Download Mechanics III

Inertia 1
Force 1
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Inertia
• Aristotle’s old belief
• Constant speed motion requires constant force
• Galileo’s law of inertia
Photo
• It is as natural for a moving object to keep moving
with a constant speed along a straight line as for a
stationary object to remain at rest.
Inertia 2
Force 1
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Inertia and Newton’s first law of motion
• Galileo’s thought experiment
Diagrams
• Newton’s first law of motion:
• Every object remains in a state of rest or uniform
speed along a straight line (constant velocity) unless
acted on by an unbalanced force.
Photo
Inertia 3
Force 1
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Inertia and Newton’s first law of motion
• Force : changes the state of rest or uniform motion
of an object (vector! Why?)
• Inertia : the resistance of an object to a change in
its state of rest or uniform motion in a straight line.
• Mass : is a measure of inertia (scalar! Why?)
• Mass can be considered as a measure of inertia
• Inertial balance Diagrams
Force 1
Force 1
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Friction
• Origin of wrong concept of Aristotle’s old belief
• Nature of friction
Diagrams
• Direction : opposite to the motion (velocity)
• Examples of smooth surface
Photo
• Demonstration of Newton’s first law on smooth
surfaces
Force 2
Force 1
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Unbalanced force (resultant force, net force)
• Falling of object in liquid (terminal velocity)
Photo
• Its relation with Newton’s first law of motion Diagram
• Friction-compensated inclined plane
• Experiment to study resultant force
• Results
Calculation
Photo
Diagram
Force 3
Force 1
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Newton’s second law of motion
• Acceleration is not zero if the resultant force
is not zero.
• Deductions from the results
1. Acceleration  force (mass is constant)
2. Acceleration  1/mass (force is constant)
3. F  ma or F = constant  ma
Force 4
Force 1
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Newton’s second law of motion
• Definition:
The acceleration of an object is directly proportional
to, and in the same direction as, the unbalanced
force acting on it, and inversely proportional to the
mass of the object
• Unit of force : Newton (N)
Force 5
Force 1
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Newton’s second law of motion
• 1 newton of force will give a mass 1 kg an
acceleration 1 m s  2
• constant in F = constant  ma becomes 1
• Mathematical form of Newton’s second law
F  ma
•Direction of resultant force = direction of acceleration
Weight & Mass 1
Force 1
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Weight and Mass
• Definition:
The force of gravity acting on the object is called
the weight of the object and is measured in newton
• Acceleration in free falling = 10 m s-2
W  mg
1 kg of mass has a weight 10 N (downwards)
Weight & Mass 2
Force 1
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Weight and Mass
• Instruments to measure weight and mass
Photo
• Weightlessness
• Discussion about the restrictions of the above
machine Calculation
Addition of Force 1
Force 1
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Addition of Forces (vectors)
•More than one force acting on an object
•Add them together to get ONE resultant force
•F = ma can only be applied for resultant force
•Tip-to-tail method (Revision)
•Example 1 Calculation
•Example 2 Calculation
Addition of Force 2
Force 1
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Addition of Forces (vectors)
•Method of resolving components
•Adding forces or vectors without drawing diagrams
•Example Calculation
•Examples for components of forces
Calculation
END of Force 1
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Force 1
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•Galileo Galilei (1564 - 1642)
Inertia 2
Force 1
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• Small bearing is released from rest on a smooth track at
A.
A
C
D
E
F
• A ball reaches point C which is of the same height
• Same situation for D and E
• If the track is infinite long, the ball will never stop.
Inertia 2
Force 1
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• Galileo’ pin-and-pendulum experiment
• Consider the swing of a simple pendulum
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Force 1
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• The bob rises to the same height as before
• Even we have a pin, the bob rises to the same height
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Force 1
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•Isaac Newton (1642-1727)
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Force 1
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• We set the platform into vibration and record the
period.
• Fix load on the platform and repeat the vibration,
we find that a longer period can be found.
• The larger the load, the longer the period.
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Force 1
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• Friction is caused by the interlocking of surface
irregularities.
Force 1
Force 1
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• A mass placed on a
thin layer of
polystyrene beads on a
glass plate
• A balloon is blown up
and attached to a short
pipe
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Force 1
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• Air-layer Ball
• Motion on a air track
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Force 1
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• An object is falling
inside liquid.
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Force 1
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• The object is falling downwards with constant velocity. Do
you know why?
• Liquid resistance is equal to the weight
• No unbalanced force
liquid resistance
weight
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Force 1
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• An inclined plane is prepared so that when we give
the trolley a hard push, it moves down with constant
velocity. It is called to be friction-compensated.
