Force and its application for k12 students

1
FORCE APPLICATION
AN INTRODUCTION
Arun Umrao
www.sites.google.com/view/arunumrao
DRAFT COPY - GPL LICENSING
Arun Umrao
https://sites.google.com/view/arunumrao

2
Contents
0.1 Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
0.1.1 Newton’s Law of Motion . . . . . . . . . . . . . . . . . . . . . . . . 2
0.1.2 Properties of Force . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
0.1.3 Force & Mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
0.1.4 Time Energy Relation . . . . . . . . . . . . . . . . . . . . . . . . . 5
0.1.5 Friction Force . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
Piled Blocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
0.2 Free Body Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
0.2.1 FBD of Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
0.2.2 FBD In Friction-less Horizontal Plane . . . . . . . . . . . . . . . . 20
Object Without External Forces . . . . . . . . . . . . . . . . . . . 20
Object With External Forces . . . . . . . . . . . . . . . . . . . . . 21
0.2.3 FBD In Frictional Horizontal Plane . . . . . . . . . . . . . . . . . 22
Object Without External Forces . . . . . . . . . . . . . . . . . . . 22
Object With External Forces . . . . . . . . . . . . . . . . . . . . . 23
Contact type Push-Pull . . . . . . . . . . . . . . . . . . . . . . . . 24
String type Pull-Pull . . . . . . . . . . . . . . . . . . . . . . . . . . 25
One Over Other . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
0.2.4 FBD In Inclined Plane . . . . . . . . . . . . . . . . . . . . . . . . . 26
0.2.5 FBD In Pulley . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
0.2.6 Acceleration of Moving Pulley . . . . . . . . . . . . . . . . . . . . 31
0.3 Stability & Equilibrium . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
0.3.1 Equilibrium of Forces . . . . . . . . . . . . . . . . . . . . . . . . . 33
0.3.2 Non-Equilibrium State of Forces . . . . . . . . . . . . . . . . . . . 39
0.3.3 Stability Under Friction & Force . . . . . . . . . . . . . . . . . . . 39
In Horizontal Plane . . . . . . . . . . . . . . . . . . . . . . . . . . 40
In Inclined Plane . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
0.4 Gravitational Stability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
0.1 Force
Force is pull or push on a body. Force changes the state of body. If body is in rest and
a force is applied on it, body came in motion. Similarly, a force bring a body to rest
from its motion if applied force is in opposite direction to the direction of momentum of
the body. Unit of force is Kg m/s2
. Second unit of force is Newton represented by N,
honoring to Sir James Newton. Mass of a body is m and force F is applied on it then
mass force relation is
F = ma (1)
While we discuss the physics’ rule, we always take ideal conditions not real one. For ex-
ample, in Newton’s force law, “body” means tiny, round, symmetrical particle of sufficient
large mass but not too much small in size. Its center of mass lies at its center. As the
“size” of body increases, the environmental phenomenon shall affect the motion of body
in several ways, by means of frictional or drag force. If body is too tiny and has sufficient
Arun
0.2.6 A
0.2.6 A
St bilit
St bilit
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celeration o
celeration o
& E ilib i
& E ilib i
0.3.1 Equilibrium of Forces . . . . . . . . . . . . . . . . . . . . . . .
0.3.1 Equilibrium of Forces . . . . . . . . . . . . . . . . . . . . . . .
0 3 2 N E ilib i St t f F
N E ilib i St t

0.1. FORCE 3
large mass then it behave like a black hole for electrons or small bodies. Irregular bodies
having finite shape start rotating about their axis due to presence of external forces.
0.1.1 Newton’s Law of Motion
Newton had gave three laws of motion. These are
1. The inertia of a body does not change until unless an external force is
not applied on it. As there is no external force acting on the body, hence its
momentum remains constant with time.
2. The force acting on a body is product of mass (m) and acceleration (a).
Therefore, mathematically,
F = ma (2)
Note that, here “mass body” or “mass object” are very small in shape (i.e. in point
particle form) but not of atomic size, so that, whole mass of body is concentrated
at center of mass point of the body, and position of center of mass of the body does
not change with time.
3. Each action has always an equal and opposite reaction. If two bodies are
colliding each other then force exerted by first body on second body is F12. From
the law of action and opposite reaction, the force acted by second body into the
first body is given by F21. Mathematically
F12 = −F21 (3)
If there is elastic collision or impact collision between two masses, then F12 = m1v1
and F21 = m2v2. From the equation (3), relation in mass and velocity form is given
by:
m1v1 = −m2v2
−ve sign represents opposite reaction.
0.1.2 Properties of Force
Force is a vector quantity and obeys all the laws of vector.
Algebraic Sum of Forces If two or more forces, acting on a body in same direction
then their resultant force is algebraic sum of all these applied forces.
x1 x2
y1
y2
x
y

F1

F2

F
θ
x1
x2
y1
y2
x
y

F1

F2

F
θ
Arun Umrao
here is elastic collision or impact collision between two masses, then
here is elastic collision or impact collision between two masses, then F
F1
1
F
F F th ti (3) l ti i d l it f
h t

4
For example if
F1 = x1î + y1ĵ and
F2 = x2î + y2ĵ are two vector forces as shown in
the first part of above figure, then their sum is given by

F = (x1 + x2)î + (y1 + y2)ĵ
Or

F = (x1î + y1ĵ) + (x2î + y2ĵ)
OR

F =
F1 +
F2 (4)
The sum of vector in case of second part of the above figure is given by

F = (x1 − x2)î + (y1 − y2)ĵ
Or

F = (x1î + y1ĵ) − (x2î + y2ĵ)
Or

F =
F1 −
F2 (5)
These two vector algebraic sum equations, i.e. (4) and (5), can be written in combined
form as

F =
F1 ±
F2
There shall be +ve sing during the sum of the two vector forces if they are in same
direction or inclined in angular domain of 0◦
≤ θ ≤ 90◦
. Similarly, there shall be −ve
sign during the sum of the two vector forces if they are in opposite direction or inclined
in the angular domain of 90◦
≤ θ ≤ 180◦
.
Product of Forces If result of two vectors being multiplied with each other is a vector
quantity (or say tries to rotate each other) then cross product or vector product of
two vectors is performed. i.e.

F =
F1 ×
F2
If result is a scalar quantity (or say tries to pull each other) then dot product of two
vectors is performed. i.e.

F =
F1 ·
F2
Resultant of Inclined Forces If magnitude of two forces are inclined with each other
at an angle θ then their resultant is given by
F =

F2
1 + F2
2 + 2F1F2 cos θ
This is a vector resultant method of forces. If both forces are parallel and opposite to
each other then resultant force is

F =
F1 −
F2
And if they are parallel and are in same direction then

F =
F1 +
F2
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ve
ve sing durin
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ng the sum of the two vector forces if they are in opposite direction or
ng the sum of the two vector forces if they are in opposite direction or
l d i f 90◦
≤
≤ θ ≤ 180
18 ◦

0.1. FORCE 5
0.1.3 Force Mass
As usual, we always compute force in terms of velocity or acceleration and it is assumed
that mass is constant variable for force. But it is not true. Take a case of road shower.
Road shower is used to sprinkle water over the road, so that dust do not ﬂy. For this
purpose, water is ﬁlled in a tank and it is sprinkled over the dusty road. Tanker moves
with constant velocity and with time, tank emptied constantly. Here, velocity is constant
but water mass reduces constantly. Question is, do the force exerted by tanker engine
change with time or not? From the law of physics, force exerted by tanker engine reduces
with the emptying tank. If velocity of varying mass remain constant them from the
momentum, p = mv, we have
d
dt
p = dF =
d
dt
mv
Now, from law of derivatives, we have
dF = m ×
dv
dt
+
dm
dt
× v
This is general relation between force, mass and velocity of the mass. In particular case
of road shower, v is constant, so dv/dt = 0. Now,
dF =
dm
dt
× v
is required change in force during the accretion or reduction of mass.
dF F
In a conveyor system, initially, above relation is followed till the mass on conveyor
varies. After sometime, when mass is accrued at one end and removed at other end of
the conveyor, force becomes constant. Then a minimum constant force is required to
maintain the constant motion of conveyor. This minimum maintaining constant force is
given by
F =
dm
dt
× v
Let the conveyor carry mass by distance x then work done is
W = F × x
And power required for conveyor motion at constant velocity is
P =
dW
dt
= F ×
dx
dt
= Fv
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in force dur
in force dur

