11th physics chapter 2

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Unique Notes Physics 1 Year
Unique Notes Physics 1“ Year
1.9: Shin* that the famous “Einstein
equation” E = me2
is dimensionally
consistent.
Data:
Given equation is E = me
To Determine:
Power of r = n
Power of v = m
To Calculate
Eli
m VECTORS A
11101033
RIUM
n = ?, m = ?
Solution:
Formula:*
a oc r"vm
y
What is the vector quantity
^
Ans. Vectors:
Definition:
Equation is dimensionally consistent.
Solution:
Since
« Ql. 11102001
I
a = constant r" vm
Using dimensional notation
[LT2
] =constant x [Ln x (LT ‘)m
]
[LT2
]
(1)
E = me2
where c is velocity of light
Dimensions of energy are that of work.
Therefore dimensions of L.H.S.
E= [ML'T2
]
Dimensions of RH.S. =[me21= (MILT1
)
2
]
Hence
“ Those physical quantities
description are calledyectors."
For example,
'
velocity; accc
The vectpl
d is ‘d'
(light face
Graphicall
end. While length
scalfy
Q2. What is meant by Rectangular Co-ordinate System or Cartesian Co-ordinate System?
Ans. Rectangular Coordinate System: 11102002
Two lines drawn at right angle to each other are known as
rdinate axes and their point of intersection is called as origin. This
lem of coordinate axes is called Cartesian or rectangular
coordinate system.
In this system one
named as y-axis. The x-axis is taken as horizontal axis with +ve
direction to the right side while y-axis is taken as vertical axis with +ve
direction upward.
uch reqinre both magnitude and direction for their complete
= constant x Ln+m
Tm
Comparing Power on both sides
m + n = 1 (2)
eratiojr' Force etc.
bold face characters such as.A, d, r, v, the magnitude of a vector
;ed
e
and -m= -2 .
From Eq. (3)
m =
~
2
(3) a vector isTepresented by a directed line segment with an arrow' head at its one
‘
thepline segment represents the magnitude of the vector according to the chosen
Dimensions of LHS =dimensions of RHS
U (MLY2
!
Sc above equation is dimensionally correct.
l.lO'
.Suppose, we are told that the
aroi icration of a particle moving in a circle
of radius r with uniform speed v is
proportional to some power of r, say r°, and
vKoe power of v, say v®
determine the
powers of r and v?
Data:
Radius of circle = r
Uniform linear speed = v
Acceleration = a
V
Put the value of m in eq. (2).
2 + n = 1,
n = -1
coo
x x
line is named as x-axis and other line is * O
11101034
Y
P (a.b)
The direction of a vector in a plane is required by the angle
which the representative line of vector makes with positive -axis in the
ami clockwise direction as shown in fig. The point P has coordinates
(a.b). This notation means that if we start at origin, we can reach point
P by moving “ a” unit, along positive x-axis and "b” units along positive
y-axis.
4 b
:
+x
O a
* The direction of a vector in space is denoted by another axis
is. This 3*axis is named as z-
which is at right angle to x-axis and y-ais
axis.
19
—18

The direction of a vector in space is requires by three angles
which are formed by representative lines with x, y and z-axis
,respectively.
The point P of vector A is thus denoted by three coordinates (a.b,c).
L Q3. Explain the addition of vectors. OR 11102003
How art* vectors added?
is reversed.
A ns.Addition of Vectors: When ‘n’ represents a tity’ dimension, then the product nA will correspond
SIC;
physical tity intensions of the resultant vector will be the same as product of
“ If A and B are two vectors acting in different directions then to a new
dimensions of two qu
When veloi
having the dimensi
Q6. Define the
(i) Position
Ans. Position Vector:
A position vector is a vector which represents the position
®of a point with respect to the origin. It is denoted by r. It is
represented by a straight line which starts from the origin and
terminates at the point P (a,b). The projections of the position
vector r on the x and y-axis are the coordinates a and b and they
rectangular components of the vector r. i.e.,
r =ai + bJ
The distance between origin O and point P (a,b) is given by:
r =
^I7bI
In three dimensional space, the position vector ot a point
jltiplied. For example:
ss the product is a new vector quantity called momentum
>ns as that the product of dimensions of mass and velocity.
ollowing
:s, are
then sum will be obtained by drawing their representative lines in such
a way that the tail of B should coincide w ith the head of vector A. Now
join the tail of A to the head of B. It will represent the vector sum A+B
m magnitude and direction. This method of vector addition is called
head to tail rule of vector addition".
This rule of vector addition can be extended to find the
sum o! any number of vectors. Also the sum of B + A is shown
by dotted lines, i.e.
ipliei
1
llKUMMi
l (ii) Null Vector (iii) Equal Vector (iv) Unit Vector
L are
A + B = B + A x
o
It means that the vector addition is commutative i.e.,
when vectors are added the result is the same for any order of
%
addition.
Resultant Vector:
* 2
P(a.b.c)
0 y
P (a,b,c) is
; The resultant of a number of similar vectors is a single
r = ai + bJ + ck
vector which has the same effect as all the original vectors 1 2 *
* 2
and its magnitude is r = va" + b~
+ c~
taken together.
* Null Vector
It is a vector of zero magnitude and arbitral direction. For example, the sum of a vector and
04. What is meant by Vector Subtraction? 11102004
* its negative vector is a null vector i.e.
Ans. Vector Subtraction:
* A + (-A) =0
The subtraction n
| a vector is equivalent to the addition
Equal Vectors:
Two vectors A and B am said to be equal if they have the
It means that parallel vectors of the same magnitude
of the same vector with its direction reversed In order to magnitude and direction.
same
t subtract B from A reverse th* direction oi U and add it to A. regardless of the position of their initial points
At i.e.. are equal to each other
A - B = A + (-B)
I

