vectors and equilibrium type equation here. 2 · 2018-02-28 · 2prof. muhammad amin chapter-2...

Prof. Muhammad Amin 1

Chapter-2 vectors and equillibrium

Type equation here.

Q.1 (a) Describe the addition of two vectors by their rectangular components

(trigonometric method).

(b)Given the various steps to find the resultant of vectors by the addition of

rectangular components?

(c)Describe the rules to find the direction of the resultant vector.

(a) VECTOR ADDITOIN BY RECTANGULAR COMPONENTS

Consider two vectors A and B which are represented by the lines OM and ON making angle 1

and 2 respectively with the x–axis as shown in figure (a).

We add the vectors A and B by head to tail rule. The resultant vector R making an angle with

x–axis is represented by the vector OP .

i.e. OP = OM + MP

or R = A + B________________(1)

Let us resolve these vectors into rectangular components.

Draw a perpendicular on the x–axis from the head of A and join ‘O’ with point Q.

xOQ A = X–component of A

yQM A = Y–component of A

Similarly, the rectangular components of the vector B are MS and SP .

x =MS B X–component of B

ySP B = Y–component of B

From figure (b), we have

xB = MS = QT

yA = QM = TS

xR = OT = X–component of R

yR = TP = y–component of R

x

Vectors and ‘equilibrium Chapter

2

Important long questions



Resultant of X–Components

OT = OQ + QT

Or Rx = Ax + Bx

In vector form,

x x xˆ ˆ ˆR i = A i + B i

x x xˆor R = A +B i _____________(2)

This shows that x–component of the resultant vector is equal to sum of x–components of two

vectors which are to be added.

Resultant of Y–Components

TP = Ts + SP

Or RY = Ay + By

In vector form,

y y yˆ ˆ ˆR j = A j + B j

Y y yˆor R = A +B j _____________(3)

This shown that y–component of the resultant vector is equal to the sum of y–components of

two vectors which are to be added.

Resultant Vector

The resultant vector is given by

x yR = R + R

Putting the values of xR and yR from equation (2) and (3)

Magnitude of the Resultant Vector

The magnitude of resultant vector R is given as

22

x yR = R = R + R

2

x x y yor R = A +B + A +B

Direction of the Resultant Vector

The direction of resultant vector R is give as

y

x

Rtan θ

R

y y

x x

A +Bor tan θ

A +B

y y-1

x x

A +Bor θ = tan

A +B

Addition of N number of Vectors

If we have large number of vectors represented by A, B, C, ………..

The magnitude of their resultant vector R is given by.

2

x x x y y yR = A +B +C +..... + A +B +C +.....

The direction of resultant vector R is given as



y y y

x x x

A +B +C +......tan θ =

A +B +C +......

y y y1

x x x

A +B +C +......or θ = tan

A +B +C +........

(b) Summary of Steps

The vector addition by rectangular components consists of the following steps.

1. Find x and y components of all given vectors.

2. Find x – component Rx of the resultant vector by adding the x – components of all the

vectors.

3. Find y-component Ry of the resultant vector by adding y – components of all the vectors.

4. Find the magnitude of the resultant vector R

using 2 2

x yR R R

5. Find the direction of the resultant vector R

by using equation. 1 y

x

Rtan

R

Where is the angle, which the resultant vector makes with positive x-axis.

(c) Rules to find the direction of resultant vector

The signs of Rx and Ry determine the quadrant in which resultant vector lies. For that purpose

proceed as given below.

Irrespective of the sign of Rx and Ry, determine the value of y-1

x

Rtan

R from the calculator or

by consulting trigonometric tables. Knowing the value of , angle is determined as follows.

If both Rx and Ry are positive, then the

resultant lies in first quadrant and direction

is

If Rx is –ive and Ry + ive, the resultant vector lies

in the second quadrant and its direction is

= 1800 -

If both Rx and Ry are negative, the resultant

lies in third quadrant and its direction is

180

If Rx is positive and Ry is –ve, the resultant

lies in the fourth quadrant and its direction is 0360

Q. 2. Define and explain scalar product of two vectors. Show that the scalar product is

commutative. Give its main properties?

