# aieee-notes_aieee-notes-physics_03 by anshu8765

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and other Entrance Exams
PHYSICS Notes & Key Point
LAWS OF MOTION                                                                                                                                   Chapter - 3

1. Newton’s laws of motion
(a) First law of motion : This law states that a body continues to remain in its state of rest or of uniform motion in a straight line, unless it is
compelled to change the state by external forces. It is also known as law of inertia.
(b) Second law of motion : This law states that the rate of change of momentum of a body is equal to the net external force acting on it and
dp d (mv)     dv   dm
=
takes place in the direction in which the force acts. i.e., F = =m +v
dt    dt      dt   dt
dm                  dv
(i) If mass of an object is constant, then    = 0 and     F=m     = ma .
dt                  dt
dm
(ii) If mass of the body into consideration is not a constant, then       v          ≠ 0.
dt
Some examples of variable mass situations are
• Rocket propulsion where the rocket moves forward due to thrust gained by ejecting exhaust gases backwards. The mass of
the rocket thus reduces continuously.
•   A truck full of sand moving with sand continuously spilling out through a hole.
• A truck moving with rain water continuously getting collected into the truck.
(c) Third law of motion : This law states that to every action, there is always an equal and opposite reaction.

2. Inertial and Non-Inertial frames
(a) Inertial frame: A frame of reference relative to which Newton’s laws of motion are valid is called an inertial frame.
in
Such a frame has a zero acceleration, i.e., such a frame is either stationary or moves with a uniform velocity.
co.
(b) Non-Inertial frame: A frame of reference relative to which Newton’s laws are not valid is called a non-inertial frame.
.
e rs
Such a frame has a non zero acceleration. A lift that is moving up or down but accelerating or retarding is a non inertial frame.

ach
e te
3. Common forces acting on a body

in
In most cases that we encounter in everyday life, some of the forces that act on the bodies are
• Weight         • Normal Reaction                • Tension              • Friction
n   l
//o
(a) Weight : It is the gravitational pull of the earth and is given by W = mg
:
t tp
The direction of the force acting on the body is vertically downwards.
h
m
(b) Normal Reaction : When two surfaces are in contact, they press each other with a certain force which is denoted by N or R .
ro
df
The direction of the normal force is always perpendicular to the surface and tends to push the body away from the surface.
(c) Tension : When a cord (string, rope or cable) is attached to a body and pulled taut, the cord is said to be under tension.
e
It pulls on the body with force T and it ‘s direction is away from the body and along the cord at the point of attachment.
nl
Unless otherwise stated, a cord is assumed to be massless and unstretchable. Under these conditions, it exists only as a connection
w
Do
between two bodies and it pulls both bodies at each end with same magnitude, T .
(d) Friction : If we move or try to move a body over the surface of another body, the motion is resisted by bonding between the surfaces
of two bodies. This resistance is called the frictional force or the force of friction.
The force of friction acts along the surface.
Let F be a variable horizontal force that attempts to slide a block of mass m kept stationary on                                F   m
a rough horizontal surface. Let f be the resulting frictional force as shown in the figure.
(i) I f F i s t oo small to cause actual sliding, then t he fri cti onal f orce t hat acts on
f
the bod y i s k nown a s sta ti c f ri cti onal forc e an d i s deno t ed b y f s .
(i i ) Static frictional force, f s is parallel to the surface and equal in magnitude but opposite to F .
(iii) Static frictional force, f s has a maximum value beyond which it cannot increase. The maximum value is given by (f s ) max = µ s N
where, µ s = coefficient of static friction N = Normal force with which the two surfaces press each other.
.
(iv) If the magnitude of F exceeds (f s ) max , then the body begins to slide along the surface.
(v) If the body actually starts moving along the surface, then the force of friction is              ( f s ) max
called the kinetic friction or dynamic friction and is denoted by f k .
Force of friction, f