Unique Notes Physics 1' Year
Unique Notes Physics 1"Year
tfl
.
Unit Vector:
I
^ 1
It is a vector whose magnitude is one and is used to represent OM
Cos0 =
the direction of a vector. OP
A unit vector in the direction of A is written as A which is
COS0 =AL
read as “ A hat” . Hence, A
A = AA A Cos0 = A*
From equations (1 ) and (2) if We
rectangular components. Vectoraforncof
Ax
^
Cos0i and AvWf
Q8. How will you Find th
Ans. Determination of a
If Ax and A4are P
then by Pythagorean theore
O P
~
>
=O t f + M
A2
=
t-(2)
A = —
I »w the vi
ttion (1) and (2) are written as:
A and angle 0 we can determine its two
or
A x
The direction along x, y and z axis are generally represented
A A A
by unit vectors i.jand k respectively.
The use of unit vector is not restricted to
Cartesian coordinate system only but unit vector may
he defined to any direction. Two frequently used unit
vectors are the vector r which represents the direction
of vector r and n represents the direction of outward
normal drawn on a specified flat surface.
Define Rectangular Components of Vector. How are they determined? Find their
expression.
Ans. Rectangular Components of a Vector:
Such components of a vector which are at right angle to each other are called rectangular
components of their resultant vector.
A component of a vector is its effective value in a given direction.
Consider a vector A is represented by a line
SinBj
nt vector if the rectangular components are given? HIOIOOH
»y rectangular Components:
igular components of a vector A
yeci
r A eci
From this.equation if we know two rectangular components
of vector then magnitude of the resultant vector can be
determined. Also
~ (3)
Q7- 11102007
MP
OM
Y
OP making an angle 0 with x-axis. If we draw projections
from point P on x-axis and on y-axis, then it will meet the x-
% /. . a' point M and .-ax: at point N As OM is along x-axis
named a A , and ON is along y-axis named as A,. Also ON
and MP are equal in magnitude and direction. So MP is also
N
From this equation we can find the direction of resultant vector by putting values of A,and A,,
Q9. Describe the method of addition of vectors by Rectangular Components.
(Board 2010)
11111200*
A,
named as A,,
Ans. Vector Addition by Rectangular Components:
If A and B are two vectors which are represented by two directed
lines OM and ON as shown in figure. The vector B is added to
vector A by head to tail rule of vector addition. Then the resultant
vector R = A + B is given in direction and magnitude, by line OI .
In the figure. A*
, B, and R* are the x-components of vectors
A. B and R and their magnitudes are given by lines OQ.MS and OR
respectively, i.e.,
OMP r. a right angled triangle in which by head to tail
rule A is the resultant vector of A, and Ay So A, and A are
components of A Because these components are at right
angle to each other are called rectangular components of A
O
In right angle tri«
.r , )c O.VI
MP
Sir.B
OP
Av
Sir.fi - -U OR = OQ+ QR
OR =OQ+ MS
R* = A, + B*
R* = RJ = (Ax + Bx) i
A
23

Which means that the sum of x-components of two vectors, which are to be added, ils equal to Rules for (he Determination of the Direction of th
the x-component of the resultant vector. Jtaiff?
Important Note: 9 is the angle from positive
Similarly y-component of the resultant vector will be: while0 is the angle from
ie vector
nearest x-axis (either positive x-axis or negati
Irrespective of the sign of Rx and R,, fi
taxis).
RP = RS +SP
alue of
Ry = Ay 4 B
ft)
y
-i
(j> = tan'
The sum of the magnitude of y-components of two vectors is equal to the magnitude of y.
component of the resultant vector.
* *
Ry = Ryj = (Ay + By)j
Then angle ‘0’ can be ca
If both Rv and Ry
i) pos resultant
vector lies in the
As. Rx and R are the rectangular components of the resultant vector R.
Hence,
R = R*i + R>i
Y
R = ( AX + BX) j + (Ay + By) j
The magnitude of the resultant vector R is thus given by
R -x/( A* + Bx)2
+ ( A> + By)2
The direction of the resultant vector R will be
-i
0 = tan
rA+ B
Ax -*- B
:)
-l
0 = tan
If we have any coplanar vectors A,B,C
Then the magnitude of their resultant will be
7
7+ (Ay + By 4 Cy +
R = /(A, 4 Bt 4 C,+
and the direction of the resultant will be
f A . 4 B C 4
)
•I
0 = tan
A . 4 B* 4 c, ^
Qlt. State the different steps and rules for addition of vectors by rectangular components.
vector will lie in fourth quadrant and its direction is:
An*. Main Steps for Addition of Vectors by Rectangular Components:
Following stept should be observed to find the resultant of vectors by the addition of rectangular
11102010
I
0 =360-(
?)
ft)
-i
components: 0 = 360-tan
Find x and y components of all the vectors to be added. f
Find x-component of the resultant vector Rt' by adding the x-components of all &
i)
>
a)
Product of Two Vectors
There are two types of vector multiplications. The product of these two vectors are known as
vectors to be added
Find y -component of the result*
*
vector R/ by adding the y-components of all &
ui )
vectors to be added (i ) Scalar Product or Dot Product
i Find the magnitude of the resultant vector R by using the formula ( ii) Vector Product or Cross Product
IV )
R = V/R2
,*R >
Similarly, find the direcn - f the resul it vector by using the relation
i
: V)
RJ
i
0 = tan
25
24

u.uUif qu/ihtii
t tnt/,
*W;ff
^ A H = ABOAB
BA* HA
.CtAh
placed between (henn
uec» * te a <Vrt 6)
['t
- AkCotM
f . Thu*
/ f d j t f t t A and B in writte
A.B = B.A
I *pi ifl .)
’ I
'f * '
The order of mu prodoct ii commutative.
n( t/ rT A to B making an angle 0 an
vt The
^
ca
^
ar pr<
TheSector prodi
~
Xl?=ABCos9Q
2. perpendicular vectors:
perpendicular from 0 on OP which r
0 A vectors A Sc B who are perpendicular to each other is:
:wi
O Bcc/
^e p
M /nof vectr/r B along A ,
(v Cos9<r=0)
Thi
A B Af Projection of vector H along vector A )
AH * A(BCa00) irs i, j and k since they are mutually perpendicular therefore
(1)
A B ABCosO
Similarly, if wc change the order of two vectors. The
perpendicular from point P is drawn on OQ lor representation
O
. !° =lxlx0 =0
of projection of A ah>ng vector B
Thus;
Similarly.
Hence
B A = B( Projection of vector A along vector B )
BA = KACMD
B.A = BACosB
_
By comparing eq. ( I ) & ( 2) we conclude that
IA.B
^BA
0
o A A A A A
i.j = j.k =Jd =0
Scalar Product of Two Parallel N ectors: . ,
The scalar product of two parallel vectors is equal to the product of their magnitudes i.e. for
parallel vectors (0 =0°).
ivB^ABCosO0
= AB
~
X
^= AB
In Case of unit vectors
(2)
*

