In the previous section we saw resolution of vectors. Now we are in a position to use analytical method to find vector sum.
• We have seen how to find the resultant using graphical method.
• But graphical method is tedious and lacks accuracy.
• So it is preferable to use analytical method. The steps are given below:
1. Consider two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$ shown in fig.4.15(a) below:
• Let the rectangular components of $\small{\vec{A}}$ be $\small{\vec{A_x}}$ and $\small{\vec{A_y}}$ respectively. This is shown in fig.4.15(b)
• Let the rectangular components of $\small{\vec{B}}$ be $\small{\vec{B_x}}$ and $\small{\vec{B_y}}$ respectively
2. Grouping of like-components:
• Group the horizontal components together. That is., group $\small{\vec{A_x}}$ and $\small{\vec{B_x}}$ together. This is shown in fig.4.15(c)
• We have seen how to find the resultant using graphical method.
• But graphical method is tedious and lacks accuracy.
• So it is preferable to use analytical method. The steps are given below:
1. Consider two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$ shown in fig.4.15(a) below:
 |
| Fig.4.15 |
• Let the rectangular components of $\small{\vec{B}}$ be $\small{\vec{B_x}}$ and $\small{\vec{B_y}}$ respectively
2. Grouping of like-components:
• Group the horizontal components together. That is., group $\small{\vec{A_x}}$ and $\small{\vec{B_x}}$ together. This is shown in fig.4.15(c)
• Group the vertical components together. That is., group $\small{\vec{A_y}}$ and $\small{\vec{B_y}}$ together
3. Add the like-components:
• Add the two horizontal components $\small{\vec{A_x}}$ and $\small{\vec{B_x}}$.
♦ For that, shift $\small{\vec{B_x}}$ so that it's tail coincide with the tip of $\small{\vec{A_x}}$
♦ We thus get a new horizonatal vector: ($\small{\vec{A_x}}$ + $\small{\vec{B_x}}$)
• Add the two vertical components $\small{\vec{A_y}}$ and $\small{\vec{B_y}}$.
♦ For that, shift $\small{\vec{B_y}}$ so that it's tail coincide with the tip of $\small{\vec{A_y}}$
♦ We thus get a new vertical vector: ($\small{\vec{A_y}}$ + $\small{\vec{B_y}}$)
4. For that, we will add the two new vectors by the triangle method.
• Let the resultant of this vector addition be $\small{\vec{R}}$. This is shown in fig.d
• We can see that: $\small{\vec{R}}$ = [($\small{\vec{A_x}}$ + $\small{\vec{B_x}}$)] + [($\small{\vec{A_y}}$ + $\small{\vec{B_y}}$)]
• This is an easy analytical method for vector addition
5. Let us check the above result graphically:
For that we add the original vectors by triangle method. This is shown in fig.e
• We find that, the resultant in fig.e is same as the resultant in fig.d
■ So we can write:
$\small{(\vec{A}\: +\: \vec{B})\: =\: [(\vec{A_x}\: +\: \vec{B_x})]\: +\: [(\vec{A_y}\: +\: \vec{B_y})]}$
6. We proved the above result graphically. But once proved, we do not need to draw graphs for doing problems. That is., while adding two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$, we do not need to use graphical methods any more.
We can straight away write:
• The horizontal component of the required resultant $\small{\vec{R}}$ is: $\small{(\vec{A_x}\: +\: \vec{B_x})}$
• The vertical component of the required resultant $\small{\vec{R}}$ is: $\small{(\vec{A_y}\: +\: \vec{B_y})}$
7. Once we find the horizontal and vertical components mentioned in (6) above, we can easily find $\small{\vec{R}}$
But how do we find those components analytically?
