Vidyamandir CCllaasssseess Introduction to Vectors & Forces
Introduction to Vectors & Forces
BASIC IDEAS Section - 1
Scalars : A quantity which has magnitude but does not have any sense of direction is called SCALAR.
For example, mass, speed, temperature, energy etc.
Scalar quantities are represented as numbers and hence can be positive and negative numbers.
Scalars are added , subtracted and multiplied like real numbers.
Vectors : A quantity which has magnitude and has some sense of direction is called VECTOR.
For example : force, velocity, momentum.
If a is a vector quantity, it must be written as a or a or in bold a.
Magnitude of vector a is a positive quantity and is represented
as | a | or simply a.
Geometrical Representation :
Geometrically, vectors are represented by directed line segments i.e., (by arrows).
Length of the arrow represents the magnitude of the vector and the direction of the arrow represents
the direction of the vector.
Vector a has magnitude of 5 units, direction is +X axis.
| b | = 6 units , direction is –X axis.
| c | = 4 units , direction is –Y axis.
| d | = 7 units , direction is +X axis.
Parallel Arrows :
Two or more parallel arrows having same length and pointing in same direction (as shown above)
represent the same vector a .
Two parallel arrows having the same length but pointing in opposite directions (as shown) represent
two opposite vectors : a and – a .
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In other words, – a is a vector having same magnitude as that of a but is opposite in direction to
vector a . i.e.
| a | = | – a |.
Multiplication of a vector by a scalar :
Consider a vector a multiplied by a scalar k.
If k > 0, vector k a is a vector in same direction as vector a but of magnitude k times that of a .
If k < 0, vector k a is a vector in opposite direction to vector a but of magnitude | k | times that of
a . For example :
Unit Vector : A vector whose magnitude is unity is called a UNIT vector.
A unit vector a can also be denoted as aˆ (pronounced as a cap).
In every direction there can be one unit vector.
The unit vector in direction of +X axis is called as ˆi (i cap).
The unit vector in direction of +Y axis is called as ˆj ( j cap).
Any vector in +X axis direction can be written as its magnitude multiplied by ˆi and in –X axis
direction as multiplied by – ˆi .
Similarly for vectors parallel to +Y and –Y
axes.
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A line perpendicular to the XY plane and passing through origin is called Z-axis. The unit vector
along +Z-axis is denoted by kˆ .
ADDITION OF VECTORS Section - 2
Collinear Vectors :
The sum of two vectors a and b is denoted as a + b and is also called the resultant of a and b .
If two vectors in same direction are added, their magnitudes are added together.
If two vectors in opposite directions are added, their magnitudes are subtracted. The resultant is
in the direction of the vector whose magnitude is greater
.
Triangle Law of Vector Addition :
If two vectors a and b lie along the two sides of a triangle in consecutive
order (as shown), the third side represents the sum of a and b .
Parallelogram Law of Addition :
If two vectors lie along two adjacent sides of parallelogram as shown, the
diagonal of the parallelogram through the common vertex represents their sum
(or resultant).
Note : The vectors must originate from the same vertex O.
Consider two vectors P and Q of magnitudes P and Q respectively
making an angle between them.
If P and Q lie along the adjacent sides of a parallelogram, the length
of the diagonal represents the magnitude of the resultant R .
R |P Q |
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R P2 Q2 2 PQ cos
The direction of R is measured by the angle between R and the vector P (angle ).
Concept :
sin = Q sin OR tan = Q sin
R P Q cos
The magnitude of resultant of two vectors a and b is minimum when they are opposite to each other
and maximum when a and b are in same direction.
R a b
Rmin | a | | b | (where | a | | b |)
Rmax | a | | b |
Illustration - 1 Two forces of magnitude 7 newton and 5 newton act on a particle at an angle to each
other which can have any value. The minimum magnitude of the resultant force is :
(A) 5 newtons (B) 8 newtons (C) 12 newtons (D) 2 newtons
SOLUTION : (D) Rmin | a | | b | 7 5 2N
for 180
Illustration - 2 Out of the following the resultant of which cannot be 4 N ?
(A) 2 N and 2 N (B) 2 N and 4 N (C) 2 N and 6 N (D) 2 N and 8 N
SOLUTION : (D) For a 2N and b 8N
Rmin 8 2 6N
Hence the resultant can’t be 4N.
Illustration - 3 Two non-collinear forces, one of 10 N and another of 6 N act upon a body. The directions of
the forces are unknown. The resultant force on the body is :
(A) between 6 and 10 N (B) between 4 and 16 N
(C) more than 6 N (D) more than 10 N
SOLUTION : (B) For F1 = 10N, F2 = 6N
F1 F2 F1 F2 F1 F2
4N F1 F2 16N.
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Concept: (a) If three collinear vectors are adding up to zero, the sum of the magnitudes of two of them
should be equal to the magnitude of the third.
a b c.
(b) If three non collinear vectors add up to zero, it should be possible to represent them along
the three sides of a triangle.
Hence sum of the magnitudes of any two should be greater than the third.
Illustration - 4 Whose resultant cannot be zero ? 10, 20, 40
(A) 10, 10, 10 (B) 10, 10, 20 (C) 10, 20, 20 (D)
SOLUTION : (D) If a 10, b 20, c 40
As a c b , a,b ,c cannot lie along the sides of a triangle
Illustration - 5 Which of the sets given below may represent the magnitudes of three vectors adding to
zero?
(A) 2, 4, 8 (B) 4, 8, 16 (C) 1, 2, 1 (D) 0.5, 1, 2
SOLUTION : (C) a 1, b 2, c 1
ac b
Illustration - 6
Find the resultant of two vectors of magnitudes 5 and 3 and making an angle 60°
between them.