(vi) Kinetic friction, f k has a constant value and is independent of the velocity or                                      fk
acceleration of the body. It is given by f k = µ k N
where, µ k = coefficient of kinetic friction
N = Normal force with which the two surfaces press each other                 Applied force, F
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(vii) The value of f k is slightly less then (f s ) max . Hence, The value of µ k is
slightly less than µ s .
(viii)Therefore, we have 0 ≤ f s ≤ µ s N and f k = µ k N
(ix) The maximum value of static friction, (f s ) max is also known as limiting friction. Hence, limiting friction is the maximum opposing
force that comes into play when one body is just at the verge of moving over the surface of other body.
(x) The force of friction is independent of the area of contact of the two surfaces. It depends upon
Normal force with which the two surfaces press each other.
Coefficient of friction.
(xi) The cocfficient of friction ( µ s or µ k ) depends upon
Material of the two surfaces in contact
Roughness of the two surfaces in contact
(xii) The force of friction on a body always acts opposite to the direction of intended motion.
(xiii) According to Newton’s third law of motion, the frictional force acting on the two bodies in contact is opposite to each other in direction.
(xiv) If the relative velocity between two surfaces is zero, then static friction acts. Otherwise, kinetic friction acts.
(xv) If in a given problem, only µ is mentioned, we may consider µ s = µ k = µ .

4. Free body diagram (FBD)
A free body diagram is a stripped down diagram (of a body under consideration) in which we show
• all forces acting on the body with due consideration to the directions in which they point.
• direction of acceleration of the body.                                                      in
. co.
5. Summary of forces to be considered while making F.B.D
ers Direction of Force
Type of Force            Magnitude of Force
ach
Gravitation                                          mg
e te
Vertically downwards
nPerpendicular to the surface.
li
Normal or Reaction
N or R
/   on
p: /
(It pushes the body away from the surface)
Tension
hTt
t                   Away from the body in line with the cord (It pulls the body)

romfAs given
Friction                                         (see note)              Parallel to surface and opposite to the direction of intended motion

d
Any other force given in the problem      f                         As given
a  dealong the surface
Note : f = µ N if the body slides
k
lodoes not slide but a slightly greater force will make it slide (limiting case)
wn
f = µ N if the body
if o
s
f < µ N Dthe body does not slide and a slightly greater force will also not make the body slide
s

6. Some examples of F.B.D
(a) Blocks in equilibrium

m                                                        f
T                                                                                           m
m                                      mg
mg N
mg                                           N

(b) Blocks moving with acceleration

T                                                                       a
T
m     µk N
a m1 m 2 a
mg      N
m1g m g
2

m1 > m 2

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7. Approach to solving problems in Mechanics
Step- I : Choose the body. A body to which Newton’s laws are applied may be a particle, a block or a combination of blocks connected
by a cord, etc.
Step- II : Identify the forces that act on the body and draw a free body diagram showing
• all forces acting on the body
• direction of acceleration of the body if any.
.
Step- III: Choose axis. It is convenient to choose one of the axis (say X − axis) along the direction in which the body is likely to have
acceleration. A direction perpendicular to this may be chosen as the other axis (say Y − axis). Resolve all forces along the
two axes.
Step-IV : Write down separate equations along the two chosen axis.
If a is the acceleration of the body, then by Newton’s second law of motion,
• along x-axis, we have Σ Fx = ma                                • along y-axis, we have Σ Fy = 0
Solve the equations to obtain the desired result.

8. Some Solved Examples                                                                                                              F.B.D
T
−2
Example 1: A body of mass 6 kg is supported by a cord. Find the tension in the cord. ( Take g = 10ms )
Solution: Since, the body is at rest, its acceleration is zero.
⇒ T − mg = 0                        ∴ T = mg = 6 × 10 = 60 N
in
co.
mg
.
Example 2: A body of mass 6 kg is supported by a cord inside an elevator                           e rs
ach                        F.B.D
which is moving up with a constant acceleration 2 m/s 2 .
n     e te                                   T

Find the tension in the cord. ( Take g = 10ms )
2
nl=i 2m / s
/o a                                                   a
T − mg = ma                        /                           2

tp:
Solution: The equation is                                                                                     m
∴ T = m(g + a ) = 6(10 + 2) = 72 N
ht                                                              mg
m
d    f ro
deacceleration 2 m/s .
Example 3: A body of mass 6 kg is supported by a cord inside an elevator which
a                                                                                            F.B.D
nlo
2
is moving down with a constant                                                                                            T
w
Find the tensiono the cord. (Take g = 10m / s )
2
D in                                                                  a = 2m / s 2          m                      a
Solution: The equation is          mg − T = ma
∴ T = m(g − a ) = 6(10 − 2) = 48 N                                                                                   mg