• Careful adjustment for the plane is needed to
achieve this situation.
Force 2
Force 1
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• Identical elastic strings are used to pull the trolley.
• At first, we use one string and then two, and three.
• We always maintain the same length for all the strings so
that each string produces the same force.
• The accelerations in each case are recorded.
trolley
Friction-compensated
inclined plane
elastic string
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Force 1
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• One elastic string is used to pull several trolleys.
• At first, we use one trolley and then two, and three.
• We always maintain the same length for the string so that
the string produces the same force in each case.
• The accelerations in each case are recorded.
elastic string
friction-compensated inclined plane
Force 2
Force 1
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• Different tape charts for different no. of strings with one
trolley are shown.
1 string
2 strings
a = 2 m s-2 a = 4 m s-2
3 strings
a = 6 m s-2
aF
if m  constant
• We find that the acceleration is directly proportional to
the no. of strings used (Force) when the mass of trolley is
kept constant.
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Force 1
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• Different tape charts for different no. of trolleys with one
string are shown.
1 trolley
2 trolleys
3 trolleys
a = 2 m s  2 a = 1 m s  2 a = 0.67 m s  2
1
a
m
if F  constant
• We find that the acceleration is inversely proportional to
the no. of trolleys used (mass) when 1 string is used
(constant force).
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Force 1
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• Beam balance
(measure mass)
• Spring balance
(measure weight)
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Force 1
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• Can we use the beam balance or spring balance on
Moon to get correct readings of mass and weight of an
object with 1 kg mass?
• The acceleration due to gravity on Moon is only
about 1.8 m s-2.
• 1 kg slot-mass is still needed to balance the object.
• The reading from the spring balance = 1  1.8 = 1.8 N!
• Mass is the same anywhere while weight depends
on position and is not a constant even for the same
object.
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Force 1
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• Two forces 3 N and 4 N are acting on an object (2 kg) as
shown below. What are the resultant force and acceleration?
3N
2 kg
4N
N
• Use a scale of 1 cm to 1 N to
draw the forces in the form of
arrows. The direction of the force
is indicated by the arrow.
3 N (3 cm
in length)
4 N (4 cm in length)
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Force 1
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• Attach the tip of an arrow to the end of another arrow.
(Tip-to-tail method)
• Draw an arrow from the
starting point to the end
5 N (5 cm)
point. It is the net force.
3
N
(3
cm
53.1°
in length) • Length of the arrow : 5 cm
• Direction : N 53.1°E
4 N (4 cm in length)
• Direction of net force : N53.1°E
• Magnitude of net force : 5 N (Why?)
• Direction of acceleration : N53.1°E (Why?)
F
• Magnitude of acceleration :  2.5 m s - 2
m
Addition of Force 2
Force 1
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• By using the concept of tip-to-tail method, one force
can also be separated into two different forces, for
example,
• Use a scale of 1 cm to 1 N
10 sin 60°N
(10 sin 60°cm)
10 N (10 cm)
60°
10 cos 60° N
(10 cos 60° cm)
Addition of Force 2
Force 1
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• We want to add the following two forces using the
method of resolving components. Scale : 1 cm to 1 N
5N
60°
6N
30°
Addition of Force 2
Force 1
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y
6N
5N
60°
30°
• We place them together on
xy-coordinate plane with
both the nails at the origin.
x
y
5 sin 60°N
6 sin 30° N
• Each force can be
represented by a force
(component) along x-axis and a
component along y-axis.
30°
x
5 cos 60°N
6 cos 30°N
Addition of Force 2
Force 1
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y
• Add the components along x-axis
together. Then add the components
along y-axis.
6 sin 30° N
+ 5 sin60° N
6 cos 30° N - 5 cos 60° N
x
• They are of the same direction and we
can add them like scalars.
Addition of Force 2
Force 1
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• Combine the components of force along each axis to
form the net force vector.
y
6 cos 30° N
- 5 cos 60° N
6 sin 30° N + 5 sin 60° N

x
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Force 1
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• Magnitude of the net force:
(6 sin 30  5 sin 60) 2  (6 cos 30  5 cos 60) 2 N
 7.81 N
• Direction of force ():
tan   6 sin 30  5 sin 60
6 cos 30  5 cos 60
   69.8
Addition of Force 2
Force 1
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• HK Convention & Exhibition Centre
• By resolving components, the roof will not
fall even no support is directly below the roof
Addition of Force 2
Force 1
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• Hong Kong Space Museum
• By resolving components, the roof will not
fall even no support is directly below the roof
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Force 1
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• The top of a tunnel