Ans. Scalar Product of Two Vectors

“If two vectors are multiplied in such a way that their resultant is a scalar quantity such

a product is known as scalar product”

x

y

R

R

0180

x

y

R

R

0180

x

y

R

R

0360

x

y

R

R

X X

Y

Y



Examples

Work is the scalar product of two vectors, force F and displacement d .

i.e. work = F • = FS Cosθd

Power is a scalar product of two vectors, force F and Velocity V .

i.e. power = F • V = FV Cosθ

Explanation

Consider two vectors A and B making an angle with each other as shown in figure.

Then scalar product of A and B is defined as

From fig(i)

A . B = (magnitude of A )(projection of B on A )

or A • B = (A)(BCosθ) ABCosθ ___________(1)

Similarly

From figure(ii)

B . A = (magnitude of B )(projection of A on B )

or B • A = (B)(A Cosθ) ABCosθ ___________(2)

From equation (1) and (2), we get

A • B = B • A

i.e. The scalar product does not change with change in the order of the vectors. Hence

scalar product of two vectors obeys commutative law.

Properties or Characteristics of Scalar Product

1- Commutative law

The scalar product obeys commutative law.

That is A • B = B • A

2- Scalar product of two perpendicular vectors

If two vectors are perpendicular to each other or any one of the vector is zero, their

scalar product is zero.

i.e. A • B = AB Cosθ = AB 90 = 0Cos In case of unit vectors

ˆ ˆ ˆ ˆ ˆ ˆi • j = j•k = k•i = 0

3- Scalar product of two parallel and anti-parallel vectors

If two vector are parallel to each other then0A• B = ABCos0 = AB (1) = AB



If two vectors are anti-parallel, then oA•B = ABCos180 ABx 1 AB

5- Associative law

The scalar product obeys associative law

i.e. mA . n B = mn A•B = A• mn B = n A• mB

6- Distributive law

The scalar product obeys distributive law i.e. A . B + C = A•B + A•C

7- Self scalar product

The scalar product of two similar vectors i.e. A•A is called the self scalar product. In this

o oCas0 Cos0 =1

o 2A•A = AA Cos0 = A

In case of unit vectors ˆ ˆ ˆ ˆ ˆ î .i = j. j = k .k = 1

8. Scalar Product in Term of Rectangular Components

Let x y z

ˆ ˆ Â = A i +A j+A k

x y zˆ ˆ ˆB = B i+B j+B k

Then

x y z x y zˆ ˆ ˆ ˆ ˆ Â • B = A i +A j+A k • B i +B j+B k

x x y y x zˆ ˆ ˆ ˆ ˆ Â • B = A i •B i+A i •B i+A i •B k y x y y y z

ˆ ˆ ˆ ˆ ˆ ˆ+ A j•B i +A j•B j+A j •B k

z x z y z zˆ ˆ ˆ ˆ ˆ ˆ+ A k •B i +A k•B j + A j •B k

x x x y x zˆ ˆ ˆ ˆ ˆ Â • B = A B i i +A B i j +A B i k y x y y x z

ˆ ˆ ˆ ˆ ˆ ˆ+ A B j.i +A B j.j +A B j.k

x y zˆ ˆ ˆ ˆ ˆ Â B i +A B k j +A B k kz z zk

But, ˆ ˆ ˆ ˆ ˆ î .i = j. j = k .k = 1

ˆ ˆ ˆ ˆ ˆ î . j = j.k = k.i = 0

x x y y z zA• B = A B +A B +A B

This is known as scalar product in terms of scalar product

Q.3 Define and explain vector product of two vectors. Show that vector product is not

commutative. Give its main properties.

Ans. Vector Product of Two Vectors

“If two vectors are multiplied in such a way that their product is a vector quantity such

a product is known as vector product”

Examples of Vector Product

1Torque

The torque about a point is defined as “the cross product of position vector r and

force F i.e. = r × F.