Example 4: The figure shows two blocks at rest on a horizontal surface.
Find the contact forces acting on each of the blocks.                                                     m1
Solution: Let the contact force between m1 and m 2 be N1 and that between                                            m2
m 2 and the horizanal surface be N 2 Since, the masses are at rest,
their accelerations are zero.
F.B.D of m1             F.B.D of m 2
⇒ m1g − N1 = 0              (from F.B.D of m1 )               ............. (i)
N1

m 2 g + N1 − N 2 = 0       (from F.B.D of m 2 )               ............ (ii)
From (i) ,      N1 = m1g
From (ii) ,      N 2 = N1 + m 2 g = ( m1 + m 2 )g                                            m1g N1                    m 2g N 2

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Example 5: A light and inextensible string passes over a light and frictionless pulley as shown.
Calculate the acceleration of the system and tension in the string
Solution: Let mass m 2 be accelerated downwards with acceleration a .
Then, the mass m1 accelerates upwards with acceleration a .
Let T be the tension in the string.                                                                          m1     m2
The equations are
F.B.D of m1                       F.B.D of m 2
T − m1g = m1a         and m 2 g − T = m 2 a
On solving the two equations, we get
T                                    T

m 2 − m1                                                                                           a                        m2         a
a=                                       2m1m 2
g             and T =               g
m1 + m 2                            m1 + m 2                                           m1g                               m 2g

Example 6: A block of mass m is lying on a smooth inclined plane,                                             m
inclined at an angle θ with the horizontal.
Find the acceleration of the block
θ
Solution: We first make a free body diagram of the
in
block. Since the acceleration is along the
F.B.D
. co.
e rs
inclined surface, let us resolve the forces

ach
along the axis as shown.                                              a                                                 Y            a
Along X − axis, mg sin θ = ma
e    te
lin
m
∴ a = g sin θ
/   on              θ
t p: /                           N                        X        mg sin θ
N
mg cos θ
ht                           mg
m
f ro
Example 7: In the above problem, if the inclined planeis rough having coefficient of kinetic friction µ k , find the acceleration of the block.
d
Solution: Along X − axis,         d   e                              F.B.D

n    loa
mg sin θ − µ k N = ma ....... (i)
Y

Along D
Y o−waxis,                                               a
µk N                                              µk N
N − mg cos θ = 0 or N = mg cos θ
Substituting, the value of N in (i) ,                               θ                                                             N
N
mg                                         mg sin θ
we have mg sin θ − µ k mg cos θ = ma                                                                 X                       mg cos θ
∴ a = g(sin θ − µ k cos θ)

m1
Example 8: Figure shows a block of mass m1 on a smooth horizontal surface pulled by a
massless string which is attached to a block of mass m 2 hanging over a light
frictionless pulley as shown. Find the acceleration of the system and tension in
the string.
m2

Solution: Note that tension in the string at both ends is same and also,                                F.B.D of m1                   F.B.D of m2
the acceleration of both blocks is same.
T
The equations are T = m1a and m 2 g − T = m 2 a                                               m1     T                       m2
On solving the two equations, we get
m2                                      m1 m 2                                          N
a=            g              and         T=            g                                    m1 g                         m2 g
m1 + m 2                                 m1 + m 2

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Example 9: In the above problem, if the horizontal surface is rough having coefficient of kinetic friction µ k , find the acceleration of the system
and tension in the string.
Solution: From F.B.D of m1 , we have
F.B.D of m1                 F.B.D of m2
N − m1g = 0               or         N = m1g
T − µ k N = m1a            or         T − µ k m1g = m1a ...... (i)                                                                   T
From F.B.D of m 2 , we have m 2 g − T = m 2 a                     ...... (ii)                         m1 T                          m2
From (i) and (ii) , we get                                                                 µk N
N
( m 2 − µ k m1 )                                  (1 + µ k ) m1m 2                                 m1 g                         m2 g
a=                  g                  and        T=                  g
m1 + m 2                                           m1 + m 2

9. Angle of the inclined plane at which a body just starts sliding
Consider a body of mass m kept on an inclined plane.
m
Let µ s be the coefficient of static friction between the body and the plane.
If we keep increasing the angle θ of the plane, at a certain value of angle θ ,
the body just starts sliding. This angle is given by θ = tan −1 (µ s ) .
θ

