Transcript Lecture8 (Equilibrium)

Engineering
Fundamentals
Session 8 (3 hours)
Motion
Distance Vs Displacement
• Distance距離 is a scalar quantity which
refers to "how much ground an object has
covered" during its motion.
• Displacement s 位移 is a vector quantity
which refers to "how far out of place an
object is"; it is the object's change in
position.
Exercise
Start here
A teacher walks 4 meters East, 2 meters
South, 4 meters West, and finally 2 meters
North.
Distance = ___________
Displacement = _____________
Exercise
The skier moves from A to B to C to D.
displacement =___________
distance traveled=____________
Answer to previous page:
distance=12;displacement=[0,0]
Speed and Velocity
•
Average speed 速率
– Average speed = total distance / total time taken
– ( or rate of change of distance, or changes in distance
per unit time)
– a scalar
• Average velocity v 速度
– v = total displacement/total time taken
v = ∆s / ∆t
– (or rate of change of displacement, or changes in
displacement per unit time)
– a vector
– Unit: ms-1
Answer to previous page:
displacement=140m,distance=420m
Example
A car moves 3 km north for 10 minutes and then 3
km east for 10 minutes. Find its average speed
and velocity.
3km
3km
Average speed = 6km / 20 minutes = 6 km / (1/3 hour)
= 18km/hr
Average velocity = √18 km/ (1/3 hour) at an angle of 45
=12.73 km/hr at 45。
。
Exercise
displacement
B
A
C
time
• Which part(s) is the car having a positive
velocity? ________
• Which part(s) is the car having a negative
velocity?_________
• What is the velocity at part B?____
Acceleration加速度
• Car speeds up  velocity increases and there is
an acceleration a
• Car slows down  velocity decreases and there is
a -ve acceleration, or deceleration.
• Average acceleration = changes in velocity / total
Average acceleration a = ∆v / ∆t
time taken
• (or rate of change of velocity, or changes of
velocity per unit time)
• SI unit: ms-2
Answer to previous page:
A, C, 0
Example
A sports car can go from rest to 100 km hr-1
within 10 seconds. What is its acceleration?
Answer
Changes in velocity = 100 - 0 km hr-1 = 100
/3600 km s-1 = 100/3.6 m s-1
Average acceleration = (10 / 3.6) /10 ms-2
Exercise
• Which part(s) of the curve shows an
acceleration? ____
• Which part(s) of the curve shows a
deceleration? _____
• Which part(s) of the curve shows a
constant (stable) velocity?_____
velocity
B
A
C
D
time
Exercise
• A car is originally at rest. It accelerates
at 2 ms-2 for 1 second. What is its velocity
afterwards? _________
• A car is originally moving at a constant
velocity of 1 ms-1. It then accelerates at 2
ms-2 for 0.5 second. What is its velocity
afterwards? _________
Instantaneous Velocity
• Average velocity 平均速度= ∆s / ∆t (average over
the time interval ∆t )
• Instantaneous velocity 瞬時速度= velocity at an
instant 瞬閒of time. (∆t  0)
• Instantaneous velocity at a time instant t1 = slope
of tangent line at t1.
instantaneous velocity (at time t) = slope of tangent at t
Displacement s
t1
t
Instantaneous Velocity vs
Average Velocity
instantaneous velocity at t=2 is 1 ms-1
Instantaneous velocity at t=3 is 0 ms-1
Instantaneous velocity at t=4 is __________
Instantaneous velocity at t=8 is __________
Instantaneous velocity at t=2 is undefined since it is
different at 2+ (slightly > 2) and 2- (slightly < 2).
Average velocity between t=0 and t=2 is 1 ms-1
Average velocity between t=0 and 7 is _____________
Average velocity between t=7 and 9 is _____________
Average velocity between t=0 and 9 is _____________
S(m)
2
2
7
9
T(s)
s
Exercise
6
4
3
1.5
v
10
13
t
Plot the v-t graph below:
1
0.5
t
Exercise
s
t
t1
t2
t3
t4
Time instants at which velocities are positive: _________
Time instants at which velocities are negative: ________
Compare velocity at t1 and velocity at t2:___________
Velocity at t3 = ______________
Realistic Displacement-Time
s
curve
Discontinuous
velocity (not realistic)
t
v
Velocity gradually increases
(realistic)
Red curves are unrealistic since the velocities
are discontinuous (implies infinite acceleration)
t
Constant Acceleration
v
t
a
Motions Equations for
Constant Acceleration
• 5 variables :
–
–
–
–
–
t
u
v
a
s
time
initial velocity
final velocity
acceleration (constant)
displacement
Motion Equations (constant
acceleration)
velocity
v
v = u + at
u
t
time
v2 = u2 + 2as
1 2
s  ut  at
2
displacement
s
t
time
Example
An object moving in a straight line with
constant acceleration takes 10 s from rest to
cover a distance of 100 m. Determine the
acceleration of the object.