2- Angular Momentum

Angular momentum is defined as “the cross product of position vector r and linear

momentum P .” i.e. L = r × P

Explanation

Consider two vectors A and B making an angle with each other as shown in fig (a).

Then cross product of A & B is defined by the relation. Â × B = AB Sin θ n ---------- (1)

Where n is a unit vector perpendicular to both A and B .

Direction

The direction of n can be found by “Right hand rule.”

Right Hand Rule

By right hand rule “rotate the fingers of right hand along the

direction of rotation. The thumb will give the direction of

vector product”

If the vector A moves towards B , then by right hand rule, the

direction of resultant vector �� will be vertically upward.

Similarly

Then cross product of B & A is defined by the relation.

ˆ× = AB Sin θ ( n )---------- (2)B A

Where n is a unit vector perpendicular to both A and B .

Then by right and rule, the direction of resultant vector �� will be

vertically downward

From (1) and (2), we get

A × B B × A Thus vector product is not commutative.

Properties or Characteristics of Vector Product

1- Commutative law

The vector product does not obey commutative law.

. . A × B B × A A × B B × Ai e but

2- Associative law

The vector product obeys associative law.

i.e. mA × nB = m n A × B = A × m n B

Where m and n are numbers

3- Vector product of two parallel or anti-parallel vectors

If two vectors are parallel or anti-parallel to each other or any one of the vector is

zero, their vector product is null vector.

o 0ˆ ˆ î.e. A × B Sin0 n = AB 180 n = 0 n 0AB Sin In case of unit vectors

ˆ ˆ ˆ ˆ ˆ î × i = j × j = k × k = 0

4- Vector product of two perpendicular vectors

If the vectors are perpendicular to each other, then vector has maximum magnitude.



oA × B = AB Sin90 = AB ×1 = AB

For unit vector ˆ ˆ,i j and ˆ,k by using right hand rule, we can write as.

ˆ ˆ ˆ ˆ ˆ î × j = k j × i = - k

ˆ ˆ ˆ ˆ ˆ ˆj × k = i k × j = - i

ˆ ˆ ˆ ˆ ˆ ˆk × j = j i × k = - j

5- Distributive property

The vector product obeys the distributive

i.e. A× B+C = A × B +A × C

6- Vector product in-terms of rectangular components

If x y zˆ ˆ Â = A i + A j + A k

x y zˆ ˆ ˆB = B i + B j + B k

x y z x y zˆ ˆ ˆ ˆ ˆ Â B = A i +A j+A k B i +B j+B k

x x x y x zˆ ˆ ˆ ˆ ˆ Â B = A B i i +A B i j +A B i k

y x y y x zˆ ˆ ˆ ˆ ˆ ˆ+ A B j i +A B j j +A B j k x y z

ˆ ˆ ˆ ˆ ˆ Â B i +A B k j +A B k kz z zk

Since, ˆ ˆ ˆ ˆ ˆ î i = j j = k k = 0

ˆ ˆ ˆ ˆ ˆ î × j = k j × i = - k

ˆ ˆ ˆ ˆ ˆ ˆj × k = i k × j = - i

ˆ ˆ ˆ ˆ ˆ ˆk × j = j i × k = - j

x x x y z zˆ ˆÂ×B = A B O +A B k +A B -j

y x y y y zˆ ˆ+ A B -k + A B o +A B i

z x z y z zˆ ˆ ˆ+ A B j + A B -i +A B o

y z z y z x x z x y y xˆ ˆ Â×B = i A B -A B +j A B -A B +k A B -A B

The resultant obtained can be also expressed for memory in determinant form as below.

x Y Z

X Y Z

ˆ ˆ î j k

A × B = A A A

B B B

Q. 4 Define torque and explain torque acting on a rigid body?

Ans. Torque

“The turning effect of a force on a body about the axis of rotation is called torque or

moment of the force.”

OR

Torque may be defined as “the physical quantity which produces angular acceleration

in a body is called torque.”

Mathematically

Torque may be defined as “the vector product of po sition vector r and force F .”

i.e. = r × F



ˆor = rF sin θ n

Where is the angle between r and F .

Magnitude

Where rF sinθ is the magnitude of the torque.

