Download unit 3 dynamics

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UNIT 3 DYNAMICS
Dynamics :
The study of forces which cause motion
( interaction between forces and matter)
Kinematics : The study of motion.
UNIVERSAL FORCES: ( 4 FORCES)
There are thousands of observable forces but
they can be categorized into 4 major
categories.
1. Gravitatonal: (Weakest)
- Forces existing between particles of mass
2. Electromagnetic: ( physical and magnetic)
- Forces that exist between charged
particles and magnets
(ex. Hand on a table is actually repulsion of
electrons)
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3. Nuclear: ( strongest ) (aka: strong force)
- Forces that hold the nucleus together.
It holds protons , neutrons together
despite the electromagnetic forces of the
protons. (holds quarks together)
4. Weak interactive: (decay of nucleus)
- These forces are responsible for
radioactive decay. β-decay ( decay of
neutrons = Proton + beta particle released
(electron)
-Both weak and strong nuclear forces act only over
a very short distance ( about the width of an atom)
-Gravitational and Electromagnetic are long range
forces: However gravitational is the only one
that can act over very long distances
( earth-sun).
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NEWTON’S LAWS OF MOTION:
1665 Isaac Newton stated three laws of
Motion.
1. Newton’s first Law: (Law of inertia)
An object continues in its state of rest or
uniform motion unless it is acted on by
some external force.
Inertia: is the tendency of an object
not to accelerate.
2. Newton’s second Law:
When an unbalanced force acts on an
object, the object will be accelerated
directly w/ the force applied and
inversely with the mass of the object.
F=ma for weight W=mg
Newton: (N) Kg*m/s2 ( SI unit of force)
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The amount of force needed to
accelerate 1kg mass @ 1m/s2.
3. Newtons 3rd Law of Motion:
( law of reciprocal action)
All forces occur in pairs, and these two
forces are equal in magnitude and
opposite in direction.
Mass:
The amount of matter in an object
Weight:
The gravitational pull of the earth on an
object. W=mg
NET FORCE:
Fnet = Applied Force - hindering forces
Fnet = ma so
a = Fnet
m
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Ex. A 1 Kg mass that is pushed w/ a 50N
force. How fast will it be accelerated if the
floor causes 15N worth of friction?
A 1.50kg ball rolls down a hill with a 40
degree incline. What is the acceleration of
the ball down the incline?
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An arrow with a .40kg mass is shot straight
up in the air by a force of 120.00Newtons.
Find the acceleration of the arrow.
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Assignment 1
Pg 100 1-10 (old book
Pg 124 Section review 1- 4
Assignment 2
Pg 128 Practice B 1-3
Pg 132 Practice C 1-5
Pg 102 old book pr 11-18
Pg 103 old
pr 19-27
Kinetic friction: The retarding frictional force on an
object in motion
Static friction: The resistance force that keeps the an
object from moving
Force Normal: is the force that is the force that is
equal to the gravitational force on an object but
opposite in direction. ( not always equal to the weight
of the object)
Which is bigger Static or kinetic friction?
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When are the equal?
Coefficient of Friction
Is defined as the ratio of the force of friction to the
normal force between two surfaces.
Coefficient of Kinetic friction is the ratio of the
force of kinetic friction to the normal
µk = Fk
Fn
Coefficient of static friction is the ratio of the
maximum value of force of static friction to
the normal force.
µs = Fsmax
Fn
Force of friction then can be calculated by..
Ff = µFn
Look at table 2 pg 138
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Ex. A 24 kg crate at rest on the floor needs a 75N
force horizontally to set it in motion.
What is the coefficient of static friction for this
system?
µs = Fsmax
Fn
What is Fsmax ?
What is Fn ?
Assignment 3
Pg 139 Practice D
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OVERCOMING FRICTION:
Spencer attaches a rope to a 50Kg weight. He
pulls with a force of 250.0N at an angle of 30o
with the horizontal. The µk is .500. Find the
acceleration of the box.
Fn
Fg
30o
1. Lets find the weight of the box:
W=mg = (50.0kg) (9.8)= 490N
30o
490N
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2. Resolve component vectors :
Cos30 = X/ 250.0N
fx =216.5N
fy = sin30 = Y/250.0N
fy = 125N
3. Find Fn = Downward force on the box
( in this case = weight of object – upward force)
Fn = 490N - 125N = 365N
4. Next we must find the force of kinetic friction
Fk
Fk= µkFn = .500 X 365N = 182.5N
F=ma acceleration = Fx - Ff
mass
= 216.5N – 182.5N = .68m/s2
50Kg
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Assignment 4
Pg 141 Practice E
Pg 150 standardized test prep 1-9
Pg 107 old 1-3