
gravity. If action is considered
to be that of the Earth pulling
down on the rock, can you
clearly identify the reaction
force?
NEXT-TIME QUESTION
reaction: the falling rock pulling up on the Earth
The recipe for action-reaction forces is simple
enough: if A exerts force on B, then in turn; B
exerts force on A. So if action is the earth
pulling down on the falling rock, reaction is
simply the falling rock pulling up on the Earth.
Does this mean that the acceleration of the
rock and the Earth should be the same? Not at
all, but only because the Earth's mass is so
much greater than that of the falling rock.
NEXT-TIME QUESTION
Two smooth balls of
exactly the same size,
one made of wood and
the other of iron, are
dropped from a high
building to the ground
below. The ball to
encounter the greater
force of air resistance
on the way down is
the
a) wooden ball
b) iron ball
c) both the same
NEXT-TIME QUESTION
b) iron ball
Air resistance depends on both the size and speed
of a falling object. Both balls have the same
size, but the heavier iron ball falls faster
through the air and encounters more air
resistance in its fall.
NEXT-TIME QUESTION
She holds the book
stationary against
the wall as shown.
Friction on the book
by the wall acts
a) upward.
b) downward.
c) can't say
NEXT-TIME QUESTION
c - we can't say unless we know how the
vertical component of her push
compares with the weight of the book,
we can't specify the direction of friction
between the book and the wall.
This defense-minded episode concerns Dr. J's dual role as a physics teacher and a
part-time CIA agent. The United States (represented by Dr. J) and Russia were both
working on a real-life nuclear-powered limb (á la Steve Austin) to be used by
government spies who might have lost or damaged their arms in some way. (You
guessed it - this is the famous nuclear arms race.) Finally, both countries finished
limbs and fitted them to suitable subjects. The Russians tested theirs by having their
agent (Big Fist) throw a ball weighing 490 N straight up into the air. They measured
the initial acceleration of the ball to be 100 m/sec2. Dr. J decided to test his limb in a
different manner. He had his agent push a large safe across the street. In order to
push the safe (which weighed 5000 N), he had to overcome a force of friction of
3500 N. In addition, members of SAP (being against any type of physics) decided to
push against the safe opposite to the bionic subject with a force of 500 N. The bionic
subject still managed to accelerate the safe forward.
At what minimum rate would the safe have to accelerate in order to demonstrate a
superior American force?
a > 2.92 m/s2
The Russian force (FR) would be the sum of the net force required to
accelerate the ball plus the weight of the ball. The mass of the ball can
be found by the fact that: Fw = mag
The net force is found by:
Fnet = ma
Fnet = (50 kg) (100 m/s)
Fnet = 5000 N
Therefore:
FR = 5000 N + 490 N
FR = 5490 N
The American force, FA, must exceed, however slightly, 5490 N. The
acceleration produced by an American force of 5490 N is found as
follows: The mass of the safe is found by:
.
The net force is equal to the American force minus the two opposing forces
provided by friction and the members of SAP.
Fnet = 5490 N - (3500 N + 500 N)
Fnet = 1490 N
The acceleration is then found by:
The minimum rate must exceed this value: a > 2.92 m/s2
This is the sad story of a lover's leap foiled by the fate of physics. The romance
centers around Dr. J's maiden aunt and her so-called boyfriend, Jilting Joe
from Buffalo. Aunt J and Jilting Joe had only known each other for a few
weeks, but she was sure it was a Coca-Cola relationship (the real thing). On
the other hand, Dr. J viewed it as a sour-milk setup (it smelled from the very
beginning). When Jilting Joe proposed marriage, Dr. J feared that he was after
her bank account more than her homely face and lack of grace. Therefore, in
final desperation, Dr. J locked his aunt in her second- story bedroom. Not to
be stopped, Jilting Joe established communication with Aunt J by sending
sines through the window with a flashlight (this guy knew all the angles). She
responded with matching sines (that's right-cosines). Together they decided
on a final plan - they would elope (check the title). Jilting Joe signaled for Aunt
J to make a rope out of hosiery and slide down to his waiting arms. She
couldn't decide which brand to use - Leggs, since they were running away, or
No-Nonsense, since this was nothing to fool around about. She finally decided
that Jilting Joe would prefer Hanes (although we know better). As luck would
have it, she came down too slowly, causing the hosiery rope to break. In the
resulting fall, she fell head over heels for Jilting Joe. However, it proved to be
the final blow to their relationship, as Joe failed to survive.
