Download Kinematics Unit Outline - Hicksville Public Schools

Dynamics Review Outline
2.1 Contact Forces
2.1.1A-C


Acceleration due to gravity can be determined by looking at the slope of a
graph of mass vs. weight.
Newton’s Laws of Motion
First Law (Inertia) – objects tend to remain in their current state of motion
(at rest of moving at a constant velocity) until acted upon by a net force
o Inertia is proportional to mass and is NOT affected by speed
Fg
Slope = g
A 10 kilogram object at rest has more inertia than a 2 kilogram object moving at 20
meters per second.

Second Law – acceleration is proportional to force and inversely
proportional to mass
A 5.0 kilogram object that is subjected to a net force of 20 newtons will accelerate
2
at a rate of 4.0 meters per second .
m
2.1.2A-C

Forces on Angles (Top/Down)
The resultant to a set of vectors is determined graphically by adding the
vectors using the ‘head-to-tail’ method.
B’
B’

Third Law – for every action there is an equal and opposite reaction.
A
If a person pushes on an object with a force of 25 newtons then the object pushes
back with a force of 25 newtons.
2.1.1D

Mass and Weight

R
A
B
A and B can be added by
drawing A’ or B’ (or both) and
drawing resultant R.
When vectors are perpendicular
the resultant can be found using
the Pythagorean Theorem.
2

A’
B
2
A +B =R
Mass is universal (object has the same mass independent of its location).
A 10 kilogram object has a mass of 10 kilogram on Earth, in orbit and on the Moon.
A’
R
2
Vectors can be broken into components that are perpendicular to one
another.
Weight is a gravitational force that depends on location.
The resultant to a set of vectors is determined algebraically by adding
the components of the vectors and using the Pythagorean Theorem on
the resultant components.
A 10 kilogram object weighs about 100 newtons on Earth and about 17 newtons on
the Moon.

Equilibrium is a state in which the net force on a system is zero.
o
o
An equilibrant is a vector that when added to a system produces a
state equilibrium within that system.
2.1.2D

When objects are pulled or pushed along surfaces using forces at angles to
the surface then acceleration does not depend on the force itself, but the
components of it.

Pulling on an upward angle reduces the normal force.

Pushing on a downward angle increases the normal force.
The equilibrant is equal in magnitude to the resultant, but
opposite in direction.
3N
R = 4.2 N
3N
E = 4.2 N
Adding these two perpendicular 3 N
vectors gives a resultant of R = 4.2 N
that points northeast. Thus the
equilibrant is also 4.2 N in magnitude,
but points southwest.
Forces on Angles (Side-view)
FN = +50N
Fg = -50N
B
The maximum resultant produced by
two vectors is obtained by aiming them
with 0° between them. Max resultant
(and equilibrant) will be A + B.
A
B
A
The minimum resultant produced by two
vectors is obtained by aiming them with
180° between them. Min resultant (and
equilibrant)will be A - B.
Any resultant (and equilibrant) between A + B and A – B can be produced!
A 10 newton vector and a 7 newton vector can be added together to
produce any resultant between 3 N and 17 N (it just depends on what angle
you choose to have between them). It is therefore true that any vector
between 3 N and 17 N could be added this system to produce equilibrium.
A 5 kilogram mass has a weight of about -50 N, so the normal force will
be +50 N if it is in equilibrium.
FN = +45N
FAY = +5N
FA = 10N
FN = +55N
30°
30°
FAY = -5N
Fg = -50N
Fg = -50N
FA = 10N
A 10 N force at 30° above
horizontal would have an
upward component of +5 N, thus
reducing the requirement for the
normal force to +45 N.
A 10 N force at 30° below
horizontal would have an
downward component of -5 N,
thus increasing the requirement
for the normal force to +55 N.
Note that FN + FAY + Fg must be
equal to 0.
Note that FN + FAY + Fg must be
equal to 0.

Pulling on an angle also changes the net force. The net force in side-view
systems depends only on the sum of horizontal forces.
FN = +45N
FAY = +5N
FA = 10N
FAX = +8.7N
30°
FAX = +8.7N
Fg = -50N
FN = +55N
30°
FAY = -5N
Fg = -50N
FA = 10N
In both cases the horizontal component of FA is +8.7 N. If there is no
friction in either system, then the net force IS 8.7 N.
We can then say that both objects will accelerate at a rate of 8.7 N / 5.0 kg
2
or 1.7 meters per second .
A 15 newton block of rubber that is sliding along an ice-covered surface will require
2.25 newtons of force to be constantly applied to keep it moving at a constant
velocity. If a force of 10.25 newtons were applied to the same block the result
2
would be a net force of 8.0 newtons and an acceleration of 0.53 meters per second .
If 40 newtons of force are required to make a 3.0 kilogram object accelerate along a
2
surface at a rate of 5.0 meters per second , then the net force on the object is 15
newtons and we can conclude that there must be a 25 newton force of kinetic
friction acting on the object. The kinetic coefficient of friction in this case must be
2
about 0.85 (arrived at using Ff = μFN where FN = 3.0 kg * 9.81 m/s .)
If a 20 newton force of friction is acting on a 4.0 kilogram object that is accelerating
2
at a rate of 3.5 meters per second then we must conclude that there is a 34 newton
force pushing on the object.
2.1.4A-B
2.1.3A-B




Friction
Friction is a force that attempts to keep objects from moving. It is the
result of electrical attraction between surfaces and can be reduced by
lubrication.
Static friction is the force that must be overcome to make an object slide.
This type of friction should be used in ALL problems where there is NO
SLIDING.
Kinetic friction is the force that acts in opposition to the motion of a sliding
object. This type of friction should be used ONLY in problems in which
objects ARE SLIDING.
Inclined Planes

When an object is placed on an incline its weight will point partially ALONG
the plane and partially INTO the plane.