6
This is required constant conveyor power.
Solved Problem 0.1 Gravel dropped in conveyor belt at rate of 75.0kg/s. The speed of
conveyor belt is 2.2m/s. What force is required to keep belt moving?
Solution Let gravel of mass m in the belt is moving with a constant velocity u.
Additional mass of gravel m
is added to the belt continuously. Now, an additional force
is required to maintain the velocity of the belt. The required force is
dF = (m + m
) × v
Here m + m
is rate of additional mass of gravel. Mathematically
dF = v × dm
If change in mass of gravel in belt is dm/dt then required force is
dF = v
dm
dt
Substituting the values
F = 75.0 × 2.2 = 165N
This is required force.
0.1.4 Time Energy Relation
In a one dimensional situation, force F(x) depends only on the mass and its acceleration
along the direction of force. Hence
F(x) = mẍ
Acceleration in terms of velocity and displacement is given by
ẍ = v
dv
dx
Hence
F(x) = mv
dv
dx
Integrating both side with respect to x
1
2
mv2
=

F(x)dx + C
Applying initial and ﬁnal conditions, when x = 0 then v = v0 and above relation give
1
2
mv2
0 = C
Now the time and energy relation becomes
1
2
mv2
−
1
2
mv2
0 =

F(x)dx (6)
Arun
Time
Time Umrao
Energy R
Energy R
dimensional situation, force
dimensional situation, force F
F(
(x
x) depends only on the mass and its acc
) depends only on the mass and its acc

0.1. FORCE 7
Right side is merely a work done by the force F(x). Similarly, the potential energy is
given by
V − V0 = −

F(x) dx (7)
Sum of the above two equation gives
T + V = T0 + V0
It gives that sum of kinetic energy and potential energy is conserved. Now
T = E − V (x)
Substituting the value of T and v, above relation becomes
t = ±

m
2
x
x0
1

E − V (x)
dx (8)
This equation is known as time energy relation.
0.1.5 Friction Force
Friction force (f) is resistible force that opposes to the relative motion between two
or more surfaces. Friction force depends on the nature of the surface, area of contact,
temperature, velocity, atmosphere. Direction of friction force is opposite to the externally
applied forces along the surfaces in contact as shown in the following figure.
M

F

f
(a)

v

ω

F

f
(b)
M
g

R

f
(c)
In first part of figure 0.1.5 (a), the direction of friction force (
f) is opposite to the
externally applied force (
F) that slides mass rightward. Similarly, in second part of above
figure 0.1.5 (b), a wheel is rolling clockwise and its center of mass moving in rightward
direction. The direction of friction force is in the same direction as the direction of velocity
is. This is because, while wheel rolling clockwise direction, it applied an external force

F leftward on the surface. Numerical value of friction force depends on the normal force
(R) acting to the surface, see figure 0.1.5 (c). In shot, direction of friction force is parallel
to the surfaces in contact. Friction force is directly proportional to normal force (R).
Mathematically
f ∝ R
= μR (9)
Where μ is the coefficient of friction force ranging between ‘0’ to ‘1’. μ depends on the
area of the surfaces in contact, roughness of surfaces, temperature, humidity etc. Rubber
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8
in contact with other surfaces has coefficient from ‘1’ to ‘2’. Occasionally it is maintained
that μ is always less than ‘1’, but this is not true. A coefficient value above ‘1’ implies
that the force required to slide an object along the surface is greater than the normal force
of the surface on the object (sticky surfaces). There is also a scope of negative coefficient
of friction in which an increase in normal force leads to an increase in friction. This
is experimentally demonstrated by using a graphene sheet in the presence of graphene-
adsorbed oxygen. Practically, friction forces between rough surfaces are very large while
between smooth surfaces are very small. The most important property of the friction
force is that, magnitude of friction force is the maximum force that may be transferred
from one body to other body. See the following arrangement of two blocks.
m2
m1
F
Force F is applied to the upper block. Upper block transfers a fraction of external
force to lower block through frictional contact. Therefore, effective force acting on the
lower block is equal to the frictional force acting between surfaces of the two blocks.
Piled Blocks
m3
m2
m1
F
μ
μ
Take a condition in which there are three blocks of masses m1, m2 and m3 respectively
and they are placed one over other as shown in above figure. An external force (F) is
applied on the mass m1 horizontally. Our problem is to find the effective forces acting
on the three masses. The surface on which three masses are placed is friction-less which
coefficients of frictions between surfaces of masses m1 and m2 is μ, and between surfaces
of masses m2 and m3 is μ
. As we aware, the force that is transferred from m1 to m2 is
equal to the friction force between their surfaces. Here, it is f1 = μm1g.
m3
m2
f1
m1
F
μ
μ
f1
Again noted here, if F f1 then f1 has upper ceiling of μm1g. Masses m1 and m2
shall behave like separate entities. Block of mass m1 will slide over the mass m2 under
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0.1. FORCE 9
effective force of F − f1. Similarly, if F f1 then mass m1 and m2 shall behave like
a single entity and both masses shall move under the effect of force F, i.e. there is no
relative motion between masses m1 and m2.
m3
f2
m2
f1
m1
F
μ
μ
f1
f2
For effective force on mass m3, we shall find the friction force between the surfaces of
mass m2 and m3. Friction force between these masses is f2 = μ
(m1 + m2)g. If f1 f2
then mass m2 and m3 shall behave like separate entities. Block of mass m2 will slide
over the mass m3 under effective force of f1 − f2. If f1 f2 then mass m2 and m3 shall
behave like single entity and both masses shall move under the effect of force f1.
Solved Problem 0.2 A block of mass m is placed over frictional surface of coefficient of
friction μ as shown in the following figure. A horizontal force F is applied on the block.
Find the fraction of force that is applied on the horizontal surface by the block.
Solution
m
F F
ff
N
w
It is friction force between surfaces of block and horizontal surface that opposes relative
motion between them. So, maximum force that block may applied on the horizontal
surface is depend on two cases:
1. F ff When applied force is larger than the friction force then force acting on
the surface by the block is equal to friction force. So, fraction of force that is transferred
from the block to the horizontal surface is ff = μmg. In this case, the block slide over
the surface under constant force F − ff .
2. F ≤ ff When applied force is equal or less than the friction force then force acting
on the surface by the block is equal to the applied force F. So, fraction of force that is
transferred from the block to the horizontal surface is F. In this case, block shall not
slide over the surface and both block and surface shall act as single entity.
Solved Problem 0.3 A block of mass m is placed over frictional surface of coefficient of
friction μ as shown in the following figure. A horizontal force F is applied on the block.
In the same time, force f is continuously applied vertically on the block throughout its
motion. Find the fraction of force that is applied on the horizontal surface by the block.
Solution
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w/arunu
ew
vie
F
F w
w/arunum
f
ff
f
f
ru

10
m
F F
ff
N
w
f
We know that, it is the friction force between surfaces of block and horizontal surface
that opposes relative motion between them. So, maximum force that block may applied
on the horizontal surface is equal to friction force. Friction force is always equal to the
product of coefficient of friction and total force applied by upper block normally on the
surface in contact. Normal force on the surface of contact is w + f. So, fraction of force
that is transferred from the block to the horizontal surface is ff = μ(mg + f).
Solved Problem 0.4 A plank of mass m and length l is held by two cables of maximum
tensile strength of mg and 4mg respectively. A block of iron is placed on the plank. Find
the maximum mass of a iron block that can be placed on plank. Also find the position of
iron block.
Solution The figure for given problem is shown below.
m
M
l
x
4mg mg
Mg
mg
If plank can hold a mass M then vertical tensions should be balanced by each other.
Now
Mg + mg = mg + 4mg
It gives us the maximum mass that can be hold by the plank system.
M = 4m
To get the position of the mass, we take moment about the line of 4mg.
mg × l = 4mg × x + mg ×
l
2
On solving it
x =
l
8
The mass should placed at l/8 unit from the line of tension of 4mg.
Arun
n Umrao
s o
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0.1. FORCE 11
Solved Problem 0.5 A flexible chain of mass m and length l is initially at rest with one
half of it resting on a smooth horizontal table, and the other half dangling over the edge.
Now it is set free to move and chain starts slide off the table. After a subsequent time t,
chain slides by x. At this time t, show that
v2
= gx +
g
l
x2
Solution
l/2
l/2
l/2 − dk
l/2 + dk
dF
A flexible chain of mass m and constant length l is initially at rest with one half of it
resting on a smooth horizontal table, and the other half dangling over the edge. Initially,
string is at rest at time t = 0. It means, weight of dangling chain is balanced by horizontal
part of chain. If ρ is mass of chain in per unit length then
ρ =
m
l
Let chain starts sliding and it slides by dk in time dt. Excess force acting on the chain as
chain is slide by dk is
dF = ρ dk × g =
mg
l
dk
This force slides whole chain. Table is friction-less, hence force dF accelerates whole chain
by
da =
dF
m
It gives
da =
g
l
dk
If chain slides from l/2 to (l/2) + x and its acceleration changes from 0 to a then
a
0
da =
(l/2)+x
l/2
g
l
dk
On solving it, we have
a =
g
l
x
It shows that acceleration is linear function of length that chain slide. Further, consider
acceleration a at any position 0 k x. Here, we have just change the parameter. Now,
we have
a =
g
l
k ⇒
dv
dt
=
g
l
k
hain. If
hain. If ρ
ρ Umrao
is mass of c
is mass of c
ρ
ρ =
=
m
om
l
l

