Download NEWTON`S LAWS OF MOTION 1 Newton`s Laws of Motion

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CHAPTER 3. NEWTONS LAWS OF MOTION
INSTITIÚID TEICNEOLAÍOCHTA CHEATHARLACH
INSTITUTE OF TECHNOLOGY CARLOW
NEWTON’S LAWS OF MOTION
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Newton’s Laws of Motion
The study of mechanics is based on three laws which were first formulated by Newton.
Every body will remain at rest or continue to move with uniform velocity unless an
external force is applied to it.
When an external force is applied to a body of constant mass the force produces an
acceleration which is directly proportional to the force.
When a body A exerts a force on a body B, B exerts an equal and opposite force on A.
Force (a vector quantity) is necessary to make an object begin to move.
Consider a body of mass m which is falling under the action of its weight only.
Weight is the gravitational force g ms−2 acting downwards.
y
y
?
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W eight
Acceleration g ms−2
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CHAPTER 3. NEWTONS LAWS OF MOTION
The weight of a body is a force. (Newtons)
Mass is a measure of the matter contained in an object. (Kg)
Before considering problems concerning the effect of a constant force on a body – we consider some
diagrams to show clearly all forces acting on a body in different situations.
Example
T − T ension
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T
W − W eight
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Example
R
D
]
J
J
3 J
J J
JJ J
JJ J
F
+
?
W
R − Reaction
W − W eight
D − Driving F orce
F − F riction
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CHAPTER 3. NEWTONS LAWS OF MOTION
1.1
Newton’s Second Law of Motion
When an external force is applied to a body of constant mass the force produces an acceleration which
is directly proportional to the force.
F = ma
The resultant force acting on a body of constant mass m (kilograms) is equal to the mass of the body
multiplied by its acceleration a (ms−2 ). Note that force and acceleration are both vector quantities
hence the equation F = ma is a vector equation. Therefore as well as the magnitudes of both sides,
force and acceleration have the same direction. The unit of force is defined to be that force which gives
a mass of 1 kg an acceleration of 1 ms−2 . This unit of force is called a newton N.
Using F = ma we have that the weight of the falling object is mg Newtons. Therefore a body of
mass m kilograms has a weight of mg Newtons.
Example A particle of mass 5Kg slides down a smooth plane inclined at 30o to the horizontal. Find
the acceleration of the particle and the reaction force between the particle and the plane.
R
]
JJ
J
J
Jy
W
30o
F orces
?
As the acceleration is down the plane, the resultant force is also down the plane. (It is the resultant
force that causes the acceleration).
If we resolve the downward force of 5gN into components that are parallel and perpendicular to the
plane we can state that the resultant force down the plane is given by 5g sin 30o . Now, from Newton’s
Second Law of Motion
∴
F
=
ma
5g sin 30o
=
5a
g
2
⇒
a =
4
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CHAPTER 3. NEWTONS LAWS OF MOTION
The acceleration of the particle is g2 ms−2 down the plane.
To determine the reaction force R between the particle and the plane we have that the forces are in
equilibrium
R − 5g cos 30o
=
0
∴
R
=
⇒
R
=
5g cos 30o
√
5g 3
2
The reaction force R is 42 · 47N perpendicular to the plane.
Example A block of mass 2Kg rests on the floor of a lift which has an acceleration of 5ms−2 upwards.
Find the reaction between the block and the lift.
R
6
F orces
W
?
There is no component of acceleration horizontally.
The resultant upward force on the block is R − 2g. Using F = ma we have
R − 2g
=
2(5)
R
=
10 + 2g
R
=
29 · 6
∴
⇒
Therefore the reaction between the block and the lift is 29 · 6N .
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CHAPTER 3. NEWTONS LAWS OF MOTION
Note For the friction force F we have
F = µR
where µ is the coefficient of friction which will be given and depends on the surfaces involved in the
problem.
6R
Direction of M otion
-
F
When a surface is described as ‘smooth’ we ignore friction at the contact, however when a surface
is described as ‘rough’ means that there is friction at the contact.
