Principle of Force Application - Physics - explained deeply by arun kumar

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FORCEAPPLICATION
ANINTRODUCTION
ArunUmrao
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DRAFT COPY-GPLLICENSING
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Contents
0.1 Force....................................... 2
0.1.1 Newton’sLawofMotion........................ 2
0.1.2 PropertiesofForce........................... 3
0.1.3 Force&Mass.............................. 4
0.1.4 TimeEnergyRelation......................... 5
0.1.5 FrictionForce.............................. 6
PiledBlocks............................... 7
0.2 Free Body Diagram . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
0.2.1 FBD of Pendulum . . . . . . . . . . . . . . . . . . . . . . . . . . . 20
0.2.2 FBDInFriction-lessHorizontalPlane ................ 20
ObjectWithoutExternalForces ................... 20
ObjectWithExternalForces ..................... 21
0.2.3 FBD In Frictional Horizontal Plane . . . . . . . . . . . . . . . . . 22
ObjectWithoutExternalForces ................... 22
ObjectWithExternalForces ..................... 23
ContacttypePush-Pull ........................ 24
StringtypePull-Pull.......................... 25
OneOverOther ............................ 25
0.2.4 FBDInInclinedPlane......................... 26
0.2.5 FBDInPulley ............................. 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
InHorizontalPlane .......................... 40
InInclinedPlane............................ 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 isKgm/s
2
. Second unit of force is Newton represented byN,
honoring to Sir James Newton. Mass of a body ismand forceFis 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
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0.3.1 Equilibrium ofForces ...................... .0.3.1 Equilibrium of Forces . . . . . . . . . . . . . . . . . . . . . . .
0 3 2 N E ilib i St t f FN 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 isF
12. From
the law of action and opposite reaction, the force acted by second body into the
ﬁrst body is given byF
21. Mathematically
F
12=−F 21 (3)
If there is elastic collision or impact collision between two masses, thenF
12=m 1v1
andF 21=m 2v2. From the equation (3), relation in mass and velocity form is given
by:
m
1v1=−m 2v2
−vesign represents opposite reaction.
0.1.2 Properties of Force
Force is a vector quantity and obeys all the laws of vector.
Algebraic Sum of ForcesIf 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
ﬃ
F
1
ﬃ
F
2
ﬃ
F
θ
x1
x2
y1
y2
x
y
ﬃ
F
1
ﬃ
F
2
ﬃ
F
θ
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For example if
ﬃ
F
1=x1
ˆi+y
1
ˆjand
ﬃ
F
2=x2
ˆi+y
2
ˆjare two vector forces as shown in
the ﬁrst part of above ﬁgure, then their sum is given by
ﬃ
F=(x
1+x2)ˆi+(y 1+y2)ˆj
Or
ﬃ
F=(x
1
ˆi+y
1
ˆj)+(x
2
ˆi+y
2
ˆj)
OR
ﬃ
F=
ﬃ
F
1+
ﬃ
F2 (4)
The sum of vector in case of second part of the above ﬁgure is given by
ﬃ
F=(x
1−x2)ˆi+(y 1−y2)ˆj
Or
ﬃ
F=(x
1
ˆi+y
1
ˆj)−(x
2
ˆi+y
2
ˆj)
Or
ﬃ
F=
ﬃ
F
1−
ﬃ
F2 (5)
These two vector algebraic sum equations, i.e. (4) and (5), can be written in combined
form as
ﬃ
F=
ﬃ
F
1±
ﬃ
F2
There shall be +vesing 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 ifthey are in opposite direction or inclined
in the angular domain of 90
◦
≤θ≤180
◦
.
Product of ForcesIf result of two vectors being multiplied with each other is a vector
quantity (or say tries to rotate each other) thencross product or vector productof
two vectors is performed. i.e.
ﬃ
F=
ﬃ
F
1×
ﬃ
F2
If result is a scalar quantity (or say tries to pull each other) thendot productof two
vectors is performed. i.e.
ﬃ
F=
ﬃ
F
1·
ﬃ
F2
Resultant of Inclined ForcesIf magnitude of two forces are inclined with each other
at an angleθthen their resultant is given by
F=
ﬃ
F
2
1
+F
2
2
+2F 1F2cosθ
This is a vector resultant method of forces. If both forces are parallel and opposite to
each other then resultant force is
ﬃ
F=
ﬃ
F
1−
ﬃ
F2
And if they are parallel and are in same direction then
ﬃ
F=
ﬃ
F
1+
ﬃ
F2
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ngthe sum of the two vector forces if they are in opposite direction orng the sum of the two vector forces if they are in opposite direction or
ld i f90
◦
≤≤θ≤18018
◦

