Question

Please answer the 3rd and 5th Questions !

My answers were coming wrong I don't why ._. ​

Answer 1

$\large \star \rm{\underline\pink{ \: Answer \: of \: Question \: no. 3 \: ) \: }}$

First , we will find the components of the vector along x-axis and y-axis. Then we will find the resultant x and y-components.

x - component of

$\: \: \: \: \: \: \sf \: \vec{A} = A \cos45 \degree = 100 \: \cos45 \degree = \dfrac{100}{ \sqrt{2} }$

x-component of

$\: \: \: \: \: \: \sf \: \vec{B} = \vec{ B }\cos135 \degree = - \dfrac{100}{ \sqrt{2} }$

x-component of

$\: \: \: \: \: \: \: \sf \vec{C} = \vec{C} \cos315 \degree = 100 \cos315 \degree$

$\: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf = 100 \cos45 \degree = \dfrac{100}{ \sqrt{2} }$

Resultant x-component

$\sf \: \: = \dfrac{100}{ \sqrt{2} } - \dfrac{100}{ \sqrt{2} } + \dfrac{100}{ \sqrt{2} } = \dfrac{100}{ \sqrt{2} }$

Now,

y-component of

$\: \: \: \: \: \: \sf \: \vec{A} = 100 \sin45 \degree = \dfrac{100}{ \sqrt{2} }$

y-component of

$\: \: \: \: \: \: \sf \: \vec{B} = 100 \sin135 \degree = \dfrac{100}{ \sqrt{2} }$

y-component of

$\: \: \: \: \: \: \: \sf \vec{C} = 100 \sin315 \degree = - \dfrac{100}{ \sqrt{2} }$

Resultant x-component

$\sf \: \: = \dfrac{100}{ \sqrt{2} } + \dfrac{100}{ \sqrt{2} } - \dfrac{100}{ \sqrt{2} } = \dfrac{100}{ \sqrt{2} }$

Magnitude of resultant

$\sf \: \: \: \: \: \: = \sqrt{ \bigg( \dfrac{100}{ \sqrt{2} } \bigg)^{2} + \bigg( \dfrac{100}{ \sqrt{2} } \bigg)^{2} }$

$\: \: \: \: \: \: \: \: \: = \sf \: \sqrt{10000}$

$\: \: \: \: \: \: \: \: \: = \sf \: \pink{100}$

Angle made by a resultant vector with the x-axis is given by

$\: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \tan \alpha = \dfrac{y - comp}{x - comp}$

$\: \: \: \: \: \sf \: = \dfrac{100 \sqrt{2} } {100 \sqrt{2} } = 1$

$\: \sf \: \: \implies \: \alpha = { \tan}^{ - 1} (1) = \green{45 \degree}$

The magnitude of the resultant vector is 100 units and it makes an angel of 45° with the x-axis.

⠀━━━━━━━━━━━━━━━━━

$\large \star \rm{\underline\pink{ \: Answer \: of \: Question \: no. 5 \: ) \: }}$

First ,Let us find the components of the vector along x-axis and y-axis. Then we will find the resultant x and y-components.

x-component of

$\: \: \: \: \: \: \sf \: \: \vec{OA} = 2 \cos30 \degree = \sqrt{3}$

x-component of $\sf \: \: \vec{BC} = 1.5\cos120 \degree$

$\: \: \: = - \dfrac{1.5}{2} = ( - 7.5)$

x-component of

$\: \: \: \: \: \: \sf \: \: \vec{DE} = 1 \cos270\degree = 0$

y-component of $\sf \: \: \vec{OA} = 2\sin30\degree = 1$

y-component of $\sf \: \: \vec{BC} = 1.5\sin120 \degree$

$\: \: \: = - \dfrac{ \sqrt{3} \times 1.5}{2} = ( 1.3)$

y-component of $\sf \: \: \vec{DE} = 1\sin270\degree =- 1$

x-component of resultant

$\: \: \sf \: R _x = \sqrt{3 } - 0.75 + 0 = 0.98 m$

y-component of resultant

$\: \: \sf \: R _y = 1 + 1.3 - 1= 1.3 m$

Now,

$\: \: \: \sf Resultant (R) \sf \: \: \: = \sqrt{ ( R _x)^{2} + ( R _y)^{2} }$

$\sf \: \: \: \: \: = \sqrt{(0.98)^{2} + (1.3) ^{2} }$

$\sf \: \: \: \: \: = \sqrt{0.96 + 1.69 }$

$\sf \: \: \: \: \: = \sqrt{2.65 }$

$\sf \: \: \: \: \: = 1.6m$

If it makes an angle α with the positive x-axis,then

$\: \: \: \: \: \: \: \: \: \: \: \sf \: \: \: \tan \alpha = \dfrac{y - comp}{x - comp}$

$\: \: \: \: \: \sf \: = \dfrac{1.3} {0.98 } = 1.332$

$\: \sf \: \: \implies \: \alpha = { \tan}^{ - 1} (1.32)$

Answer 2

-: Question 3 :-

Given :-

Vector A , Vector B and Vector C all have magnitude of 100 units and are inclined at angle of 45° , 135° and 315° .

To Find :-

The magnitude of the resultant vector .

Solution :-

For the diagram refers to the attachment 1 .

