Chapter 5.10 - Equilibrium of a particle

In the previous section we saw conservation of momentum. In this section we will see equilibrium of a particle

We will write the steps:
1. ‘Equilibrium of a particle’ refers to that state of the particle in which net force on it is zero
2. Consider the following situation:
• A person points to a particle and says that, ‘it is in equilibrium’.
• If his statement is verified, we can say this: Net force acting on that particle is zero
3. This can mean any one of the following two:
(i) No force is acting on the particle at all
(ii) Forces are acting, but they cancel each other out
4. Further, it may be noted that, ‘zero net force’ need not imply ‘no motion’
• The particle can be in motion even when there is no net force.
• In such cases, the particle will be in ‘uniform motion’. We have seen such cases in previous sections.
■ So, a particle in ‘uniform motion’ can also be said to be in equilibrium.
5. If there are only two forces $\mathbf\small{\vec{F_1}\: and \: \vec{F_2}}$ acting on a particle which is in equilibrium, we can write:
■ The resultant of the two forces is a null vector
• That is., the vector sum of the two vectors is zero 
• That is., $\mathbf\small{\vec{F_1}+\vec{F_2}=0}$
• We have already seen how to calculate vector sum in a previous section
6. Similarly, if a particle is in equilibrium under the action of 3 forces, we can write:
$\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}=0}$
7. The above equation for 3 forces can be rearranged in 3 different ways:
(i) $\mathbf\small{\vec{F_1}+\vec{F_2}=-\vec{F_3}}$
(ii) $\mathbf\small{\vec{F_1}+\vec{F_3}=-\vec{F_2}}$
(iii) $\mathbf\small{\vec{F_2}+\vec{F_3}=-\vec{F_1}}$
This means that, we can group 'any two' together. The vector sum of those two, should be equal and opposite to the third
8. This can be explained using an example
• Consider fig.5.27(a) below:

Fig.5.27

• Three forces are acting on a particle. If the particle is in equilibrium, then: 
We can represent the forces by the sides of a triangle ABC. This is shown in fig.b
[If the particle is not in equilibrium, then we will not be able to form the triangle ABC]
• Also note that, strict order should be maintained while forming the triangle
    ♦ That is., the head of a vector should coincide with the tail of the next vector. 
    ♦ Then only we will be able to form the triangle.
    ♦ See triangle method of vector addition 
9. Note that, in fig.c, the resultant of $\mathbf\small{\vec{F_1}}$ and $\mathbf\small{\vec{F_2}}$ is obtained
    ♦ This resultant is exactly equal in magnitude but opposite in direction to $\mathbf\small{\vec{F_3}}$
    ♦ This is same as the analytical result in 7(i) above
• In fig.d, the resultant of $\mathbf\small{\vec{F_1}}$ and $\mathbf\small{\vec{F_3}}$ is obtained
    ♦ This resultant is exactly equal in magnitude but opposite in direction to $\mathbf\small{\vec{F_2}}$
    ♦ This is same as the analytical result in 7(ii) above
• In fig.e, the resultant of $\mathbf\small{\vec{F_2}}$ and $\mathbf\small{\vec{F_3}}$ is obtained
    ♦ This resultant is exactly equal in magnitude but opposite in direction to $\mathbf\small{\vec{F_1}}$
    ♦ This is same as the analytical result in 7(iii) above

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So how do we put the above information to practical use?

We will write the steps:
1. Consider the following scenario. See fig.5.28(a) below:

Fig.5.28

• We see a particle in equilibrium
    ♦ Three forces are acting on it
    ♦ We know the details (magnitude and direction) about 2 forces
    ♦ We want the third force.
2. We can use either one of the two methods given below:
Method 1: Analytical method
Simply use the relation: $\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}=0}$
• We have already seen how to calculate vector sum in a previous section.
Here we will write it again:
• For the above vector sum to be equal to zero, the following two conditions must be satisfied:
(i) $\mathbf\small{\vec{F_{1x}}+\vec{F_{2x}}+\vec{F_{3x}}=0}$ 
(ii) $\mathbf\small{\vec{F_{1y}}+\vec{F_{2y}}+\vec{F_{3y}}=0}$
• In (i), the only unknown is: $\mathbf\small{\vec{F_{3x}}}$
    ♦ So it can be easily calculated
• In (ii), the only unknown is: $\mathbf\small{\vec{F_{3y}}}$
    ♦ So it can be easily calculated
• The resultant of $\mathbf\small{\vec{F_{3x}}}$ and $\mathbf\small{\vec{F_{3y}}}$ will give the required $\mathbf\small{\vec{F_{3}}}$

