1. Mechanics - 2
For more information about Newton’s
laws and applications search the
presentation- Mechanics-1 By Aditya
Abeysinghe
Mechanics-2 By Aditya Abeysinghe
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2. Tension of a string
Tension is the external force that act on objects
connected to the string or the internal forces that
develop inside the string.
F (internal force to the
T
string)
F (External force
to the object)
mg
Mechanics-2 By Aditya Abeysinghe
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3. Different types of tension
1. Tension in accelerated environments
2. Tension in non-accelerated environments
a. Normal tension
b. Tension in inclined planes
c. Tension in pulley systems
Mechanics-2 By Aditya Abeysinghe
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4. Tension in non-accelerated environments
Normal tensionSingle object
T
F
T > F for the
object to move
Mg
* Note: For the next examples, the friction on the
object(s) is/are not considered.
Mechanics-2 By Aditya Abeysinghe
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5. Two objects
For M1 ,
F = ma
T = M1 a
a
T
T
For M2,
F = ma
F – T = M2a
①
②
F
①+②,
M1g
F = (M1 + M2 )a
M2g
Therefore, T = M1F/ (M1 + M2 )
Three or more objects
T2
T1
M3g
5
Mechanics-2 By Aditya Abeysinghe

6. For M1,
For M2,
For M3 ,
F = ma
F = ma
F = ma
F – T1 = M1a ① T1 – T2 = M2a ② T2 = M3a ③
①+②+③
F = (M1 + M2 + M3 )a
Therefore, T1 = (M2 + M3 )F/ (M1 + M2 + M3 )
T2 = M3F/ (M1 + M2 + M3 )
Mechanics-2 By Aditya Abeysinghe
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7. Tension in pulley systems
T”
a
T
M1g
M2g
For M1, F =ma
M1g – T = M1a ①
For M2, F = ma
T– M2g= M1a ②
①+②
a = (M1 - M2) g / (M1 + M2)
Therefore,
T = 2M1 M2 g / (M1 + M2)
T” = 2T
T” = 4M1 M2 g / (M1 + M2)
Mechanics-2 By Aditya Abeysinghe
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8. Tension in inclined planes
F = ma
Mg Sinθ = Ma
a = g Sinθ
F = ma
R – Mg Cosθ = M × 0
R = Mg Cosθ
θ
Mg
Mechanics-2 By Aditya Abeysinghe
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9. Tension in accelerated environments
T
θ
a
mg
For m,
F = ma
T Sinθ = ma ①
F = ma
T Cosθ –mg = m× 0
T Cosθ = mg ②
①/② , tanθ = a/g
Mechanics-2 By Aditya Abeysinghe
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10. T
For m,
F = ma
T-mg = ma
a
mg
T = m(g + a)
Pulley Systemsa
T
m1 a
T
m1 g
F
m2 a
Mg
M’g

11. For m,
F = ma
T = m1a
For M,
F = ma
T = Mg
Thus, m1a = Mg
Therefore, a = (M/m1)g
By considering the whole system,
F =ma
F = (M + M’ + m1) a = (M + M’ + m1) (M/m1)g
Inclined planes –
a
maCosθ
maCosθ = mgSinθ
ma
θ
mg
mgSinθ
a = g tanθ
Thus, a α θ
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12. Apparent weight in a lift
Consider the following occasions:
N
1. The lift is at rest or moving with constant velocity
For man, F=ma , N – Mg = 0 OR N = Mg
(Apparent weight equals true weight)
2. The lift accelerates upwards
For man, F = ma , N – Mg = Ma OR N = M (g+a)
(Apparent weight is more than true weight)
Mg
3. The lift accelerated downwards
For man, F =ma , Mg – N = Ma OR N = M(g-a)
(Apparent weight is less than true weight, for g > a )
4. The lift falls under gravity
For man, F = ma , Mg – N = Mg OR N = 0
(Apparent weight is zero)
Mechanics-2 By Aditya Abeysinghe
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13. Contact Force
(Masses in contact)
a
M
F
m
For M, F = ma
F – R = Ma
For m, F = ma
R = ma
①+②
①
②
Thus,F = (M + m)a
And, a = F / (M +m)
M
R
Internal force
, retards the
object
R
m
R – Contact
Force
Internal
force, accelerat
es the object
Method II:
By applying F = ma to the whole system, the contact force
can be ignored.
Thus, F = (M + m)a
Mechanics-2 By Aditya Abeysinghe
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14. Motion of a body on a frictionless
inclined plane
Mgsinθ = Ma
Therefore, a = gsinθ
R
R = MgCosθ
a
θ
Mg
Mechanics-2 By Aditya Abeysinghe
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15. A body moving down a rough inclined
plane
When a body is moving down a rough inclined plane, the frictional force,
which opposes the motion, acts upwards.
μR
R
a
θ
R = MgCosθ
However, F = μR
Therefore, F = μ MgCosθ
By applying F = ma,
MgSinθ – μMgCosθ = Ma
OR
a = g(Sinθ – μCosθ)
Mg
Mechanics-2 By Aditya Abeysinghe
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16. A body moving up a rough inclined
plane
When a body is moving up a rough inclined plane, the frictional force,which
opposes the motion, acts downwards.
R
a
θ
Mg
μR
R = MgCosθ
However, F = μR
Therefore, F = μ MgCosθ
By applying F = ma,
MgSinθ+ μMgCosθ = Ma
OR
a = g(Sinθ + μCosθ)
Acceleration also lies downwards as the object
retards and becomes stationary on its way up.
Mechanics-2 By Aditya Abeysinghe
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17. Flying of birds
A bird flies by displacing air below its wings.
F
The force due to the change of
momentum is the force given to
air by the wings of the bird.
If F > Mg , it flies up.
Similarly, if F = Mg , it stays
stationary.
And, if F < mg, it flies down.
A
V
mg
If V is the speed of the bird, the
distance it travels within a unit
amount of time is also V.
V
However, force = rate of change of
momentum.
Therefore, F = mv /1 = ρAV × V = ρAV2
F = ρAV2
Then, the volume of air displaced by
the bird within a unit time is AV.
However, m = ρ × Volume (ρ – density
of air).
Thus, m = ρAV
Mechanics-2 By Aditya Abeysinghe
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18. Movement of a bicycle
Movement of a bicycle is due to the movement of
the body of a bicycle on wheels.
When applying brakes frictional
forces are arranged as follows
F1
First a frictional force occurs
when the whel tries to move
backward relative to Earth
F2
F
Due to this F1 force the body moves
forward which causes the front wheels
to develop a F2 frictional force
Mechanics-2 By Aditya Abeysinghe
F
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19. The momentum or the speed of an object changes only
when an external unbalanced force acts on it
Mass of trolley – m
u
v
Mass of man - M
F
v
F
Although when considered as a whole, there’s no unbalanced
force, the frictional force F which acts on the trolley decelarates
the trolley while the frictional force which acts on the object
accelerates the object
Mechanics-2 By Aditya Abeysinghe
Since there’s no
unbalanced force
within the system,
Applying Principle of
Conservation of Linear
Momentum to the
system,
mu = (m + M)V
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20. But if a person leaves the trolley perependicularly
while the trolley is moving, trolley’s speed will not
be affected, since there’s no generation of
frictional forces.
Rod
Rod
v
u
From Principle of Conservation of Linear Momentum,
mu + Mu = Mu + mv
Solving this equation we get,
u=v
Mechanics-2 By Aditya Abeysinghe
Thus the speed is
unaffected when
there’re no frictional
force in both the
objects
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