SOLUTION :
Here P = 5
Q=3
= 60°
R P2 Q2 2 PQcos 52 32 2.5.3 cos 60 7units
sin = Q sin 3 33
= sin 60° =
R7 14
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COMPONENTS OF A VECTOR Section - 3
Consider a given reference of X and Y axes and a given vector a (as shown).
We can split a in two parts (known as components of a ), so that one part is parallel to X-axis
and the other part is parallel to Y-axis.
According to parallelogram law of addition of vectors we can see that
a = PA + PB
PA is parallel to X-axis and is known as X-component of a .
PB is parallel to Y-axis and is known as Y-component of a .
If magnitude of a = | a | = a and is the angle between a and the X-axis :
X-component of a = ax = a cos
Y-component of a = ay = a sin
In vector notation :
a = PA + PB
a = ax ˆi + ay ˆj
as PA = ax ˆi , PB = ay ˆj
a = a cos ˆi + a sin ˆj
ax and ay can be negative or positive according to the direction of components.
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Illustration - 7 Find the vectors :
a ˆi ˆj b 3ˆi 2ˆj c 2ˆi ˆj
Note : If components of a are ax and ay, then
| a | = ax2 a2y
Illustration - 8 Resolve into components :
(i)
(ii)
(iii)
Note : (i) The components can be resolved along any two perpendicular axes. They may not be horizontal
(ii) and vertical always.
Observe that the given angle always comes between given vector and its cosine component.
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ADDITION OF VECTORS USING COMPONENTS Section - 4
PROBLEM SOLVING TECHNIQUE
To add vectors using components :
Resolve each vector into components.
Add the X components and Y components separately.
If the components along an axis are in same direction, magnitudes are added (like vector addition of
parallel vectors)
If the components along an axis are in opposite direction, magnitudes are subtracted.
Rx = X-component of resultant = (X-components of all given vectors)
Ry = Y-component of resultant = (Y-components of all given vectors)
The magnitude of resultant R = (Rx2 + Ry2)1/2
Illustration - 9 Find the resultant of vectors.
SOLUTION : Resolve both the vectors.
Rx = X-component
= 8 cos 60° + 10 cos 30°
= 4 + 53 towards +X axis.
RY = Y-component
= (8 sin 60° – 10 sin 30°)
= (43 – 5) towards – Y-axis.
Magnitude of R = (Rx2 + Ry2)1/2 = 2 41 units.
R makes an angle below + X-axis
where tan 4 3 5 (As is acute, tan Ry
45 3 Rx
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Illustration - 10 Find the magnitude and direction of the resultant of given vectors.
SOLUTION : Resolving the vectors :
Rx = 2 – 2 sin 30° = 1 towards + X axis
Ry = 2 cos 30° = 3 towards + Y axis
R = [12 + (3)2]1/2 = 2 units
The angle of R with X-axis =
tan = 3 = 60°
NOW ATTEMPT IN-CHAPTER EXERCISE-A BEFORE PROCEEDING AHEAD IN THIS EBOOK
FORCES IN EQUILIBRIUM Section - 5
Aparticle is said to be in equilibrium if it is at rest or moving with uniform velocity.
According to Newton’s First Law, the net force on a particle in equilibrium is zero.
PROBLEM SOLVING TECHNIQUE
If the sum of given forces is zero , we say that forces balance each other. In that case:
(magnitudes of X-components in +X direction) will be equal to (magnitudes of X-components
in -X direction)
(magnitudes of Y-components in +Y direction) will be equal to (magnitudes of Y-components
in - Y direction).
Illustration - 11 Find a and b if the sum of given vectors is zero.
SOLUTION : Resolving the vectors :
Since the sum of the given vectors
is zero, so vectors balance each other.
i.e., a sin 30° = b
a cos 30° = 10
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a = 20/(3)
b = 10/(3)
Illustration - 12 If the sum of three given vectors is zero, find a and .
SOLUTION : Resolve into components :
Balance X-components
a cos = 5 cos 60° + 5
Balance Y-components
a sin = 5 sin 60°
a sin = 53/2
a cos = 15/2
Divide to get : tan = 1/3
= 30°
a sin 30° = 53/2
Hence a = 53
Illustration - 13 Three forces of equal magnitudes and making an angle of 120° with each other act on
a body. Show that the resultant force is zero.
SOLUTION : Consider one force along + X axis and subsequently draw rest. Let F = magnitude of
each force.
Net Y-component = F sin 60° – F sin 60° = 0
Net X-component = F – (F cos 60° + F cos 60°)
= F – 2 F cos 60° = 0
Hence the resultant force is zero.
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Illustration - 14 A particle is in equilibrium under the action of three forces as
shown. Find the magnitude of forces F1 and F2.
SOLUTION :
As the angle between F1 and F2 is 90°, we will take the X-axis along
F1 and Y-axis along F2. Then resolve the force of 102 N along the axes.
As the particle is in equilibrium, net force = 0
so balancing X-components :
F1 = 102 cos 45° = 10 N
and balancing Y-components :
F2 = 102 sin 45° = 10 N
ANALYSIS OF A PARTICLE IN EQUILIBRIUM Section - 6
There are mainly three types of forces which are being considered in the next few examples :
9.1 Weight of a body :
It is the gravitational attraction of earth and is denoted by W or mg (m = mass of body, g = 9.8 m/
s2). It is directed vertically downward.
9.2 Tension in a string :
Whenever a body is connected to a tight string, it is pulled by the string with a force which will
be called as the tension in the string (T).
Unless stated otherwise, strings will be assumed massless and hence the tension in a string will
be same throughout its length.