Direction

The direction of torque is represented by a unit vector n . The direction of the torque is

perpendicular to the plane formed by r and F given by right hand rule for the vector

product of two vectors.

Torque acting on a rigid body

Consider the torque due to force F acting on the rigid body. Suppose the force F is

applied on the particle at point P.

Its position vector from the origin ‘o’ (i.e. axis of rotation) is OP = r the magnitude of

torque acting on the particle can be calculated in two ways.

Method 1

We can resolve the force into two rectangular components as shown in figure.

1. Fx = F Cos which acts in the direction of r .

2. Fy = F Sin which acts in the direction perpendicular to r .The force (F Cos) can pull

the particle but cannot rotate it. The force (F Sin) produces rotation.

So, Magnitude of torque T = r (F Sin )

or = r F Sin θ__________(i)

Method 2

Resolve r into two rectangular components as shown in figure.

1. rx = r Cos which acts in the direction of F .

2. ry = r Sin which acts in the direction perpendicular to F as shown in figure

So, Magnitude of torque T =(r sin) F



or = r F Sin θ__________(ii)

Hence magnitude of torque is given by = r F Sin θ

S.I Unit The SI unit of torque is N m.

Dimensions

Torque = rF = r m a

Torque = Displacement Mass

2

Displacement×

Time

Torque2

L= L×M

T

Torque = 2 -2ML T

Factors upon Which Torque Depends

The applied force F .

The position vector r

The direction of F and r .

Special Cases

(1) Magnitude of torque will be zero if

When the line of action of force passes through the origin. Then r = 0

When the angle between F and r is 0o or 180o.

Then 𝜏 = r Fsin0o = r F sin180o = r F 0 = 0

When a body is at rest or moving with uniform angular velocity.

(2) When = 90o. Then magnitude of the torque will be maximum

𝜏 = r F sin90o = r F 1 = rF (Maximum torque)

(3) If we reverse the direction of r or F , the direction of torque T is reversed.

(4) If we reverse the direction of both r and F , the direction of T remains the same.

Convention

Positive Torque

Anticlockwise torque is taken as positive.

Negative Torque

Clockwise torque is taken as negative.



Q. 2.1 Define the term (i) unit vector (ii) Position vector and (iii) Components of a

vector?

Ans. (i) Unit Vector

“A vector whose magnitude is one in a given direction and use to represent the

direction of a vector is known as unit vector”

e.g vector 𝐴 can be written as

A = A A

A

A =A

A unit vector in the direction of A is A , which we read as ‘A hat’

The direction along x, y and z axes are generally represented by

unit vectors i , j and k respectively.

(ii) Position Vector

“The position vector r is a vector that describes the location of a point with respect to the

origin”

It is represented by a straight line drawn in such a way that its tail

coincides with the origin and the head with point P (a,b) as

shown in Figure. The projections of position vector r on the x and

y axes are the coordinates a and b and they are the rectangular

components of the vector r .

Hence.

2 2ˆ ˆr = a i + b j and r = a + b

In three dimensional space, the position vector of a point P (a, b, c) is shown

in Figure and is represented by

2 2 2ˆ ˆ ˆr = a i + b j+ ck and r = a + b c

(iii) Component of a Vector

A component of a vector is its effective value in a given direction. a vector can be split up into

two or more than two components.

Rectangular Components of A Vector

Components of a vector which are perpendicular to each other are

called rectangular components of vector.

Example

Let there be a vector A represented by OP making angle with

the x-axis

A = Ax i + Ay j

Thus Ax i and Ay j are the components of vector A . Since these are at right angle to each other,

hence, they are called rectangular components of A .

Important short questions

r

Y

N

a

P(a,b)

Xo

b

x

y

z

( , , )P a b c

r

O



Q. 2.2 The vector sum of three vectors gives a zero resultant. What can be the orientation

of the vectors?

Ans. Yes.

When the three vectors represented sides a triangle by head

to tail rule then their resultant is zero.

By adding head to tail rule, we can get zero resultant.

i.e. R = A+ +C= 0B

Q. 2.3 If one of the components of a vector is not zero, can its magnitude be zero?

Explain?

