Kinetics Consider the forces and their relation with motion Kinematics Eq Newton 2 nd Law 1 Math diff Integral 2 Graph xy nt r q Particles Solving Differential Equation ID: 391296
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Slide1
1
Kinetics of Particles
Kinetics
Consider
the forces
and their relation with motion.
Kinematics Eq
:
Newton 2
nd Law
1) Math (diff, Integral)
2) Graph,
x-y n-t
r-
q
Particles
Solving Differential Equation
Kinetics
Eq.
No FBD (+KD): no score!!!
Constrain Eq.Slide2
Basic Principles of Mechanics
1.
The Parallelogram Law
2.
The Principle of Transmissibility
3. Newton’s First Law
4. Newton’s Second Law
5. Newton’s Third Law
Some principles that governs the world of Mechanics:
6. Newton’s Law of GravitationSlide3
Newton’s Law of Motion
3
Philosophiæ
Naturalis
Principia Mathematica
(first published 5 July 1687)In Newton’s own word,“Every body perseveres in its state of rest, or of uniform motion in a right line, unless it is compelled to change that state by forces impressed thereon”
“The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed.”“To every action there is always opposed an equal reaction; or the mutual actions of two bodies upon each other are always equal, and directed to contrary parts.”Slide4
Newton’s Law of Motion
4
First Law
Second Law
Third Law
Action = Reaction
There are many misleading or misunderstanding these Three Law, so we will examine it one by one and, detail by detailSlide5
1
st
Law:
An object at rest tends to stay at rest and an object in motion tends to stay in motion with the same speed and in the same direction, unless acted upon by
an unbalanced force.
no motion(at rest)
straight line
Under Balanced Force
no change
in speed
When measuring in the “Inertia Frame”
Sometimes, it is call “
Law of
Inertia”v
orSlide6
2
nd
Law: The acceleration of a particle is
proportional to the vector sum of forces acting on it, and is in the direction of this vector sum
.
mSlide7
3
rd
Law: The mutual forces of
action and reaction between two particles are equal in magnitude,
opposite in direction,
and collinear.
Confusing?
Concept of
FBD
(
Free Body Diagram
)
Point: Isolate the body
Forces always occur in pairs – equal and opposite action-reaction force pairs.Slide8
8Slide9
9
Kinetics of Particles
Kinetics
Consider
the forces
and their relation with motion.
Kinematics Eq
:
Newton 2
nd Law
1) Math (diff, Integral)
2) Graph,
x-y n-t
r-
q
Particles
Solving Differential Equation
Kinetics
Eq.
Constrain Eq.Slide10
10
Rectilinear Motion
In general
Use
x-axis
as
its moving direction
straight line
v , a
A
Even we consider only 2D motion, we may need to establish
3 dimensional equation
Since
No
FBD
:
no score!!!
x
y
+ KD
inertia
FBD
(Free Body Diagram)
KD
(Kinetics Diagram)
ma
One FBD(+KD) :
At most 3 unknowns Slide11
11
SP3/4 The design model for a new ship has a mass of 10 kg and is tested in an experimental towing tank to determine its resistance to motion through the water at various speeds. The test results are plotted on the accompanying graph, and the resistance R may be closely approximated by the dashed
parabolic curve
shown.
If the model is released when it has a speed of 2m/s
, determine the time t required for it to reduce its speed to 1 m/s and the corresponding
travel distance x.
x
y
inertia
At any time
tSlide12
12
SP3/5 The collar of mass
m
slides up the vertical shaft under the action of a
force
F
of constant magnitude but variable direction. If
q =
kt where k = const. and
if the collar starts from rest with q=0, determine the magnitude F of the force which will result in the collar coming to rest as q=p/2
.
x
y
0
F
N
Not depend on k
x
y
At any time
t
Unknown:
N, F, a
F must be certain value
that make velocity = 0 when
q=p/2
. Slide13
13
SP3/3 The block A is released from rest, and pulls the 200-kg log up. Determine the velocity of the block A when it hits the ground
Find constrained motion
measured from a
convenient
“fixed”
reference
+
+Slide14
14
measured from a convenient
fixed referenceSlide15
15
3/43 The sliders A and B are connected by a light rigid bar of length L=0.5 m and move with negligible friction in the horizontal slots shown. For the position where Xa= 0.4 m, the velocity of A is 0.9 m/s to the right. Determine the acceleration of each slider and the force in the bar at this instant.Slide16
16
-(3)
x
A
y
B
q
W
B
T
q
N
B
q
P
N
A
T
A
B
x
y
0.5m
T
T
At this instant
-(1)
-(2)
(2)
(3)
(1)
Some constraints
Why? - 2 force member (result from statics)
No, this is dynamics!?!
OK (we ignore mass of bar)
0
Eq
: 2x2
+1
(constraint)
Fixed
point
Fixed
pointSlide17
17
The inclined block A is given a constant rightward acceleration
a
.