Thus a massless string will pull the bodies connected to its two ends with equal forces (tensions).
9.3 Normal Reaction Between two bodies :
Whenever, two surfaces are in contact, they press (or push) against each other with equal and opposite
forces. These forces are perpendicular to the surfaces in contact and are known as the force of
Normal Reaction. This is denoted by N or R.
To analyse the equilibrium of a particle :
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PROBLEM SOLVING STRATEGY
1. Draw a diagram showing all the forces acting on the particle. This diagram is known as force
diagram or free body diagram.
2. Select a convenient X and Y-axes (so that number of components is minimum).
3. Balance the X-components and Y-components of all the forces to get two equations from each
force diagram.
Illustration - 15 Consider two blocks of weight 50 N and 20 N suspended
one below the other as shown with help of two strings A and B. Find the
tensions in the strings A and B.
SOLUTION : Analysing forces on W2
as block is at rest, TB = W2 = 50 N (balancing forces in vertical direction)
Analysing forces on W1
TA : pulling force (tension) on block by string A
TB : pulling force on block by string B
W1 : block weight
Balancing forces : TA = TB + W1
TA = (50 + 20) N = 70 N.
Illustration - 16 A block of weight 100 N is suspended as shown with
the help of three strings. Find the tension in all the three strings.
SOLUTION : T0 = tension in the string connected to block
Let T1 = tension in the string OA
T2 = tension in string OB
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Force diagram of block
Force diagram for tensions at O
* balance forces in horizontal direction to get,
T1 = T2 sin 60°
* balance forces in vertical direction to get,
T0 = T2 cos 60°
On solving, we get :
T2 = 2T0 = 200 N
T1 = 100 3 N
Illustration - 17 Two blocks A and B of weights 20 N and 15 N are placed on a
horizontal surface one on top of the other as shown. Find the normal reaction at
all contact surfaces.
SOLUTION :
Surfaces of A and B are in contact,
Let N1 = normal reaction between A and B
A is in contact with horizontal surface
Let N2 = normal reaction between A and surface below it.
Force diagram of B
N1 = normal force exerted by A on B
as B is at rest, N1 = 15 N
Force diagram of A
N1 = normal force exerted by B on A
N2 = normal forces exerted by horizontal surface on A
On balancing the forces, we get :
net downward force = net upward force
N1 + 20 = N2 N2 = 35 N
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Note : In the force diagram of A, we show forces acting on A and not forces exerted by A.
Illustration - 18 A block of mass m is lying on a smooth inclined
plane and is tied to a string fixed parallel to the plane as shown.
Analyse forces on the block.
SOLUTION : Let = inclination of plane with horizontal
angle between a line to plane and the vertical =
Note : In case of a block on an inclined plane, one axis will be taken parallel
to the plane and the other axis to the plane.
Force diagram of block
mg = weight of the block
mg cos = component of weight to the plane
mg sin = component of weight || to the plane
N = normal reaction force on the block exerted by the inclined plane
T = tension (force) exerted by the string
As the block is at rest,
T = mg sin (balancing forces || to the plane)
N = mg cos (balancing forces to the plane)
Note : You should remember that the components of the weights of a body on an inclined plane are mg
cos perpendicular to the plane and mg sin parallel (and downwards) to the plane. (if =
angle between inclined plane and horizontal.)
Illustration - 19 In the diagram shown, the blocks A and B are connected
together by a string and placed on a smooth inclined plane. B is connected
to C (which is suspended vertically) by another string which passes over
a smooth pulley fixed to the plane. The system is at rest. If WA = 10 N,
WB = 20 N, find WC and analyse the force diagrams of A, B and C.
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SOLUTION :
Force diagram of A :
T1 : tension in string between A and B
NA :
normal reaction between A and inclined
plane.
Balancing forces on A :
NA = WA cos = 10 cos 30° = 53 N
T1 = WA sin = 10 sin 30° = 5 N
Force diagram of B :
As B is connected to both strings, two tensions T1 and T2 will act on it.
T2 = tension (force) of string between B and Cacting upwards
T1 = tension of string between A and B acting downwards
WB sin , WB cos are components of weigth WB.
Balancing forces on B :
NB = WB cos = 20 cos 30° = 103 N
T2 = T1 + WB sin = 5 + 20 sin 30° = 15 N
Force diagram of C :
T2 = pulling force by the string on the block C
WC = T2
Hence WC = 15 N.
Illustration - 20 Blocks A and B rest on a horizontal surface
in contact with each other. Pushing forces F1 and F2 are F1
applied as shown. Weight of A is 30 N and of B is 20 N.
30° AB F2 = 15 N
Find the force F1 and the normal reactions between all
the contact surfaces.
SOLUTION :
Force diagram of B :
NB = normal reaction exerted by floor on B
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R = normal reaction exerted by A on B
R = F2 = 15 N & NB = 20 N
Force diagram of A :
R = normal force exerted by B on A.
NA = normal reaction exerted by floor on A.
balancing forces :
F1 cos 30° = R (horizontal direction)
F1 sin 30° + 30 = NA (vertical direction)
F1 = 2R 30
= = 10 3 N
3 3
NA = 10 1 + 30 = (30 + 5 3) N
3.
2
Note : The two normal reaction forces : R by A on B and R by B on A
are of equal magnitude as they form an action reaction pair. This will be discussed later in detail
under Newton’s III law (Chapter 3 : Newton’s Laws)
FORCE OF FRICTION Section - 7
Whenever two rough surfaces are in contact, sliding between the surfaces is opposed by the force of
friction which the surfaces exert on each other. The force of friction acts parallel to the surfaces in contact
and on both the surfaces.