Ans. No.

If one of the components is not zero, then magnitude of vector can never be zero.

If Ax = 0, then magnitude is

2 2

x yA = A +A

2 2 2

y yA = O +A = A = A

If Ay = 0, then A = Ax

Hence the magnitude of a vector can only be zero if all of its rectangular components

are zero.

Q. 2.4 Can a vector have a component greater than the vector’s magnitude?

Ans. No.

The rectangular component of a vector can never be greater than the vector’s

magnitude. The maximum value of magnitude of component can be equal to the

magnitude of the vector.

The magnitude of vector 𝐴 is given by

2 2

x yA = A +A

⇒ 𝐴𝑥 ≤ 𝐴 and 𝐴𝑦 ≤ 𝐴

Q. 2.5 Can the magnitude of a vector have a negative value?

Ans. No. The magnitude of a vector cannot be negative.

The magnitude of vector 𝐴 is given by

2 2

x yA = A +A

Since 2 2

x yA +A always gives a positive value. If any of its components is negative the

square will make it positive.

Hence the magnitude of a vector always has a positive value

Q. 2.6 If A+B = 0 , what can you say about the components of the two vectors?

Ans. let 𝐴 = 𝐴𝑥𝑖 + 𝐴𝑦𝑗 and �� = 𝐵𝑥𝑖 + 𝐵𝑦𝑗

A + B = 0

( 𝐴𝑥𝑖 + 𝐴𝑦𝑗 ) + ( 𝐵𝑥𝑖 + 𝐵𝑦𝑗 ) = 0

(𝐴𝑥 + 𝐵𝑥 )𝑖 + (𝐴𝑦 + 𝐵𝑦 )𝑗 = 0

𝐴𝑥 + 𝐵𝑦 = 0 and 𝐴𝑦 + 𝐵𝑦 = 0

⇒ 𝐴𝑥 = −𝐵𝑥 𝑎𝑛𝑑 𝐴𝑦 = −𝐵𝑦

Hence we can conclude that A+B = 0 if the respective components of vector 𝐴 𝑎𝑛𝑑 ��

will be equal in magnitude and opposite in direction.

𝐴

�� 𝐶



Q. 2.7 Under what circumstances would a vector have components that are equal in

magnitude?

Ans. When a vector makes and angle of 45 o with x–axis (horizontal), then the components of

that vector are equal in magnitude.

Proof:-

Let 𝐴𝑥 𝑎𝑛𝑑 𝐴𝑦 are rectangular components of vector 𝐴

xA = A y

A Cos = ASin

Sin Cos

sin= 1

Cos

tan 1

1

0

tan (1)

45

Q. 2.8 Is it possible to add a vector quantity to a scalar quantity? Explain?

Ans. No.

Since Only similar type of physical quantities can be added or subtracted hence A

vector cannot be added to a scalar because both are different quantities by nature.

The scalar obeys the rules of arithmetic and ordinary algebra. Vectors have magnitude

as well as direction so obey the special rules of vector algebra.

Q. 2.9 Can you add zero to a null vector?

Ans. No.

Reason

Zero is a scalar and null vector is a vector quantity. As scalar cannot be added to the

vector. Hence zero cannot be added to a null vector. Physical quantities of the same

nature can be added.

Q. 2.10 Two vectors have unequal magnitudes. Can their sum be zero? Explain?

Ans. No. The sum of two vectors having unequal magnitudes cannot be zero.

Reason

Because the sum of two vectors will be zero only when their magnitudes are equal and

they act in opposite direction.

Q. 2.11 Show that the sum and difference of two perpendicular vectors of equal lengths

are also perpendicular and of the same length?

Ans. Consider two vectors A and B of equal lengths are perpendicular to each other, then their

A+B sum is equal in length to their difference A - B . Both A+B and A B are also

perpendicular to each other A+B makes an angle of 45o with A and A B makes and

angle of –45o with A .