[vCosO°= l]
U4 A A
i.i = ixiCosO0
V* (M pwdud Fmot
a 4COIM(
juo/rttTtu
O'j P= ?~)
U
^^ftedud- J JU
^
a, •
'
= IxiCosO
0
=1x1x1= I
A A A A A A
i.i = j.j =kvK = l
When two vectors one anti parallel then
•or mitipanilld vectors 0 = 1K0°
AIK’osI HO*1
All
The Dot Product with Itidf:
The sell dot product o! a vector A IM
"
^.7? ; AA ( os0 A
/ •e
‘
Similarly
!vCosl80°
4.
•ijual U> square of its magnitude i.e.
27
I
*
20

lVLf Q#
_
m
Unique Notes Physics I Yeat
Unique Notes Physics 1“ Year
Scalar product of two vectors A and B in terms of their rectangular components
A=A,i+Afj+A,k
B=B,i+B, j+BJk
A.B=A,B.+A > BJ+A,B,
In order to find the angle between two vectors
A.B=AB CosB^A.B.+A.Bj+A
^,
-> -9
~~
C&ixSL - A - B_
5
Right Hand Rule:
According to this rule, the vector A is rotated
vector B through small angle. Curling the fingers
in the direction of smaller rotation, the exi
iiA x 8
M
it hana
^
indicates the direction of A x B. The direeti
perpendicular to the plane containing vectfc A
If n is a unit vector in th
the vector product A x B is com
tion
AxB= IAXB
A x B =AB sini d)
A6 Similarly, if we chi then
ie o]
N
Bx A = IBxA!(-nj
B x A = BA sin0 f-n
Comparing (1) andj(2) it is con
*
0
(2) O
A
that
B x A
-h
r product is non-commutative.
v
a = )
r:
AB “ The turning effect of the body is called torque "
Mathematical Expression/Formula:
Mathematically, it can be defined as the vector product of force and moment arm.
T =rxF
The S.I. unit of torque is Nm and dimensions are ML T
J.
Angular Momentum:
“ Angular Momentum can be defined as '‘the cross product of pos.uon vector and linear
^
nomentum." Mathematically
product Describe its important characteristics.
^ 11102012 T.
< )2 Define and explain vector
An*. Definition:
When two vectors are multiplied and their product results in a ve
product n uiui to he vector product or cross product.”
It tr»e product of two vectors u a vector quantity, the product is known a sector product.
The vector product i also called cross product because a x" is placed in between
i
Jnlt:
tity then tli
*
ctor
i S L = rxP
f characteristics of vector product:
^ Vector product is not commutative.
K I vectors
Magnitude of Vector Product:
Consider two vector* A and B, make y an angle 0 with ach other
then magnitude of vector product is defined atf
IA x B I * AB Sir.fi
lr»e direction of A x B is determined right h *nd rule.
.(i)
A x B =ABSinQn
Xj
By rule, the direction on vectot n is upward.
Bx A =-ABSin0n
Negative sign shows ihai
The direction of n is downwarii while the magnitude is same
29
—
28
-

7 {
' ) Nor-Cfirr
*n*Xfi&»C i
'
As AXB - AB5h<a O
«4- * -A& 2
A KV^
<H "
ixB*
T
, Unique Notes Physics I " Year
For unit vectors T.j and k
ixi =ixiSinO=)xlxO=0
-&J0 .
Uonce j Vect&C 0
^CAos
*p0(iM6f
^Iwo
J<
^cko
li
GrrnrYOjJzoJti
^ .
Similarly,
nw
- jxj= kxk =0
Hence
V ^ A A A
^
ixi = jxj=kxk
^
2. ( TOSS product of perpendicular vectors:
The cross product of two perpendicular vectors has maximum magnitude i.e
[v Sin90° =jl
In case of unit vector which are mutually perpendicular,
ix j=ijSin90°k = Ixlxlk
A A A A
ixj = lxk = k
jxi =-jiSin90°k
A A A
jxi =-k
5. If A and B are t'
the magnitude
parallelogram.
SIi f a parallelogram then
act will be area of
*
A
tor pi t
A x B = AB sin 90 n = ABn
B
Bsin 0
!
0
»
A
ector Product in terms of their Rectangular Components:
If consider two vectors A and B in Cartesian space, which are given by:
A=Axi + Ayj + Azk
Bt=Bxi + Byj+ Bzk
Then AxB = (Axi + Aj + A2fc) x ( Bxi+ Bv j+ Bzk)
AxDxixi i AxDyi)dj AkBjvk,
*
= AyR Jvi * A} B jvl * A ,Bjvk +
— *1 n f 11 A
.P H i yp ^
A x B
-AxBvH0 I A B;( J ) i AyBd k) i AyB,i i A,BJ i AjBd R
A x B = i(AyBz - ByAz) - j(AxBz - BXAZ) + k(AxBy - BxAy)
The above equation is written in determinant form:
6. V
and
Similarly,
jxi =-k
ft ft A
k x j =— i
A * *
ixk =-j
Cross Product of Parallel Vectors:
The cross product of two parallel vectors is null vector i.e.
AxB = ABSin0°n = AB(0)n =0
Also If 0 = 180° then
AxB = AB sin 180°/? =0
ix j = k
Aft ft
jxk = i
i
kxi = j
3.
° =o] j k
SinO I
Ax B = Ax Ay Az
Bx By Bz
, Ay A2
J A, A,
By B,
‘
B, B,
=/( AyB.- AlBy )- j( A,B.- A,.Bx) +«ArDy ~ AyB< >
ri
[vSinl80° =0]
1
i
A, A,
*
AxB = i
A x B = AB (0) n
A x B =0
Self Cross Product:
Self cross product is also equal to null vector i.e.
Ax A = A A Sin0°n = AA (0) =0
AxA =0
3^ 4. v
1
component.
1
* of AxB-
A A
^
. A .B.I
Thu
( A v H I - ( A I L
( A x B )y
( A x B h
( AAB,
( AnBy -AjBJ
*
30
/