If the tendency to slide against each other is too small to cause actual sliding motion, the force of
friction is called the force of static friction. The magnitude of this force balances the net applied
force and can be anything between zero and s N. Hence if there is no sliding between the surfaces,
Force of static friction = Net applied force parallel to the surfaces.
If the sliding between the surfaces is about to begin, the static friction is at its maximum value
which is equal to sN, where N = normal reaction between the surfaces and s = coefficient of
static friction. In this situation, we say that the surfaces are at their point of sliding and are exerting
a force sN on each other so as to oppose sliding.
If actual sliding is taking place between the surfaces, the force of friction is called as force of
kinetic friction or the force of sliding friction (fk).
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fk = k N where k = coefficient of kinetic friciton (k s)
Note : Force of friction on a body always acts against the sliding tendency.
Illustration - 21 A block lying on a horizontal surface is pulled by a force of
0.1 N but the block does not move i.e., it remains at rest. Analyse the frictional
force on the block.
SOLUTION :
As the block remains at rest, the force of static friction is balancing
the 0.1 N force (the applied force). So the frictional force = 0.1 N
Illustration - 22 A block of weight 100 N lying on a horizontal surface just begins to move when a
horizontal force of 25 N acts on it. Determine the coefficient of static friction.
SOLUTION :
Illustration - 5
As the 25 N force brings the block to the point of sliding,
the frictional force = sN.
From the force diagram :
N = 100
sN = 25 s = 0.25.
Illustration - 23 A block lying on an inclined plane has a weight of 50 N. It just begins to slide down
when the inclination of plane with the horizontal is 30°. Find s.
SOLUTION :
The block reaches the point of sliding when the plane makes an angle of 30° with the horizontal.
Hence in this situation, frictional force = sN
Balancing the forces :
N = W cos 30°
Note : sN = W sin 30°
sW cos 30° =W sin 30°
1
s = tan 30° = 3
Minimum angle for which a block starts sliding down an inclined plane is known as the angle of
repose.
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Illustration - 24 A block of weight 100N lying on a horizontal surface is
pulled by a force F acting at an angle 30° with horizontal. For what value
of F will bethe block begin to move if s = 0.25 ?
SOLUTION :
Consider the force diagram of the block at the moment when it is just to start moving.
Balancing the forces :
N + F sin 30° = mg
F cos 30° = sN
F cos 30° = s (mg – F sin 30°)
F= s mg 0.25 (100 ) 2
cos 30 s sin 30 =
3 0.25
F = 25.2 N
So the block begin to move for F > 25.2 Newtons.
Illustration - 25 Two blocks of equal weights are connected by a
string as shown.If the blocks slide with constant velocity, find the
coefficient of kinetic friction (k).
SOLUTION :
As the surfaces are sliding against each other, frictional force will be k N.
The blocks are sliding with constant velocities, the net force on each block will be zero.
Force diagrams :
W sin 30° = T + k N1 T = k N2
N1 = W cos 30° N2 = W
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sin 30 1/ 2 1
k = 1 cos 30 = 1 = =2– 3
3/2 2 3
Illustration - 26 The resultant of four vector a, b , c and d as
shown in the figure is zero. Find the magnitude of b and d if
the magnitude of a and c are 5 2 and 13 respectively
(Given : sin 22.6° = 5/13, sin 37° = 3/5)
SOLUTION :
Resolving the given vector into components
c cos 22.6° + d cos 53° = a cos 45° + b cos 37°
and a sin 45° + d sin 53° = b sin 37° + c sin 22.6°
12 3 14
13 d 5 2 b
13 5 25
1 43 5
and 5 2 d b 13
2 5 5 13
4b 3d 3b 4d
7 and 0
55 55
d = 15 units and b = 20 units
SUBTRACTION OF VECTORS & RELATIVE VELOCITY Section - 8
Subtraction of Vectors
The idea of subtracting one vector from another is based on the concept of “negative of a vector”. We have
already discussed that negative of a vector a is another vector which is of same magnitude as a but in a
direction opposite to that of a .
The vector difference b a can be thought of as vector addition of the vectors b and a .
b a b a
This amounts to adding the negative of a to the vector b . In order to subtract one vector a from another
vector b , we add the reversed arrow of a to b .
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Relative Velocity
Imagine yourself standing by the side of a road observing two
cars : A and B, both moving towards right with a speed of 10 m/
s. Let us consider car A for a moment. According to you its
velocity is + 10 m/s. But an observer in car B will measure the
speed of car A as 0 m/s. His observations about the velocities
are different. According to him, car A is at rest and the road
itself is moving with 10 m/s (i.e. backwards).
Hence whenever we speak about the velocity of any moving
object, we must also specify the reference from where we
are observing it.
Velocities observed from the ground will be called as velocities relative to the ground. Velocities observed
from the car B will be called as velocities relative to the car B.
Hence, velocity of car A relative to the ground = 10 m/s.
velocity of car A relative to the car B = 0 m/s.
Note : In symbolic notation we write :
vAG = velocity of A relative to the ground (G) = 10 m/s.
vAB = velocity of A relative to the car B = 0 m/s.
In general we have the relation :
vAB vAG vBG
When specifying the velocities relative to the ground, we usually write :
vA instead of vAG and vB instead of vBG. In that case the formula for the relative velocity should also be
remembered as : vAB vA vB vA vB
(i) To find velocity ofArelative to a reference B, we impose (add) the reversed velocity of reference (B).
(ii) The formula : vAB vA vB
involves vectors and hence velocities should be added or subtracted like vectors while using this relation.
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Illustrating the Concept :
A particle A is moving towards +X-axis with speed 7 m/s.