Magnitude of 2 2A+B = A +B

Magnitude of 2 2A-B = A+ -B = A +B

Hence the sum and difference of two perpendicular

vectors of equal lengths are also perpendicular and of

the same length



Q. 2.12 How would the two vectors of the same magnitude have to be oriented, if they

were to be combined to give a resultant equal to a vector of the same magnitude?

Ans. The two vectors of the same magnitudes are combined to

give a resultant equal to a vector o f the same magnitude if

the two vectors are oriented such that an equilateral triangle

is formed. In this case the angle b/w the vectors is 120o as

shown in figure.

C = A+B

C = A B

Q. 2.13 suppose the sides of a closed polygon represent vector arranged head to tail.

What is the sum of these vectors?

Ans. The sum of these vectors will be zero.

Reason

In this case, the tail of the first vector meets with the head of last vector according to

head to tail rule as shown in figure. So, resultant will be zero

R = A + B + C + D + E + F = 0

Q. 2.14 If all the components of the vector 1A and 2A were revers, how would this alter

1 2A ×A ?

Ans. 1 2A ×A will not change

Reason

If all the components of the vectors 1A and 2A are reversed, then both of the vectors will

become 1A and 2A .

1 2 1 2A A A A

1 2So A ×A will not alter (change).

Q. 2.15 Name the three different conditions that could make 1 2A A 0

Ans. CONDITIONS

1- Either 1 2A = 0 or A = 0

1 2 1 2ˆA ×A = A A Sinθn

1 2A ×A =0

2- 1A is parallel to 2A i.e. = 0o

o O1 2 1 2

ˆA ×A = A A Sin 0 n Sin0 = 0

1 2ˆ= A A ×o× n =0



3- 1A is anti-parallel to 2A i.e. = 180o

o O1 2 1 2

ˆA ×A = A A Sin180 n Sin180 = 0

1 2

ˆ= A A ×o× n =0

Q. 2.16 Can a body rotate about its centre of gravity under the action of its weight?

Ans. No, a body cannot rotate under the action of its weight.

Reason

The body cannot rotate about its centre of gravity under the action

of its weight because in this case the line of action of weight passes

through centre of gravity. Therefore, moment arm will be zero. As a

result torque is also zero.

i.e

𝜏 = ℓ 𝐹 = (0)𝐹 = 0

Q. 2.17 What is meant by Null vector?

Ans: A vector whose magnitude is zero and has an arbitrary direction is called Null vector. It

is represented by O .

We can obtain the null vector by adding a vector into its negative vector.

0A A

Q. 2.18 If force of magnitude 20N makes an angle of 30o with x – axis then find its y –

components?

Ans: F = 20N

θ = 30o

Fy =?

Fy = F Sin θ

= 20 Sin 30o

120

2

Fy = 10N

Q. 2.19 If force F of magnitude 10N makes an angle of 30o with y – axis then find its x –

component.

Ans: F = 10N

Angle of F with x-axis

90 30 60o o o

Fx =?

Fx = F Cosθ

= 10 Cos (60o)

110

2

Fx = 5N

Q. 2.20 Prove that 2 3A i j k and 4 5B i j k are mutually perpendicular.

Ans: 2 3A i j k

4 5B i j k

. (2 3 ).(4 5 )A B i j k i j k

. 2 4 3 1 1 5A B

30o

60o

F

x axis

y axis

O

W



= 8 3 5

= 0

Since dot product of two vectors &A B is equal to zero. So they are perpendicular to

each other.

Q. 2.21 What is the physical significance of cross product of two vectors?

Or prove that magnitude of A B ia equal to the area of parallelogram…..?

Ans: Magnitude of A B is equal to the area of the parallelogram formed with Aand B as two

adjacent sides.

Area of parallelogram = (length) (height)

sinA B

sinAB

A B

Q. 2.22 Define the moment arm

Ans. The perpendicular distance between the line of action of force and axis of rotation is

called moment arm. In the given figure moment arm = OP = r

Q. 2.23 How do we add and subtract the vectors?

Ans: Addition of vectors

We add the vectors by head to tail rule.