This can be obtained by making a FBD .
in
co.F.B.D
Since, it is a limiting case,
.
friction is equal to µ s N   and acceleration of the body is zero.
e rs
From FBD , we get                                                                       m
achN
µ                                      m         µs N
e te
s

µ s N = mg sin θ              and        N = mg cos θ
nθ
µ N mg sin θ
/o n li         N                  mg sin θ
mg cos θ
N

p: /
⇒ s =                               or        µ s = tan θ                           mg
mg cos θ
ht t
N

∴ θ = tan −1 (µ s )
ro     m
e     df
10. Uniform Circular Motion

nl
(a) In the last chapter, we have seen that a body moving in a circle (of radius r ) with uniform speed v is accelerated towards the centre
given by a = v 2 r .       w
Do
(b) This acceleration is known as centripetal acceleration and the force responsible for this is known as centripetal force and is given by
F = ma = mv 2 r .
(c) For a stone rotating in a circle by a string, the centripetal force is provided by the tension in the string.
(d) For a planet revolving around the sun, the centripetal force is provided by the gravitational force on the planet due to the sun.
(e) For a car taking a circular turn on a horizontal road, the centripetal force is provided by the force of friction.

11. Circular motion of a vehicle on a level road
(a) Consider a vehicle of mass m moving at a speed v and making a turn on
Axis of rotation

a circular path of radius r . The external forces acting on the vehicle are                                                             v2
a=
• Weight, mg                                                                                                                             r
• Normal force, N                                                                                                             fs
N
• Friction, f s
m v2
The equations from F.B.D are N − mg = 0 and f s = ma =
r
(b) The only horizontal force that acts towards the centre is the friction, f s . This is static friction and is self adjustable.
(c) The tyres have a tendency to skid outwards and the frictional force, f s apposes the skidding tendency and acts towards the centre.
m v2
(d) Thus, for a safe turn,     f s ≤ µs N            ⇒      ≤ µ s mg                                   ∴ v 2 < µ s rg
r
Hence, the maximum safe speed an a level road so that vehicle do not skid is given by                                    v max = µ s rg
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12. Circular motion of a vehicle on a banked road
(a) We have seen that on a level road, the maximum safe speed is given by v max = µ s rg
(b) To avoid dependence on friction, the roads are banked at the turn so that the outer part of the road is somewhat lifted up as compared
to the inner part. This increases the maximum safe speed.
(c) Let the surface of road make an angle θ with the horizontal.
fs
(d) Clearly, if the speed of vehicle is zero or very low, the tendency of the vehicle
is to skid downwards and the direction of friction f s will be upwards as shown.                                     θ

(e) If the speed of vehicle is very high, the tendency of the vehicle is to skid
upwards and the direction of friction, f s will be downwards as shown.                             fs
θ
(f) At a certain value of speed, the frictional force is neither upwards nor downwards (its value is zero).
Such a speed is known as the speed for which there is no dependence on friction

a = v2 / r
2
v
a=                                                                                                        N sin θ
in
co.
r

θ
θ N
rs   .           θ
N cos θ
c he
te a
Axis of
rotation
e
n   lin                            mv 2
//o
From F.B.D , the equations are            N cos θ − mg = 0           and         N sin θ =
tp:
r
v2
Eliminating N ,      we get       tan θ =      ht
m
rg

d f ro
Hence, the speed for which there is no dependence on friction is given by,                     v = rg tan θ ,
de
avalue of speed, the frictional force, f
lo                                                   = µ s N acts downwards.
wn
(g) To find the maximum safe                                              s

Do

mv 2
From F.B.D , the equations are          N cos θ − f s sin θ − mg = 0            and        N sin θ + f s cos θ =
r
Since, for     v max ,    fs = µs N ,       we have
mv 2
N cos θ − µ s N sin θ = mg                  and                    N sin θ + µ s N cos θ =             max
r
Dividing 2nd equation by 1st equation, we get

sin θ + µ s cos θ v 2                                   tan θ + µ s 
= max                                 1 − µ tan θ 
⇒ v 2 = rg             
cos θ − µ s sin θ
max
rg                                       s      
 µ + tan θ 
Hence, the maximum safe speed an a banked road is given by                                              1 − µ tan θ 
v max = rg s           
      s      
(h) Therefore, the maximum safe speed on a banked road is more than that on a level road.
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