Transcript Newton`s Laws

Newton’s 1st Law of Inertia
Any object continues in its state of rest or in its
uniform velocity unless it is made to change that
state by an unbalanced force is acting upon it.
 An object does not accelerate itself and it
wants to retain a state of zero
acceleration.
Every object possesses inertia and the amount
depends on the amount of matter or mass.
 The greater the mass, the greater the inertia
or resistance to acceleration.
 Mass is a measure of the inertia.
Mass is the amount of matter contained in an
object.
 Mass is a scalar quantity meaning it has
magnitude only.
 Mass is measured in g, kg, or slugs.
 For nonrelativistic speeds, the mass of an
object remains constant.
Do not use mass and weight interchangeably!
 Weight is a measure of the gravitational
attraction between an object and the earth.
 Weight is a vector quantity because it has
both magnitude and direction (the direction
is always assumed to be towards the center
of the earth).
 The weight of an object varies with location
as it is dependent on the distance from the
center of the earth.
Newton’s 2nd Law of Acceleration
The acceleration of an object is directly
proportional to the net force acting on the object
and is inversely proportional to its mass.
Inertia is the tendency to resist changes in
motion and Newton’s 2nd law expresses this
mathematically.
a is directly proportional to the Fnet.
 By whatever factor Fnet changes, a
changes by the same factor.
 a α Fnet.
If you double the force, you double the
acceleration.
If you decrease the Fnet by 1/3, you
decrease the acceleration by 1/3.
A graph of Acceleration vs Force would be a
straight line passing through the origin.
a is inversely proportional to m.
 a α 1/m
 If you double the mass, the acceleration is
reduced by ½.
 If you decrease the mass by a factor of 1/3,
you would triple the acceleration.
 A graph of Acceleration vs Mass would be a
hyperbola.
An object always accelerates in the direction of
the net force.
 If the net force is applied in the direction of
the object’s motion (velocity), the object
accelerates positively (speeds up).
 If the net force is applied in the opposite
direction of the object motion (velocity), the
object decelerates.
Mathematically, Newton’s 2nd Law is
 Fnet = ma
where m is the mass of the object in kg, a is
the acceleration in m/s2, and Fnet is the net
force in N.
 We now can formally define 1 N of force.
If you have a mass of 1.0 kg and the net
force causes it to accelerate at 1.0 m/s/s,
then it is by definition 1.0 N of force.
Which Is It?
Which is true?
 a = Δv/Δt or a = Fnet/m?
 Acceleration was previously defined to be
the rate at which the velocity changes.
 Now we are defining acceleration to be
Fnet/m.
 Why the difference?
Both are true!
Previously, we looked at kinematics or how do we
describe motion?
 Do objects move at a constant velocity or a
constant acceleration?
Newton’s Laws describe the dynamics or why do
objects move as they do?
 Is the net force equal to greater than zero?
Newton’s 2nd Law Problems
A car traveling at 32 m/s slows down to a stop
and travels a distance of 52 m. If the mass of
the car is 1375 kg, what net force acted on the
car?
vi = 32 m/s
m = 1375 kg
vf = 0
Δx = 52 m
vf2 = vi2 + 2aΔx
0 = (32 m/s)2 + 2 × a × 52 m
a = -9.8 m/s2
Fnet = ma = 1375 kg × -9.8 m/s2 = -1.4 × 104 N
The negative values for a and F make sense
because the car decelerated in coming to a stop
requiring a force in the opposite direction to its
motion.
A stone weighs 7.4 N. What force must be
applied to make it accelerate upward at 4.2 m/s2?
Fw = 7.4 N
a = 4.2 m/s2
FT
•
Fw
g = 9.80 m/s2
FT = Fnet + Fw
Fnet = ma
Fw = mg
m = 7.4 N/9.80 m/s2 = 0.76 kg
Fnet = 0.76 kg × 4.2 m/s2 = 3.2 N
FT = 3.2 N + 7.4 N = 11 N
Some notes from the previous problem:
 If the acceleration of the stone is upward,
then the Fnet must also act upwards.
 This implies that FT > Fw because the rope
must provide the total force to support the
weight of the object and also provide the net
force.
Free Fall
Free fall exists when an object’s weight is the
only force acting on it (straight down, towards
the center of the earth).