0.1. FORCE 5
0.1.3 Force & Mass
As usual, we always compute force in terms ofvelocity 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,wehave
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,vis constant, sodv/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 distancexthen 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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This is required constant conveyor power.
Solved Problem 0.1Gravel 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?
SolutionLet gravel of massmin the belt is moving with a constant velocityu.
Additional mass of gravelm
ﬀ
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
ﬀ
)×v
Herem+m
ﬀ
is rate of additional mass of gravel. Mathematically
dF=v×dm
If change in mass of gravel in belt isdm/dtthen 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, forceF(x) depends only on the mass and its acceleration
along the direction of force. Hence
F(x)=m¨x
Acceleration in terms of velocity and displacement is given by
¨x=v
dv
dx
Hence
F(x)=mv
dv
dx
Integrating both side with respect tox
1
2
mv
2
=
ﬀ
F(x)dx+C
Applying initial and ﬁnal conditions, whenx=0thenv=v
0and above relation give
1
2
mv
2
0
=C
Now the time and energy relation becomes
1
2
mv
2
−
1
2
mv
2
0
=
ﬀ
F(x)dx (6)
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0.1. FORCE 7
Right side is merely a work done by the forceF(x). Similarly, the potential energy is
given by
V−V
0=−
θ
F(x)dx (7)
Sum of the above two equation gives
T+V=T
0+V0
It gives that sum of kinetic energy and potential energy is conserved. Now
T=E−V(x)
Substituting the value ofTandv, above relation becomes
t=±
α
m
2
θ
x
x
0
1
ρ
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
ﬃ
F
ﬃ
f
(a)
ﬃv
ﬃ ω
ﬃ
F
ﬃ
f
(b)
Mﬃg
ﬃ
R
ﬃ
f
(c)
In ﬁrst part of ﬁgure 0.1.5 (a), the direction of friction force (
ﬃ
f) is opposite to the
externally applied force (
ﬃ
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
ﬃ
Fleftward 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)
Whereμis the coeﬃcient of friction force ranging between ‘0’ to ‘1’.μdepends on the
area of the surfaces in contact, roughness of surfaces, temperature, humidity etc. Rubber
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in contact with other surfaces has coeﬃcient from ‘1’ to ‘2’. Occasionally it is maintained
thatμis 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
m1F
ForceFis 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
m1F
μ
μ
ﬀ
Take a condition in which there are three blocks of massesm 1,m2andm 3respectively
and they are placed one over other as shown in above ﬁgure. An external force (F)is
appliedonthemassm
1horizontally. 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 massesm
1andm 2isμ, and between surfaces
of massesm
2andm 3isμ
ﬀ
. As we aware, the force that is transferred fromm 1tom 2is
equal to the friction force between their surfaces. Here, it isf
1=μm 1g.
m3
m2f1
m1F
μ
μ
ﬀ
f1
Again noted here, ifF>f 1thenf 1has upper ceiling ofμm 1g. Massesm 1andm 2
shall behave like separate entities. Block of massm 1will slide over the massm 2under
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0.1. FORCE 9
eﬀectiveforceofF−f
1. Similarly, ifF<f 1then massm 1andm 2shall behave like
a single entity and both masses shall move under the eﬀect of forceF, i.e. there is no
relative motion between massesm
1andm 2.
m3f2
m2f1
m1F
μ
μ
θ
f1
f2
For eﬀective force on massm 3, we shall ﬁnd the friction force between the surfaces of
massm
2andm 3. Friction force between these masses isf 2=μ
θ
(m1+m2)g.Iff 1>f2
then massm 2andm 3shall behave like separate entities. Block of massm 2will slide
over the massm
3under eﬀective force off 1−f2.Iff 1<f2then massm 2andm 3shall
behave like single entity and both masses shall move under the eﬀect of forcef
1.
Solved Problem 0.2Ablockofmassmis placed over frictional surface of coeﬃcient of
frictionμas shown in the following ﬁgure. A horizontal forceFis applied on the block.
Find the fraction of force that is applied on the horizontal surface by the block.
Solution
mF 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>f
fWhen 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 isf
f=μmg. In this case, the block slide over
the surface under constant forceF−f
f.
2.F≤f
fWhen applied force is equal or less than the friction force then force acting
on the surface by the block is equal to the applied forceF. So, fraction of force that is
transferred from the block to the horizontal surface isF. In this case, block shall not
slide over the surface and both block and surface shall act as single entity.
Solved Problem 0.3Ablockofmassmis placed over frictional surface of coeﬃcient of
frictionμas shown in the following ﬁgure. A horizontal forceFis applied on the block.
In the same time, forcefis 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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10
mF 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 isw+f. So, fraction of force
that is transferred from the block to the horizontal surface isf
f=μ(mg+f).
Solved Problem 0.4Aplankofmassmand lengthlis held by two cables of maximum
tensile strength ofmgand 4mgrespectively. 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.
SolutionThe ﬁgure for given problem is shown below.
m
M
l
x
4mg mg
Mg
mg
If plank can hold a massMthen 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 atl/8 unit from the line of tension of 4mg.
ArunnUmrao
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0.1. FORCE 11
Solved Problem 0.5A ﬂexible chain of massmand lengthlis 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 timet,
chain slides byx. At this timet, show that
v
2
=gx+
g
l
x
2
Solution
l/2
l/2
l/2−dk
l/2+dk
dF
A ﬂexible chain of massmand constant lengthlis 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 timet= 0. It means, weight of dangling chain is balanced by horizontal
part of chain. Ifρis mass of chain in per unit length then
ρ=
m
l
Let chain starts sliding and it slides bydkin timedt. Excess force acting on the chain as
chain is slide bydkis
dF=ρdk×g=
mg
l
dk
This force slides whole chain. Table is friction-less, hence forcedFaccelerates whole chain
by
da=
dF
m
It gives
da=
g
l
dk
If chain slides froml/2to(l/2) +xand its acceleration changes from 0 toathen
θ
a
0
da=
θ
(l/2)+x
l/2
g
l
dk
On solving it, we have
a=
g
l
x
It shows that acceleration is linear function oflength that chain slide. Further, consider
accelerationaat 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. Ifhain. IfρρUmrao
is mass ofcis mass of c
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ρρ==
m
om
ll

12
As velocity also changes with distancek, 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
θ
v
0
v
dv
dk
=
g
l
k
Or
θ
v
0
vdv=
θ
(l/2)+x
l/2
g
l
kdk
On solving it, we have
v
2
=gx+
g
l
x
2
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 byxchange in its potential
energy is converted into its kinetic energy. Taking table level as zero potential line.
Velocity of chain isvwhen it slides byx.So,
−
m
l
×
l
2
×g×
l
4
+
m
l
×
α
l
2
+x
π
×g×
τ
l
2
+x
2

=
1
2
mv
2
On simpliﬁcation, we have
v
2
=gx+
g
l
x
2
This is proof required here.
Momentum Method
Umrao mrao
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0.1. FORCE 13
Solved Problem 0.6A uniform rod of length 2lrests on a table, with a lengthl−ain
contact with the table. The remainingl+asticking over the edge. Hereais 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 angleθwith the
horizontal before it slip. Where
tanθ=
2μ
2+9(a/l)
2
Whereμis 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 2lrests on a table as shown in above ﬁgure. Its mass ism
that concentratedat its center of mass. Its center of mass is at distanceafrom the edge of
the table (i.e. from pointA). ForceFis applied at right end of the rod so that it remains
horizontal. When forceFis removed rod ﬁrst starts rotating aboutAtill it attains
inclination of angleθ. After that, it starts sliding. The rod follows: (i) ﬁrst it starts
rotating about pointAdue to its own weight (w=mg) till it does not attain inclination
of angleθand then (ii) static forces and momentum generated due to rotation of rod help
it in sliding. At angleθ, weight of rod has two resoluted components,w
h=mgsinθand
w
v=mgcosθ. As mass of the rod is concentrated at its center of mass, so, at angleθ,
tangential acceleration (a
t) of center of mass isa t=aα,whereais distance of center of
mass of the rod fromAandαis angular acceleration.The diﬀerence of forcesmgcosθ
andNare responsible for the tangential acceleration (a
t=rα). Therefore,
ma
t=mgcosθ−N (10)
The rod rotates about the pointA. Before sliding, it rotates under the eﬀect of centripetal
force aboutA. The net forcef−w
his responsible for the balance of centrifugal force of
the rod. So, applying law of forces along the length of rod.
ma
r=−mgsinθ+f (11)
Centripetal force is given by
F
cp=m
v
2
t
a
Therefore, radial acceleration is equal tov
2
t
/a. Similarly,v t=aω. As The rod rotates
aboutA, therefore it has torque too. Therefore,
I
Aα=mg×acosα (12)
Arun
centratedentratedUmrao
atits centerat its center
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(i.e.from point(i.e. from pointA).Force) Fis applied at right end of the rod so that its applied at right end of the rod so that
al.Whenforceal. When forceFFis removed rodﬁrst starts rotatingaboutis removed rod ﬁrst starts rotating aboutAAtill itill i