Another particle B is going towards -X-axis with speed 3
m/s. Find the velocity B relative to A.
VBA = vB vA
= (3 m/s) (+7 m/s) = 10 m/s.
i.e. B appears to move towards - X-axis with 10 m/s if
observed from A.
VBA = 10 m/s towards ve X-axis
Illustration - 27 A boat is moving with a velocity 3i + 4j with respect to ground. The water in the river is moving
with velocity 3i 4 j with respect to ground. The relative velocity of the boat with respect to water is :
(A) 8i (B) 6i 8 j (C) 6i + 8j (D) 5 2
SOLUTION : (C)
vbw vb vw 3ˆi 4ˆj 3ˆj 4ˆj = 6ˆi 8ˆj .
Illustration - 28 A man sitting in a bus travelling in a direction from west to east with a speed of 40 km/h
observes that the rain-drops are falling vertically down. To the another man standing on ground the rain will
appear
(A) To fall vertically down (B) To fall at an angle going from west to east
(C) To fall at an angle going from east to west
(D) The information given is insufficient to decide the direction of rain
SOLUTION : (B)
vr vrb vb
Vr makes an angle with vertical
The rain falls at an angle from west to east.
Illustration - 29 A balloon is moving horizontally in air with speed of 5 m/s towards north. A car is moving
with 5 m/s towards east. If a person sitting inside the car sees the balloon, what will he observe ?
SOLUTION :
Let us take Xaxis along east-west and Yaxis along northsouth.
vbc = vb vc = vb + (vc)
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To calculate the balloon’s velocity as seen from the car, we
have to view the situation by placing ourselves inside the
car.
The equation : v = v + (v ) gives an idea about how toc
bc b
do this.
For calculation relative to car, we will impose the reversed
velocity of car (vc) on the balloon velocity (vb). The net
effect of these gives the velocity of the balloon as seen from
the car.
Hence the velocity of balloon as seen from the car is 52 m/
s towards North West.
Illustration - 30 A man standing on a road observes that raindrops are lifting his back with speed 5 m/s are
making an angle of 45 with vertical. With what velocity should he run so that the raindrops hit his head
vertically ?
SOLUTION :
When the man is at rest, what he observes is the velocity of rain with respect to ground.
Vr = 5m/s at 45with vertical.
When he runs, what he observes is the velocity of rain relative to him.
Vrm is vertical.
Using Vrm Vr Vm
From right triange OAB, sin 45 AB |V m |
OA |V r |
|V m | |V r | sin 45 5 5 2 m/ s.
m/s
22
Swimmer in A Flowing River :
When a person swims in a flowing river, his net velocity is :
vs vSR vR
vS = net velocity of swimmer (As seen from ground)
vR = velocity of river vS
vSR = velocity of swimmer relative to river = velocity of swimmer in still water
Hence we have vS = swimmer velocity in still water + velocity of river
Note that the RHS involves a vector addition of two velocities.
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Illustration - 31 A person can swim in still water at the rate of 1.0 Km/hr. He tries to cross a river by
swimming perpendicular to the river flowing at the rate of 2 Km/hr. If the width of the river is 10 m, find the
location of the point where he lands on the other side of the river. Also find the time taken by him to cross the
river.
SOLUTION :
Assume that the person starts from point A and tries to swim towards B (AB is perpendicular to river flow).
VS VSR VR
VSR = velocity of swimmer relative to river = 1 ˆj
VR = velocity of river = 2ˆj
Hence the net velocity of the swimmer is directed
from A towards C at an angle with the river.
tan VSR 1 tan1 1
VR 2 2
sy AB vyt d=V t
SR
d 10 0.01 Hr V S R 1 km / hr 103
t 103 m / h r
VSR
BC sx Vxt VRt 2 0.01 0.02km 20m
The swimmer lands on the other side of the river at a point C which is 20 m from the point B, towards
which he was trying to swim.
Illustration - 32 A man who can swim with speed of 1 Km/hr in still water wants to cross a river. He starts
from a point A from the river bank and wants to reach the point B which is directly opposite to A. In what
direction should he try to swim ? Speed of the river flow is 0.5 Km/hr. Also find the time taken to cross the
river if width is 10 m.
SOLUTION :
The swimmer should direct his velocity in such a direction that his net velocity should be perpendicular to the
river flow.
Let vs = net velocity of swimmer.
= angle between his velocity and line AB
By parallelogram law, the vector addition of velocities VSR and VR should be perpendicular to riverflow.
We have sin vR 0.5km / hr 1 (from APQ )
vSR 1km / hr 2
= 30
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Hence he should direct his velocity at an angle of 30 with the line AB against the flow.
Swimmer’s net velocity (as seen from ground)
vs vS2R VR2 (from APQ )
Time taken = t sy d d 10 / 1000 km 1
hr 41.57 sec
vy vS vS2R vR2 12 0.52 km / hr 50 3
Note : This situation is possible only if vSR > vR.
SCALAR OR DOT PRODUCT Section - 9
Angle between Two Vectors
Angle between any two vectors is taken as the angle between their direction.To find the angle between two
vectors a and b , we should first make (imagine) their tails at the same point. : an angle between a and
b.