12
As velocity also changes with distance k, hence changing the parameters in left hand side
from time domain to distance domain as explained below:
dv
dt
=
dv
dt
×
dk
dk
=
dv
dk
×
dk
dt
= v
dv
dk
Substituting it in above relation v
0
v
dv
dk
=
g
l
k
Or v
0
v dv =
(l/2)+x
l/2
g
l
k dk
v2
= gx +
g
l
x2
This was to be proved here. This problem may also be solved by using conservation of
enery and momentum method.
Energy Method
l/2 l/4 l/2 − x l
2
+ x
2
dF
Chain slides under the eﬀect of its own weight, and there is no friction forces, hence
energy of the chain remains conserved. So, when chain slides by x change in its potential
energy is converted into its kinetic energy. Taking table level as zero potential line.
Velocity of chain is v when it slides by x. So,
−
m
l
×
l
2
× g ×
l
4
+
m
l
×

l
2
+ x

× g ×

l
2 + x
2

=
1
2
mv2
On simpliﬁcation, we have
v2
= gx +
g
l
x2
This is proof required here.
Momentum Method
Umrao
mrao
://sites.g g w/arunu a

0.1. FORCE 13
Solved Problem 0.6 A uniform rod of length 2l rests on a table, with a length l − a in
contact with the table. The remaining l + a sticking over the edge. Here a is the distance
of mid point of rod from the edge of the table. Initially rod is placed horizontally in rest
by applying vertical force on its one end that is over the table. When rod is free to turn
about the edge of the table by removing force, show that it makes an angle θ with the
horizontal before it slip. Where
tan θ =
2μ
2 + 9(a/l)2
Where μ is the coefficient of limiting static friction at the edge.
Solution
F
l
a l − a
A
f
N
ar
at
wv
w
wh
θ
A
A uniform rod of length 2l rests on a table as shown in above figure. Its mass is m
that concentrated at its center of mass. Its center of mass is at distance a from the edge of
the table (i.e. from point A). Force F is applied at right end of the rod so that it remains
horizontal. When force F is removed rod first starts rotating about A till it attains
inclination of angle θ. After that, it starts sliding. The rod follows: (i) first it starts
rotating about point A due to its own weight (w = mg) till it does not attain inclination
of angle θ and then (ii) static forces and momentum generated due to rotation of rod help
it in sliding. At angle θ, weight of rod has two resoluted components, wh = mg sin θ and
wv = mg cos θ. As mass of the rod is concentrated at its center of mass, so, at angle θ,
tangential acceleration (at) of center of mass is at = aα, where a is distance of center of
mass of the rod from A and α is angular acceleration. The difference of forces mg cos θ
and N are responsible for the tangential acceleration (at = rα). Therefore,
mat = mg cos θ − N (10)
The rod rotates about the point A. Before sliding, it rotates under the effect of centripetal
force about A. The net force f − wh is responsible for the balance of centrifugal force of
the rod. So, applying law of forces along the length of rod.
mar = −mg sin θ + f (11)
Centripetal force is given by
Fcp = m
v2
t
a
Therefore, radial acceleration is equal to v2
t /a. Similarly, vt = aω. As The rod rotates
about A, therefore it has torque too. Therefore,
IAα = mg × a cosα (12)
Arun
centrated
entrated Umrao
at its center
at its center
(i.e. from point
(i.e. from point A). Force
) F is applied at right end of the rod so that it
s applied at right end of the rod so that
al. When force
al. When force F
F is removed rod first starts rotating about
is removed rod first starts rotating about A
A till i
till i

14
The inertia of rod about A is given by
IA = m

l2
3
+ a2

(13)
Eliminating α from relations 10, 11 and 12, we have
N = mg cos θ ×

l2
l2 + 3a2

(14)
Till rod rotates about point A, there is no external forces that aﬀect its rotation, therefore,
potential energy at initial state shall be equal to the rotational kinetic energy before sliding
of the rod. Hence
mg × a sin θ =
1
2
Iω2
Here, vt = aω. Solving, it we have
mg × a sin θ =
1
2

m

l2
3
+ a2

×
v2
t
a2
=
1
2

m

l2
3
+ a2

×
ar
a
(15)
It gives
ar =
2ga2
sin θ
1
3 l2 + a2
(16)
Substituting it in 11, we get
f = mg sin θ

l2
+ 9a2
l2 + 3a2

(17)
Now, from f = μN, we get
tan θ =
2μ
2 + 9(a/l)2
(18)
This is required answer.
Solved Problem 0.7 A yo-yo is of mass M and rotational inertia I. The radius of its axle
is r, and is falls in the usual way with a length of string wrapped around the axle. Find
that its linear acceleration downward is
a =
Mr2
I + Mr2
× g
and tension in the string is
T =
I
I + Mr2
× Mg
Solution
Arun
ting it in
ting it in Umrao
11, we get
11, we get
f
f = mg
m sin
i θ
θ

l
l2
2
+ 9
+ 9a
a2
2

0.1. FORCE 15
r C
C
A
W
W
T
C
r
A
W
a
Let a yo-yo has mass M, rotational inertia I and radius of its axle is r. Here, yo-yo is
like a ring. A mass-less, tensile and flexible string is wrapped at its axes and it is allowed
to fall freely as usual. Due to its mass the yo-yo tries to rotate about point A where
unspung string is in contact with yo-yo axle. Its rotation is such that, its center of mass
moves from C to C
. But at the same time, string is unspung such that, center of mass,
i.e. C remains a distance of r from point A. Length of unspung string is sufficient large,
hence it appears vertical as shown in the second and third part of above figure.
1. Now, rotation of yo-yo about A comes from the torque applied by mass of yo-yo.
So,
IA × α = W × r
Inertia of yo-yo about A (IA) is given by
IA = I + Mr2
So,

I + Mr2

× α = Mg × r
Again, if tangential acceleration is at and angular acceleration is α then at = rα. So

I + Mr2

×
at
r
= Mg × r
In case of yo-yo, tangential acceleration shall be equal to the acceleration of yo-yo in
downward direction (a say).
a =
Mr2
I + Mr2
× g
This is ﬁrst answer.
Second Method We can solve this problem by using conservation of energy method.
Let yo-yo falls by a distance x from rest (u = 0) at initial. So, change in potential energy
of center of mass of the yo-yo shall be equal to the change in kinetic energy of the center
of mass of yo-yo, when it covers a distance x in vertical plane. So,
Mg × x =
1
2
mv2
+
1
2
Iω2
Here, yo-yo has translational and rotational (about C) kinetic energies. From the relation,
velocity of center of mass v = rω, so above relation becomes
Mg × x =
1
2
mv2
+
1
2
I
v
r
2
Arun
f yo-yo ab
yo-yo a
Umrao
out
out A (
(IA
A
I )
)
IA
A
I I + Mr

16
v2
= 2 ×
Mr2
g
I + Mr2
× x
This is distance-velocity relation of motion in vertical plane. So, vertical acceleration is
a =
Mr2
I + Mr2
× g
Third Method There may be a difference concept of solution. Let yo-yo falls by a
distance x from rest (u = 0) at initial. So, change in potential energy of yo-yo shall be
converted into the rotational kinetic energy of the yo-yo about A. This is, as yo-yo tries
to rotate about point A when it falls usually. So
Mg × x =
1
2
IAω2
Substituting the values, we have
Mg × x =
1
2
(I + Mr2
)
v
r
2
v2
= 2 ×
Mr2
g
I + Mr2
× x
This is distance-velocity relation of motion in verical plane. So, vertical acceleration is
a =
Mr2
I + Mr2
× g
2. To find the tension in the string, apply force rule for vertical direction. From the
third part of above figure
W − T = Ma
Or
T = W − Ma = Mg − M ×
Mr2
I + Mr2
On simplification, we have
T =
I
I + Mr2
× Mg
This is second part of the answer.
Arun Umrao
istance-velocity relation of motion in verical plane. So, vertical acceler
istance-velocity relation of motion in verical plane. So, vertical acceler

0.1. FORCE 17
Solved Problem 0.8 A yo-yo is of mass M and rotational inertia I. The radius of its axle
is r and outer radius is R. It is rests on a horizontal table. The string is wrapped around
the axle and held vertically by applying a force P. The coeﬃcient of yo-yo and surface is
μ. Show that if
μ
MrRP
(Mg − P)(I + MR2)
the initial motion of the yo-yo will be to roll to the left without slipping, with an initial
linear acceleration
rRP
I + MR2
. But if
μ
MrRP
(Mg − P)(I + MR2)
then the yo-yo will rotate counter-clock wise without rolling with an initial angular ac-
celeration about its center
P(r + μR) − μMg
I
.
Solution
P
r
R
N
f
P
C
C