14
The inertia of rod aboutAis given by
I
A=m
α
l
2
3
+a
2
π
(13)
Eliminatingαfrom relations 10, 11 and 12, we have
N=mgcosθ×
α
l
2
l
2
+3a
2
π
(14)
Till rod rotates about pointA, 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×asinθ=
1
2
Iω
2
Here,v t=aω.Solving,itwehave
mg×asinθ=
1
2

m
α
l
2
3
+a
2

×
v
2
t
a
2
=
1
2

m
α
l
2
3
+a
2

×
a
r
a
(15)
It gives
a
r=
2ga
2
sinθ
1
3
l
2
+a
2
(16)
Substituting it in 11, we get
f=mgsinθ

l
2
+9a
2
l
2
+3a
2

(17)
Now, fromf=μN,weget
tanθ=
2μ
2+9(a/l)
2
(18)
This is required answer.
Solved Problem 0.7A yo-yo is of massMand rotational inertiaI. The radius of its axle
isr, and is falls in the usual way with a length of string wrapped around the axle. Find
that its linear acceleration downward is
a=
Mr
2
I+Mr
2
×g
and tension in the string is
T=
I
I+Mr
2
×Mg
Solution
Aruntingit inting it inUmrao 11,weget11, we get
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ff=mgmsiniθθ

ll
22
+9+9aa
22

0.1. FORCE 15
r
C
C
ﬃ
A
W
W
T
C
r
A
W
a
Let a yo-yo has massM, rotational inertiaIand radius of its axle isr. 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 pointAwhere
unspung string is in contact with yo-yo axle. Its rotation is such that, its center of mass
moves fromCtoC
θ
. But at the same time, string is unspung such that, center of mass,
i.e.Cremains a distance ofrfrom pointA. 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 aboutAcomes from the torque applied by mass of yo-yo.
So,
I
A×α=W×r
Inertia of yo-yo aboutA(I
A)isgivenby
I
A=I+Mr
2
So,

I+Mr
2

×α=Mg×r
Again, if tangential acceleration isa
tand angular acceleration isαthena t=rα.So

I+Mr
2

×
a
t
r
=Mg×r
In case of yo-yo, tangential acceleration shall be equal to the acceleration of yo-yo in
downward direction (asay).
a=
Mr
2
I+Mr
2
×g
This is ﬁrst answer.
Second MethodWe can solve this problem by using conservation of energy method.
Let yo-yo falls by a distancexfrom 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 distancexin vertical plane. So,
Mg×x=
1
2
mv
2
+
1
2
Iω
2
Here, yo-yo has translational and rotational (aboutC) kinetic energies. From the relation,
velocity of center of massv=rω, so above relation becomes
Mg×x=
1
2
mv
2
+
1
2
I

v
r

2
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f yo-yo abyo-yo a
Umrao
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IAAII+Mr

16
On solving it, we have
v
2
=2×
Mr
2
g
I+Mr
2
×x
This is distance-velocity relation of motion invertical plane. So, vertical acceleration is
a=
Mr
2
I+Mr
2
×g
Third MethodThere may be a diﬀerence concept of solution. Let yo-yo falls by a
distancexfrom 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 aboutA. This is, as yo-yo tries
to rotate about pointAwhen it falls usually. So
Mg×x=
1
2
I
Aω
2
Substituting the values, we have
Mg×x=
1
2
(I+Mr
2
)

v
r

2
On solving it, we have
v
2
=2×
Mr
2
g
I+Mr
2
×x
This is distance-velocity relation of motion inverical plane. So, vertical acceleration is
a=
Mr
2
I+Mr
2
×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×
Mr
2
I+Mr
2
On simpliﬁcation, we have
T=
I
I+Mr
2
×Mg
This is second part of the answer.
ArunUmrao
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istance-velocity relation ofmotion in verical plane.So, vertical acceleristance-velocity relation of motionin verical plane. So, vertical acceler

0.1. FORCE 17
Solved Problem 0.8A yo-yo is of massMand rotational inertiaI. The radius of its axle
isrand outer radius isR. It is rests on a horizontal table. The string is wrapped around
the axle and held vertically by applying a forceP. The coeﬃcient of yo-yo and surface is
μ. Show that if
μ>
MrRP
(Mg−P)(I+MR
2
)
the initial motion of the yo-yo will be to roll to the left without slipping, with an initial
linear acceleration
rRP
I+MR
2
.Butif
μ<
MrRP
(Mg−P)(I+MR
2
)
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
ﬃ
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 ofa(acceleration of
center of mass of yo-yo). In this case, yo-yo tries to rotate aboutAto move leftward. So
taking torque about pointA:
P×r−f×0=I
A×α
Here,αis angular acceleration, that is related with linear acceleration asa=Rαfor the
pointA. Inertia of yo-yo aboutAis given by
I
A=I+MR
2
So,
P×r=

I+MR
2

×
a
R
On Solving it, we have
a=
PrR
I+MR
2
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https://sites.google.com/view/arunumraogeses/sites/sitestt
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gomome.come.comAAAA.c unun/arunw/arunrrNNNN
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18
This is acceleration, when there is no slipping of yo-yo. Yo-yo shall not slip iff>Ma.
So
μN > M×
PrR
I+MR
2
Again,N=Mg−PSo,
μ>
MrR×P
(I+MR
2
)(Mg−P)
Again, yo-yo shall slip atAwithout rolling whenf<Ma, i.e.
μ<
MrR×P
(I+MR
2
)(Mg−P)
The yo-yo when slip atA, it shall rotate aboutC. In this case, net moment by all linear
forces shall be equal to torque of the yo-yo. So, taking force moment about pointC
P×r−f×R=Iα
Here,Iis 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 0.9A 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 equalsμ. 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 theμand 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.
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α
II
l l ti f th h it t t t l k i ithlti fth

0.1. FORCE 19
mg
mgsin60
◦
mgcos60
◦
R
μR
mg
R
μR
Now the work done by friction force when ball moves from P to Q along the ramp.
W
1=μmgsin60×PQ
Similarly the work done by friction force while ball moves from Q to R is
W
2=μmg×QR
From the question, energy losses due to friction force are equal, hence
μmgsin60×PQ=μmg×QR
Again PQ is equal to 2/sin30
◦
,soQRis2
√
3m. The ball moves under the gravitational
force but presence of frictional force makesa control ﬂow. The resultant acceleration of
the ball isa=gcos60
◦
−μgsin60
◦
. Velocity at the bottom of the sloped ramp is given
by
v
2
=8(gcos60
◦
−μgsin60
◦
)
Kinetic energy at the bottom of the sloped ramp is lost in the frictional motion of ball in
horizontal part of ramp.
KE=4m(gcos60
◦
−μgsin60
◦
)
Now equating this kinetic energy toμmg×QR
μmg QR=4m(gcos60
◦
−μgsin60
◦
)
On solving it,μis approximately 0.29.
Solved Problem 0.10Two massmandMare placed at a ﬁnite distance over a friction-
less surface. BlockMis moving towards the massmwith constant velocity ofv.Wehave
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
Mv
m
l
x
Mv
m
l−vt
x
ﬀ
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 massm.Andthis
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( )
nergyat the bottom of the sloped rampis lost in the frictional motionnergy 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 massMis moving with constant
velocity. So,
x=
m×0−M×l
m+M
and
x
ﬀ
=
m×0−M×(l−v)
m+M
Here, displacement of massMin one second isv×1=v. Now, change in the position of
center of mass in one second isx−x
ﬀ
and it is also equal to the velocity of the center of
mass.
v
ﬀ
=
−M×l
m+M
−
−M×(l−v)
m+M
Or
v
ﬀ
=
−Mv
m+M
The center of mass shall shift leftward at the speedv
ﬀ
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 massMin the same direction of its velocity. It shall
increase the velocity of massM, therefore, center of mass shall move leftward faster than
v
ﬀ
.
2.If spring is compressing then there is retarding force acting on the massMin
the opposite direction of its velocity. It shall decrease the velocity of massM, therefore,
center of mass shall move leftward slower thanv
ﬀ
.
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.
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f spring is compressing then there is retarding force acting on the mf spring is compressing then there is retarding force acting on the m
site direction of its velocity. It shall decrease the velocityof masssite direction of its velocity. It shall decrease the velocity of massMM,tt