Given two vectors Aand B . Their sum is obtained by drawing their representative line in

such a way that the tail of vector B coincides with the head of vector A . Now if we join

the tail of A to the head of B . This line will represent the vector sum A B in

magnitude and direction. The vector sum is called resultant vector and is indicated by R .

Subtraction of vectors

Ans: The subtraction of a vector B from A is equivalent to the addition of the negative of

vector B and vector A . Thus, to subtract vector B from A , reverse the direction of B and

add it to A as shown in the figure.

A B A B where B is negative vector of B

Q. 2.24 what is dot product? Write the formula of K.E in terms of self dot product of

velocity vectors.

Ans: The scalar product of two vectors Aand B is a scalar quantity, defined as . cosAB AB

As kinetic energy of object of mass m and speed v is given by the relation

O

r

F

P

F

B Sin h

A

B

B A

BA



21.

2K E mv

As we know square of magnitude of vector is equal to its self dot product. So

2 .v v v

1

. .2

K E m v v

Q. 2.25 What does sin

A B

AB

represent?

Ans: sin

A B

AB

represent the unit vector. Which shows the direction of A B .

By definition

sinA B AB n

sin

A Bn

AB

It is unit vector perpendicular to plane containing Aand B .

Q. 2.26 Show that vector addition is commutative?

Ans: When A is added to B then resultant is R A B ------ (i)

when B is added to A then the resultant is R B A ------ (ii)

as shown in fig

So from equation (i) and (ii) it is clear,

A B B A

It shows that vector addition is commutative.

Q.2.27 Define dynamic and static equilibrium.

Ans: Dynamic equilibrium

If the body is moving with uniform velocity, it is said to be in dynamic equilibrium.

Example

A rain drop moving in air in downward direction with constant terminal velocity is an

example of dynamic equilibrium.

Static equilibrium

If a body is at rest, it is said to be in static equilibrium.

Example A book lying on a table.

Q.2.28 what is the unit vector in the direction of the vector A= ˆ ˆ4i + 3 j?

Given: A=4i+3j

AA=

A

Where A is the magnitude of vector A which is expressed as

2 2

x yA= A +A

Putting the values

2 2

A= 4 + 3

= 16+9= 25

or A=5

Putting the values in formula A=A/A

Hence,

4i + 3jA =

5 Ans.



2.1 Two particles are located r1 = 2

ˆ ˆ ˆ ˆ3i +7 jand r =-2i +3jrespectively. Find both the magnitude of

the vector (r2--r1) and its orientation with respect to the x-axis.

Data:

Location of first particle = 1r =3i+7 j

Location of second particle 2r = -2i+3j

To Find:

Magnitude of the vector r = 2 1r r = ?

Direction of the vector r = = ?

Calculation:

As we know that

2 1r=r -r

Putting these values we get

r= -2i+3j - 3i+7 j

= -2i+3j-3i-7 j

r =-5i-4j

Magnitude of position vector r is given by 2 2

x yr r

2 2r = -5 + - 4

= 25+16 = 41

r = 6.4 Ans.

x-component = x = -5 and y – component = y = -4

1 1 4tan tan

5

y

x

r

r

1tan 0.8

39 o

As both components are –ve, so the vector lies in the 3rd quadrant. Therefore,

1180 tan

y

x

ro

r

180 39 o o

= 219o Ans.

2.2 If vector B is added to vector A , the result is ˆ ˆ6 i j . If B is subtracted from A , the result is

ˆ ˆ4i 7 j . What is the magnitude of vector A ?

Solutions:

Data:

A+B=6i+j __________ (i)

A-B=-4i+7j __________ (ii)

To Find:

Magnitude of A=?

Numerical Problems



Calculation:

Adding equation (i) and (ii), we get

A+B=6i+ j

A-B=-4i+7 j

__________

2A=2i+8j

A=i+4j

Using the formula of magnitude,

2 2

x yA= A +A

2 2

A= 1 + 4 = 1+16= 17

A = 4.1 units Ans.