A
P
C
N
f
A
W
The yo-yo is placed in a surface and a string is unspun so that, it rotate in counter
clockwise direction as shown in the above ﬁgure. As there is no vertical acceleration,
hence
P + N = Mg
As there is no slipping, hence yo-yo move leftward with acceleration of a (acceleration of
center of mass of yo-yo). In this case, yo-yo tries to rotate about A to move leftward. So
taking torque about point A:
P × r − f × 0 = IA × α
Here, α is angular acceleration, that is related with linear acceleration as a = Rα for the
point A. Inertia of yo-yo about A is given by
IA = I + MR2
So,
P × r =

I + MR2

×
a
R
On Solving it, we have
a =
PrR
I + MR2
Arun
n
n
n
n Umrao
o
o
Umrao
Umrao
mra
mr
mr
Um
U
g
es
es
/sites
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t
t
N
f
f g om
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e.com
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A
A
A
A
.c un
un
/arun
w/arun
r
r
N
N
N
N
f
f
ar

18
This is acceleration, when there is no slipping of yo-yo. Yo-yo shall not slip if f Ma.
So
μN M ×
PrR
I + MR2
Again,N = Mg − P So,
μ
MrR × P
(I + MR2) (Mg − P)
Again, yo-yo shall slip at A without rolling when f Ma, i.e.
μ
MrR × P
(I + MR2) (Mg − P)
The yo-yo when slip at A, it shall rotate about C. In this case, net moment by all linear
forces shall be equal to torque of the yo-yo. So, taking force moment about point C
P × r − f × R = Iα
Here, I is inertia of the yo-yo about its center of mass. So
P × r − μN × R = Iα
P × r − μ(Mg − P) × R = Iα
It gives
α =
P(r + μR) − μMg
I
This is angular acceleration of the yo-yo when it rotates counterclockwise with slipping.
Solved Problem 0.9 A point particle of mass m, moves along the uniformly rough track
PQR as shown in the ﬁgure. The coeﬃcient of friction, between the particle and the
rough track equals μ. The particle is released, from rest, from the point P and it comes
to rest at a point R. The energies, lost by the ball, over the parts, PQ and QR, of the
track, are equal to each other, and no energy is lost when particle changes direction from
PQ to QR. Find the μ and distance of QR.
Solution
2m
30◦
P
Q R
When a ball released from the point P, it starts moving under the gravitational force.
There is a friction force that acts opposite to the motion of ball. There are no other forces
hence energy loss is due to friction force only.
Arun Umrao
α
I
I
l l ti f th h it t t t l k i ith
l ti f th

0.1. FORCE 19
mg
mg sin 60◦
mg cos 60◦
R
μR
mg
R
μR
Now the work done by friction force when ball moves from P to Q along the ramp.
W1 = μmg sin 60 × PQ
Similarly the work done by friction force while ball moves from Q to R is
W2 = μmg × QR
From the question, energy losses due to friction force are equal, hence
μmg sin 60 × PQ = μmg × QR
Again PQ is equal to 2/ sin30◦
, so QR is 2
√
3m. The ball moves under the gravitational
force but presence of frictional force makes a control flow. The resultant acceleration of
the ball is a = g cos 60◦
− μg sin 60◦
. Velocity at the bottom of the sloped ramp is given
by
v2
= 8(g cos 60◦
− μg sin 60◦
)
Kinetic energy at the bottom of the sloped ramp is lost in the frictional motion of ball in
horizontal part of ramp.
KE = 4m(g cos 60◦
− μg sin 60◦
)
Now equating this kinetic energy to μmg × QR
μmg QR = 4m(g cos 60◦
− μg sin 60◦
)
On solving it, μ is approximately 0.29.
Solved Problem 0.10 Two mass m and M are placed at a finite distance over a friction-
less surface. Block M is moving towards the mass m with constant velocity of v. We have
to find the velocity of center of mass of the mass system. Would the velocity of center of
mass be change if both masses are attached with mass-less spring?
Solution
M
v
m
l
x
M
v
m
l − vt
x
To get the velocity of center of mass of the mass system, we shall compute the change
in position of center of mass of the mass system in one second about mass m. And this
Arun Umrao
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( )
nergy at the bottom of the sloped ramp is lost in the frictional motion
nergy at the bottom of the sloped ramp is lost in the frictional motion

20
change will equal to the velocity of center of mass as mass M is moving with constant
velocity. So,
x =
m × 0 − M × l
m + M
and
x
=
m × 0 − M × (l − v)
m + M
Here, displacement of mass M in one second is v × 1 = v. Now, change in the position of
center of mass in one second is x − x
and it is also equal to the velocity of the center of
mass.
v
=
−M × l
m + M
−
−M × (l − v)
m + M
Or
v
=
−Mv
m + M
The center of mass shall shift leftward at the speed v
given by above relation. If both
masses are attached with mass-less spring then there is no eﬀect of mass of the spring
but due its elasticity, there may be two conditions:
1. If spring is elongated and it is contracting to its original length, then there is
additional force acting on the mass M in the same direction of its velocity. It shall
increase the velocity of mass M, therefore, center of mass shall move leftward faster than
v
.
2. If spring is compressing then there is retarding force acting on the mass M in
the opposite direction of its velocity. It shall decrease the velocity of mass M, therefore,
center of mass shall move leftward slower than v
.
0.2 Free Body Diagram
Free body diagram is graphic representation of all forces, either dynamic or static acting
in a body.
0.2.1 FBD of Pendulum
A simple pendulum consists a heavy bob and strong, elastic and non stretchable string.
Pendulum is hanged from fulcrum. Bob weight is vertically downward while tension in
string acts vertically upward balancing to each other.
T
mg
Figure 1: FBD of a simple pendulum.
Arun Umrao
f spring is compressing then there is retarding force acting on the m
f spring is compressing then there is retarding force acting on the m
site direction of its velocity. It shall decrease the velocity of mass
site direction of its velocity. It shall decrease the velocity of mass M
M, t
t

0.2. FREE BODY DIAGRAM 21
Pendulum does not displace in vertical direction, hence
T − mg = 0 (19)
0.2.2 FBD In Friction-less Horizontal Plane
In friction-less surfaces, it is assumed that there is no resisting forces between two surfaces
when they slide against each other. The coeﬃcient of friction for friction-less surface is
zero.
Object Without External Forces
An object is in rest over a horizontal plane, its weight force acts normally downward,
applying an action force to the surface of plane.
m mg
R
Figure 2: FBD of a mass in friction-less horizontal plane surface at rest condition.
Plane surface applied a reaction vertically upward on the object. These two forces
are in vertical direction or say normal to the surface and net displacement in vertical
direction is zero. hence
R − mg = 0 (20)
There is not any forces along the horizontal direction or say along the surface.
Object With External Forces
An object is placed in friction-less plane and a force (F) is applied on it parallel to the
surface. Due to its weight, a weigh force is acted on the surface by this mass, and equal
and opposite reaction force (R) is acted by surface on the body.
m
F F
mg
R
Figure 3: FBD of a mass in frictionless horizontal plane surface at forced condition.
As there is no vertical displacement of the object, hence algebraic sum of normal forces
shall be zero. Hence
R − mg = 0 (21)
Arun
surface
surface Umrao
applied a rea
applied a re
rtical direction or say normal to the surface and net displacement in
rtical direction or say normal to the surface and net displacement in
is zero. hence
is zero. hence

22
As the surface is friction-less, and only external force (F) is applied on it. Therefore,
there shall be a net acting force on the body in horizontal direction. It shall displace this
body according to the Newton’s second law. So,
F = ma (22)
Where a is the acceleration of the object.
Solved Problem 0.11 An unknown mass, m1 hangs from a mass-less string and descends
with an acceleration g/2. The other end is attached to a mass m2 which slides on a
friction-less horizontal table. The string goes over a uniform cylinder of mass m2/2 and
radius r. The cylinder rotates about a horizontal axis without friction and the string does
not slip on the cylinder. Express the results in terms of g, m2 and r. (a) Draw free-body
diagrams for the cylinder and the two masses. (b) What is the tension in the horizontal
section of the string? (c) What is the tension in the vertical section of the string and (d)
What is the value of the unknown mass m1?
Solution The ﬁgure for given problem is shown below.
m2
m1
g/2
m2/2
a Free body diagram of masses and cylinder are
T
m2g
R
Tv
m1g
Tv
Th
m2
2
g
R
b The accelerations of both masses are g/2. Hence tension on the horizontal portion
of the string is Th = m2g/2.
c The accelerations of both masses are g/2. Hence tension on the vertical portion of
the string is Tv = m1g/2.
d Using the free body diagrams, we have three force relations:
m1g − Tv = m1 ×
g
2
(23)
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Tv − Th = I × α =
m2rg
8
(24)
Here, I is inertia of the cylinder about its axis along its length passes through its center
of base. α is angular acceleration. Its relation with linear acceleration is given by a = rα.
Th = m2 ×
g
2
(25)
Solving equation 23, 24 and 25 simultaneously, we have
m1
g
2
=
m2g(r + 4)
8
On simpliﬁcation, it gives
m1 =
m2(r + 4)
4
This is value of unknown mass.
0.2.3 FBD In Frictional Horizontal Plane
In rough surfaces, there is always a non-zero force which acts opposite to the external
applied force. This force is called friction force. The coeﬃcient of friction for rough
surfaces ranges from ‘0’ to ‘1’. In presence of friction forces, a fraction of input power
lost in form of heat.
Object Without External Forces
A body is in rest in friction-surface if there is no external forces acting on it. Its free body
diagram will be constructed similarly as in the case of body placed in rest in friction-less
surface.
m mg
R
For this state of body, only vertical forces are balancing to each other. i.e.
R − mg = 0 (26)
Object With External Forces
A body is in rest over a friction-surface and an external force (F) is applied parallel to
the surface. Weight of the body is acting a force on the surface normally downward (mg)
while surface applied a reaction force (R) on the body vertically upward. We shall apply
conservation of force relation for (a) horizontal direction and (b) for vertical direction.
There is no vertical displacement of the body as it remains in contact with surface, hence
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Without External Forces
Without External Forces
f f f f
f f