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)
ArunsurfacesurfaceUmraoapplied a reaapplied a re
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rtical direction or say normal to the surface and net displacement inrtical direction or say normal to the surface and net displacement in
is zero. henceis 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)
Whereais the acceleration of the object.
Solved Problem 0.11An unknown mass,m 1hangs from a mass-less string and descends
with an accelerationg/2. The other end is attached to a massm
2which slides on a
friction-less horizontal table. The string goes over a uniform cylinder of massm
2/2and
radiusr. The cylinder rotates about a horizontal axis without friction and the string does
not slip on the cylinder. Express the results in terms ofg,m
2andr. (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 tensionin the vertical section of the string and (d)
What is the value of the unknown massm
1?
SolutionThe ﬁgure for given problem is shown below.
m2
m1
g/2
m2/2
aFree body diagram of masses and cylinder are
T
m2g
R
Tv
m1g
Tv
Th
m2
2
g
R
bThe accelerations of both masses areg/2. Hence tension on the horizontal portion
of the string isT
h=m 2g/2.
cThe accelerations of both masses areg/2. Hence tension on the vertical portion of
the string isT
v=m 1g/2.
dUsing the free body diagrams, we have three force relations:
m
1g−T v=m 1×
g
2
(23)
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0.2. FREE BODY DIAGRAM 23
T
v−Th=I×α=
m
2rg
8
(24)
Here,Iis inertia of the cylinder about its axis along its length passes through its center
of base.αis angular acceleration.Its relation with linear acceleration is given bya=rα.
T
h=m 2×
g
2
(25)
Solving equation 23, 24 and 25 simultaneously, we have
m
1
g
2
=
m
2g(r+4)
8
On simpliﬁcation, it gives
m
1=
m
2(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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WithoutExternalForcesWithout External Forces
fff f ff

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 forceF.μis 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
ﬃ
Rand it gives the friction force
f=μR (28)
Net force along the the surface is given byF−μR. Note that friction force between pair
of surfaces depends on the net load/forceacting normal to the sufaces in contactor
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 ismg+f. It gives net reaction by the horizontal surfaceR=mg+f.
Consequently the friction force between block and horizontal surface isμR, i.e.μ(mg+f).
Now, there are three cases for relation between apllied force and friction force:
WhenF<fIf 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 calledstatic coeﬃcient of the friction(μ
k) for the given pair of surfaces.
WhenF=fIf 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 calledterminal coeﬃcient of the friction(μ
T) for the given pair of
surfaces.
ArunUmrao
https://sites.google.com/view/arunumraoc
RR

0.2. FREE BODY DIAGRAM 25
WhenF>fIf 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 calleddynamic coeﬃcient
of the friction(μ
d) for the given pair of surfaces. Body acquires an accelerationaand
the net force is given by product of mass of object (m) and its accelerationa. 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
m
1. The coeﬃcients of friction are diﬀerent for two masses and depend on the surfaces in
contact of the two blocks. In blockm
1, mass forcem 1gis balanced by reactionR 1.The
net force along the surface of plane is
F−μ
1R1 (33)
m1 m2F
μ1
μ2
R1
m1g
F
μ1R1
F−μ 1R1
R2
m2g
F−μ 1R1
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 massm
2as action reaction forces between the
surfaces of massm
1andm 2.Inmassm 2, mass forcem 2gis balanced by reactionR 2.
Net horizontal force is
F−μ
1R1−μ2R2 (34)
This is the net force that will displaced bothobjects right ward. If so, the acceleration
of the system is
F−μ
1R1−μ2R2=(m 1+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 massm
2. The coeﬃcients of frictions are diﬀerent for two masses
ArunUmrao
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μμ2
coo
μ11
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μμ11
unnuμμ22RR22