Example A cyclist is riding up a hill inclined at 20o to the horizontal. His speed at the foot of the hill
is 10ms−1 but after 30 seconds it has dropped to 4ms−1 . The total mass of the cyclist and his machine
is 100kg and there is a wind of strength 15N down the slope. Find, correct to 3 significant figures, the
constant driving force exerted by thee cyclist up the slope.
BM R
B
B
B
B
B
B
B
D
B
B
B
:
B F
9
o
20
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CHAPTER 3. NEWTONS LAWS OF MOTION
As the acceleration is up the plane, the resultant force D is also up the plane. Firstly to the
acceleration of the cyclist up the slope. We have
v
= u + at
⇒
4
=
⇒
a
= −0 · 2 ms−2
10 + 30a
The cyclist has an acceleration of −0 · 2ms−2 up the plane.
If we resolve the downward force of 100gN into components that are parallel and perpendicular to
the plane we can state that the resultant force up the plane is given by
D − 15 − 100g sin 20o
Using F = ma, we have
D − 15 − 100g sin 20o
=
100a
D
=
330 · 1
The cyclist driving force is 330N up the plane.
Example A particle of mass 5Kg is pulled along a rough horizontal surface by a string which is
inclined at 60o to the horizontal. If the acceleration of the particle is
friction between the particle and the plane is
2
3,
g
−2
3 ms
and the coefficient of
find the tension in the string.
R
T
6
F
60o
?
W
As there is no component of acceleration vertically there is no resultant force vertically.
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CHAPTER 3. NEWTONS LAWS OF MOTION
Therefore, the vertical forces are in equilibrium, i.e.,
R + T sin 60o − 5g = 0
(1)
A friction is limiting, we have
F =
2
R
3
(2)
The horizontal forces are as follows
T cos 60o − F
Using F = ma we have
T cos 60o − F =
5g
3
(3)
Eliminating F and R from (1),(2) and (3) yields
2
5g
T cos 60o − (5g − T sin 60o ) =
3
3
Solving for T yields
√
√
T = 10g 3(2 − 3)N
Example A car of mass 1000Kg is brought to rest from a speed of 40ms−1 in a distance of 80m. Find
the breaking force B of the car assuming that it is constant and that there is a constant resistance to
motion of 100N .
6R
Res
B
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CHAPTER 3. NEWTONS LAWS OF MOTION
Now u = 40ms−1 , v = 0ms−1 and s = 80m.
v2
= u2 + 2as
⇒
0
=
∴
a
= −10ms−2
1600 + 2a(80)
Therefore the car has an acceleration (deceleration) of −10ms−2 . The horizontal force is B + Res.
Using F = ma we have
B + Res = 1000(10)
But Res = 100N , therefore
B = 10, 000 − 100 = 9900N
Therefore the breaking force of the car is 9900N .
Exercise A car is being pulled along a horizontal track by two cables, against resistances totalling
1100N , with an acceleration of 0 · 8ms−2 . One cable is horizontal and the other is inclined at 40o to
the track. The tensions in the cables are shown in the diagram. Taking g = 9 · 81ms−2 , find
i the mass of the car,
ii the reaction force R exerted by the track on the car.
6R
1000N
3
1100N
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800N
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CHAPTER 3. NEWTONS LAWS OF MOTION
Exercise A wooden box of mass 8Kg is slides down a rough horizontal surface which is inclined at
α = sin−1
1
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to the horizontal. If the acceleration of the particle is 1ms−2 find the coefficient of friction
µ between the particle and the plane.
R
KA
F
A
A
*
A
A W
α
?
[Solution: µ=0.0656]
Exercise A crate of weight 8N is resting in rough contact with a plane inclined at an angle α to
the horizontal where tan α =
3
4.
The coefficient of friction between the crate and the plane is µ. A
horizontal force P newtons is applied to the crate. When P = 16N the crate is on the point of slipping
up the plane. Find the coefficient of friction µ.
R
KA
A
A
P
A
- A
F
W
α
?
[Solution: µ = 0 · 5]