If Aunt J weighed 750 N, and if the hosiery rope could only withstand a maximum
tension of 600 N, what was the minimum acceleration that Aunt J could have
had in order not to break the rope?
1.96 m/sec2
Aunt J's mass can be determined from her weight by:
If the force of gravity on Aunt J was 750 N downward and the force
of the rope (tension) exerted on Aunt J was 600 N upward, the net
force exerted on Aunt J was 150 N downward. The resulting
acceleration is then found by:
Dr. J is working hard to make Labor Day a memorable holiday for all of his
relatives. It is his turn to host the annual family reunion, and he has
decided to have a picnic in the park. He is especially interested in
providing a good time for all of his nieces and nephews. To insure this, he
assigns someone to be in charge of each area. For example, Tripod will
help with the three-legged race, while Dr. J's henpecked brother-in-law will
barbecue the chicken. Uncle Darwin will supervise play on the monkey
bars while his thrice-divorced sister is a natural choice for the merry-goround. His uncle, a retired carpenter who made a fortune building
shorefront condominiums, will manage the see-saws, while a playboy
cousin will watch the swingers. That only leaves the sliding board, which
Dr. J eagerly assigns to himself. As he once said, "Isn't it just great to
watch kids accelerate?“
If the sliding board makes an angle of 60º with the ground and the coefficient
of sliding friction between the kids and the sliding board is 0.15, at what
rate will they accelerate?
7.75 m/sec2
The weight of the children, Fw, does not affect the rate
of acceleration. The components of the weight are
found by:
The force of friction, FF, is found using the coefficient
of friction:
The net force down the slide is equal to the difference
between the component parallel to the slide, FP and
friction:
The mass of the children can be found using their
weight:
The acceleration is determined using Newton's
Second Law:
Dr. J was so displeased with his season tickets for local football games
that he invented his own version of a box seat. It was designed to let
Dr. J and his dog Tripod view the games from a different angle. One
night, while watching a game from his favorite angle of 25º, Dr. J,
Tripod, and the box seat were taken for a ride. A former member of
SAP named Cy had left the organization in order to have himself
cloned nine times. He and his clones had formed a new group. (Yes,
the Cyclones - I hope I didn't blow you away with that one.) Cy
attached a rope of negligible weight to the back of the box and then
ran it up the incline and over a frictionless pulley. Cy and the
Cyclones were then able to climb into an attached 50 N box, causing
Dr. J, Tripod, and the box seat to accelerate up the incline at a rate
of 8.00 m/sec2.
If the box seat and its contents weighed 1000 N, and the coefficient of
kinetic friction was 0.466, how much did Cy weigh?
899 N
The problem involves no less than five simultaneous forces. Dr. J's box
and contents exert a force normal to the plane, FN, and parallel to the
plane, FP, as shown. These are components of the weight, FW1.
The force of friction, FF, is found by using the coefficient of friction:
The mass of Dr. J's box and contents, M1, can be
determined from its weight:
The net force can be expressed in two different
ways: Knowing the acceleration of the system, we
can use Newton's Second Law where M2 represents
the total mass of Cy's box and contents.
By finding the vector sum of all the forces we get:
By setting these two expressions for net force equal to each other, M2 can
be calculated:
The weight of Cy's box and contents, FW2, can then be determined:
Cy's weight, FWc, can finally be calculated by subtracting the 50 N weight
of the box and dividing by 10 (remember, Cy was joined in the box by 9
identical clones):