• We have already seen the method in the previous section. We will write it again:
(i) If $\theta_A$ is the angle made by $\small{\vec{A}}$ with the x axis, then:
• $\small{\vec{A_x}}$ = $\small{\left [\left | \vec{A} \right |\, cos\, \theta_A \right ]\hat{i}}$
♦ That means: magnitude of $\small{\vec{A_x}}$ is: $\small{\left [\left | \vec{A} \right |\, cos\, \theta_A \right ]}$
♦ Direction of $\small{\vec{A_x}}$ is same as the direction of the unit vector $\small{\hat{i}}$
• $\small{\vec{A_y}}$ = $\small{\left [\left | \vec{A} \right |\, sin\, \theta_A \right ]\hat{j}}$
♦ That means: magnitude of $\small{\vec{A_y}}$ is: $\small{\left [\left | \vec{A} \right |\, sin\, \theta_A \right ]}$
♦ Direction of $\small{\vec{A_y}}$ is same as the direction of the unit vector $\small{\hat{j}}$
(ii) If $\theta_B$ is the angle made by $\small{\vec{B}}$ with the x axis, then:
• $\small{\vec{B_x}}$ = $\small{\left [\left | \vec{B} \right |\, cos\, \theta_B \right ]\hat{i}}$
♦ That means: magnitude of $\small{\vec{B_x}}$ is: $\small{\left [\left | \vec{B} \right |\, cos\, \theta_B \right ]}$
♦ Direction of $\small{\vec{B_x}}$ is same as the direction of the unit vector $\small{\hat{i}}$
• This can be written as: $\small{\left ( \left | \vec{A_x} \right |+\left | \vec{B_x} \right | \right )\hat{i}}$
(ii) New vector in the y direction is: $\small{\left | \vec{A_y} \right |\hat{j}+\left | \vec{B_y}\right |\hat{j}}$
• This can be written as: $\small{\left ( \left | \vec{A_y} \right |+\left | \vec{B_y} \right | \right )\hat{j}}$
3. Add the like-components:
• Add the two horizontal components $\small{\vec{A_x}}$ and $\small{\vec{B_x}}$.
♦ For that, shift $\small{\vec{B_x}}$ so that it's tail coincide with the tip of $\small{\vec{A_x}}$
♦ We thus get a new horizonatal vector: ($\small{\vec{A_x}}$ + $\small{\vec{B_x}}$)
• Add the two vertical components $\small{\vec{A_y}}$ and $\small{\vec{B_y}}$.
♦ For that, shift $\small{\vec{B_y}}$ so that it's tail coincide with the tip of $\small{\vec{A_y}}$
♦ We thus get a new vertical vector: ($\small{\vec{A_y}}$ + $\small{\vec{B_y}}$)
4. For that, we will add the two new vectors by the triangle method.
• Let the resultant of this vector addition be $\small{\vec{R}}$. This is shown in fig.d
• We can see that: $\small{\vec{R}}$ = [($\small{\vec{A_x}}$ + $\small{\vec{B_x}}$)] + [($\small{\vec{A_y}}$ + $\small{\vec{B_y}}$)]
• This is an easy analytical method for vector addition
5. Let us check the above result graphically:
For that we add the original vectors by triangle method. This is shown in fig.e
• We find that, the resultant in fig.e is same as the resultant in fig.d
■ So we can write:
$\small{(\vec{A}\: +\: \vec{B})\: =\: [(\vec{A_x}\: +\: \vec{B_x})]\: +\: [(\vec{A_y}\: +\: \vec{B_y})]}$
6. We proved the above result graphically. But once proved, we do not need to draw graphs for doing problems. That is., while adding two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$, we do not need to use graphical methods any more.
We can straight away write:
• The horizontal component of the required resultant $\small{\vec{R}}$ is: $\small{(\vec{A_x}\: +\: \vec{B_x})}$
• The vertical component of the required resultant $\small{\vec{R}}$ is: $\small{(\vec{A_y}\: +\: \vec{B_y})}$
7. Once we find the horizontal and vertical components mentioned in (6) above, we can easily find $\small{\vec{R}}$
But how do we find those components analytically?