24
m F
μ mg
R
F
f
Figure 4: FBD of a mass in frictionless horizontal plane surface at forced condition.
R − mg = 0 (27)
Reaction of friction force is always in opposite direction to the applied force F. μ is the
coefficient of friction for the pair of surfaces, i.e. surface of the body and surface of the
plane on which body is placed. Maximum contact force is
R and it gives the friction force
f = μR (28)
Net force along the the surface is given by F − μR. Note that friction force between pair
of surfaces depends on the net load/force acting normal to the sufaces in contact or
on net reaction acting normal to the sufaces in contact.
F
ff
R
w
f
From the above figure, load at surface on which block is placed is equal to the sum of
weight of the block and applied normal force in downward direction. So net load at the
horizontal surface is mg + f. It gives net reaction by the horizontal surface R = mg + f.
Consequently the friction force between block and horizontal surface is μR, i.e. μ(mg+f).
Now, there are three cases for relation between apllied force and friction force:
When F f If externally applied force (F) is less than friction force (f) then
F − μR 0 (29)
And in this case body remain in stationary state (body and surface shall act like a single
entity) even if there is an external force acting on it. For this condition, coefficient of
friction is called static coefficient of the friction (μk) for the given pair of surfaces.
When F = f If externally applied force (F) is exactly equal to the friction force (f)
then
F − μR = 0 (30)
And in this case body is in terminal condition. It means body is in just to move state. In
this case too body and surface shall act like a single entity. For this condition, coefficient
of friction is called terminal coefficient of the friction (μT ) for the given pair of
surfaces.
Arun Umrao
c
R
R

When F f If externally applied force (F) is greater than the friction force (f) then
F − μR 0 (31)
And in this case body is in dynamic state. It means body is moving relatively to the
horizontal surface. For this condition, coefficient of friction is called dynamic coefficient
of the friction (μd) for the given pair of surfaces. Body acquires an acceleration a and
the net force is given by product of mass of object (m) and its acceleration a. i.e.
F − μR = ma (32)
Contact type Push-Pull
Two blocks are placed in contact over a plane. A horizontal force is applied on the mass
m1. The coefficients of friction are different for two masses and depend on the surfaces in
contact of the two blocks. In block m1, mass force m1g is balanced by reaction R1. The
net force along the surface of plane is
F − μ1R1 (33)
m1 m2
F
μ1
μ2
R1
m1g
F
μ1R1
F − μ1R1
R2
m2g
F − μ1R1
μ2R2
Figure 5: FBD of two body mass system in frictional horizontal plane surface at forced
state in Push-Pull condition.
This net force is passes to second mass m2 as action reaction forces between the
surfaces of mass m1 and m2. In mass m2, mass force m2g is balanced by reaction R2.
Net horizontal force is
F − μ1R1 − μ2R2 (34)
This is the net force that will displaced both objects right ward. If so, the acceleration
of the system is
F − μ1R1 − μ2R2 = (m1 + m2) × a (35)
String type Pull-Pull
Two blocks are placed over a plane and they are connected with a string. A horizontal
force is applied on the mass m2. The coefficients of frictions are different for two masses
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μ
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un
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μ
μ2
2R
R2
2

26
depend on the surfaces of the two blocks. In block m2, mass force m2g is balanced by
reaction R2. The net force along the surface of plane is
F − μ2R2 (36)
m1 m2 F
μ1 μ2
R1
m1g
μ1R1
F − μ2R2
R2
m2g
F − μ2R2
μ2R2
F
state In Pull-Pull condition.
This net force passes to string as tension. This tension acts on the mass m1. In mass
m1, mass force m1g is balanced by reaction R1. Net horizontal force on the mass m1 is
F − μ2R2 − μ1R1 (37)
This is the net force that may displaced both objects right ward. If so, the acceleration
of the system is
F − μ1R1 − μ2R2 = (m1 + m2) × a (38)
One Over Other
A mass m2 is placed over another mass m1. These mass system is placed over a plane
surface. All the surface in contact have frictions. The coeﬃcient of friction between m1
and m2 is μ2 and m1 plane surface is μ1. The friction force between mass m2 and mass
m1 is μ2R2. This friction force tends to displace mass m1.
m1
m2
F
μ1
μ2
R1 + R2
m1g R2
μ1(R1 + R2)
μ2R2
R2
m2g
F
μ2R2
state. One body is over to other.
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Umrao
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stem is
stem is
( )

In mass m1, total reaction is (R1 + R2) hence friction force between the mass surface
and plane surface is μ1(R1 + R2). Now the net horizontal force on the mass m2 is
F − μ2R2 (39)
and mass m1 is
μ2R2 − μ1(R1 + R2) (40)
If both masses displaced rightward then acceleration of mass m2 is
F − μ2R2 = m2a2 (41)
and acceleration of mass m1 is
μ2R2 − μ1(R1 + R2) = m1a1 (42)
0.2.4 FBD In Inclined Plane
A mass of m is placed over an inclined plane. The angle of inclination is θ. This mass
is attached with a string and it passes through a pulley. The mass is pulled by applying
force F on the string. This force passes to the mass in form of tension T .
μ
m
T
θ
i
j
T
R
μR
mg
θ
Figure 8: FBD of body in frictional inclined plane surface.
The mass force mg acts vertically downward. As the body moves parallel to the plane
surface, friction force acts downward and parallel to the plane. Reaction always acts
normal to the surface. Taking forces parallel and normal to the surface of plane, reaction
force is
R = mg cos θ (43)
There is no displacement along the normal to the plane surface. Net force parallel to the
plane surface is
T − μR − mg sin θ (44)
It displaces the mass along the surface of the plane. If there is an acceleration (a) of the
object then T − μR − mg sin θ shall be equal to the product of mass and acceleration of
the object.
T − μR − mg sin θ = ma (45)
Arun
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Umrao
google.
ps:
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T
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28
Solved Problem 0.12 Two blocks of masses 50kg and 30kg are placed later over first.
Static coefficient of friction between two blocks is μ2 = 0.36, Static coefficient of friction
between 50kg body and the surface is μ1. Two different forces of strength 500N and
100N are applied on the 30km body horizontally to get two values of μ1. Assume the
system is in ‘just move’ state and there is no sliding between two blocks. Calculate the
value of coefficient of static friction μ1 for two given forces. Also described the motion if
static coefficient of friction between bodies system and surface is zero.
Solution
50kg
30kg
F
μ1
μ2
R1 + R2
m1g
R2
μ1(R1 + R2)
μ2R2
R2
m2g
F
μ2R2
It is clear that force applied on 30kg body is horizontal. Due to inter-body interaction,
only friction force (μ2R2) between two body surface is transformed from 30kg body to
50kg body in form of shear force. This force let the move two objects against the friction
force between 50kg body and surface (μ1(R1 + R2)). To prevent the sliding between two
blocks, force should be less than friction force between two blocks. So
500N μ2R2
Or
500N 0.36 × 30 × 10 108N (46)
which is not possible, hence whole system never came in motion without sliding between
two blocks. In second case, when 100N force is applied, motion of the system is possible
if applied force is less than μ2R2 = 108N. In this case whole force can be transferred to
body of 50kg. In this case two blocks would behaves like single body and force 100N is
considered for whole system. For body system and surface
μ1(R1 + R2) = 100N
This gives
μ1 =
100
50 × 10 + 30 × 10
=
100
800
= 0.125 (47)
This is coefficient of static friction between body system and surface. If μ1 is zero, the
frictional force μ2R2, transferred to 50kg body let it came to move. For 500N force, body
system has two type of motion, one is sliding between two blocks and surface and other
is sliding between two blocks. For 100N force, body system has only sliding of whole
system over the surface.
Arun
lear that
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Umrao
orce applied
orce applied
(
( R
R ) b t
) b t
(
( )
y in form of shear force. This force let the move two objects against th
y in form of shear force. This force let the move two objects against th
50k b d d f (
f ( (R + R )) T t th lidi b t