In the absence air resistance, all objects
accelerate at the same rate.
 a = Fnet/m = Fw/m
 a = g = 9.80 m/s2 = 980 cm/s2 = 32 ft/s2
Air resistance can usually be ignored for small
dense objects that travel short distances but there
can be exceptions:
 Every baseball fan has heard the expression
that the ball was headed out but the wind
knocked it down.
 A ping pong ball will never be confused with
a small dense object.
Sometimes air resistance is not what you want:
 When throwing a football for distance, a tight
spiral minimizes air resistance.
 Long range rifles have grooves in the barrel
so the bullets come out spiraling.
 If either the football or the bullet started to
topple end over end, well …
 The reason will be explained in another set
of slides discussing angular momentum.
Sometimes you want air resistance.
 Just ask any parachutist.
 Air resistance depends on both velocity and
surface area.
a = g = Fnet/m = Fw – R/m
At t = 0, R = 0.
As an object accelerates downward, R
increases.
a = g = Fnet/m = Fw – R/m
At t = 0, R = 0.
As an object accelerates downward, R
increases.
When Fw = R, Fnet = Fw – FR = 0, and a = 0.
When the acceleration equals zero, the
object is said to be moving with a terminal
velocity.
Two brothers, Pete and Repeat, jump from the
same helicopter and their parachutes are initially
opened. The parachutes are the same size and
Pete weighs 500 N and Repeat weighs 450 N.
Who hits the ground first?
True Weight vs Apparent Weight
A man stands on a bathroom scale in an
elevator. The scale reads 917 N.
(a) What is the man’s weight?
Fup
•
Fw
Fup is the force that the bathroom scale
pushes up on the man.
Fw = Fup = 917 N and the man appears
to weigh 917 N.
(b) What is the man’s mass?
Fw = 917 N
g = 9.80 m/s2
Fw = mg
m = 917 N/9.80 m/s2 = 93.6 kg
(c) As the elevator moves up, the scale reading
increases to 1017 N. Determine the upward
acceleration of the elevator.
Fup
•
Fw
Fnet = ma = Fup - Fw
Fnet = 1017 N – 917 N = 100. N
a = Fnet/m = 100. N/93.6 kg
a = 1.1 m/s2 straight up
(d) As the elevator approaches the 13th floor, the
scale reading decreases to 798 N. What is the
acceleration of the elevator?
•
Fup
Fw
Fnet = ma = Fw - Fup
Fnet = 917 N – 798 N = 119 N
a = Fnet/m = 119 N/93.6 kg
a = 1.3 m/s2 straight down
(e) When the elevator reaches the 13th floor it
stops. After about 5 sec the man looks down
and the scale is reading 0. What is going on?
•
Fnet = Fw = mg
Fw
The guy is in a heap of trouble as
he is in a state of free fall!
Thoughts To Ponder
If the elevator was sound proof and there was
no visible connection to the outside world, there
is nothing the man could do to detect uniform
motion.
 When the acceleration of a system is
zero, there is no experiment that
distinguishes between an object at rest
(∑F = 0) or an object moving in a straight
line at constant speed (∑F = 0).
There are no relativistic speeds involved, so
that the mass of the man remains constant.
Whenever there is an acceleration involved, the
net force will always be in the same direction as
the acceleration.
Friction
Friction is a force that resists the relative motion
of solid objects that are in contact with each
other.
 If the solid is in a fluid (a liquid or a gas),
then it is called viscosity.
Friction is caused by uneven surfaces of
touching objects.
Six Principles of Friction
Friction acts parallel to the surfaces that are in
contact and always opposes motion.
Friction depends on the nature or composition
of the solid surfaces in contact.
Rolling Friction < Sliding Friction < Starting
Friction
Sliding friction is practically independent of
surface area for a given object.
Sliding friction is practically independent of
medium speeds.
Sliding friction is directly proportional to the
force pressing the two surfaces together.
Coefficient of Friction
The formula for friction is given by
Ff = µFN
where Ff is the frictional force in N (newtons)
and FN is the normal (perpendicular) force
pressing the two surfaces together.
 The normal force, FN, will not always equal
the weight of the object!