Unique Notes Physics 1'
Year
^Unique Notes Physics V' Year
013. Define Torque, name its unit and hence describe the factors on which torque depend Direction of Torque:
Ans. Torque:
The direction of the torque is represented b
Uii)
Definition:
direction of plane containing r and F and is de
When a fan t it applied ona body and body rotates about a fixed axis then turning effect vector form is written as:
force produced in the body which is called torque." OR x = (r FSin0) n
Mathematically, it can be defined as the torque is product of force and moment arm.
" Special Cases of Torque:
Unit: Its SI units are N m and dimension ML2
T 2]. Casel:
The magnitude of the
Moment Arm: given torce and for given point of its application depends
upon the angle ‘0.
’- between ft e an positionVector with respect to axis of rotation.
‘
It is a perpendicular distance between line of action of force and axis of rotation"
. When 0 = 90°
Unit: T = rF Sin90'
Which is
The SI unit of torque is Nm and dimensions are ML T 2
.
m le torque.
Case 2:
Expression of Torque on a Rigid Body: When force F and po vector r are in the same direction. In this case 0 = 0°.
( onxider a rigid body which is pivoted at the point O. Consider the ore
in0°
tore*- I acting on the rigid body at point P whose position vector with
x =0
respect to a x i s of rotation is r. The force F can be resolved into two
minimum value of the torque.
ie
components, one in the direction of r and the other perpendicular to r. The
be noted that anticlockwise torque is taken as positive whereas clockwise torque is
component ol the force in the direction of r and perpendicular to it are,
taken as negative. Force determines linear acceleration in a body and torque acting on a body
H o .0 and I Sin0 This is as shown in figure. The torque due to FCos0 is determines its angular acceleration. Torque is the analogous of force in rotational motion.
014. Define equilibrium. W hat are its type? State and explain two conditions of equilibrium.
/cro !»« •( ausc it line of action is passing through point of rotation O. The
Ans. Equilibrium: II 102U14
torqm due to the component I Sin0 about point O is given by:
Definition:
Torque Force moment arm “ Body is said to be in state of equilibrium if it is at rest or moving with uniform velocity.
"
Types of Equilibrium:
x FSinB OP
Static Equilibrium:
When- OP is the moment arm and is equal to the magnitude of position
When the body is at rest it is in static equilibrium.
vector r Therefore Translational Equilibrium:
When a body is moving with uniform linear velocity v. it is said to be Translational
lI
X FSmOXi
Equilibrium.
x rPSinO I I )
Rotational Equilibrium:
In this equilibrium state, the body is moving with constant angular velocity w
Where 0 r the angle hoi ween force 1 and position vector
Complete Equilibrium:
When a body is in Translational Equilibrium as well as in rotational equilibrium then this
Alternative Method of Calculation of Torque:
Alternatively torque cun al be d<* tunned by resolving r into two
'•late is known as a complete equilibrium.
In static ant.dynamic equilibrium the accelerate of the body
al« »ng the direct ion of h*
,t e P and the other
one
components. is zero i .e..
perpendicular to it . I be component of / pcrpendicul.u to the force F is r a = 0 and u = 0
Sinfi and it is also the loment .nm hy definition of torque is zero i.e. u = ()
In rotational equilibrium the angular acceleration
Equilibrium of Forces or First Condition of Equilibrium:
"It states that a body will be in translational equilibrium if the vector sum of all the lorees
t Force x ( M w nt arm)
X F ( rSinO )
S
acting on a body is zero. i.e. IF = 0
* rF Sin0
• X =

Unique Notes Physics 1* Year
In case of co-planar forces, right side force is equal to left side force then
acting along x -axis i.e. I F =0
Also when upward force is equal to downward force then no resultant force is
y -axis and X Fy =0.
no resultant f0rce]
acting a|
, r 5
2.1. Define the terms (i) unit vector (ii) f 2.PVector A li
Position vector (iii) Components of a vector. orientations wi
Ift =0
lFy=0
Z F=0
the xy plane. For what
•th of its rectangular
components are negative and for what
orientation will its components have
opposite signs?
Ans.: If vector A lies in the 3rd
quadrant then
its both components A* and A> will be
negative and If vector A lies in second and
As
And (Board 2010, 14) 111020
Ans. (i) Unit Vector:
Hence
Second Condition of Equilibrium:
It states that for body to he in rotational
equilibrium the sum of all the torques acting on a body
about the same axis, should be zero i.e.
I T =0.
A vector whose magnitude is one and
used to represent the direction <
Thus a unit vector is foundry
^ vector by its magnit
vector in the directio
hat
1
thus
11102017
a vector,
ividing a
n of unit .
A
n
fourth quadrant then its both components Ax
and Ay will have opposite signs.
2.4 If one of the components of a vector is
not zero, can its magnitude be zero?
Explain.
If two equal and opposite force are acting on a
body but in the same straight line then first condition of
equilibrium is satisfied i.e. I Fx =0 I Fy =0
( no Force). But this body can rotate clockwise.
If these forces are acting in such a way that forces
are parallel but in opposite direction then these forces
will not rotate the body and body will be in rotational
equilibrium i.e. anti clockwise torque will be balanced by
clockwise torque and
XT = 0
It means that sum of all the torques acting on the body is
zero. And it is called second condition of equilibrium.
A = IAIA
( Board 2009)
Ans.: The magnitude of the vector in term of
its rectangular components is written as
11102018
(ii)
position vector r is a vector that describes
location of a particle with respect to the
i-ordinates of point P(a,b) then
tor of point “ P” in Cartesian
/ 2 2 2
VA*
+ Ay + Az
A =
origin. If __
position vect
coordinates is
F
C If Ax = 0.Ay =0 and Az*0
-
Jo+ o+ A“
A =
: ai + bj (In two dimensions)
r
I A = Az
'cioww* r = ai + bJ + ck (ln three dimensions)
Anti Clockwise
A *0
(iii) Components of a Vector:
It means that if one of the components of a
Such parts of a vector whose combined effect
vector is not zero then its magnitude will not
called
the resultant
gives vector are
zero.
components of a vector. 2.5. Can a vector have a component
2.2 The vector sum of three vectors gives a
greater than the vector's magnitude?
t zero resultant. What can be the orientation ( Board 2010) 11102019
of vectors? Ans. If components of vector are rectangular
11102016
Ans.: If three vectors A, B, C are represented then
by the sides of triangle taken in cyclic order in
A = >/AX + A;
such a way that head of last vector coincides A
with the tail of first vector then their resultant Which implies A>
A > Ax
will be zero i.e.
A > A A
A + B =-C >
Hence the magnitudes ol the components of
the vector can never be greater than magnitude
A + B + C =
0
of the vector, if they are rectangular
not the
But if they are
components,
rectangular components then components may
h means that vector sum of vectors A, B and C
be greater than the resultant vector.
,s equa! to zero.
34