Scalar (Dot) Product a . b
The scalar or dot product of two vectors a and b is defined as the product of their magnitudes and the
cosines of the angle between them. It is represented by a dot (.) between a and b . The product itself is a
scalar quantity.
a . b ab cos 0
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IMPORTANT POINTS
1. If a and b are in same direction 0 , then a . b ab
2. If a and b are in opposite direction 180 , then a . b ab cos180 ab
3. If a is perpendicular to be b 90 , then a . b 0
4. If is acute, a . b is positive.
5. If is obtuse, a . b is negative. Section 9 25
6. a . a aa cos 0 a2 a2 a a .a
7. iˆ . ˆj ˆj . kˆ kˆ . ˆi 0
ˆi . ˆi ˆj . ˆj kˆ . kˆ 1
8. Commutative and Distributive properties :
a.b b.a
a . b c a . b a . c
9. For a axˆi ay ˆj azkˆ
and b bxˆi by ˆj bzkˆ
a . b axbx ayby azbz
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IMPORTANT POINTS
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Work Done by a Force
Work done by a force F over a displacement s :
W F . s Fs cos
(a) Work done is itself a scalar quantity while it is obtained by multiplying two vector quantities :
Force and displacement
(b) W = 0 if F is perpendicular to s .
(c) W is negative if F makes an obtuse angle with displacement s .
(d) W is positive if F makes acute angle with s .
(e) The instantaneous rate of doing work is defined as power.
P F .v {where v velocity}
Illustration - 33 A point of application of a force F 5ˆi 3ˆj 2kˆ is moved from r1 2ˆi 7ˆj 4kˆ to
r2 5ˆi 2ˆj 3kˆ . The work done is :
(A) 22 (B) 22 (C) 11 (D) 0
SOLUTION : (A)
W F . s F . AB
W F . r2 r1
r2 r1 5ˆi 2ˆj 3kˆ 2ˆi 7ˆj 4kˆ
7ˆi 5ˆj kˆ
W 5ˆi 3ˆj 2kˆ . 7ˆi 5ˆj kˆ
35 15 2 22 units
Illustration - 34 A particle moves with a velocity v 6ˆi 4ˆj 3kˆ under the influence of a force
F 20ˆi 15ˆj 5kˆ . The instantaneous power applied to the particle is :
(A) 35 umits (B) 45 units (C) 25 units (D) 195 units
SOLUTION : (B)
P F . v 6ˆi 4ˆj 3kˆ . 20ˆi 15ˆj 5kˆ
120601545 units
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Illustration - 35 In the previous situation (llustration - 34), the angle between the force F and the velocity
v is :
(A) cos1 45 (B) cos1 45 (C) cos1 45 (D) cos1 9
61 26 26 61
26 61
SOLUTION : (D)
cos F . v 45
Fv 62 42 32 202 152 52
cos1 45 cos1 9
61 650 26 61
Illustration - 36 If R is the resultant of two vectors a and b , show that R a2 b2 2abcos where is
the angle between a and b .
SOLUTION : R2 a2 b2 2a . b
R a b R2 a2 b2 2 ab cos
R . R a b. a b R a2 b2 2ab cos
R2 a . a a . b b . a b . b
Illustration - 37 Express the equations of motion for uniformly accelerated motion in terms of vectors.
SOLUTION : Scalar form for Straight Line Motion General Vector form
V = u + at V u at
s ut 1 at2 S u t 1 at2
2 2
v2 u2 2as v . v u .u 2a . s or v 2 u 2 2a . s
s 1 v at s 1 v u t
2 2
Note : If the particle goes fromA to B during the interval from t = 0 to time t.
(i) u can also be written as v 0
(ii) v can be written as v t
(iii) s can be written as s r t r 0
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VECTOR OR CROSS PRODUCT Section -10
The vector product of two vectors a and b is defined as a vector whose
(a) magnitude is equal to the product of their magnitudes and the sine of the angle between them.
(b) direction is perpendicular to the plane containing the vectors a and b and is given by the right hand
thumb rule.
It is represented by a cross () between a and b .
The product itself is a vector quantity.
a b | | a | |b | sin
ab sin
Right Hand Thumb Rule
If we bend the figures of the right hand in such a way that they rotate the vector a towards b through the
angle , then the thumb gives the direction of vector a b .
Note that a b = (b a ) their magnitudes are equal and directions are opposite.
IMPORTANT POINTS
1. Vector product is not commutative.
Since b a a b
2. Distributive property :
a b c a b a c
3. If a and b are in same or in opposite directions i.e., if they are collinear, 0 or 180 ,
|a b | ab sin 0 or ab sin180
=0
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Hence a b 0 for collinear vectors.
4. If a is perpendicular to b
a b ab sin 90 ab
5. a a 0
6. ˆi ˆi ˆj ˆj kˆ kˆ 0
ˆi ˆj kˆ ˆj kˆ ˆi kˆ ˆi ˆj
7. a b gives the area of the parellelogram formed with a and b as adjacent sides.
area = base height
= ap
= ab sin | a b |
Torque Due to a Force
When a force F acts on a body of finite size and shape, its capability to rotate that body is measured by a
quantity called torque .
If the force F acts at a point P as the body, then its torque about the point of rotation O is defined as :
OP F
r F
r F Fr sin
F r gives the magnitude of torque.
OM = r dropped on F vector from O.
= r sin
where r = length of perpendicular dropped from centre of rotation O to the line through the force vector
F.
Note : (i) If the force F tries to give an anticlockwise rotation to the body, we say that it is creating an anticlockwise
torque.
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(ii) The arrow representing the torque vector lies along the axis of rotation and is perpendicular to the
plane of rotation (which contains r and F ).
Illustration - 38 The area of the parallelogram formed by the vectors a ˆi 2ˆj and b 3ˆi 2ˆj as
adjacent sides is:
(A) 4 (B) 6 (C) 8 (D) 0
SOLUTION : (A)
Area of parallelogram = a b
a b ˆi 2ˆj 3ˆi 2ˆj
2ˆi ˆj 6ˆj ˆi
2kˆ 6kˆ 4kˆ
Area = | a b | 4 units
Illustration - 39 A force F 4ˆi 10ˆj N acts at a point P whose coordinate are 5, 4. Find the
torque of the force about origin O.