26
depend on the surfaces of the two blocks. In blockm
2, mass forcem 2gis balanced by
reactionR
2. The net force along the surface of plane is
F−μ
2R2 (36)
m1 m2 F
μ1 μ2
R1
m1g
μ1R1
F−μ 2R2
R2
m2g
F−μ 2R2
μ2R2
F
Figure 6: FBD of two body mass system in frictional horizontal plane surface at forced
state In Pull-Pull condition.
This net force passes to string as tension. This tension acts on the massm
1.Inmass
m
1, mass forcem 1gis balanced by reactionR 1. Net horizontal force on the massm 1is
F−μ
2R2−μ1R1 (37)
This is the net force that may displaced bothobjects right ward. If so, the acceleration
of the system is
F−μ
1R1−μ2R2=(m 1+m2)×a (38)
One Over Other
Amassm
2is placed over another massm 1. These mass system is placed over a plane
surface. All the surface in contact have frictions. The coeﬃcient of friction betweenm
1
andm 2isμ2andm 1& plane surface isμ 1. The friction force between massm 2and mass
m
1isμ2R2. This friction force tends to displace massm 1.
m1
m2F
μ1
μ2
R1+R2
m1g
R2
μ1(R1+R2)
μ2R2
R2
m2g
F
μ2R2
Figure 7: FBD of two body mass system in frictional horizontal plane surface at forced
state. One body is over to other.
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https://sites.google.com/view/arunumrao
yp j g ,yp j g
stem isstem is
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0.2. FREE BODY DIAGRAM 27
In massm
1, total reaction is (R 1+R2) hence friction force between the mass surface
and plane surface isμ
1(R1+R2). Now the net horizontal force on the massm 2is
F−μ
2R2 (39)
and massm
1is
μ
2R2−μ1(R1+R2) (40)
If both masses displaced rightward then acceleration of massm
2is
F−μ
2R2=m 2a2 (41)
and acceleration of massm
1is
μ
2R2−μ1(R1+R2)=m 1a1 (42)
0.2.4 FBD In Inclined Plane
Amassofmis placed over an inclined plane. The angle of inclination isθ.Thismass
is attached with a string and it passes through a pulley. The mass is pulled by applying
forceFon the string. This force passes to the mass in form of tensionT.
μ
m
T
θ
i
j
T
R
μR
mg
θ
Figure 8: FBD of body in frictional inclined plane surface.
The mass forcemgacts 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=mgcosθ (43)
There is no displacement along the normal to the plane surface. Net force parallel to the
plane surface is
T−μR−mgsinθ (44)
It displaces the mass along the surface ofthe plane. If there is an acceleration (a)ofthe
object thenT−μR−mgsinθshall be equal to the product of mass and acceleration of
the object.
T−μR−mgsinθ=ma (45)
Arunuu
jj
Umrao
https://sites.google.com/view/arunumraogoogle.ps:ps
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28
Solved Problem 0.12Two blocks of masses 50kgand 30kgare placed later over ﬁrst.
Static coeﬃcient of friction between two blocks isμ
2=0.36, Static coeﬃcient of friction
between 50kgbody and the surface isμ
1. Two diﬀerent forces of strength 500Nand
100Nare applied on the 30kmbody horizontally to get two values ofμ
1. Assume the
system is in ‘just move’ state and there is no sliding between two blocks. Calculate the
value of coeﬃcient of static frictionμ
1for two given forces. Also described the motion if
static coeﬃcient of friction between bodies system and surface is zero.
Solution50kg
30kgF
μ1
μ2
R1+R2
m1g
R2
μ1(R1+R2)
μ2R2
R2
m2g
F
μ2R2
It is clear that force applied on 30kgbody is horizontal. Due to inter-body interaction,
only friction force (μ
2R2) between two body surface is transformed from 30kgbody to
50kgbody in form of shear force. This force let the move two objects against the friction
force between 50kgbody and surface (μ
1(R1+R2)). To prevent the sliding between two
blocks, force should be less than friction force between two blocks. So
500N<μ
2R2
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 100Nforce is applied, motion of the system is possible
if applied force is less thanμ
2R2= 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 100Nis
considered for whole system. For body system and surface
μ
1(R1+R2) = 100N
This gives
μ
1=
100
50×10 + 30×10
=
100
800
=0.125 (47)
This is coeﬃcient of static friction between body system and surface. Ifμ
1is zero, the
frictional forceμ
2R2, transferred to 50kgbody let it came to move. For 500Nforce, body
system has two type of motion, one is sliding between two blocks and surface and other
is sliding between two blocks. For 100Nforce, body system has only sliding of whole
system over the surface.
Arun
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https://sites.google.com/view/arunumrao
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yinform ofshearforce. Thisforce let the move two objects against thy in form of shear force. This force let the move two objects against th
50kbd d f (f( (R+R)) T t th lidi b t

0.2. FREE BODY DIAGRAM 29
Solved Problem 0.13Twoblocksofmasses30kgand 50kgare placed ﬁrst over later
respectively in friction-lesssurface. Static coeﬃcient of friction between two blocks is
μ=0.36. A force of strength 500Nis applied horizontally on the 30kmblock that is at
left edge of 50kgblock. If length of 50kgblock is 5m, then ﬁnd the time after which 30kg
block leaves the surface of 50kgblock.
Solution
50kg
30kgF
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
F
f=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
a
1=
500−108
30
=13.07m/s
2
a2=
108
50
=2.16m/s
2
The relative sliding accelerationaisa 1−a2(both accelerations are in same direction) and
its numerical value isa=10.91m/s
2
. Now the time taken by the block of 30kgto leave
the surface of length of 5mof lower block is
5=0+
1
2
×10.91×t
2
(49)
on solving it
t=0.96s (50)
After 0.96s, upper block would leave the surface of lower block.
Solved Problem 0.14Two blocks of massm 1andm 2are 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
ﬃ
Facts on the top blockm
2. When the force
ﬃ
Fis small, the two blocks move
together. Find the magnitude of the force
ﬃ
Fabove which the blockm
2starts to slide
relative to the blockm
1.
Solution
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https://sites.google.com/view/arunumrao
aa11==
oglec3030
==1313..0707m/sm/s
2

30
m1
m2F
μ
R1+R2
m1g
R2
μR2
R2
m2g
F
μR2
If force
ﬃ
Fis 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
accelerationa. 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 ofm
2. Now in ‘just slide’ state, let applied force is
ﬃ
Fand acceleration
of both blocks isathen
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
μm
2g=m 1a=m 1
F
m1+m2
(52)
On simplifying
F=
μm
2g(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.
ArunUmrao
https://sites.google.com/view/arunumrao
pplied force is larger than the force on lower body due to acceleratiopplied force is larger than the force on lower body due to acceleratio
orce)then upper block starts sliding. Soorce) then upper block starts sliding. So

0.2. FREE BODY DIAGRAM 31
m1 m2
m1g
T1
m2g
T2
T
T1T2
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
T
1=T2 (54)
Ifm
1>m2then massm 1will move downward and there is an acceleration in the direction
of motion of the massm
1. Now the acceleration is
m
1g−T=m 1a (55)
Similarly for massm
2, acceleration is
T−m
2g=m 2a (56)
Solved Problem 0.1530kgand 50kgbodies are placed in two diﬀerent surfaces of diﬀerent
coeﬃcients of frictions 0.5andμ. Both surfaces are in the same elevation and distance
between two object is 10m. Assume 50kgbody is at the edge of beginning of second
surface. A force of strength 225Nis applied horizontally on the 30kgbody. The body
moves and collides elastically with 50kgbody and came to rest. Find the initial velocity
of the 50kgbody.
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 externallyapplied force. The net
force towards rightward shall accelerate 30kg body with accelerationa.So,
F−μmg=ma
Arun
for massfor mas
Umrao
mm22,accelera, acceler
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TT−−mm22gg==mm22aa

32
Substituting the values, we have
225−0.5×30×10 = 30×a
It givesa=2.5 meter per square second. When this mass covers a distance of 10 meter,
its velocity increases fromu=0tou
1. So, from velocity-distance relation
u
2
1
=u
2
+2as
Or
u
1=
√
2×2.5×10
It givesu
1=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
m
1u1+m2u2=m 1v1+m2v2
Or
30×7.07 + 50×0=30×0+50×v
2
It givesv 2=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 distancex,
then mass moves a distance of 2x(i.e. twice of the distance covered by pulley).
m
ﬃ
F
m
am
ﬃ
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 twicetime derivative of the distance covered by
it.
a
p=
d
dt

d
dt
x

=x
θ
(57)
ArunAcceleAcceleUmraoration ofration of
https://sites.google.com/view/arunumraobelow ﬁgure, it has seen that when pulley is pulled outward upto a dibelow ﬁgure, it has seen that when pulley is pulled outward upto a di
di t f 2di t f 2 (i t i f th di t d b ll )