By using the equation
1 2
s  ut  at
2
• u=0
• t = 10 sec
• s = 100 m
(Ans) a = 2 m/s2
Example
A particle with u = 80 m/s and zero
acceleration for first 5 sec. The particle is
then slowed down with acceleration of -15
m/s2. How far it has travelled after 5 sec
more? Find its velocity at that time.
1 2
s  ut  at
2
• In last 5 sec,
• s = 80(5)+0.5(-15)(5)2 = 212.5
m
• But it has travelled 80x5 = 400
m in the first 5 sec
Example (continued)
• Total distance travelled = 400 + 212.5
=612.5 m
• Velocity at that time
• v = u + at
• v = 80 + (-15)5
• = 5 m/s
Motion under the action of gravity
• The acceleration due to gravity引力 g is
the acceleration of a freely falling object
as a result of a gravitational force. For
most practical purpose is taken as being
9.81 m/s2 at the surface of the earth.
Why is there Gravity?
(this slide will not be tested)
Newton hit by an apple. Why
does the apple fall downwards?
Law of Gravity萬有引力
m1
r
There is gravitational force
between 2 masses
m2
Example
An object is thrown vertically upwards with a
velocity of 8 m/s. What will be the maximum
height it reaches and the time taken for it to
Positive
reach that height ?
• u = 8 m/s
• a = g = -9.81 m/s2
• v = 0
at the
maximum height
direction
g
Example (Continued)
Apply
v2 = u2 + 2as
0 = 82 + 2(-9.81) s
s = 3.26 m
Also
( The maximum height )
v = u + at
0 = 8 + (-9.81) t
t = 0.82 sec.
Force
• A force cannot be seen, only the
effect of a force on a body may be
seen.
• Force Units: S.I. Unit ,Newton, (N)
or (kN)
• Force is a vector quantity. It has
both magnitude and direction.
Newton
•Born 1643
•Newton’s Laws
•Gravitational Force
•Calculus
Newton’s First Law
(Law of Inertia)
• First Law – Every body will remain at
rest or continue in uniform motion in a
straight line until acted upon by an
external force.
• Inertia慣性: tendency for a body
maintains its state of rest or move at
constant speed
• The greater the mass質量, the larger
is the resistance to change.
Life Examples
Once the dummy is in motion,
it will be in motion
Astronauts in Space Shuttle.
Observe instances of Law of
Inertia in the following video clip.
http://spaceflight.nasa.gov/gallery/video/living/net56/fun_destiny_56.asf
Newton’s Second Law
• When an external force is applied to a
body of constant mass. It produces an
acceleration which is directly proportional
to the force
• The large the mass, the more force it
takes to accelerate it.
• The large the force, the larger the
acceleration.
• Force (F)= mass (m) x acceleration (a)
Video Example
• When a given tension (force) is given to a
slider without friction (with air track).
View 1-2 video of acceleration from
http://www.doane.edu/Dept_Pages/PHY/P
hysicsVideoLibrary/videolibrary.html
(Use the flash version if you do not have
quicktime.)
Constant force provided by
falling object
Example
A net force of 200 N accelerates an
object with a mass of 100 kg. What is
the acceleration?
F = 200 N
m = 100 kg
F=ma
a=F/m = 2 m/s2
mass
meter
Newton’s Third Law
• Every action produces an equal and
opposite reaction.
• Action and Reaction
Action Force = Reaction Force
Life Example
Life Example (continued)
Concept map
Gravity=9.81
Newton's
third law
m/s2
Uniformly accelerated motion
v = u + at
1
s  ut  at 2
2
v2 = u2 + 2as
Action = Reaction
Force
Motion
Linear
Motion
Scalars and Vectors
Newton's law
Newton'first law
Law of Inertia
Newton's
second law
F=ma