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Mechanics - 2
For more information about Newton’s
laws and applications search the
presentation- Mechanics-1 By Aditya
Abeysinghe
Mechanics-2 By Aditya Abeysinghe

1
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

2
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

3
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

4
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
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

6
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

7
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

8
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

9
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
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
11
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

12
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

13
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

14
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

15
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

16
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

17
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

18
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
19
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

20

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Mechanics 2

  • 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 1
  • 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 2
  • 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 3
  • 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 4
  • 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 6
  • 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 7
  • 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 8
  • 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 9
  • 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 11
  • 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 12
  • 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 13
  • 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 14
  • 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 15
  • 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 16
  • 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 17
  • 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 18
  • 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 19
  • 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 20