0.2. FREE BODY DIAGRAM 33
Similarly, acceleration of mass is twice timederivative of the distance covered by it.
a
m=
d
dt

d
dt
2x

=2x
θθ
(58)
The ratio of accelerations is
a
p
am
=
x
θθ
2x
θθ
=
1
2
(59)
It shows that, acceleration of mass is double to the acceleration of pulley.
Solved Problem 0.16Assume 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 40kgby a chord. This mass is pulled by a force of 300Nhorizontally.
Find (a) accelerations of the masses,(b) Tensions in each of the chords.
Solution
m2
a2
T2
T2
T1
ﬃ
F
m
1
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
ﬃ
F
m
1
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−T
1=m 1a1 (60)
T
1=2T 2 (61)
T
2=m 2a2 (62)
From equations 60 and 61, we have
F−2T
2=m 1a1
ArunUmrao U
https://sites.google.com/view/arunumrao//sitesgooglecom/view/arunum

34
Or
F−2m
2a2=m 1a1 (63)
We know that, acceleration of dynamic pulley is half to the acceleration of massm
2.
Therefore for this problem,a
2=2a 1and equation 63 becomes
F−4m
2a1=m 1a1
Substituting the numerical values as given in the problem:
300 = 40a
1+80a 1
On solving it, we havea 1=2.5m/s
2
.Nowa 2=2a 1and it givesa 2=5m/s
2
. From
equation 62,T
2=m 2a2and it givesT 2= 100Nand from relation 61T 1= 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 ifF
1,F2,...,F narenforces acting on a body at equilibrium
then
F
1+F2+F3+...+F n=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 forcesF
1,
F
2,...,F nareF 1cosθ,F 2cosθ,...,F ncosθand at equilibrium their algebraic sum
F
1cosθ+F 2cosθ+...+F ncosθ=0
Similarly at equilibrium, algebraic sum of vertical components must be zero. ie
F
1sinθ+F 2sinθ+...+F nsinθ=0
Solved Problem 0.17Ablockofmassmis tied to two strings as shown in ﬁgure. Length
of each string isL. The angle between string and horizontal is 30
0
. 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.
SolutionThe free body diagram of the mass when it hangs with string is shown in
below ﬁgure.
Arun
aybesevay be sev
Umrao
eral forcesaeral forces a
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https://sites.google.com/view/arunumrao
pp 1,2,,n gyq

0.3. STABILITY & EQUILIBRIUM 35
m
T T
θ
T T
Tsin30
◦
Tsin30
◦
Tcos30
◦
Tcos30
◦
mg
b.Balancing horizontal and vertical forces
2Tsin30
◦
=mg
It gives that the tension in each string is
T=mg
c.If length of string islthen total vertical height themass covers during the fall is
h=l−lsin30
◦
. 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
mv
2
=mg(l−lsin30
◦
)
Simpliﬁcation and solving, it gives
v=
ρ
lg
T
θ
mg
mv
2
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
v
2
l
+mg
Substituting the value ofvand simplifying it, we have
T
θ
=2mg
It is the maximum tension when mass will be its maximum depth after cutting any of the
two string.
Umrao
https://sites.google.com/view/arunumrao
ation and solving, it givesation and solvin
vv==
ρρ
lglg

36
Solved Problem 0.18A mass-less plank of lengthl=1mis supported by two ropes, those
can bear maximum tensions ofT
1= 600NandT 2= 400Nrespectively. 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 isT
1+T2which is 1000N.Theweightof
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 distancexfromT
1. Now, the torque by both ropes should be
balanced. So,
T
1×x=T 2×(l−x)
Substituting the values, we have
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 0.19A plank of lengthl=1mand massm=10kgis supported by two
ropes, those can bear maximum tensions ofT
1= 600NandT 2= 400Nrespectively. 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
https://sites.google.com/view/arunumrao
( )
ingthe values,we haveing the values, we have

0.3. STABILITY & EQUILIBRIUM 37
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 isT
1+T2which
is 1000N. So the maximum weight shall be
M×9.8+10×9.8 = 1000N
It givesM=92.04kg. This mass can be placed in the plank at suitable location, i.e. at
distancexfrom the rope that can bear tension ofT
1. 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 massM,wehave
T
1×x=−m×
α
l
2
−x
π
+T
2×(l−x)
Substituting the values
600×x=−10×(0.5−x) + 400×(1−x)
On simpliﬁcation, we shall getx=0.403 meter from rope that can support maximum
tension upto 600N.
Solved Problem 0.20A mass-less plank of lengthl=1mis supported by two ropes, those
can bear maximum tensions ofT
1= 600NandT 2= 400Nrespectively. A body is put
on the plank and whole system is lifted by a constant acceleration ofa=2m/s
2
. 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 accelerationa=2m/s
2
, 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 isxfrom rope that can support
maximum tension uptoT
1. In this arrangement, upward tensions shall be equal to the
downward tension. Now, using law of forces
(T
1+T2)−Mg=Ma
Substitute the values, we have
600 + 400 =M(2.0+9.8)
It gives maximum massM=84.75 kilogram approximately.
m
plank andUmrao
m
whole syste
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imum mass of the body that can be lifted without breaking of ropes.

38
Solved Problem 0.21AcylinderofmassMand radiusRis lying on the street against
the side-walk. The height of side-walk ish. A rope is attached to the axis of cylinder
and force is applied by pulling it with an angleαwith 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 ofα,θ,Mandg,whereθis the angle between line joining
side-walk contact with axis of cylinder. (b) At what angle ofαis the ratio as described
under question (a) is minimum or is maximum.
SolutionA cylinder of massmis 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 pointPshould
be balanced by force and mass force. Now
F
mg
α
θ
P
F
mg
α
θ
P
Rsin(θ+α)
Rcosθ
F×Rsin(θ+α)=mg×Rcosθ
On simpliﬁcation, the ratio between force and mass weight should be
F
mg
=
cosθ
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 pointPare equalθ=45
◦
andα= 0. For minimum ratio, normal on force
should be minimum in comparison to the normal on the mass force drawn from the point
P,ieifθ<45
0
moment by forceFwill be lesser than the moment by mass force and
ratio ofF/mgwill be minimum. Again at this caseα=0
◦
. For maximum ratio, normal
on force should be maximum in comparison to the normal on the mass force drawn from
ArunUmrao FF
https://sites.google.com/view/arunumrao
iﬁcation, the ratio betweenforce and mass weight should beiﬁcation, the ratio between force and mass weight should be

0.3. STABILITY & EQUILIBRIUM 39
the pointP.Atthiscaseθ>45
0
andα>0
0
. The normal on the force line will be
maximum whenα=45
◦
.
Solved Problem 0.22An object subjected to three equal forces. Of them two are perpen-
dicular to each other (one along−xaxis and other along−yaxis) while third is making
angleθwith horizontal (+xaxis) 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 angleθand (d) It is not possible to ﬁnd the answer without known forces
and angle.
Solution
ﬃ
F
ﬃ
F
ﬃ
F
θ
x
y
ﬃ
F
ﬃ
F
√
2
ﬃ
F
ﬃ
F
θ
From the question, two forces are perpendicular to each other, in which one is along
−xaxis and other is along−yaxis. The resultant of these two forces is
√
2
ﬃ
Fwhose
direction shall be 45
◦
from the−yaxis 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
toF. 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
ﬃ
Fwhatever value ofθis.
b.It is not possible for this object to remain at rest. This statement is true, as third
forceisalwayslessthan
√
2
ﬃ
F.
c.This statement is false. We can give correct answer without knowing the value of
the angleθas 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 0.23Amassofmkg is hanging with a spring of spring constant ofk.Mass
is undergoing simple harmonic motion in vertical plane. At a certain point its velocity
isvand mass is moving upward direction. Find the total work done by the spring when
mass moves fromx
atoxbfrom the rigid platform to which spring is attached. Also ﬁnd
the velocity at the second position.
ArunUmrao
https://sites.google.com/view/arunumraothe question, twoforces are perpendicular to each other, in which onethe question, two forces are perpendicular to each other, in which one
√