• We have already seen the method in the previous section. We will write it again:
(i) If $\theta_A$ is the angle made by $\small{\vec{A}}$ with the x axis, then:
• $\small{\vec{A_x}}$ = $\small{\left [\left | \vec{A} \right |\, cos\, \theta_A \right ]\hat{i}}$
♦ That means: magnitude of $\small{\vec{A_x}}$ is: $\small{\left [\left | \vec{A} \right |\, cos\, \theta_A \right ]}$
♦ Direction of $\small{\vec{A_x}}$ is same as the direction of the unit vector $\small{\hat{i}}$
• $\small{\vec{A_y}}$ = $\small{\left [\left | \vec{A} \right |\, sin\, \theta_A \right ]\hat{j}}$
♦ That means: magnitude of $\small{\vec{A_y}}$ is: $\small{\left [\left | \vec{A} \right |\, sin\, \theta_A \right ]}$
♦ Direction of $\small{\vec{A_y}}$ is same as the direction of the unit vector $\small{\hat{j}}$
(ii) If $\theta_B$ is the angle made by $\small{\vec{B}}$ with the x axis, then:
• $\small{\vec{B_x}}$ = $\small{\left [\left | \vec{B} \right |\, cos\, \theta_B \right ]\hat{i}}$
♦ That means: magnitude of $\small{\vec{B_x}}$ is: $\small{\left [\left | \vec{B} \right |\, cos\, \theta_B \right ]}$
♦ Direction of $\small{\vec{B_x}}$ is same as the direction of the unit vector $\small{\hat{i}}$
• $\small{\vec{B_y}}$ = $\small{\left [\left | \vec{B} \right |\, sin\, \theta_B \right ]\hat{j}}$
♦ That means: magnitude of $\small{\vec{B_y}}$ is: $\small{\left [\left | \vec{B} \right |\, sin\, \theta_B \right ]}$
♦ Direction of $\small{\vec{B_y}}$ is same as the direction of the unit vector $\small{\hat{j}}$
♦ Direction of $\small{\vec{B_y}}$ is same as the direction of the unit vector $\small{\hat{j}}$
• In the above example, we simply added the like-components
• We may encounter problems in which we will have to do subtraction. One such example is given below:
1. Consider two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$ shown in fig.4.16(a) below:
• Let the rectangular components of $\small{\vec{A}}$ be $\small{\vec{A_x}}$ and $\small{\vec{A_y}}$ respectively. This is shown in fig.4.16(b)
• Let the rectangular components of $\small{\vec{B}}$ be $\small{\vec{B_x}}$ and $\small{\vec{B_y}}$ respectively
2. Grouping of like-components:
• Group the horizontal components together. That is., group $\small{\vec{A_x}}$ and $\small{\vec{B_x}}$ together. This is shown in fig.4.15(c)
• We may encounter problems in which we will have to do subtraction. One such example is given below:
1. Consider two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$ shown in fig.4.16(a) below:
 |
| Fig.4.16 |
• Let the rectangular components of $\small{\vec{B}}$ be $\small{\vec{B_x}}$ and $\small{\vec{B_y}}$ respectively
2. Grouping of like-components:
• Group the horizontal components together. That is., group $\small{\vec{A_x}}$ and $\small{\vec{B_x}}$ together. This is shown in fig.4.15(c)
• Group the vertical components together. That is., group $\small{\vec{A_y}}$ and $\small{\vec{B_y}}$ together
3. Add the like-components:
• Add the two horizontal components $\small{\vec{A_x}}$ and $\small{\vec{B_x}}$.
♦ For that, shift $\small{\vec{B_x}}$ so that it's tail coincide with the tip of $\small{\vec{A_x}}$
♦ We thus get a new horizonatal vector: ($\small{\vec{A_x}}$ + $\small{\vec{B_x}}$)
♦ Note that, in the fig.d, $\small{\vec{B_x}}$ will come above $\small{\vec{A_x}}$. It is shown slightly out of position. This is for clarity only.
♦ From fig.d, we see that, ($\small{\vec{A_x}}$ + $\small{\vec{B_x}}$) is in effect, a subtraction
• Add the two vertical components $\small{\vec{A_y}}$ and $\small{\vec{B_y}}$.
♦ For that, shift $\small{\vec{B_y}}$ so that it's tail coincide with the tip of $\small{\vec{A_y}}$
♦ We thus get a new vertical vector: ($\small{\vec{A_y}}$ + $\small{\vec{B_y}}$)
4. Add the two new vectors by the triangle method.
• Let the resultant of this vector addition be $\small{\vec{R}}$. This is shown in fig.d
• We can see that: $\small{\vec{R}}$ = [($\small{\vec{A_x}}$ + $\small{\vec{B_x}}$)] + [($\small{\vec{A_y}}$ + $\small{\vec{B_y}}$)]
5. Addition of the original vectors by triangle method. This is shown in fig.e
• We find that, the resultant in fig.e is same as the resultant in fig.d
■ The above points will be clear when we see a solved example. Link to the pdf file is given below:
3. Add the like-components:
• Add the two horizontal components $\small{\vec{A_x}}$ and $\small{\vec{B_x}}$.
♦ For that, shift $\small{\vec{B_x}}$ so that it's tail coincide with the tip of $\small{\vec{A_x}}$
♦ We thus get a new horizonatal vector: ($\small{\vec{A_x}}$ + $\small{\vec{B_x}}$)
♦ Note that, in the fig.d, $\small{\vec{B_x}}$ will come above $\small{\vec{A_x}}$. It is shown slightly out of position. This is for clarity only.