<
' 1
'AM
u * 4 a *
*
***
* t
*
*+
* a
*“* %
AM1M*
mrur %
*9**+
*m
4 *
$
*
*A 0
V
**4 r
*4
*k
1 M* y
*«W+
4 M
r+
/' M
'+
' ** & ' l/ ' <
* M s* > •
^
**ft 9
+fitt
* 4 « Vfr/* y.> w>/ A tArt'
m
*vv/ tml v/ V jM&f m*v <w
|
mmC
^ MMM*
*v 4**r ft i*
SE
** Jr**#-**
//A
-fc/M
P- /,
%*&
Hence'A + ftp t* a^ »vt
The *n0e ft a- <j
111 H<w» < wM
*#
44V«** v*******
*****
*" vW V V2
^
rtr *4 Sf
^SfAf 40f t V
*«MV V*>«r
-s*
«fefefe the vil*; yvveMfM 4
f -fl 04901(
SpnniMi r £
* f
'
> A
fc* '*
4* i*v
^ V 2
.M.
/
**#
repreywif wttr
Cafe Ufa*fc
nwm* (4
'
1fc<v« r4 /
*** ::4HK
Aooi.K fw*y vttj<
*
tarvt tmeqtral magmrtMfe
fhea their MM eao never be zero »heo * •
fwo /ettor* of equal magnitude are aetm*1
tppfrtite drrect*r*» then their wm wifi he equa
to zero
2.12. Show that the *om and difference t
1
*
r Mrt
W A A 1
1 *
iTthey AIK. When mare
cnmhrned .n wh a way Thai die head of Last
vector oomcides with the tad of fint vector
Then they form a dosed polygon and cheir
vector on is zero i.e
Form Fig
p
*w> wc«r» we
were to he o
eqoaf to a vect
H * B 1 # fc a resultant
the
fee* A * ft
A » *
ft
sen Motor A and B of
120° then their resultant
mitude as A and B. This
A_+ B +C#D = AE
AE = -E
A + B + C + D =-E
A + B + C + D + E =0
For any closed path if initial and final points
are same then the resultant is equal to zero
'B* * j)
A
Ay l two perpendicular vectors of equal length*
are aho perpendicular and of the sarw
length.
A, i > Av j= - B, i - By j
ft comparing the coefficients
A, =-ft, and Ay = -By
It means that components are equal in
magnitude hut opposite in direction.
2.H. l nder what circumstances would a
vector have component that are equal in
magnitude?
I1IO202A
A ns.
Form Fig.
(OQ)
2
= (OP)2
+ <PQ,2
As R = A = B =F
OQ = yl(OP )
2
* (PQ)
:
F =A/F2
+ F2
+ 2F:Cos6
F
2
= 2F2
+ 2F2
COS0
-F = 2F2
Cose
Cos0 = -
11102022
OQ = JT7F- M
IA + Bl =JA~ + B2
( I )
(OR ,’
= (OP)2
+ ( PR)2
OR =V'OP)
'
+ (PR)2
OR =VAT+ B
IA - Bl = VAJ+ B2
IA + Bl* IA Bl
Ans. If a vector A is making an angle of 45°
with horizontal then its both rectangular
components arc equal in magnitude i.e.
A, = ACos 45° = A ( .707) = 0.707A
Av = A Sin 45° = A ( 707)
=0.707A
It means that A, and A . are equal in
magnitude
I he other orientations for which components
have same magnitude are 115*.225°.115*
2.9. Is it possible to add a vector quantity
Hence, the resultant of all the vectors along
the side of the closed polygon from head to
tail rule is zero.
F2 1
2F2 2 2.16.Identify the correct answer.
( i ) Two ships ami 1 are travelling in
different directions at equal speeds. The
actual direction of motion of is due north
observer on y, the apparent
I
or 0 =Cosl(-
^) = 120°
2.14.The two vectors to lie combined have
magnitudes 60N and 35N. Pick the correct
answer from those given below and tell
"In
is it the only one
0) IIHIN
Ar»s4
acting in the same direction then tht u r
‘'"han
will in*
(2)
hut to an
direction of motion of X is north Fast.The
actual direction of motion of V as observed
from the shore will be:
1, 1 S + ft ) ( A ft ) A A A ft 4 ft.A ft ft
of the three that is correct.
nnmux
A' A ft* ft A ft
tit scalar quantity? Explain. r Il 102010
( iili 20N
( A 4 B).(A -It ) * A1
- 11^ fil) 70N
II two vectors ol 60N and 15N arc
.iiM *11 ( It1 West
(A) Fast
V A It
Aox.It »s not possible to a/k. v« tor quam > y (|
» South west
(( *
) North east
Ans. The relative velocity of ship *x’ can be
of velocity of
>n(r> A sc alar quantity be< ause t. quantities » »l Ok
Form diagram
.<rne nature < an he arkled mu/ frr»e '»ther
obtained by adding negative
Id ) 15 95N
37

7
i
ship V to the velocity of ship X as shown in
fig
If y is moving towards west, then v{= v, +(-v)
is North east.
Unique Notes Physics ln Year
i J k
- A, x C-A2) = 2.20. A picture is suspended from a wall by
two strings. Show by diagram ,he
Comparing (1) Md
' of „,« s,rings »Wth
that
( ) 11 ls
Proved ,ens,on ,n the stnn8s will be minimum.
A| x A2 = - A, x (-A2)
AI* Aly A,z (2)
5in^(l) =90'1
A2, A2y A22
N -v
w
V
ran - *
2x1
V,
W
Ans.:
w<
E ann
y
2
Therefore for
Tension in the
8
(II) A horizontal force F applied
object P of mass
an angle 0 to the horizontal as shown in fig.
The magnitude of the resultant force acting
up and along the surface of the plane
(a) F Cos 0 - mg Sin 0
(b) F Sin 0 - mg Cos 0
(c) F Cos 0 + mg Cos 0
F Sin 0 -» mg Sin 0
<e) mg Tan 0
The net upward force
along the inclined plane is:
R = F Cos 6 - W Sin 0
As W = mg
. R = F Cos0-mg Sin 0
2.17. If all the components of the vector Aj
and
i were reversed, how would this alter
Aj x A2?
Aas. Let A = A , i -At> j -r A!z k and
AZ= A2I i ^ A;) j + A-, k
i k
A
^
A„
A A2# A2Z
When components are reversed
- A, x (-A2 ) =
minimum
to a small
m at rest on inclined plane at
strings, the
picture is suspended asshown.
2.21. Can
. A
0 a body rotate
gravity
under the action of its weight?
about its centre of
A
W=mg
. 1110203!
(Board 2009) 11102036
Ans.: The body can’
t rotate about the centre
of gravity under the action of its weight
because in this case the line of action i.e.
weight of body is passing through the
of gravity and its moment arm is zero i.e. r=0.
T =rx F
By reversing the components of Ai and A2.
-Ai x -A 2 = A] x A2
The vector product will
F P.
->
<d )
remain same as
e Az
Aas. centre
2.18.Name the three different
that could make Aj x A2=0
Condition are:
Ans. The Product A
(i) Vectors are parallel ( 0 =
(ii) Vectors are anti-parallel
( iii) If one of the
vector.
= 0 x w
T = 0
i x A 2 =
vectors Aj and A2 is a nui
11102032
2.19.Identify true or false statements an-
explain the reason.
» 111020«
( a) A body in nuilibrium implies that it ;
not mo> ing nor rotating.
Ans. The first statement is false because <
moving body cm also be in equilibrium wh*
it is mov mg with uniform velocity.
(b) If coplanar forces
A . x A. = A ., ( ,)
s
k
i i
- A ,
.
- A>#
A >i
- A
*
- A„
- A?,
acting on a lx**
forii closed polygon, then the body is S3
*
to be ii equilibrium.
Taking common negative from second and
ns. This statement i
third row
Aj x ( A2 ) = ( ) (-) )
I