Determine the range of values of
q
for which block B will not slip relative to block A, regardless of how large the acceleration a is.
N
q
mg
f
a
x
ySlide18
18
P
1: A,B at rest
2: A moving along with B
4 unknowns,
2 Eq.
3: A moving with lower acc compared with B
3/46
(ignore tipping)
4 unknowns,
2
Eqs
+
2
Eqs
(2 more eqs)
(1 more eq)
(1 more eq)
(1 more eq)
(1 more eq)Slide19
19
What will happen?Slide20
20
Recommended Problems
3/39 3/31
3/26
3/16 3/11 3/44 3/23 3/28
3/46Slide21
21
From eq.(1)-(4), we get
-(3)
-(4)Slide22
22Slide23
23
P
1: A,B at rest
2: A moving along with B
4 unknowns,
2 Eq.
3: A moving with lower acc compared with B
3/46
(ignore tipping)
4 unknowns,
2
Eqs
+
2
Eqs
(2 more eqs)
(1 more eq)
(1 more eq)
(1 more eq)
(1 more eq)Slide24
24Slide25
25
The car P starts from rest at the position O and slides along the curve path shown in the figure. The time rate of change of its speed is constant along its motion, where
t
is the time in seconds, and starts from zero when P is at O. If , determine the following 4 quantities: ,when the block P is at the position :
x-y
coord
r-
q
coordSlide26
At the instant shown the 50-kg block A is moving down the plane at 2 m/s while being attached to the 25-kg block B. if the coefficient of kinetic friction is 0.2, determine the acceleration of A at this instant and the distance A slides before its stops. Neglect the mass of pulleys and cables.
x
y’
x
y
x’
ySlide27
27Slide28
28
3/5 Curvilinear Motion
Rectangular
Coordinates
x
y
path
Normal Tangential
Coordinates
Polar
Coordinates
Use appropriate
coordSlide29
29
n
SP3/8 A 1500-kg car enters a section of curved road in the horizontal plane and
slows down at a uniform rate
from a speed of 100 km/h at A to a speed of 50 km/h as it passes C
. Determine the force acted on the car at A, B and C. Point B is the inflection point.n-t
t
n
t
t
n
t
n
t
tSlide30
30
Find
b
which locates the point where the vehicle leaves the path.
mg
N
t
n
q
general
position
Time when vehicle leaves path
N = 0Slide31
31Slide32
32
3/78 What is the
maximum speed
at which the car A can go
over a hump
and still maintain contact with the road? If the car maintains this critical speed, what is the total reaction N under its wheel at the bottom of a dip? Car’s mass is m.
N
t
n
mg
0
N
t
mg
n
At top
At
bottomSlide33
33
If the disk starts rotating from rest and is given a constant angular acceleration
a
, determine an expression for the number of revolutions N through which the disk turns before the coin slips.
f
0
0
n
t
friction
No relative movement in this direction
direction of friction?Slide34
34
z
n
Find the range of
w
which make the box remain on the dish without slipping.
Assume that speed changes are made slowly so that any angular acceleration may be neglected.
N
mg
f
f
w
:small
f ‘
w
: large
f
Direction
of friction?
(front view)
(top view)
w
may change direction of friction
andSlide35
35
slider A (180 gram) moves
without friction
in the tube, which rotates in a horizontal plane with a
constant speed
W=7 rad/s. The slider is launch with an initial state relative to the tube at the inertial coordinates x=150 mm and y=0.
Determine the magnitude P of the horizontal force exerted on the slider by the tube just before the slider exits the tube.
N
q
r
0
only + makes sense
no force in this direction
oSlide36
36
Since N is negative, the normal force should be from side B.
Ans
Ans
r
q
N
T
FBDSlide37
37
The smooth 2-kg cylinder C has a peg P through its center which passes through the slot in art OA. If the arm rotates in the vertical plane at a constant rate , determine the force that arm exerts on the peg at the
instatnt
q
=60.
kinematics eq.
constrain motion eq.
Solve for
Solve for
C moves as rectilinearSlide38
38
The smooth 2-kg cylinder C has a peg P through its center which passes through the slot in art OA. If the arm rotates in the vertical plane at a constant rate , determine the force that arm exerts on the peg at the
instatnt
q
=60.
r
and q
are related in somehowr=r(q )
kinematics eq.
constrain motion eq.Slide39
39
m = 2 kg
= 0.4 m
Determine the force of the arm OA on the ball when
q
= 30
.
.
..
r
and
q
are related in somehow
r=r(
q
)
kinematics eq.
constrain motion eq.Slide40
40Slide41
41
SP 3/7
SP 3/9
SP 3/10Slide42
42
Recommended Problems
SP3 3/85 3/75
3/84 3/89 3/78 3/93 3/98
3/102