Solved Problem 0.13 Two blocks of masses 30kg and 50kg are placed first over later
respectively in friction-less surface. Static coefficient of friction between two blocks is
μ = 0.36. A force of strength 500N is applied horizontally on the 30km block that is at
left edge of 50kg block. If length of 50kg block is 5m, then find the time after which 30kg
block leaves the surface of 50kg block.
Solution
50kg
30kg
F
μ
l
50kg
30kg
a1
a2
The force that is transferred to lower block is the friction force between two blocks.
The maximum friction force that can be transferred to lower block is
Ff = 30 × 10 × 0.36 = 108N (48)
As this force is less than the applied force, there is sliding between two blocks. The
friction force tries to accelerate to lower block and rest of force will accelerate to upper
block. So Accelerations of upper and lower blocks are
a1 =
500 − 108
30
= 13.07m/s2
a2 =
108
50
= 2.16m/s2
The relative sliding acceleration a is a1 −a2(both accelerations are in same direction) and
its numerical value is a = 10.91m/s2
. Now the time taken by the block of 30kg to leave
the surface of length of 5m of lower block is
5 = 0 +
1
2
× 10.91 × t2
(49)
on solving it
t = 0.96s (50)
After 0.96s, upper block would leave the surface of lower block.
Solved Problem 0.14 Two blocks of mass m1 and m2 are put on a friction-less level surface
as shown in the figure below. The static coefficient of friction between the two blocks is
μ. A force
F acts on the top block m2. When the force
F is small, the two blocks move
together. Find the magnitude of the force
F above which the block m2 starts to slide
relative to the block m1.
Solution
o Acceler
A
Umrao
tions of upp
tio
a
a1
1 =
=
ogle c
30
30
=
= 13
13.
.07
07m/s
m/s2

30
m1
m2
F
μ
R1 + R2
m1g
R2
μR2
R2
m2g
F
μR2
If force
F is small two blocks behave like single block. If force is too large, two blocks
starts sliding against each other. If applied force is small, it is directly transferred to
the lower block in form of friction force. In this case both blocks move together with an
acceleration a. As the force increases, acceleration increases. If applied force is larger than
frictional force between two blocks, upper block starts sliding over the lower block. In
case of ‘just slide’ state friction force between two blocks should be equal or greater than
the forced motion of m2. Now in ‘just slide’ state, let applied force is
F and acceleration
of both blocks is a then
a =
F
m1 + m2
(51)
Now, if applied force is larger than the force on lower body due to acceleration (linked
friction force) then upper block starts sliding. So
μm2g = m1a = m1
F
m1 + m2
(52)
On simplifying
F =
μm2g(m1 + m2)
m1
(53)
This is the critical force.
0.2.5 FBD In Pulley
In a pulley system, a string is pass through it and one end of the string has a weight and
force is applied at second end of the string to lift the object.
Arun Umrao
pplied force is larger than the force on lower body due to acceleratio
pplied force is larger than the force on lower body due to acceleratio
orce) then upper block starts sliding. So
orce) then upper block starts sliding. So

m1 m2
m1g
T1
m2g
T2
T
T1 T2
Figure 9: FBD of pulley system.
At equilibrium, tension balances the weight of the masses via string, hence tension
throughout the string is same and mathematically
T1 = T2 (54)
If m1 m2 then mass m1 will move downward and there is an acceleration in the direction
of motion of the mass m1. Now the acceleration is
m1g − T = m1a (55)
Similarly for mass m2, acceleration is
T − m2g = m2a (56)
Solved Problem 0.15 30kg and 50kg bodies are placed in two different surfaces of different
coefficients of frictions 0.5 and μ. Both surfaces are in the same elevation and distance
between two object is 10m. Assume 50kg body is at the edge of beginning of second
surface. A force of strength 225N is applied horizontally on the 30kg body. The body
moves and collides elastically with 50kg body and came to rest. Find the initial velocity
of the 50kg body.
Solution
μ = 0.5 μ
30kg
200N
50kg
f
w
10m
30kg body is placed in frictional surface. A friction force will act between the surfaces
in contact in the opposite direction to the direction of externally applied force. The net
force towards rightward shall accelerate 30kg body with acceleration a. So,
F − μmg = ma
Arun
for mass
for mas Umrao
m
m2
2, accelera
, acceler
T
T −
− m
m2
2g
g =
= m
m2
2a
a

32
225 − 0.5 × 30 × 10 = 30 × a
It gives a = 2.5 meter per square second. When this mass covers a distance of 10 meter,
its velocity increases from u = 0 to u1. So, from velocity-distance relation
u2
1 = u2
+ 2as
Or
u1 =
√
2 × 2.5 × 10
It gives u1 = 7.07 meter per second. Now, it collides with mass 50kg with speed of 7.07
meter per second. Collision between the two masses is ellastic, hence during the course
of collision, momentum remains conserve. So
m1u1 + m2u2 = m1v1 + m2v2
Or
30 × 7.07 + 50 × 0 = 30 × 0 + 50 × v2
It gives v2 = 4.24 meter per second. It means after collision, body of mass 50kg shall
move with initial speed of 4.24 meter per second.
0.2.6 Acceleration of Moving Pulley
From the below figure, it has seen that when pulley is pulled outward upto a distance x,
then mass moves a distance of 2x (i.e. twice of the distance covered by pulley).
m

F
m am

F
ap
2x x
Therefore, the accelerations of the pulley and mass are not equal but they are different
for each. Now, acceleration of pulley is twice time derivative of the distance covered by
it.
ap =
d
dt

d
dt
x

= x
(57)
Arun
Accele
AcceleUmrao
ration of
ration of
below figure, it has seen that when pulley is pulled outward upto a di
below figure, it has seen that when pulley is pulled outward upto a di
di t f 2
di t f 2 (i t i f th di t d b ll )

Similarly, acceleration of mass is twice time derivative of the distance covered by it.
am =
d
dt

d
dt
2x

= 2x
(58)
The ratio of accelerations is
ap
am
=
x
2x
=
1
2
(59)
It shows that, acceleration of mass is double to the acceleration of pulley.
Solved Problem 0.16 Assume a mass-less and friction-less pulley that can rotate about
its axis as shown in below ﬁgure. A mass-less chord is ﬁxed at its one end and it is passed
over to the pulley and connected to a mass of 20kg. The hook of pulley is connected to
another mass of 40kg by a chord. This mass is pulled by a force of 300N horizontally.
Find (a) accelerations of the masses, (b) Tensions in each of the chords.
Solution
m2 a2
T2
T2
T1

F
m1 a1
In this problem, there are three free body diagrams for (a) large mass, (b) small mass
and (c) pulley as shown below:
m2 a2
T2
T2
T2
T1 T1

F
m1 a1
Using the free body diagrams as shown above, and applying the relations of force-
equilibrium conditions, for large mass, pulley and small mass respectively. We have
F − T1 = m1a1 (60)
T1 = 2T2 (61)
T2 = m2a2 (62)
From equations 60 and 61, we have
F − 2T2 = m1a1
Arun Umrao
U
//sites google com/view/arunum

34
Or
F − 2m2a2 = m1a1 (63)
We know that, acceleration of dynamic pulley is half to the acceleration of mass m2.
Therefore for this problem, a2 = 2a1 and equation 63 becomes
F − 4m2a1 = m1a1
Substituting the numerical values as given in the problem:
300 = 40a1 + 80a1
On solving it, we have a1 = 2.5m/s2
. Now a2 = 2a1 and it gives a2 = 5m/s2
. From
equation 62, T2 = m2a2 and it gives T2 = 100N and from relation 61 T1 = 200N.
0.3 Stability Equilibrium
Stability of a body is its state when it resists external disturbance upto a critical extent.
If a body regains its original state after releasing external disturbance, then it is said that
body is in stable state otherwise it is said that body is in unstable state.
0.3.1 Equilibrium of Forces
There may be several forces acting on a body. At equilibrium, vector sum of all forces
must be zero. For example if F1, F2, . . ., Fn are n forces acting on a body at equilibrium
then
F1 + F2 + F3 + . . . + Fn = 0
n

i=0
Fi = 0
If components of all forces can be resolved, then at equilibrium, algebraic sum of horizontal
and vertical components must be zero. For example, horizontal components of forces F1,
F2, . . ., Fn are F1 cos θ, F2 cos θ, . . ., Fn cos θ and at equilibrium their algebraic sum
F1 cos θ + F2 cos θ + . . . + Fn cos θ = 0
Similarly at equilibrium, algebraic sum of vertical components must be zero. ie
F1 sin θ + F2 sin θ + . . . + Fn sin θ = 0
Solved Problem 0.17 A block of mass m is tied to two strings as shown in ﬁgure. Length
of each string is L. The angle between string and horizontal is 300
. Find (a) Draw the
free-body diagram of the block. (b) Tension in each string. (c) Velocity of block at the
lowest point if one of the string is cut and (d) Tension on the string at lowest position of
the block.
Solution The free body diagram of the mass when it hangs with string is shown in
below ﬁgure.
Arun
ay be sev
ay be sevUmrao
eral forces a
eral forces a
l f
l
p
p 1, 2, , n g y q