SOLUTION : 20ˆi ˆi 50ˆi ˆj 16 ˆj ˆi 40 ˆj ˆj
r F OP F 0 50kˆ 16kˆ 0 66kˆ
OP 5ˆi 4ˆj and F 4ˆi 10ˆj
5ˆi 4ˆj 4ˆi 10ˆj
.
Illustration - 40 If a 3ˆi ˆj and b pˆi qˆj which of the following can possibly represent a b ?
(A) 5ˆj (B) 7ˆi kˆ (C) ˆi 4kˆ (D) None of these
SOLUTION : (D)
As a b is perpendicular to both a and b , it is also perpendicular to the plane containing a and b .
Here a and b have only X and Y components. They both lie in the XY plane. Hence a b must lie
along +Z or Z axis which is perpendicular to the XY plane.
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Hence a b must have only kˆ in its representation. Hence none of the choices (A), (B) or (C) is correct.
NOW ATTEMPT IN-CHAPTER EXERCISE-B BEFORE PROCEEDING AHEAD IN THIS EBOOK
SUBJECTIVE SOLVED EXAMPLES
Example - 1 The pulleys and strings shown in figure are of negligible
mass. For the system to remain in equilibrium, find the value of .
SOLUTION : The arrangement is symmetrical, hence the tension is
same in both the strings attached to block of mass 2 m.
2 mg cos = 2 mg
1
cos = = 45°
2
Example - 2 Block A of weight W slides down on inclined plane S of slope
37° at constant velocity while the plank B also weighing W rests on the top
of A. The plankis attached by a cord to the top of the inclined plane.
(a) Draw a diagram of all forces acting on A.
(b) If the coefficient of kinetic friction is same between A and B and
between S and A, determine its value.
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SOLUTION :
From force diagram of B From force diagram of A
R + W sin 37° = T ...(i) R + R1 = W sin 37° ...(iii)
R = W cos 37° ...(ii) R1 = R + W cos 37° ...(iv)
Substitute the value of R in (iv) : R1 = 2 W cos 37°
1 3
(W cos 37° + 2 W cos 37°) = W sin 37° = 3 4 = 0.25
Example - 3 Weights of B and C are 10 N and 10 3 N respectively.
Find the weight of A if the system is in equilibrium. Also find the angle
which T2 makes with horizontal.
SOLUTION :
From the force diagram of B and C, it can be seen that tensions T1
and T2 in the two strings tied to B and C are 10 N and 10 3
respectively.
Force diagram at O :
Let = angle between T2 and horizontal
= angle between T1 and vertical
(since T1 is perpendicular to T2)
Resolving the tensions
T2 cos = T1 sin
T2 sin + T1 cos = T0
tan = T2 / T1 = 3
= 60°
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also T0 = 10 3 sin 60° + 10 cos 60°
T0 = 20 N
From force diagram of block A
T0 = WA WA = 20 N
Alternative Approach :
As T1 and T2 are perpendicular to each other, we can take the two
perpendicular axes along these forces.
Balancing forces along X-axis :
T2 T0 sin
Balancing forces along Y-axis :
T1 T0 cos
T02 T12 T22 and tan T2
T1
T0 102 10 3 2 20 N and tan 3
Hence WA T0 20 N and 60
Example - 4 The weight W1 in the given figure is of 300 N.
Find T1 , T2 , T3 and W2 .
SOLUTION :
T3 sin 53 W2 T3 T2
T3 cos 53 T2 cos 53
T1 cos 37 T2 5
T1 W1 3 500N
T1 sin 37 W1 T2 T1 cos 37 500 4 400N
5
T3 T2 400 2000
cos 53 3/ 5 N
3
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W2 T3 sin 53 2000 4 1600 N
3 5 3
Example - 5 In the given figure, let the weight of the hanging block be 50 N.
(a) Find the tension in each chord
(b) If the angle 60° is changed to 45°, find the
tension in each chord.
SOLUTION :
100
T3 3
(a) T3 cos 60 T2 50
T3 sin 60 T1 W
50N T2 3
T3 cos 45 T2 T3 50 2 N
T2 50N
(b) T3 sin 45 T1 W 50N
Example - 6 In each of the following arrangement, find tensions in each cord
and the weight of the suspended body, if the indicated tension T is 10 N
SOLUTION :
T1 T2 sin 30 [T2 = W]
T T2 cos 30
T 20 20
T2 cos 30 3
33
T1 T2 sin 30 10 3 20 3
3 W T2 3 N
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Example - 7 In the setup of the figure, the mobile pulley and the fixed
pulley are associated with equal weights W. Find the angle .
SOLUTION :
2T sin = W T=W
W 1 30
sin
2T 2
Example - 8 The system in the given figure remains at rest when the hanging
weight w is 220 N. What are the magnitude and the direction of the frictional
on the 200 N block ?
SOLUTION : As T > 200 sin37 , friction acts down
T = 220 N, T = f + 200 sin37
f = 220 – 200 sin37
Example - 9 The block A in the given figure weighs 100 N. The coefficient
of friction between the block and the surface on which it rests is 0.30. The
weight W is 20 N and the system is in equilibrium. Find the frictional force
exerted on block A.
SOLUTION :
f T1 T0 sin 45 T2 T2 = W = 20N
T0 cos 45 T1
N 100
f = frictional force = T1 = T2 cot45 = 20 N (Verify that f < sN)
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Example -10 A force of 50 N is applied on a 40 N block which is in contact with the wall. Assuming
there is no slipping, find the minimum coefficient of friction between the block and the wall.