L'
|
Unique Notes Physics 1" Year
Solved Examples Step (ill)
The magnitude of y component Fy of the resultant’
fora,
Fy = F|y +F2y
Fy = 5 N + 17.32 N = 22.32 N
Example 1: The positions of two aeroplanes at any instant are
( 2.3, 4) and B<5, 6.7) from an origin O in kin as shown in P ig. 2.8.
(i) What are their position vectors?
(ii) Calculate the distance between the two aeroplanes.
Solution: (i) A position vector r is given by
A A A
r = ai + bj + ck
Thus position vector of first aeroplane A is
OA = 2 i + 3 j+ 4 k
And position vector of the second aeroplane B is
OB = 5i + 6 j+ 7 k
By head and tail rule
OA +AB = OB
Therefore, the distance between two aeroplanes is given by
AB =OB -OA =(5i + 6 j+ 7 k )-(2i + 3 j+ 4k )
= (3i + 3 j+ 3k )
Magnitude of vector AB is the distance between the position of two aeroplanes which is then:
AB = yj( 3km )
2
+(3km)
2
+(3km)2
=5.2km
represented by two points
11102037
Step (iv)
The magnitude F of the resultant foj
F = JFX + Fy2
= /(18
^66 22.32 NT =29 N
-fnakes an angle (
Step (v)
If the resultant force F the x-axis then
0= tan
'
=t
1.19i
=t;
Example 3: Fi
A
their resultant is
Solution: Let 0 be the angle between two forces F| and F2, where Fj is along x-axis. Then x-
component of their resultant will be:
Rx =Fi cosO0
+ F2 COS 0
;le between two forces of equal magnitude when the magnitude of
to the magnitude of either of these forces. 11102039
Rx = Fj + F2 COS 0
Any y-component of their resultant is:
Ry = F| sin 0° + F2 sin 0
Ry = F2 sin 0
R2
=R? + Ry
Example 2: Two forces of magnitude 10 N and 20 N act on a body in directions making
angles 30° and 60° respectively with x-axis.Find the resultant force. 11102038
Solution:
The resultant R is given by
R = Fj = F2 = F
F2
= (F + F cos 0)2
+ (F sin 0)2
0 = 2 F2
cos 0 + F2
(cos2
0 + sin2
0)
0 = 2 F2
cos 0 + F2
2F2
Cos 0 =-F2
As
Hence
.
Step (i) x-components
The x-component of the first force = Flx = Fj cos 30°
= 10 Nx 0.866 =8.66 N
The x-component of second force = F2x = F2 cos 60°
= 20 x 0.5 = 10 N
Or
L Or
-F2
:
i
Cos 0 = 2p* = 2
1
=-0.5
y-components
The y-component of the first force = F) y = Fj sin 30°
= 10 N x 0.5 = 5N
The y-component of second force = E = F2 sin60
=20N x 0.866 - 17.32 N
Cos 0 ='
2 “
1
Or cos 0 =-0.5
0 = cos"
'(-0.5) = 120°
Or
A force F = 2i + 3j «
*,
-»> P»"' ‘
T
Fxanipie 4:
Step (ii)
The magnitude of x component Fx ol tin resuliant force F

Unique Notes Physics 1” Year
Example 7: A load is suspended by two cords
maximum load that can be suspended at P, if maxim
A yy
^
Displacement = d = rB - rA = (5 - 1) i + (7 - 3) j = 4 i + 4 j
Work done = F.d = (2 i + 3 j ).(4 i + 4 j )
= 8 + 12 = 20 units
- Fig. 2.15. Determine the
ision of the cord used is 50
11102043
breal
N.
Find the projection of vector A = 2 i - 8 j+ k in the direction of the
B =3i — 4 j— 12k .
Solution: If 0 is the angle between A and B, then, A cos0 is the required projection.
By direction
Example 5: Solution: For using conditions of equili
point P are shown by a force diagram aaii
assumed to be the maximum weight
inclined forces can now be easil
Applying lg
^= 0
vector e forced acting at
ig. 2.16 where w is
Suspended. The
and y directions.
trated i
11102041 can
al
A.B = AB cos 0
A.B O >
y
T2 0
A cos B = A. B T,
Or
B rig.zie
Tj has thej'maximu?,
Ti»N, therlT
^26.6 N
T,
As
W here B is the unit vector in the direction of B
sion
B =<
y32 + ( ~4)2
+ (12)
2
= 13
, (3i — 4j— 12k )
Ra! L
Now If
No y
2sin20°- w =0
, sin
-
Thcreforc. B le v;
13
50J x 0.866 + 26.6 N x 0.34 = w
*W52 N
(3) -4j-,2k| Or
' t projection of A on B = (21-8j+
kJ.
(2)(3)^(-B)(-4)ilf-12) 26
13
"
13
13
xample 8; A uniform beam of 200N Is supported horizontally
is shown. If the breaking tension of the rope is 400N, how far can
he man of weight 400N walk from point A on the beam as shown
n Fig. 2.18?
400 N
f
11M/2444
The line of action of a force f passes through a point P of a body who c posftk*
Exampie 6:
-ertor in metre, is . 2 j 4 k .l f F* 2 i ~ 3 j+ 4 k r i n newton;, determine th« torque about Solution: Ijzt breaking point be at a distance d from the pivot A The A r
—; T
" 1 "
"
"
"
wee diagram of the situation is given in Fig 2.19. By applying 2
JOON
condition of equilibrium about point A. < *
'
Zt =0
400 N x 6m -400 Nxd -200 N x 3m =0
Or 400 N x d = 2400 Nm -600 Nm = 1800 Nm
jx>int <
*ho*e position vector On metre; is 2 i -t j+ k 11102/142
' r
400 N
Solution:
# A
The pmcUon vectorof point A * rj * 2i + J + k
A A A
The position vector of pmnt P = r2 = i - 2 j+ k relt Oi,
The position vcct/ jf of P relative to A is;
AF = r = r2 - r,
A P = fi 2 j+ k ) - (2i + j + k ; = i 3 j
/ >0 21»
s d =4,5 m
*
i
A
Example 9: A boy weighing 300 N is standing at the edge of a *
uniform diving board 4.0m in length. The weight of the boar u
^
200 N. (Fig. 2.20 a). Find the forces exerted by pedestals on e
hoard.
the torque about A = r x F A
- (~ i 3 j; x (2 l 3 j k ;
t m 2 TOta)
« -H i + 4 j+ 9k N m
J*
—We i
=be diving hoard which
,in
lhe force diagram ( Fig. 2.20 b). Note that the wc.ght 200 N ot
t0 act at point C. the
D
2.15
43
42