0.3. STABILITY EQUILIBRIUM 35
m
T T
θ
T T
T sin 30◦
T sin 30◦
T cos 30◦
T cos 30◦
mg
b. Balancing horizontal and vertical forces
2 T sin 30◦
= mg
It gives that the tension in each string is
T = mg
c. If length of string is l then total vertical height the mass covers during the fall is
h = l − l sin 30◦
. Total potential energy of the mass will convert into the kinetic energy
when it reaches to the maximum depth as shown in the below ﬁgure. Now
1
2
mv2
= mg(l − l sin 30◦
)
Simpliﬁcation and solving, it gives
v =

lg
T
mg
mv2
r
h
It is the velocity of the mass at its lowest point. (d) The maximum tension at this
point will be sum of centripetal force and the mass force of the body.
T = m
v2
l
+ mg
Substituting the value of v and simplifying it, we have
T
= 2mg
It is the maximum tension when mass will be its maximum depth after cutting any of the
two string.
Umrao
ation and solving, it gives
ation and solvin
v
v =
=

lg
lg

36
Solved Problem 0.18 A mass-less plank of length l = 1m is supported by two ropes, those
can bear maximum tensions of T1 = 600N and T2 = 400N respectively. A mass of 100kg
is to be placed on the plank. Find the position of the mass where it should put without
breaking of ropes.
Solution
l
T1 T2
x l − x
W
First we check whether the two ropes may support the weight of mass or not. The
maximum tension support by the both ropes is T1 + T2 which is 1000N. The weight of
the mass is 100 × 9.8 = 980N 1000N. It means by suitable positioning of mass can be
supported by the ropes without breaking. We ﬁnd a suitable position for the mass in the
plank, so that the weight of the mass is distributed rationally between the two rope and
fraction of weight does not exceed the maximum tension supported by the rope. Assume
that the mass is placed at distance x from T1. Now, the torque by both ropes should be
balanced. So,
T1 × x = T2 × (l − x)
600 × x = 400 × (1 − x)
Or
x =
400
1000
= 0.4
It means, the mass should placed at 0.4 meter from rope that can support tension upto
600N.
Solved Problem 0.19 A plank of length l = 1m and mass m = 10kg is supported by two
ropes, those can bear maximum tensions of T1 = 600N and T2 = 400N respectively. Find
the maximum mass (M) that can be placed on the plank without breaking of ropes. Also
ﬁnd the position of the mass where it should be put without breaking of ropes.
Solution
l
T1 T2
x l − x
W
w
Arun
. So,
So,
Umrao
( )
ing the values, we have
ing the values, we have

To put a mass into the plank, maximum weight of all masses should be not exceed the
tensions of the ropes. The maximum tension support by the both ropes is T1 + T2 which
is 1000N. So the maximum weight shall be
M × 9.8 + 10 × 9.8 = 1000N
It gives M = 92.04kg. This mass can be placed in the plank at suitable location, i.e. at
distance x from the rope that can bear tension of T1. We ﬁnd a suitable position for the
mass in the plank, so that the weight of the mass is distributed rationally between the
two rope and fraction of weight does not exceed the maximum tension supported by the
rope. So, taking moment of force about the mass M, we have
T1 × x = −m ×

l
2
− x

+ T2 × (l − x)
600 × x = −10 × (0.5 − x) + 400 × (1 − x)
On simpliﬁcation, we shall get x = 0.403 meter from rope that can support maximum
tension upto 600N.
Solved Problem 0.20 A mass-less plank of length l = 1m is supported by two ropes, those
can bear maximum tensions of T1 = 600N and T2 = 400N respectively. A body is put
on the plank and whole system is lifted by a constant acceleration of a = 2m/s2
. Find
the maximum mass of the body that can be lifted without breaking of ropes.
Solution
a
l
T1 T2
a
x l − x
W
The whole system is moving upward with a constant acceleration a = 2m/s2
, hence
the mass should be placed at certain location, so that its fractional weight on the rope
shall not exceed ropes maximum bearable tension. Let it is x from rope that can support
maximum tension upto T1. In this arrangement, upward tensions shall be equal to the
downward tension. Now, using law of forces
(T1 + T2) − Mg = Ma
Substitute the values, we have
600 + 400 = M(2.0 + 9.8)
It gives maximum mass M = 84.75 kilogram approximately.
m
plank and Umrao
m
whole syste
imum mass of the body that can be lifted without breaking of ropes.

38
Solved Problem 0.21 A cylinder of mass M and radius R is lying on the street against
the side-walk. The height of side-walk is h. A rope is attached to the axis of cylinder
and force is applied by pulling it with an angle α with horizontal such that it is just lift
cylinder off the street. (a) What is the ratio of this force to the weight of the cylinder.
Express your answer in terms of α, θ, M and g, where θ is the angle between line joining
side-walk contact with axis of cylinder. (b) At what angle of α is the ratio as described
under question (a) is minimum or is maximum.
Solution A cylinder of mass m is lying on the street against the side-walk as shown
in the figure below. If cylinder is just to lift off then moment about the point P should
be balanced by force and mass force. Now
F
mg
α
θ P
F
mg
α
θ
P
R sin(θ + α)
R cos θ
F × R sin(θ + α) = mg × R cos θ
On simplification, the ratio between force and mass weight should be
F
mg
=
cos θ
sin(θ + α)
This is first part of the answer. Assume that the height of side walk will never be zero.
Direction of force can be changed from 0 to 90◦
in first quadrant. Height of the side wall
ranges 0 ≤ h ≤ R.
F
mg
M
N
45
◦
P
F
mg
M
N
30
◦
P
F
mg
M
N
30
◦
60
◦
P
When force equals to the mass force, cylinder is lift off. The normal on force and mass
force from point P are equal θ = 45◦
and α = 0. For minimum ratio, normal on force
should be minimum in comparison to the normal on the mass force drawn from the point
P, ie if θ 450
moment by force F will be lesser than the moment by mass force and
ratio of F/mg will be minimum. Again at this case α = 0◦
. For maximum ratio, normal
on force should be maximum in comparison to the normal on the mass force drawn from
Arun Umrao
F
F
ification, the ratio between force and mass weight should be
ification, the ratio between force and mass weight should be

the point P. At this case θ 450
and α 00
. The normal on the force line will be
maximum when α = 45◦
.
Solved Problem 0.22 An object subjected to three equal forces. Of them two are perpen-
dicular to each other (one along −x axis and other along −y axis) while third is making
angle θ with horizontal (+x axis) in first quadrant. Which of the following statement is
true and explain your answer. (a) It is possible for this object to remain at rest. (b) It is
not possible for this object to remain at rest. (c) Answer can not be given without known
the value of the angle θ and (d) It is not possible to find the answer without known forces
and angle.
Solution

F

F

F
θ
x
y

F

F
√
2
F

F
θ
From the question, two forces are perpendicular to each other, in which one is along
−x axis and other is along −y axis. The resultant of these two forces is
√
2
F whose
direction shall be 45◦
from the −y axis in counter-clockwise direction at third quadrant.
This resultant force shall be fixed in magnitude as well as in direction. Only third force
may change its direction only within 0 ≥ θ ≤ 90◦
and its magnitude shall be always equal
to F. Now, for the given statements:
a. It is possible for this object to remain at rest. This statement is false. Object can
not remain in rest as the third force shall never be equal to
√
2
F whatever value of θ is.
b. It is not possible for this object to remain at rest. This statement is true, as third
force is always less than
√
2
F.
c. This statement is false. We can give correct answer without knowing the value of
the angle θ as given in the part (a) and (b)
d. This statement is false. We can explain the answer well as given in part (a) and
part (b).
Solved Problem 0.23 A mass of mkg is hanging with a spring of spring constant of k. Mass
is undergoing simple harmonic motion in vertical plane. At a certain point its velocity
is v and mass is moving upward direction. Find the total work done by the spring when
mass moves from xa to xb from the rigid platform to which spring is attached. Also find
the velocity at the second position.
Arun Umrao
the question, two forces are perpendicular to each other, in which one
the question, two forces are perpendicular to each other, in which one
√