Method 2: Graphical method
(i) Draw the two known force vectors on a piece of paper. This is shown in fig.5.28(b) above
• Head of one should coincide with the tail of the other
(Method of drawing vectors can be seen here) 
• This will give two sides of a triangle
(ii) Once two sides are obtained, the third side can be easily drawn
This is shown in red color in fig.c
• This third side is called the 'closing side' 
    ♦ The length of the closing side will give the magnitude of the unknown force
    ♦ The direction of the closing side will give the direction of the unknown force.  
(iii) The red vector represent the required force
• We can copy it 'as such' and place it in fig.a. The result is shown in fig.d
• The particle in fig.d is in equilibrium under the action of the three forces
3. In the above example, three forces were acting. The third force was unknown. We successfully calculated it. Let us see another case related to the same example: 
• Two forces $\mathbf\small{\vec{F_{1}}}$ and $\mathbf\small{\vec{F_{2}}}$ are acting on the particle. See fig.5.29(a) below:

Fig.5.29

• The particle is not in equilibrium
• We want that force which will enable us to keep the particle in equilibrium
4. We can use either one of the two methods given below:
Method 1: Analytical method
• Let $\mathbf\small{\vec{R}}$ be the resultant of $\mathbf\small{\vec{F_{1}}}$ and $\mathbf\small{\vec{F_{2}}}$
• If we calculate this $\mathbf\small{\vec{R}}$ and apply it in it's reverse direction, the particle will obtain equilibrium  
• So, for finding $\mathbf\small{\vec{R}}$, we simply use the relation: $\mathbf\small{\vec{F_1}+\vec{F_2}-\vec{R}=0}$
• We have already seen how to calculate vector sum in a previous section.
Here we will write it again:
• For the above vector sum to be equal to zero, the following two conditions must be satisfied:
(i) $\mathbf\small{\vec{F_{1x}}+\vec{F_{2x}}-\vec{R_{x}}=0}$ 
(ii) $\mathbf\small{\vec{F_{1y}}+\vec{F_{2y}}-\vec{R_{y}}=0}$
• In (i), the only unknown is: $\mathbf\small{\vec{R_{x}}}$
    ♦ So it can be easily calculated
• In (ii), the only unknown is: $\mathbf\small{\vec{R_{y}}}$
    ♦ So it can be easily calculated
• The resultant of $\mathbf\small{\vec{R_{x}}}$ and $\mathbf\small{\vec{R_{y}}}$ will give the required $\mathbf\small{\vec{R}}$

Method 2: Graphical method
(i) Draw the two known force vectors on a piece of paper. This is shown in fig.5.29(b) above
• Head of one should coincide with the tail of the other
(Method of drawing vectors can be seen here) 
• This will give two sides of a triangle
(ii) Once two sides are obtained, the third side (the closing side) can be easily drawn
• But this time, draw the closing side in the reverse direction 
• This is shown in orange color in fig.c
    ♦ The length of the orange vector will give the magnitude of the resultant
    ♦ The direction of the orange vector will give the direction of the resultant
(iii) The orange vector represents the resultant
• We can reverse it and place it in fig.a. The result is shown in fig.d
• The particle in fig.d is in equilibrium under the action of the three forces
■ Comparing figs.5.28 and 5.29, we find that: $\mathbf\small{\vec{F_{3}}=-\vec{R}}$

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Another example:
1. Consider the following scenario. See fig.5.30(a) below:

Fig.5.30

• We see a particle in equilibrium
    ♦ Four forces are acting on it
    ♦ We know the details (magnitude and direction) about 3 forces
    ♦ We want the fourth force
2. We can use either one of the two methods given below:
Method 1: Analytical method
Simply use the relation: $\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}+\vec{F_4}=0}$
• We have already seen how to calculate vector sum in a previous section.
Here we will write it again:
• For the above vector sum to be equal to zero, the following two conditions must be satisfied:
(i) $\mathbf\small{\vec{F_{1x}}+\vec{F_{2x}}+\vec{F_{3x}}+\vec{F_{4x}}=0}$
(ii) $\mathbf\small{\vec{F_{1y}}+\vec{F_{2y}}+\vec{F_{3y}}+\vec{F_{4y}}=0}$
• In (i), the only unknown is: $\mathbf\small{\vec{F_{4x}}}$
    ♦ So it can be easily calculated
• In (ii), the only unknown is: $\mathbf\small{\vec{F_{4y}}}$
    ♦ So it can be easily calculated
• The resultant of $\mathbf\small{\vec{F_{4x}}}$ and $\mathbf\small{\vec{F_{4y}}}$ will give the required $\mathbf\small{\vec{F_{4}}}$

Method 2: Graphical method
(i) Draw the three known force vectors on a piece of paper. This is shown in fig.5.30(b) above
• Head of one should coincide with the tail of the other
(Method of drawing vectors can be seen here) 
• This will give three sides of a quadrilateral
(ii) Once three sides are obtained, the fourth side can be easily drawn
This is shown in red color in fig.c
• This fourth side is called the 'closing side' 
    ♦ The length of the closing side will give the magnitude of the unknown force
    ♦ The direction of the closing side will give the direction of the unknown force.  
(iii) The red vector represent the required force
• We can copy it 'as such' and place it in fig.a. The result is shown in fig.d
• The particle in fig.d is in equilibrium under the action of the four forces
3. In the above example, four forces were acting. The fourth force was unknown. We successfully calculated it. Let us see another case related to the same example: 
• Three forces $\mathbf\small{\vec{F_{1}}}$,  $\mathbf\small{\vec{F_{2}}}$  and $\mathbf\small{\vec{F_{3}}}$ are acting on the particle. See fig.5.31(a) below:

Fig.5.31

• The particle is not in equilibrium
• We want that force which will enable us to keep the particle in equilibrium
4. We can use either one of the two methods given below:
Method 1: Analytical method
• Let $\mathbf\small{\vec{R}}$ be the resultant of $\mathbf\small{\vec{F_{1}}}$,  $\mathbf\small{\vec{F_{2}}}$  and $\mathbf\small{\vec{F_{3}}}$
• If we calculate this $\mathbf\small{\vec{R}}$ and apply it in it's reverse direction, the particle will obtain equilibrium  
• So for finding $\mathbf\small{\vec{R}}$, we simply use the relation: $\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}-\vec{R}=0}$
• We have already seen how to calculate vector sum in a previous section.
Here we will write it again:
• For the above vector sum to be equal to zero, the following two conditions must be satisfied:
(i) $\mathbf\small{\vec{F_{1x}}+\vec{F_{2x}}+\vec{F_{3x}}-\vec{R_{x}}=0}$ 
(ii) $\mathbf\small{\vec{F_{1y}}+\vec{F_{2y}}+\vec{F_{3y}}-\vec{R_{y}}=0}$
• In (i), the only unknown is: $\mathbf\small{\vec{R_{x}}}$
    ♦ So it can be easily calculated
• In (ii), the only unknown is: $\mathbf\small{\vec{R_{y}}}$
    ♦ So it can be easily calculated
• The resultant of $\mathbf\small{\vec{R_{x}}}$ and $\mathbf\small{\vec{R_{y}}}$ will give the required $\mathbf\small{\vec{R}}$

Method 2: Graphical method
(i) Draw the three known force vectors on a piece of paper. This is shown in fig.5.31(b) above
• Head of one should coincide with the tail of the other
(Method of drawing vectors can be seen here) 
• This will give three sides of a quadrilateral
(ii) Once three sides are obtained, the fourth side (the closing side) can be easily drawn
• But this time, draw the closing side in the reverse direction 
• This is shown in orange color in fig.c
    ♦ The length of the orange vector will give the magnitude of the resultant
    ♦ The direction of the orange vector will give the direction of the resultant
(iii) The orange vector represents the resultant
• We can reverse it and place it in fig.a. The result is shown in fig.d
• The particle in fig.d is in equilibrium under the action of the four forces
■ Comparing figs.5.30 and 5.31, we find that: $\mathbf\small{\vec{F_{4}}=-\vec{R}}$