I
Unique Notes Physics T Year
gUnique Notes Physics P
1
Year
centre of gravity which is taken as the mid-point of the board, R| and R2 are the taction fa,
exerted by the pedestals on the board. A little consideration will show that R, is in the
direction, because the board must be actually pressed down in order to keep it in equilibriu^
shall see that this assumption will be automatically corrected by calculations. ^
Let us now apply conditions of equilibrium
(No x-directed forces)
R, + R2 - 300 - 200 = 0
Numerical!
?
2.1. Suppose in a rectangular coordinate
system, a vector A has its tail at the point
P(-2, -3) and its tip at Q (3,9). Detj
the distance between these two points
2.2.% certain c<
as the origin ol
system. If an insect Is crawling on an
adjacent wall at a point having co-ordinates
(2,1). Where the units are in metres, what is
the distance of the insect from the corner of
the room?
Data:
ler of a room is selected
rectangular coordinate
LFx = 0
!
Fy =0
Ri + Ri = 500 N
ine
J x «
.
(1)
r
1 0 m
«— 4«-
lT =0 (Pivot at point D)
- R, x AD - 300 N x DB - 200 N x DC = 0
P (-2, 3)
Q (3, 9r
Data:
3.0 m
IlI02047
1.0 mX
- Ri x 1m - 300 N x 3 m- 200 N x 1m = 0
R, =- 1100 N = - 1.1 kN
Substituting the value of K inEq. (i). we have
- 1100 + R2 = 500
R2 = 1600 N = 1.6 kN
The negative sign of R( shows that it is directed downward.
Distance between Point = P = (2,1)
Origin = O = (0.0)
200 N
300N
1
2.20
(b)
Solution:
Solution:
Formula:
Q
r = ai + bj
r = 2i +J
Thus the result has corrected the mistake of our initial assumption.
V(2) + (D
r =
r = 2.2m|
P(-2,-3)
Fromhead to tail rule.
/0(0,0)
n + A = r 2
2.3. What is the unit vector in the direction
A =r2 - ri
of the vector A = 4i + 3J? 11102048
= (3, 9)- (-2,- 3)
Solution:
= 3i +9j- (-2i- 3 j ) Formula:
A = 4i + 3j
= 3i +9j +2i +3 j
A ^
A =
IAI
5i +12 j
4i + 3J
Magnitude of A A -N/(4r + or
A = -J(5 )2
+(12)2
t>"
. 4i + 3J
A _
^fl^+ 9
= ,/25 +144
4i + 3J
A =
. 4i + 3i 4: 3;
= Vl69
13 units
=
i

Unique Notes Physics 1*
Year
14. Two particles located at
are Dividing bv 2
r,= to + 7J and r2 = 2i + 3J respectively.Find 2 7 Find the angle between the two vectors
As5i + j and B =2i +4j

A = i + 4j
—8. Find the done when the point of
>rce 3i + 2J moves in a
both tht of the vector and its A =Vo)i
+ (4r U102052 appi
** rtsPf(
1 to the x axis, nitan Data: straight li
A =VTTl6 point (2,-1) to the
A = 5i + J
A =^f n
r =3l li 11102*53
ta:
B = 2l + 4J.
A = 4.12
f: =-2» + 3J Point <1-1 2i - j
e=?
-r = int B =(6, 4) = 6i +4j
16: Given that A 2i + 3j
^
Formula:
= F=3i + 2j
B = 3i -4J Find the magnitude and dine
=( -2t + ii)-(3i + 7i)
e =Cos '
f:-r, of
--:^ 3i - 3i -7j
Hit
* ormuia:
rz-tj Solution:
<a) C = A + B (b) D =3A •21 W =F.d.
A. B = ASBX + A,BV + A-
=-5i - 41 Solution: Displacement is given by
d =(X2 -Xj)i + ( yz -yi>i
=(5K2) + (1) (4
(a) C = A + B
tZ ~ T )
=10 + 4 + 0 d =(6 -2)I + (4 - M)j
Substitute the values:-
= 25 + 16
= 14
C = 2i + 3J + 3i -4J d =4i + 5J
=
W = F.d =(3i + 2lx(4i + 5J)
= 64 C =5i -J
As TI r, fees LS the 3^quadrant therefore the = 12 + 10
c =Vt5r + (-i >2
|W
)