Breaking the weight into components will cause all forces to be referred to
in reference to the plane – either parallel or perpendicular to it.

The normal force will be equal to the perpendicular component of weight.

Whether or not the object slides along the plane (and its direction)
depends on the sum of the forces directed along the plane.
FN = Fg⊥
Coefficient of friction tells about the relative “stickiness” that two surfaces
have when in contact with each other – this a dimensionless quantity (has
no units.)
If a 20 newton block of wood is placed on a wooden surface then the normal force
between the block and surface is 20 newtons. It will take slightly more than 8.4
newtons of force to get the block to begin sliding along the surface. Once the block
begins to move, the force of friction will immediately drop to 6 newtons.
Since the box weighs 80 newtons, there
are 69 newtons of force directed into
the plane – this means the normal
force is 69 newtons.
Fg
30°
Fg⊥
Fg = 80N
If the surface of the plane is frictionless
then the net force is 40 newtons.
2.2 Auxiliary Forces
Ff
FN = Fg⊥
Fg
2.2.1A-B
If the block slides down at a constant
speed, then the force of friction must
be 40 newtons.
-11
2
N·m /kg
2
Gravity is a force that attracts objects with mass to one another.

The force of gravity between two objects is directly proportional to the
masses of each object.
Fg = 80N
The force of gravity acting between two
objects with masses of 3000 kilograms that
-2
are separated by 2 x 10 meter is 150
newtons.
Fg
FN = Fg⊥
Ff
Fg
30°
Fg⊥
m
If the force of friction is 30 newtons
then a net force of 10 newtons is
causing the block to accelerate down
the plane.
If the mass of one object were doubled, the
force would become 300 newtons.
If the masses of both objects were halved, the
force would become 37.5 newtons.
Fg = 80N

F
The force of gravity between two objects is inversely proportional to the
distance between the CENTERS of the objects.
FN = Fg⊥
Ff
30°
Fg⊥
G = 6.67 x 10

30°
Fg⊥
Universal Gravitation
Fg
Fg = 80N
If the block is being pulled up the plane
at a constant speed and the force of
friction is 15 newtons, then force F
must be 55 newtons.
The force of gravity acting between two
objects with masses of 3000 kilograms that
-2
are separated by 2 x 10 meter is 150
newtons.
Fg
r
If the distance between the objects is doubled,
the force becomes 37.5 newtons.
rd
If the distance between them is cut to 1/3 of
its original value the force becomes 1350
newtons .
2.2.2A-C

Uniform Circular Motion
Objects in circular motion are being pulled into a circular path by
centripetal force – this is an unbalanced, net force that causes centripetal
acceleration.
o Both centripetal force and centripetal acceleration must point
toward the center of the object’s circular path of travel.
o


The velocity of the object will be tangent to its circular path
The force needed to bend the path of an object depends on:
o How massive the object is (m)
o The object’s centripetal acceleration (aC)
FC
FC
The minimum radius with which a 70 kilogram bike could make a 12 meter per
second turn with rubber tires on asphalt is 17 meters.
v
m
aC
aC
aC
When a car (or other object) makes a turn on a flat surface it is STATIC
friction between the object and surface that allows the turn to occur.
The maximum turning speed on a 30 meter radius turn for a 1500 kilogram car with
rubber tires on asphalt is 15.8 meters per second.
aC
FC
A 5.0 kilogram object that is moving around a circular path with a speed of 6.0
meters per second will be subjected to a centripetal force of 60 newtons if the
path has a radius of 3.0 meters and 45 newtons if the path has a radius of 4.0
meters.

The object’s centripetal acceleration depends on:
o How much speed the object has (v)
o How tight the turn needs to be (r)
v
An object that is moving around a circular path with a radius of 10 meters while
moving at a speed of 4.0 meters per second will have a centripetal acceleration
2
of 1.6 meters per second .
r
An object that makes one complete trip around a path with a circumference of
4.0 meters every 0.5 seconds has an average speed of 8.0 meters per second.
An object that moves around a circular path with a radius of 5.0 meters at a
speed of 20 meters per second will take about 1.6 seconds to make one
complete trip around the circle.
The minimum coefficient of friction needed for a 1000 kilogram vehicle to make a
safe turn with a 10 meter per second radius while moving at 5.0 meters per second
is 0.25
2.2.3
Hooke’s Law

Force is required to change the length (compress or elongate) of a spring.

The change in the spring’s length will be:
o Directly proportional to force applied (FS)
o Inversely proportional to the spring constant of the spring (k)
A force of 200 newtons will cause a spring with a spring constant of 5000 newtons
per meter to stretch a distance of 4 centimeters.
FS
The slope of the graph of FS vs. x is the spring
constant of the spring.
x