SOLUTION : When is minimum, the block is about to slip.
R = 50 N
R = 40 40
0.8
50
Example -11 A man, using a 70 kg garden roller on a level surface exerts a pushing force of 200 N at
45° to the ground. What is the vertical force of the roller on the ground ?
SOLUTION :
N = F sin45 + 70g
1
= 200 686
2
= 141.4 + 686
= 827.4 N
Example -12 Five forces of magnitude 10N each are applied at one point and all are lying in one plane.
If the angles between them are equal, the resultant of these forces will be :
(A) Zero (B) 10N (C) 20N (D) 10 2N
SOLUTION :
Along y-axis component of b cancels the component of e and
component of c
cancels with the component of d .
Along positive x-axis, net vector is 2r cos 36° in magnitude i.e.
equal to
5 1 5 1
2.r r when r is magnitude of each vector and
42
solving
negative x-axis, net vector in magnitude = 2 2r cos 72
5 1 5 1
r 2r r
42
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resultant is zero
[Important Note :Any number of coplanar vectors of equal magnitudes have their always equal to zero if
they are equally incided to each other. Also note cos 36° = sin 54° = 5 1 and cos 72° = sin 18° =
4
5 1
]
4
Example -13 In the figure, the tension in the diagonal string is 60N
(A) Find the magnitude of the horizontal forces F1 and F2 that must be
applied to hold the system in the position shown
(B) What is the weight of the suspended block
SOLUTION :
T1 = T2 = w = 36 N and F1 = F2 = 49 N.
Example -14 In the adjacent figure, angle of the incident plane is
adjustable. A block kept on the incline, and is increased slowly from
0° and it is observed that block starts sliding down the incline when is
just greater than 37°. The coefficient of static friction is :
(A) 0.5 (B) 0.6 (C) 0.75 (D) 1 / 2
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SOLUTION :
When the block is about to start sliding then limiting force of static friction acts
N = mg cos
Ns = mg sin
3
0.75
s = tan 37° = 4
Example -15 A car is moving towards south with a speed of 20 m/s. A motorcylist is moving towards
east with a speed of 15 m/s. At a certain instant, the motorcylist is due south of the car and is at a
distance of 50 m from the car. Find the shortest distance between the motorcylist and the car. Also find
the time after which they are closest to each other.
SOLUTION :
Taking N + Y axis and E as + X-axis.
Imagine yourself as an observer sitting inside the car. You will regard the car as being at rest (at C).
Relative to you, the speed of hte motorcylist is obtained by imposting the reversed velocity of car on
motorcylist as shonw in the figure.
vmc 152 202 25m / s
tan1 20 53
15 with x-axis
The motorcylist appears to move along the line MP with speed 25 m/s.
The shortest distance= perpendicular distance of MP from C = d
d = 50 cos 53° = 30 m.
Time taken to come cloest
= time taken by motorcylist to reach B
MB 50 sin 53 1.6 sec
t
vmC 25
Example -16 A swimmer at point A on one side of river wants to reach
at a point B on the other side by swimming only. Line AB makes angle
30° with the river flow. Velocity of swimmer is equal to velocity of river
flow in magnitude. In what direction with the line A should he try to swim
to reach at B.
SOLUTION : (C)
resultant should be along AB
u sin 30
tan 30
u cos 30 u
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3 sin 30 cos 20 1
3 sin 30 1 cos 30 1
22 2
cos 30 sin 30 sin 30 cos 30 1
2
sin 30 30 1
2
1 Note : sin A B sin A cos B cos A sin B
sin 30
2
ATTEMPT OBJECTIVE WORKSHEET TO COMPLETE THIS EBOOK
THINGS TO REMEMBER
1. Scalars : A quantity which has magnitude but is not related to any direction is called scalar for example,
mass, speed, temperature, energy etc.
2. Vectors : A quantity which has magnitude and is also related to some definite direction is called
vector, for example, force, velocity , momentum.
3. If k > 0, vector k a is a vector in same direction as vector a but of magnitude k times that of a .
If k < 0, vector k a is a vector in opposite direction to vector a but of magnitude k times that of a .
4. Triangle Law of Vector Addition :
If two vectors a and b lie along the two sides of a triangle in consecutive
order (as shown), the third side represents the sum of a and b .
If P and Q lie along the adjacent sides of a parallelogram,
the length of the diagonal represents the magnitude of the
resultant R .
R |P Q |
R P2 Q2 2PQ cos
5. If a number of forces acting on a particle are in equilibrium, then
Fx 0 and Fy 0
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6. Force of static friction f acts when surfaces have slipping tendency but are not slipping against each other.
0 f sN
7. Force of kinetic friction fk acts when the surfaces are slipping against each other.
fk k N k s
8. Net contact force between two surfaces is F N 2 force of friction2
9. (i) To find velocity of A relatively to a reference B, we impose (add) the reversed velocity of reference
(B).
(ii) The formula : vAB vA vB
involves vectors and hence velocities should be added or subtracted like vectors while using this
relation.
(iii) The velocity of swimmer as seen fromground vs = velocity of swimmer in still water + velocity of
river.
10. The scalar or dot product of two vectors a and b is defined as the product of their magnitudes and
the cosines of the angle between them. It is represented by a dot (.) between a and b . The product
itself is a scalar quantity.
a b ab cos (0 )
11. The vector product of two vectors a and b is defined as a vector whose :
(i) Magnitude is equal to the product of their magnitudes and the sine of the angle 9 between them.
(ii) Producct itself is a vector quantity.
a b a b sin
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Illustration - 1
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