mioue Notes Physics 1*
Year
= (4)(3) sin 90°
= 12 x1
= 12 units
This shows that these vectors are
mutually perpendicular.
2.10 Given that A = i - 2J + 3k and B = 3i -
4k. Find the length of the projection of A on
11102055
i j k
7 3 1
x =
N
1 5
3
A = 4
B. 1
-l 7
5 -2
1
+ fc
7 3
-3 1
3
Solution:
Data:
A = i - 2i + 3i
B = 3i - 4k
Find the projection of
A on B
t = i
-3 5
x =i [(15) -_(!)] - J [35—(—3)] + k [(7) -(-9)]
=1(14) -1(35 + 3) + fc (7 + 9)
E 1
C=8
n> about O' is:
*
« y i k
I
The direction of AxB according to rightharx =(l4i-38J + 16k) N m
rule is vertically upward.
-1 -1 0
F - -3i+j+5k
(2.-1) (6.4) 1 -2 0
N
A cos 0 = ?
Formula:
A. B = AB cos 0
A = 4
A W
B = 3
A.B 0
A cos 0 = ~
^ (b) IA x Cl = AC sin 0
= (4)(8) sin 90°
= 32 x 1
= 32 units ,
a)
Using right hand rule direction of A x C
^
vertically downward.
*
Acos0 B 1.13. The line of action of force F = I - 2
masses through ( he point whose position
is<4-hk).>ind
e moment of F above the origin,
e moment of F about the point
vhich the position vector is i + k.
-)
A.B = AxBx + AyBy + AZB/
= (l)(3) + (-2)(0) + (3)(-4)
= 3 - 0 - 1 2
/ector 11102058
= -9
0 J -1 -1
= fa+ B*
= x/9 + 16
= y[25
+ k
B
1 -2
F= i -2j
A A
r = - j + k
Formula:
A = 4
=i (0 - 0) - J(0 - 0) + k (2 - (-l )
|
=0 -0 + 3k
t =3k
*
C = 8
(c) IB x Cl = BC sin 180°
= (3)(8) sin 0
IB x Cl = 24 x 0
IB x Cl = 0
B x C will be a null vector.
x =rxF
Solution: 2.14.The magnitude of dot and cross
6-s/3 and 6
=5
Thus torque about O is: products of two vectors are
respectively. Find the angle between the
vectors?
Solution: A. B = 6A/3
|A x B| =6
Find 0 =?
A.B
Hence ACos0 = j k
i
B
( Board 2009) 11102050
Put the values x = r x F = 0 -1 1
1 -2 0
-9
180°
A cos0 = y
4
o M 1
-2 0
f 0 1 L
°
0
+
ki
- 1
2.11.Vectors A, B and C are 4 units north, 3
units west and 8 units east, respectively.
11102056
(c) B x C
- J
T = |
»
<r
1 ( I )
- 2 A.B =AB cos 0
I A x Bl =AB sin 0
Dividing eq.(2) by ( I )
I A x B I
A.B
"
AB cos 0
C = 8
B = 3 _ .
2.12. The torque or turning effect of t
°r
= i [(0>- (-2)] - j(0 — 1 ) + fc(() - (- 1 )]
about a given point given by r x F = 2i + j + k
is the vector from the given point to ©ata: (Part b)
point of application of F. Consider a foftl F= j - 2]
= -3i + i + 5k N acting on the point 7i + •
r, =- j + £
k (m). What is the torque in Nm about 1
11102051
1
(2)
Describe carefully
(a) A x B (b) A x C
Data:
- AB sin 0
Vector A = 4 units north
— = tan 0
Vector B = 3 units west
Vector C =8 units east j r2 = i + k
3orque about O'
= x = r x F = ?
Mow
i I
origin?
Solution:
-7= = tan 0
Solution:
*-
using head to tail rule, we have
(a) 1A x Bl = AB sin 0
} 49
x =rxF
48

1 Unique Notes Physics T Year
Unique Notes Physics T Year
bridge supports when the rear wheel ^
middle of the bridge span.
Solution:
tan 0 =0.57
0 = tan
1
0.57
0 = 30°
2.15. A load of ION is suspended from a
clothes lines. This distorts the line so that it
makes an angle of 15 with the horizontal at
each end. Find the tension in the clothes
11102060
2.17. A spherical ball of weight 50N i
lifted over the step as shown in fig.
Calculate the minimum force needed just to
lift it about the floor. ,
11102062
ts to be Putting the ED, ECand W in eq. 3
^ J>0 x 13.2
25
s 4
Solution:
F 26.4N
B
F *?
i.m
1; '
orm sphere of weight ION is held
; attached to the frictionless wall
; string makes an angle of 30°
vail as shown in fig. Find the
tension in the string and the force
on the sphere by the wall.
Solution:
Ri
by a st
D C
A B that
line.
3m "
0
th th
Data: D,
exerted
11102063
T = ?
W = ION
Wi* 5000 N W =18000 N
A!
Solution:
Using first condition of equilibrium
Ri =?
0.5 m
RJ=? Ty » Ttm»
Takin
Anii-Cldckwise 1
que about Poi.
0.5 m J
= T! Ty
'h’j
«cft
£
x.=J
20m T* » T cosO R
clockwise torque x2 W a 10 N
X Wx
21 m
XFs =0
Using first condition of Equilibrium se over the step R - T* =0
If x
XFy = 0 R - T cos 0 =0
=T2
T cos15°
T cos15° 1
R = T cos 0
Ri + R2 - W, - W2 = 0 x E x ED
F. «ii
R = T cos 60°
W = 10N
R, + R:- 5000 - 18000 = 0
R = 0.5T ~ d )
(3)
(1)
R, + R2 = 23000 EC !Fy = 0
W
using second condition of equilii Tv - W = 0
XFx = 0 rom fig EC = EO + OC
T sin0 -W =0
taking C as axis of rotation
T cos 15 - T cos 15C
=0
T sin 60° - W =0
= 15 + 10 = 25 cm
R2 x BC + W, x CD R x AC =0
iFy = 0
0.86T = W
In right angle triangle AEOD
+ Tv - w = 0 R2 X 10 + 5000 x 3 - Ri x 10 =0
OD
2
= OE
2
+ DE
2 T"
0.86
T sin 15° + T sin 15 - 10 = 0 IOR2 + 15000 - 1OR , =0
10
0 258T + 0.258T = 10
DE
2
=OD
2
-oS2
— (2)
-R , + R2 = -1500 T"
0.866
0.516T = 10
DE =V(20)2
- ( 15)2
Adding eq. ( 1) in (2) T = 11.54N
10
Putting the value of T in eq. (1)
=
^400 - 225
T 0.516 Ri + Rz = 23000
R =0.5 x 11.54
DE = OD2
- OE-
-Ri + R:= -1500
T = I9.3.V R = 5.8N]
=^/400 - 225
2R2 = 21 3(
^1
2.16. A tractor of weight 15.000N crosses a
= V175
21500(
j)
single span bridge of weight I8000N and of R:= DE = 13.2cm
length 21.0m. The bridge span Ls supported
= I 0750N
half a metre from either end. I he ( actors R2 = 10.75kN
front wheels take 1/3 of the total we; i of the
Ri + R2 = 23000
Ri = 23000 - 10750
R , = 12250N

11th physics chapter 2

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11th physics chapter 2