µ (mu) is the coefficient of friction which is
determined by what the two solid surfaces
consist of (glass on glass, wood on wood, etc.).
µ = Ff/FN is the ratio of the frictional force to the
normal force.
 µs > µ k
where µs is the coefficient of starting
friction and µk is the coefficient of sliding
friction.
Friction Problems
A crate weighing 475 N is pulled along a level
floor at a uniform speed by a rope which makes
an angle of 30.0° with the floor. The applied
force on the rope is 232 N.
(a) Draw a free-body diagram of the box.
FN
Ff
F
Fv
θ F
H
Fw
(b) Determine the coefficient of friction.
µ =
Ff
FN
=
FH
FW - FV
=
F × cosθ
Fw - F × sinθ
232 N × 0.866
µ =
475 N – 232 N × 0.500
= 0.560
(c) How much force is needed to pull the box?
FH = F × cosθ = 232 N × 0.866 = 201 N
(d) Compare the force in (c) to the weight of the
box.
It is easier to pull the crate, 201 N, than it is
to lift the crate, 475 N.
An Inclined Plane Problem
A roller coaster reaches the top of a steep hill
with a speed of 7.0 km/h. It then descends the
hill, which is at an angle of 45° and is 35.0 m
long. If µk is 0.12, what is the speed when it
reaches the bottom?
Vi = 7.0 km/h
θ = 45°
Δx = 35.0 m
µk = 0.12
FN
.
Fp
•
θ
θ F
w
Ff
FN
Fnet = Fp - Ff = Fw sin θ - µk FN
Fnet = mg sinθ - µk mg cos θ
Fnet = ma
.
a=
mgsinθ - µmgcosθ
m
a = 9.80 m/s2 × 0.707 – 0.12 × 9.80 m/s2 × 0.707
a = 6.10 m/s2
3 m
km
10
1h
vi = 7.0
× 1 km × 3600 s = 1.9 m/s
h
vf2 = vi2 + 2a(x-xi)
vf2 = (1.9 m/s)2 + 2 × 6.10 m/s2 × 35.0 m
vf = 21 m/s
Another Inclined Plane Problem
A block is given an initial speed of 4.2 m/s up a
24.0° inclined plane. Ignoring frictional effects,
calculate:
(a) How far up the inclined plane will the block
travel?
FN
Fp
•
θ FN
θ Fw
vi = 4.2 m/s
µ=0
θ = 24.0°
vf = 0
Fnet = ma
Fnet
-Fp
a= m = m
-Fw sinθ
=
=
m
-mg sinθ
m
a = -9.80 m/s2 × 0.407 = - 3.99 m/s2
vf2 = vi2 + 2a(x-xi)
0 = (4.2 m/s)2 + 2 × (- 3.99 m/s2) × Δx
Δx = 2.2 m
(b) How long does it take before the block returns to
its starting point?
vf = vi + aΔt
0 = 4.2 m/s/-3.99 m/s2
Δt = 1.1 s
ΔtT = 2 × 1.1 s = 2.2 s
Newton’s 3rd Law
When one object exerts a force on a second
object, the second object exerts a force on the
first that is equal in magnitude but opposite in
direction.
 These two forces are called an
action-reaction pair of forces.
 F1 = - F2
To apply Newton’s 3rd Law, you must
distinguish between forces acting on an object
and forces exerted by the object.
 When using Newton’s 3rd Law, you must
have two different objects!
Examples of Newton’s 3rd Law
Consider a 10. N ball falling freely in a vacuum
where there is no air resistance.
 What is the action force?
 What is the reaction force?
The action force could be the earth pulling down
on the ball with a force of 10. N.
•
Fa = Fw = 10. N
The reaction force would be the 10. N ball
pulling up on the earth.
Fr = Fw = 10. N
•
It is easy to see why the ball falls toward the
center of the earth.
From Newton’s 2nd Law:
Fb
Fw
mbg
=
=
= g = 9.80 m/s2
ab =
mb
mb
mb
straight down
It is easy to see why the earth remains
stationary.
10. N
Fb
Fw
=
=
ae =
me
me
5.96 x 1024 kg
ae = 1.68 x 10-24 m/s2 straight up
At The Firing Range
What happens when you fire a long range rifle?
 The action force can be considered to be
the force the gun exerts on the bullet.