2.3 Find the angle between the two vectors, ˆ ˆ ˆ ˆ5 and B 2A = i + j i + 4 j

Solution:

Data:

A=5i+j __________ (i)

B=2i+4j _________ (ii)

To Find:

Angle 'θ' between A and B = ?

Calculation:

We know that A.B=ABcosθ __________ (a)

Cosθ=A.B/AB

A.B= 5i+j . 2i+4j

=10i.i+20i.j+2j.i+4j.j

. 10 0 0 4 14AB

Because, i.i= j.j=1

And i. j= j i=0.

A.B=14 ---------------- (iii)

2 2

A= 5 + 1 = 26

2 2

B= 2 + 4 = 4+16= 20

Putting the values in Eq (a), we get



o

-1

14 14Cosθ= =

5.1×4.526× 20

14=

22.95

=0.61

θ = Cos 0.61

oθ = 52 Ans.

2.4 Find the work done when the point of application of force ˆ ˆ3 i 2 j moves in a straight line

from the point (2,-1) to the point (6,4).

Solution:

Data:

The force = F=3i+2j

Point A (2, -1) and point B (6, 4)

To Find:

Work done = W = ?

Calculation:

Position vector of these points are

Ar =2i- j ………………. (i)

Br =6i+4j ----------------- (ii)

Displacement = B Ad=r -r

Putting the values of rA and rB

ˆd= 6i+4j - 2i-j

=6i+4j-2i+j

d=4i+5j

Using the formula of work

Work done = W=F.d

Putting the values we get

W= 3i+2j . 4i+5j ---------------------- (iii)

= 12 i.i 10 j.j Because i.i j.j 1

= 12 + 10 = 22 J

W = 22 Units Ans

2.5 Show that the three vectors ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆi j k, 2 i 3j k and 4i j 5k are mutually perpendicular.

Solution:

Data:

The three vectors are given as



A=i+j+k

B=2i-3j+k

C=4i+j-5k

To Find:

If the vectors are mutually perpendicular to each other, then we have to show that

(i) A.B=0 (ii) B.C=0 (iii) C.A=0

Calculation:

(i) ˆA.B= i+j+k . 2i-3j+k

A.B 2 i.i 3 j.j 1 k.k

= 2 - 3 +1

A.B= 0....................(a)

(ii) B.C = 2i-3j+k . 4i+j-5k

= 8 i.i 3 j.j 5 k.k

Because i.i j.j k.k 1

= 8-3-5

Hence B.C=0......................(b)

(iii) C.A= 4i+j-5k . i+j+k

C.A 4 i.i 1 j.j 5 k.k

Because i.i j.j k.k 1

= 4+1-5

Hence C.A=0---------(c)

Equations (a), (b) and (c) show that their mutual products are zero therefore all the three vectors are

mutually perpendicular to each other.

2.6 Given that ˆ ˆ ˆA = i - 2j + 3k and ˆ ˆ3B = i - 4k , find the projection of A on B .

Solution:

Data:

A=i-2j+3k

B=3i-4k

To Find:

Length of Projection of A on B = Acos = ?



Calculation:

We know that

A.B =ABCosθ

or A.B

Acosθ = B

__________ (i)

2 2

B = 3 + -4 = 9+16 = 25

or B = 5......................(ii)

Now A.B = i-2j+3k . 3i-4k

A.B 3 i.i 2 0 12 k.k

3 0 12

or A.B = -9------------(iii)

Putting the values in equation (i) we get Acosθ = - 9 / 5 Ans.

2.7 The magnitude of dot and cross products of two vectors are 6 3 and 6 respectively. Find the

angle between them.

Solution:

Data:

Magnitude of dot product of two vectors A and B = ABcos = 6 3 ________ (i)

Magnitude of cross product of two vectors A and B = ABsin = 6 __________ (ii)

To Find: Angle = = ?

Calculation:

Dividing Eq. (2) by Eq. (1), we have;

1

A Bsinθ 6

A Bcosθ 6 3

sin θ 1

cosθ 3

1tan θ

3

1θ tan

3

oθ= 30 Ans.

vectors and equilibrium type equation here. 2 · 2018-02-28 · 2prof. muhammad amin chapter-2...

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