♦ From fig.d, we see that, ($\small{\vec{A_x}}$ + $\small{\vec{B_x}}$) is in effect, a subtraction
• Add the two vertical components $\small{\vec{A_y}}$ and $\small{\vec{B_y}}$.
♦ For that, shift $\small{\vec{B_y}}$ so that it's tail coincide with the tip of $\small{\vec{A_y}}$
♦ We thus get a new vertical vector: ($\small{\vec{A_y}}$ + $\small{\vec{B_y}}$)
4. Add the two new vectors by the triangle method.
• Let the resultant of this vector addition be $\small{\vec{R}}$. This is shown in fig.d
• We can see that: $\small{\vec{R}}$ = [($\small{\vec{A_x}}$ + $\small{\vec{B_x}}$)] + [($\small{\vec{A_y}}$ + $\small{\vec{B_y}}$)]
5. Addition of the original vectors by triangle method. This is shown in fig.e
• We find that, the resultant in fig.e is same as the resultant in fig.d
Based on the above discussions, we can write a general form for finding the resultant of two vectors.
We will write it in steps:
1. Given two vectors:
(i) $\small{\vec{A}=\left | \vec{A_x} \right |\hat{i}+\left | \vec{A_y} \right |\hat{j}}$
(ii) $\small{\vec{B}=\left | \vec{B_x} \right |\hat{i}+\left | \vec{B_y} \right |\hat{j}}$
• We want ($\small{\vec{A}}$ + $\small{\vec{B}}$)
2. Add like-components to form new vectors:
(i) New vector in the x direction is: $\small{\left | \vec{A_x} \right |\hat{i}+\left | \vec{B_x}\right |\hat{i}}$• This can be written as: $\small{\left ( \left | \vec{A_x} \right |+\left | \vec{B_x} \right | \right )\hat{i}}$
(ii) New vector in the y direction is: $\small{\left | \vec{A_y} \right |\hat{j}+\left | \vec{B_y}\right |\hat{j}}$
• This can be written as: $\small{\left ( \left | \vec{A_y} \right |+\left | \vec{B_y} \right | \right )\hat{j}}$
3. The vector in 2(i) is the x component of ($\small{\vec{A}}$ + $\small{\vec{B}}$)
The vector in 2(ii) is the y component of ($\small{\vec{A}}$ + $\small{\vec{B}}$)
■ So we can write:
■ So we can write:
$\small{\left ( \vec{A}+\vec{B} \right )=\left ( \left | \vec{A_x} \right |+\left | \vec{B_x} \right | \right )\hat{i}+\left ( \left | \vec{A_y} \right |+\left | \vec{B_y} \right | \right )\hat{j}}$
$\theta_R$ = $\tan^{-1}\frac{73.87}{20.28}$ = 74.61o
■ So the resultant has a magnitude of 76.60 units. It makes an angle of 74.61o with the x axis
An example:
Find the resultant of two vectors given below:
(i) -17 $\small{\hat{i}}$ + 29.44 $\small{\hat{j}}$
(ii) 37.28 $\small{\hat{i}}$ + 44.43 $\small{\hat{j}}$
Solution:
• Add like components to form new vectors:
♦ The new vector in the x direction is: (-17+37.28)$\small{\hat{i}}$ = 20.28$\small{\hat{i}}$
♦ The new vector in the y direction is: (29.44+44.43)$\small{\hat{j}}$ = 73.87$\small{\hat{j}}$
• So the resultant vector is: 20.28 $\small{\hat{i}}$ + 73.87$\small{\hat{j}}$. See fig.4.17 below.
• Note that, fig.4.17 shows rough sketches. For analytical method, we do not need accurate drawings
■ Magnitude of the resultant:
• Note that, fig.4.17 shows rough sketches. For analytical method, we do not need accurate drawings
 |
| Fig.4.17 |
|$\small{\vec{R}}$| = √[20.282 + 73.872] = 76.60 units
■ Direction of the resultant:$\theta_R$ = $\tan^{-1}\frac{73.87}{20.28}$ = 74.61o
■ So the resultant has a magnitude of 76.60 units. It makes an angle of 74.61o with the x axis
In the next section, we will see a formula to obtain resultant.
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