 The reaction force would be the force the
bullet exerts on the gun.
 The acceleration of the bullet and the gun
would be given by:
Fb
ab =
mb
- Fb
ag =
mg
mg > mb, therefore, ab > ag.
This accounts for the “kickback” or recoil
velocity of the gun.
Universal Law of Gravitation
The mutual force of attraction between two
objects is directly proportional to the product of
their masses and inversely proportional to the
square of the distance between their centers.
 F α m1 × m2
 F α 1/d2
 F α (m1 × m2)/d2
The proportionality sign can be replaced with an
equals sign and a constant.
F =
G × m1 × m2
r2
where G = 6.67 x 10-11 n•m2/kg2.
Newton’s Universal Law of Gravitation is an
example the inverse square law.
 If you double the distance, the force
decreases by the factor of ¼.
Universal Law of Gravitation Problem
What is the mutual force of attraction of a 1.0 kg
mass and the earth if the 1.0 kg mass is resting
on the ground?
m1 = 1.0 kg
re = 6.37 x 106 m
F =
G × m1 × m2
r2
m2 = me = 5.96 x 1024 kg
G = 6.67 x 10-11 Nm2/kg2
.
F =
F =
G × m1 × m2
r2
6.67 × 10-11 Nm2/kg2 × 5.96 × 1024 kg × 1.0 kg
F = 9.8 N
(6.37 × 106 m)2
Everything Fits!
What is the acceleration due to gravity in the
previous problem?
re = 6.37 x 106 m
me = 5.96 x 1024 kg
G = 6.67 x 10-11 Nm2/kg2
F =
G × m1 × m2
r2
.
Fw =
mb g =
G × mb × me
r e2
G × mb × me
r e2
G × me
g =
r e2
=
6.67 x 10-11 Nm2/kg2 × 5.96 x 1024 kg
(6.37 x 106 m)2
g = 9.80 m/s2
.
Sound familiar?
g is a constant for a given location!
Relation of Gravity to Weight
Gravity describes the force of gravitational
attraction on or near the surface of a planet.
 Objects at higher altitudes will weigh less
than at sea level.
Masses weigh a little more at either pole than at
the equator.
Going inside the surface of the earth decreases
the acceleration due to gravity.
Once below the surface of the earth, the
attraction of the earth above the object causes
the object to weigh less.
What makes Newton’s Three Laws and the
Universal Law of Gravitation so beautiful is that
they work anywhere in the universe.
 All of Newton’s Laws are mass dependent.
 The only time they break down is at
relativistic speeds.
Wrap Up Questions
Assess the following statement:
When an object is at rest, there are no external
forces acting on it.
This statement is false because when an object
is at rest, there is no resultant force. The vector
sum of the forces, ΣF = 0.
Two boys pull on a 5.0 m rope each with a
horizontal force of 225 N. If each boy increases
their applied force by the same amount, can the
rope ever be horizontal?
No, because of the weight of the rope. No
matter how much force each boy exerts, there
is no vertical force to cancel the weight of the
rope.
You can reduce the force of friction (i.e. sanding
or polishing the surfaces) only so much, before
it increases again. Why?
By smoothing the surfaces as much as
possible, the separation distance of the atoms
or molecules decreases. This makes for a
stronger attraction.
If the two surfaces are the same material, the
force is cohesion, otherwise the force is
adhesion.
Assuming the earth is a perfect sphere and its
mass is evenly distributed, how much would a
225 N person weigh at the center of the earth?
The person would weigh 0 N. This answer can
be arrived at either qualitatively or
quantitatively.
Qualitatively, the person would an experience a
force of attraction from all directions. But the
attractive force would not be downward toward
the center of the earth but rather radially away
from the center of the earth.
Quantitatively, one could start with Newton’s
Universal Law of Gravitation.
F=G
m1 × m2
r2
This formula is applicable when you are on the
surface of earth or above it. However, once you
go below the surface of the earth, the formula
has to be modified.
Using the assumptions given in the question,
the mass of the earth is given by:
me = De × Ve = De × 4/3 × π × re3
mp × De × 4/3 × π × re3
F=G
r e2
F = G × mp × De × 4/3 × π × r
where mp is the mass of the person and r is the
distance from the center of the earth.