40
SolutionAssume at a certain depthxfrom the platform, net force on the mass is
F
net=mg−kx (64)
If mass undergoes a vertical displacement bydxthen work done is
dW=F
netdx
Substituting the value of net force and integrating it for the vertical displacement ofx
a
toxb.
θ
dW=
θ
xb
xa
(mg−kx)dx (65)
W
xa→xb
=mg(x a−xb)−
1
2

x
2
b
−x
2
a

(66)
We know that if external forces or negligible and change in energy of mass is due to only
spring forces thenwork done is equal to the change in energy of the mass.So,
W=
1
2
mv
2
b
−
1
2
mv
2
a
(67)
On simpliﬁcation
v
2
b
=v
2
a
+
2W
m
(68)
This is required answer.
0.3.2 Non-Equilibrium State of Forces
Ifnforces 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
F
1cosθ+F 2cosθ+...+F ncosθ=F H
and resultant vertical component is
F
1sinθ+F 2sinθ+...+F nsinθ=F V
The resultant force on the body is
R=
ﬃ
F
2
H
+F
2
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 0.24A ball of mass 2kgis hit by a force of 20N. If friction force is 14.2N
then ﬁnd the acceleration of the ball.
Arun
equiredaequired a
Umrao
nswer.nswer.
https://sites.google.com/view/arunumraoNon-Equilibrium State ofForcesNon-Equilibrium State of Forces

0.3. STABILITY & EQUILIBRIUM 41
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 givesa=2.9m/s
2
.
In Horizontal Plane
Solved Problem 0.25Two blocks of massm 1andm 2are put on a friction-less horizontal
surface. Here,m
1is placed top of the massm 2. The static co-eﬃcient of friction between
the two blocks isμ. A forceFis applied on the top blockm
1. Find (a) When the force
Fis 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 forceFabove which the blockm
1starts to slide relative to the block
m
2.
Solution
aThe free body diagram of the two blocks is given below:
m2
m1F
ff
F
N1
w1
ff
N1
N2
w2
bIf both blocks move together on friction-less surface under forceF, then acceleration
of the block system is
a=
F
m1+m2
cForce (F)onblockm 1is 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 massm
2. So maximum force
that can be transferred to lower block by upper block isμm
1gthat is frictional force (f s)
between the block surfaces. Till the external forceFis lesser thanf
s, both block shall
move together. WhenFis larger thanf
s, upper block shall starts sliding over the lower
block. So,
F≥f
s=μm 1g
Arun
de
Umrao
for
https://sites.google.com/view/arunumraoonon

42
These are the answers.
Solved Problem 0.26AblockofmassMis placed in horizontal friction-less surface. Its
one end is attached with a spring of force constantkand lengthl. Another block of mass
mis placed over the block. ForceFis applied on the upper block to move it rightward
as shown in the following ﬁgure. The coeﬃcient of friction between block surfaces isμ.
Find the elongation of the spring.
Solution
M
m F
μ
l
M
m F
l+dl
Block of massMis placed over friction-less surface. It is connected with a spring
of force constantkand lengthl. The force applied on upper block of massmisF
(F>f
f) and it is moving rightward with constant acceleration. The friction force
between surfaces of blocks isμmg. This force shall be transferred to lower block. As
lower block is placed over friction-less surface, therefore friction force (f
f) shall elongate
the spring. Spring is continuously under constant forcef
firrespective of position of upper
block. At equilibrium, length of spring isl+dl. Now, from Hook’s law
f
f=kdl
It gives
dl=
μmg
k
This is elongation of the spring.
In Inclined Plane
θ
m
F
w
hff
ww
v
R
θ
m
F
w
h
ff
ww
v
R
Take a ramp of inclinationθ,onwhichamass mis placed as shown in above ﬁgure.
An external forceFis 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
w
h=mgsinθandw v=mgcosθ. Note that, in physics, horizontal and vertical axes mean
axis-line parallel to the surface and perpendicular to the surface respectively. Friction
ArunUmrao
g
https://sites.google.com/view/arunumrao
fff

0.3. STABILITY & EQUILIBRIUM 43
force between pair of surfaces depends on the net load/forceacting perpendicular
to the surfaces in contactor on net reaction acting perpendicular to the surfaces in
contact. Therefore, friction force is
f
f=μR=μw v=μmgcosθ
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.
w
h=μmgsinθ
Now, there are three cases:
Whenw
h<FIn 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.
Whenw
h>FIn 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.
Whenw
h=FIn this case all forces along the surface of slope are balanced, hence
block shall remain in rest.
Solved Problem 0.27a10gball rolls down a 1.2mhigh 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
W
fric=

mgh 1+
1
2
mv
2
1

−

mgh
2+
12
mv
2
2

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

0.01×10×1.2+
1
2
×0.01×0
2

−

0.01×10×0+
1
2
×0.01×4
2

On solving
W
fric=0.04J
Arun
Problem
H
Umrao
.27a10gba
hkid
https://sites.google.com/view/arunumraohtt//it l /i/
on

44
Solved Problem 0.28An electric train is powered on a 30kVpower supply, where the
current is 200A. If train is traveling at 90km/hthen ﬁnd the net force exerted on it in
forwards direction.
SolutionThe electric power supplied by electric train motor is
P=VI= 30000×200=6×10
6
Watt
Speed of train is 90km/hor 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
6×10
6
=F×25
On solving
F= 240kN
Solved Problem 0.29A car engine can deliver 90kWof power. The mass of car is 1000kg.
Find (a) Assume the total resistible force is proportional to the velocity.F
fric=αv.The
drag coeﬃcientαis 100Ns/m. How fast can the car move on a level Road? Express the
speed in the units ofm/s. (b) How fast can the car travel up a slope if we ignore all
friction? The slope of plane isθ=sin
−1
(3/5). Express the speed in the units ofm/s.
Solution
v
Ffric
P
v
h
θ
vcosθ
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
Substituting the values
v=30m/s
Arun
ﬃcient
the uniUmrao
i /
sofm/s.(
https://sites.google.com/view/arunumraohttps://sitesgooglecom/view/arunumrao
The slope of plane isθ=sin(/5). Express the speed in the units o

