1.5 Newton`s Law of Motion

```Newton's law of motion
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Newton’s second law ( law of inertia )
Every body remains in a state of rest or uniform
velocity unless acted on by a resultant / net /
unbalanced force.
i.e. F = 0  Dv = 0
Inertia is the tendency of a body to maintain its state
of rest or of constant velocity.
Mass is a measure of inertia. A large mass requires
a large force to produce a particular acceleration.
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Newton’s second law ( law of dynamics )
The rate of change of momentum of a particle is
directly proportional to the resultant force acting on it.
i.e.
F ∝ d(mv)/dt or F = k d(mv)/dt
If the mass is constant, F = kmd(v)/dt = k ma
i.e. The acceleration of an object is directly
proportional to the unbalanced force acting on it.
1 N is defined as the force which gives a mass of 1 kg
an acceleration of 1 ms-2 .
∴ k = 1 and F = ma
For F = 0  a = 0 if m ≠ 0  Dv = 0  v is constant.
i.e.
First law is a special case of second law.
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Sand drops vertically with 2 kgs-1 on to a conveyor belt
moving horizontally with 0.1 ms-1.
Calculate the external force required to keep belt
moving with 0.1 ms-1.
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By F = d(mv)/dt = v(dm/dt)
(v is constant)
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External force required
= 0.1 x 2 = 0.2 N
v = 0.1 ms-1
F
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A jet of water emerges from a hose of cross-section
area 5 x 10-3 m2 with a speed of 0.3 ms-1 and strikes a
wall at right angles.
Calculate the force exerted on the water by the wall
assuming the water in brought to rest and does not
rebound. (Density of water = 1000 kgm-3)
0.3 m s -1
5 x 10-3 m2
water
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0.3 m
In 1 s
5 x 10-3 m2
water
In 1 second:
 Volume of water striking the wall
= 5 x 10-3 x 0.3 = 0.0015 m3
 Mass of water striking the wall
= 1000 x 0.0015 = 1.5 kg (density = mass / volume)
 The force exerted on the water
= rate of change of momentum of the water
 = (mv – mu) / t = (0 – 1.5 x 0.3) = -0.45 N
 Negative sign means the force is opposite in direction to
the speed of water hitting the wall.
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Mass
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The mass m of an object is a measure of its
inertia.
Inertia is the resistance of an object to change
its state of rest or uniform motion.
An object with the greater mass is more difficult
to start or stop moving.
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 Weight
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The weight W of an object is the gravitational
force acting on it towards the centre of the earth.
Weight is a force and W = mg.
The weight of a body is proportional to its mass.
Thus, we can find the mass of a body m by
comparing its weight with that of a standard
masse m0 on a beam balance (天秤).
In the same place, W = mg and W0 = m0g.
If the beam balance is in equilibrium, W = W0 ⇒
m = m0.
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Mass
Weight
varies with position
on the earth’s surface
independent of the presence decreases if the object becomes
or absence of the earth
far away from the earth
scalar
vector
constant
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Newton’s third law ( Action and Reaction )
If body A exerts a force on body B, then B will
exert an equal and opposite force on A.
i.e. FAB = FBA
Forces never occur alone but always in pairs.
Action and reaction act on different bodies.
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A worker of mass 60 kg standing on a lift pulls down a
rope with a force. The mass of the lift is 20 kg and the
acceleration of the lift is 0.5 ms-2.
(a)
Draw free body diagrams for the man and the lift.
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Free body diagrams
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(a) Free body diagrams
T
R T
R: reaction from the lift
R’: from acting on the lift
by the man
T: tension
Mg: weight of the man
mg: weight of the lift
Mg
mg
Any action and
reaction pair?
R’
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(b) Find the tension of the rope and normal reaction
on the man.
R = R’ (Action &amp; Reaction)
T
For the man,
R T
R + T – Mg = Ma
R + T – 600 = 60 x 0.5
R + T = 630 --- (1)
For the lift,
T – mg – R’ = ma
T – 200 – R = 20 x 0.5
T – R = 210 --- (2)
Mg
(1) + (2):
mg
2T = 840 ⇒ T = 420 N
(1) – (2):
R’
2R = 420 ⇒ R = 210 N
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Example 4
A long trolley of mass M with a flat top is placed on a smooth
horizontal bench. A block of mass m is projected horizontally
on the top of the trolley with speed u and the coefficient of
kinetic friction is m.
u
(a)
(b)
Draw free body diagrams for the block and the trolley.
(i)
the deceleration of the block, the acceleration
of the trolley and the distance traveled
by the
block before it stops moving on
the trolley.
(ii)
the final common velocity of the system.
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Example 4
(a)
Draw free body diagrams for the block and the trolley.
u
R
f
f’
mg
R’
f: frictional force acting on the block
f’: frictional force acting on the trolley
mg: weight of the block
Mg: weight of the trolley
R: reaction on the block
R’: force acting on the trolley from
the block
S S: reaction on the trolley from the
floor
Any action and
reaction pair?
Mg
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(b) (i)
Find the deceleration of the block and the
acceleration of the trolley before the block stop moving.
R
f
S
f’
mg
R’
For the block,
f = mR and R = mg
f = mmg
Deceleration = f/m = mg
Mg
For the trolley,
f’ = f (action and reaction)
Acceleration = f’/M = mmg/M
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Find the distance traveled by the block before it stops
moving on the trolley.
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For the block,
Initial velocity = u
Final velocity = 0 (relative to the trolley)
Acceleration = -mg
By 2as = v2 – u2
2(-mg)s = 02 – u2
s = u2/(2mg)
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(b)
(ii)
the final common velocity of the system.
u
Common velocity v
At rest
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By conservation of momentum,
mu + 0 = (m + M)v
v = mu/(m + M)
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