It is clear from the diagram that we want the values of Sin 315° , Sin 135° , Cos 315° and Cos 135° . And we knows the value of Cos 45° and Sin 45° .

:- Sin 315°

=> Sin ( 90 × 3 + 45 )°

=> - Cos 45°

{ Sin in 4th Quadrant }

=> -1/√2

{ Cos 45° = 1/√2 }

:- Sin 135°

=> Sin ( 90 × 1 + 45 )°

=> Cos 45°

{ Sin in II Quadrant }

=> 1/√2

{ Cos 45° = 1/√2 }

:- Cos 315°

=> Cos ( 90 × 3 + 45 )°

=> Sin 45°

{ Cos in IV Quadrant }

=> 1/√2

{ Sin 45° = 1/√2 }

:- Cos 135°

=> Cos ( 90 × 1 + 45 )°

=> - Sin 45°

{ Cos in II Quadrant }

=> -1/√2

{ Sin 45° = 1/√2 }

Hence , Sin 315° = -1/√2

=> Sin 45° = 1/√2

=> Sin 135° = 1/√2

=> Cos 45° = 1/√2

=> Cos 315° = 1/√2

=> Cos 135° = -1/√2

Let , The resultant Vector = Vector R

Now. ,we need the components x , y of All the vectors . So ,

Magnitude of Vector A = 100 units

Magnitude of Vector B = 100 units

Magnitude of Vector C = 100 units

X - Component :-

x - component of Vector A :-

100 Cos 45°

=> 100 × 1/√2 = 100/√2

x - component of Vector B :-

100 Cos 135°

=> 100 × -1/√2 = -100/√2

x - component of Vector C :-

100 Cos 315°

=> 100 × 1/√2 = 100/√2

Y - Component :-

y - component of Vector A :-

100 Sin 45°

=> 100 × 1/√2 = 100/√2

y - component of Vector B :-

100 Sin 135°

=> 100 × 1/√2 = 100/√2

y - component of Vector C :-

100 Sin 315°

=> 100 × -1/√2 = -100/√2

Now , Let us calculate the x and y component of the resultant i.e Vector R :-

x - component of Vector R :-

=> 100/√2 + 100/√2 + ( -100/√2 )

=> 100/√2

y - component of Vector R :-

=> 100/√2 + 100/√2 + ( -100/√2 )

=> 100/√2

Now , The magnitude of the resultant is given by :-

√ ( 100/√2 )² + ( 100/√2 )²

=> √ 10000/2 + 10000/2

=> √ 5000 + 5000

=> √10000

=> 100

Now , we knows that , For an angle "x" maked by a vector along x - axis :-

tan x = y component/x - component

=> tan x = 100/√2 ÷ 100/√2

=> tan x = 100/√2 × √2/100

=> tan x = 1

=> x = tan^-1 ( 1 )

=> x = 45°

Therefore ,The magnitude of Vector R is 100 unit at angle of 45° to X - axis.

Hence , Done :)

-: Question 5 :-

For diagram kindly see the attachment 2 .

Here. , We need the values of Sin 120° , Sin 270° , Cos 120° and that of Cos 270° .

:- Sin 120°

=> Sin ( 90 × 1 + 30 )°

=> Cos 30°

{ Sin in II Quadrant }

=> √3/2

{ Cos 30° = √3/2 }

:- Sin 270°

=> Sin ( 90 × 3 + 0 )°

=> - Cos 0°

{ Sin in III Quadrant }

=> -1

{ Cos 0° = 1 }

:- Cos 120°

=> Cos ( 90 × 1 + 30 )°

=> - Sin 30°

{ Cos in II Quadrant }

=> -1/2

{ Sin 30° = 1/2 }

:- Cos 270°

=> Cos ( 90 × 3 + 0 )°

=> - Sin 0°

{ Cos in III Quadrant }

=> -0 = 0

{ Sin 0° = 0 }

Let , The resultant Vector = Vector R

Magnitude of Vector OA = 2 m

Magnitude of Vector BC = 1.5 m

Magnitude of Vector DE = 1m

X - Component :-

x - component of OA :-

2 Cos 30°

=> 2 × √3/2 = √3

x - component of BC :-

1.5 Cos 120°

=> 1.5 × -1/2 = -0.75

x - component of DE :-

1 Cos 270°

=> 1 × 0 = 0

Y - Component :-

y - component of OA :-

2 Sin 30°

=> 2 × 1/2 = 1

y - component of BC :-

1.5 Sin 120°

=> 1.5 × √3/2 = 1.3

y - component of DE :-

1 Sin 270°

=> 1 × -1 = -1

b) :-

The x and y component of Vector R are as follows :-

x - component = √3 + 0 + ( - 0.75 )

=> 1.73 - 0.75 = 0.98

y - component = 1 + 1.3 + ( -1 ) = 1.3

a) :-

The magnitude of Vector A is :-

√ ( 0.98 )² + ( 1.3 )²

=> √ 0.9604 + 1.69

=> √2.6504

=> 1.62 m ( approx )

c) :-

Now , as we knows that for an angle "x" maked by a vector along x - axis :-

tan x = y - component/x - component

=> tan x = 1.3/0.98

=> tan x = 1300/980

=> tan x = 130/98

=> tan x = 1.32

=> x = tan^-1 ( 1.32 )

Hence,. We are Done :) .

I hope your Answers got correct now .