40
Solution Assume at a certain depth x from the platform, net force on the mass is
Fnet = mg − kx (64)
If mass undergoes a vertical displacement by dx then work done is
dW = Fnetdx
Substituting the value of net force and integrating it for the vertical displacement of xa
to xb.
dW =
xb
xa
(mg − kx)dx (65)
Wxa→xb
= mg(xa − xb) −
1
2

x2
b − x2
a

(66)
We know that if external forces or negligible and change in energy of mass is due to only
spring forces then work done is equal to the change in energy of the mass. So,
W =
1
2
mv2
b −
1
2
mv2
a (67)
On simplification
v2
b = v2
a +
2W
m
(68)
This is required answer.
0.3.2 Non-Equilibrium State of Forces
If n forces are acting on a body and body is not in equilibrium position then, the resultant
force on the body is the square root of sum of square of resultant horizontal component
and vertical components. For example, resultant horizontal component is
F1 cos θ + F2 cos θ + . . . + Fn cos θ = FH
and resultant vertical component is
F1 sin θ + F2 sin θ + . . . + Fn sin θ = FV
The resultant force on the body is
R =

F2
H + F2
V
0.3.3 Stability Under Friction Force
Frictional force always oppose the relative motion of the surfaces in contact. Therefore,
while solving the stability problems, frictional forces are always taken in negative direction
to the applied force and they are subtracted from the external force to get the effective
force.
Solved Problem 0.24 A ball of mass 2kg is hit by a force of 20N. If friction force is 14.2N
then find the acceleration of the ball.
Arun
equired a
equired a Umrao
nswer.
nswer.
Non-Equilibrium State of Forces
Non-Equilibrium State of Forces

Solution
F a
ff
The effective force that causes the motion of the ball is difference of force applied and
restriction forces, i.e. frictional forces. It means, effective force is 20 − 14.2 = 5.8N. Now
the acceleration of the ball is
5.8 = 2 × a
It gives a = 2.9m/s2
.
In Horizontal Plane
Solved Problem 0.25 Two blocks of mass m1 and m2 are put on a friction-less horizontal
surface. Here, m1 is placed top of the mass m2. The static co-efficient of friction between
the two blocks is μ. A force F is applied on the top block m1. Find (a) When the force
F is small, the two blocks move together. For this case, draw the free body diagram of
both blocks. (b) Find the acceleration when both block moves together. (c) Find the
magnitude of the force F above which the block m1 starts to slide relative to the block
m2.
Solution
a The free body diagram of the two blocks is given below:
m2
m1
F
ff
F
N1
w1
ff
N1
N2
w2
b If both blocks move together on friction-less surface under force F, then acceleration
of the block system is
a =
F
m1 + m2
c Force (F) on block m1 is applied to let the two blocks move together. System
of blocks is placed in friction-less surface. There is no force, other than friction force
between the surfaces of two blocks, acting on the block of mass m2. So maximum force
that can be transferred to lower block by upper block is μm1g that is frictional force (fs)
between the block surfaces. Till the external force F is lesser than fs, both block shall
move together. When F is larger than fs, upper block shall starts sliding over the lower
block. So,
F ≥ fs = μm1g
Arun
de
Umrao
for
on
on

42
These are the answers.
Solved Problem 0.26 A block of mass M is placed in horizontal friction-less surface. Its
one end is attached with a spring of force constant k and length l. Another block of mass
m is placed over the block. Force F is applied on the upper block to move it rightward
as shown in the following figure. The coefficient of friction between block surfaces is μ.
Find the elongation of the spring.
Solution
M
m F
μ
l
M
m F
l + dl
Block of mass M is placed over friction-less surface. It is connected with a spring
of force constant k and length l. The force applied on upper block of mass m is F
(F ff ) and it is moving rightward with constant acceleration. The friction force
between surfaces of blocks is μmg. This force shall be transferred to lower block. As
lower block is placed over friction-less surface, therefore friction force (ff ) shall elongate
the spring. Spring is continuously under constant force ff irrespective of position of upper
block. At equilibrium, length of spring is l + dl. Now, from Hook’s law
ff = k dl
It gives
dl =
μmg
k
This is elongation of the spring.
In Inclined Plane
θ
m
F
wh ff
wwv
R
θ
m
F
wh
ff
wwv
R
Take a ramp of inclination θ, on which a mass m is placed as shown in above figure.
An external force F is applied on it so that the block moves upward along the ramp
surface. Surface of ramp and block are not friction-less, therefore, there is a friction force
between the surfaces and it will oppose the relative motion between the block and ramp.
Weight of the mass is acting vertically downward. Its two components can be resolute
parallel to the surfaces in contact and perpendicular to the surfaces in contact. These are
wh = mg sin θ and wv = mg cos θ. Note that, in physics, horizontal and vertical axes mean
axis-line parallel to the surface and perpendicular to the surface respectively. Friction
Arun Umrao
g
ff
f

force between pair of surfaces depends on the net load/force acting perpendicular
to the surfaces in contact or on net reaction acting perpendicular to the surfaces in
contact. Therefore, friction force is
ff = μR = μwv = μmg cos θ
The weight force that may move the block in the ramp surface is acting on the mass,
parallel to the surface of the ramp and leftward as shown in the above ﬁgure.
wh = μmg sin θ
Now, there are three cases:
When wh F In this case block shall move in upward direction and parallel to
the ramp surface. Friction force will be in downward direction and parallel to the ramp
surface. Note that friction force is always opposite to the direction of acting force.
When wh F In this case block shall move in downward direction and parallle to the
ramp surface. Friction force will be in upward direction and parallel to the ramp surface.
Note that friction force is always opposite to the direction of acting force.
When wh = F In this case all forces along the surface of slope are balanced, hence
block shall remain in rest.
Solved Problem 0.27 a 10g ball rolls down a 1.2m high slope and leaves it with a velocity
of 4m/s. How much work is done by the friction.
Solution
v1
v2
m
h
The work done by friction force is change in total energy of the body between two
states. Now for the ball
Wfric =

mgh1 +
1
2
mv2
1

−

mgh2 +
1
2
mv2
2

Taking upper height as state ‘1’ and lower base as state ‘2’. So
Wfric =

0.01 × 10 × 1.2 +
1
2
× 0.01 × 02

−

0.01 × 10 × 0 +
1
2
× 0.01 × 42

On solving
Wfric = 0.04J
Arun
Problem
H
Umrao
.27 a 10g ba
h k i d
htt // it l / i /
on

44
Solved Problem 0.28 An electric train is powered on a 30kV power supply, where the
current is 200A. If train is traveling at 90km/h then find the net force exerted on it in
forwards direction.
Solution The electric power supplied by electric train motor is
P = V I = 30000 × 200 = 6 × 106
Watt
Speed of train is 90km/h or 25m/s. This distance is traveled by train in one second.
Hence work done in one second is
W = F × 25
Work done in one second is power. So for one second
6 × 106
= F × 25
On solving
F = 240kN
Solved Problem 0.29 A car engine can deliver 90kW of power. The mass of car is 1000kg.
Find (a) Assume the total resistible force is proportional to the velocity. Ffric = αv. The
drag coefficient α is 100Ns/m. How fast can the car move on a level Road? Express the
speed in the units of m/s. (b) How fast can the car travel up a slope if we ignore all
friction? The slope of plane is θ = sin−1
(3/5). Express the speed in the units of m/s.
Solution
v
Ffric
P
v
h
θ
v cos θ
v
θ
v
a. A car moves ahead when it exerts force on ground. This force applied on the
ground should not be more than frictional forces otherwise wheels of car will skid. So
maximum force is equivalent to friction force. Work delivered by the engine is
W = F × d
For one second, work is equivalent to power and distance is equivalent to velocity. Hence
90000 = αv × v
v = 30m/s
Arun
fficient
the uni Umrao
i /
s of m/s. (
The slope of plane is θ = sin ( /5). Express the speed in the units o

b. Let the car travels in inclined plane and there is no frictions. So power delivered
by engine will change the total energy of the car in vertical. So
P = mgv sin θ
Substituting the value, we have
v = 15m/s
Solved Problem 0.30 A square of side L has center of mass at its center ‘C’ (assume it as
origin). A part of area L2
/4 is cut out from left bottom corner of the square. Find the
new coordinate of the center of mass of remaining part of the square.
Solution
L
C
L/2
C
L
C
C̄
Take the center of mass of complete square (C) as origin. A square of area L2
/4 is
removed from left bottom corner as shown in second part of above ﬁgure. If ρ is density
per unit area of the square, then mass of complete square is M = ρL2
and removed
square is m = ρL2
4. Center of mass of complete square is at origin while removed square
is at (−L/4, −L/4). As a square area is removed from the complete square, hence center
of mass of the remaining portion shall be shifted from C to new coordinate (x̄, ȳ). So,
applying the laws of center of mass for horizontal distance (i.e. x−axis).
x̄ =
M × 0 − m × −L
4
M − m
=
0 + ρL2
4 × L
4
ρL2 − ρL2
4
=
L
12
Similarly, applying the laws of center of mass for vertical distance (i.e. y−axis).
ȳ =
M × 0 − m × −L
4
M − m
=
0 + ρL2
4 × L
4
ρL2 − ρL2
4
=
L
12
So, new center of mass of the remaining portion of the square is
C̄ =

L
12
,
L
12

from the origin C. This is required answer.
Arun
A Umrao
t
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