0.3. STABILITY & EQUILIBRIUM 45
b.Let the car travels in inclined plane andthere is no frictions. So power delivered
by engine will change the total energy of the car in vertical. So
P=mgvsinθ
Substituting the value, we have
v=15m/s
Solved Problem 0.30A square of sideLhas center of mass at its center ‘C’ (assume it as
origin). A part of areaL
2
/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
θ
¯
C
Take the center of mass of complete square (C) as origin. A square of areaL
2
/4is
removed from left bottom corner as shown in second part of above ﬁgure. Ifρis density
per unit area of the square, then mass of complete square isM=ρL
2
and removed
square ism=ρL
2
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 fromCto new coordinate (¯x,¯y). So,
applying the laws of center of mass for horizontal distance (i.e.x−axis).
¯x=
M×0−m×
−L
4
M−m
=
0+ρ
L
2
4
×
L
4
ρL
2
−ρ
L
2
4
=
L
12
Similarly, applying the laws of center of mass for vertical distance (i.e.y−axis).
¯y=
M×0−m×
−L
4
M−m
=
0+ρ
L
2
4
×
L
4
ρL
2
−ρ
L
2
4
=
L
12
So, new center of mass of the remaining portion of the square is
¯
C=
α
L
12
,
L
12
π
from the originC. This is required answer.
ArunAUmrao
https://sites.google.com/view/arunumraottt gogtps://sites.googite
L //v rarview/arew/
L/2

46
Solved Problem 0.31A spool of wire of massMand radiusRis unwound along a
horizontal surface under a constant force
ﬃ
F. Assume the spool is uniform solid cylinder
that does not slip. The coeﬃcient of static friction isμ
s. Assume that the radius of the
spool does not decrease signiﬁcantly while the spool is rolling. Find (a) Moment of inertia
of the cylinder about its central axis. (b) What is the acceleration of the center-of-mass?
(c) What is the force of friction? (d) What is the total kinetic energy of the spool when
center of mass is rolled through a distanceL.
Solution
aMoment of inertia of cylinder of radiusr, about its axis is
I=
1
2
mr
2
(69)
bTorque and moment of inertia relation is
T=Iα (70)
Whereαis the angular acceleration. Takingtorque about the point of contact
T=F×2r (71)
And moment of inertia about the contact point is
I=
1
2
mr
2
+mr
2
=
3
2
mr
2
(72)
This gives angular acceleration
α=
4
3
F
mr
(73)
Linear acceleration is given bya=rαso, acceleration of center of mass is
a=
4
3
F
m
m/s
2
(74)
For unwound of spool,a>F/m and from ﬁgure, force and friction forces are in same
direction. So
f+F=ma=
4
3
F (75)
That gives
f=
1
3
F (76)
As the center of mass does not move upward hence total work done is converted into the
total kinetic energy (potential energy is zero). As the center of mass rolled byL, string
is pulled by 2L,So
KE=F×2Ljoule (77)
It is the kinetic energy of the spool of wire.
Arunment of inment of iUmrao ertia aboutertia about
https://sites.google.com/view/arunumrao
II==
11g
mmrr
22
++mrmr
22
==
33
mmrr
22

0.3. STABILITY & EQUILIBRIUM 47
Solved Problem 0.32A string is hanging over a pulley of radius 0.25masshowninbelow
ﬁgure. If torsional friction of pulley and its axle is 20Nthen ﬁnd the lengthl
1when
pulley starts rotation. Given, length of string is 10mand mass of string per unit length
is 2kg/m.
Solution
μR
ﬃ
R
ﬃ
F
1
ﬃ
F
2
l1
r
ˆj
ˆi
The torsional friction of pulley isμR. This torsional friction is due to weight of pulley
and that part of the string which is wound semicircle over the pulley. The remaining
portion of the string is hanging both-side of the pulley as shown in above ﬁgure. Now,l
1
is the length of string that hangs right side of the pulley. And 10−l 1−πris the length
of string that hangs left side of the pulley. Now, mass of hanging strings is
m
1=2l1; m 2=2×(10−l 1−πr)
It gives
m
2=2l1; m 2=2×(9.22−l 1)
Thesem
1andm 2shall try to rotate the pulley clockwise and counter clockwise respec-
tively. According the question, when pulley is in just rotation state, then clockwise torque
should be balanced by counter clockwise torque. So,
m
1g×r=μR+m 2g×r
Substituting the values, we have
2l
1g×0.25 = 20 + 2×(9.22−l 1)×g×0.25
On solving it, we givel
1=6.61 meter.
Solved Problem 0.33A particle of unit mass is moving along the +x-axis under the
inﬂuence of force while its total energy is conserved. The potential energy of the particle
is given by
U(x)=
U
0
2

1−

x
α

2

2
Assuming thatU 0andαconstant, ﬁnd the position where force is zero.
Arun
thathanthat han
Umrao
gsleftsideogs left side o
https://sites.google.com/view/arunumrao
mm11=2=ll11;; mm22=2=2××(10(10−−ll11−−πrπr))

48
SolutionAs the total energy is conserved, force on the particle is given by
F=
dU
dx
=
d
dx

U
0
2

1−

x
α

2

2

So,
F=
U
0
2
×2×

1−

x
α

2

×

0−2×
x
α
×
1
α

Position of particle when force is zero, i.e.F=0
0=
U
0
2
×2×

1−

x
α

2

×

0−2×
x
α
×
1
α

It gives two cases:
1.When
1−

x
α

2

=0
In this casex=±α.
2.When
0−2×
x
α
×
1
α

=0
In this casex= 0. It is not acceptable. Hence, force on particle shall be zero when
x=±α.
0.4 Gravitational Stability
Assume a T-shaped frame in which two heavy masses are ﬁtted as shown in ﬁgure below.
This object is hinged to a point O. This object is free to rotate and bend about its hinge.
Assuming the mass of frame negligible, centers of masses are at the center of the two
heavy masses. This object is placed vertically.
Hmgmg
O
θ
H
ﬀ
O
h
O
θ
h
ﬀ
O
The height of the masses from hinge O isH>h. When this system is bend leftward
by an angleθ, center of mass of left side mass moves away from O horizontally and center
of mass of right side mass moves towards the O. Net torque is given by
τ=τ
1±τ2
aseasexxUmrao
0Itisnot0Itisnot
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0.4. GRAVITATIONAL STABILITY 49
If heightHis suﬃciently large then gravity center of right hand size object is overpasses
to the hinge O and acts at left hand side of the O. In this case net torque becomes
τ
ﬀ
=τ1+τ2
That tries to the bent object in counter clock wise direction. Again if the height of the
masses ish<Hthen the center of gravity of right hand size object remains right side of
the hinge O. In this case net torque becomes
τ
ﬀﬀ
=τ1−τ2
Now, it is clear thatτ
ﬀﬀ
<τ
ﬀ
, i.e. magnitude of bending of mass system is less in second
case, i.e. the system is more stable when center of gravity is at lower height from the axis
or point of rotation or bending. In practical, it is easy to carry heavy mass by hanging
both side of the pellian seat in a bike rather than ﬁxing it over the pellian seat.
ArunUmrao
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Principle of Force Application - Physics - explained deeply by arun kumar

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