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• We saw the case when 3 forces are acting on a particle
• We saw the case when 4 forces are acting on a particle
■ Now we can write the general case:
A. To find the unknown force when n forces are acting
Method 1: Analytical method
Use the relation: $\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}+ .\: .\: .+\vec{F_n}=0}$
Method 2: Graphical method
Draw the n-sided polygon. The closing side will give the unknown force
B. To find the resultant when (n-1) forces are acting
Method 1: Analytical method
Use the relation: $\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}+ .\: .\: .+\vec{F_{(n-1)}}-\vec{R}=0}$
Method 2: Graphical method
Draw the n-sided polygon. The 'reverse of the closing side' will give the resultant force
■ This graphical method is known as: Polygon law of vector addition 
It states that, if a number of vectors can be represented in magnitude and direction by the sides of a polygon taken in the same order, then their resultant is represented in magnitude and direction by the closing side of the polygon taken in the opposite order.

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Solved example 5.14

Three forces are acting on a body as shown in fig.5.32(a) below. If the body is in equilibrium, find the magnitude of $\mathbf\small{\vec{F_2}\: \: \text{and}\,\,\vec{F_3}}$

Fig.5.32

Solution:
1. Since the body is in equilibrium, we have:
$\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}=0}$
• For the above vector sum to be equal to zero, the following two conditions must be satisfied:
(i) $\mathbf\small{\vec{F_{1x}}+\vec{F_{2x}}+\vec{F_{3x}}=0}$ 
(ii) $\mathbf\small{\vec{F_{1y}}+\vec{F_{2y}}+\vec{F_{3y}}=0}$
■ We will use the usual sign convention:
• Forces towards right are positive
• Forces towards left are negative
• Forces in the upward direction are positive
• Forces in the downward direction are negative
2. Substituting the values, we get:
(i) $\mathbf\small{0-(|\vec{F_2}|\cos 60)\hat{i}+(|\vec{F_3}|\cos 30)\hat{i}=0}$
$\mathbf\small{\Rightarrow |\vec{F_2}|=\sqrt{3}\left [|\vec{F_3}|\right ]}$ 
(ii) $\mathbf\small{-100+(|\vec{F_2}|\sin 60)\hat{j}+(|\vec{F_3}|\sin 30)\hat{j}=0}$
$\mathbf\small{\Rightarrow \sqrt{3}\left [|\vec{F_2}|\right ]+|\vec{F_3}|=200}$
3. Solving 2(i) and 2(ii), we get:
$\mathbf\small{|\vec{F_2}|=50\sqrt{3}\: N \,\,\text{and}\,\,|\vec{F_3}|=50\: N}$

Solved example 5.15
Three forces acting on a body are shown in the fig.5.32(b) above. What is the minimum additional force to be applied so that, the resultant of the four forces will be along the vertical direction only?
Solution:
■ We will use the usual sign convention:
• Forces towards right are positive
• Forces towards left are negative
• Forces in the upward direction are positive

• Forces in the downward direction are negative
1. Vector sum of the horizontal components of the 3 forces already present:
$\mathbf\small{\vec{F_{1x}}+\vec{F_{2x}}+\vec{F_{3x}}}$
= $\mathbf\small{-(4 \times \cos 60)\hat{i}+(2 \times \cos 60)\hat{i}+(1 \times \cos 60)\hat{i}}$
= $\mathbf\small{-0.5 \hat{i}}$
2. So the available resultant in the horizontal direction is $\mathbf\small{-0.5 \hat{i}}$
That is., in the horizontal direction, a force of magnitude 0.5 N is acting towards the left
3. If we apply an additional force of magnitude 0.5 N in the horizontal direction, towards the right, then there will be no net force in the horizontal direction
• The resultant of the 4 forces will then be in the vertical direction only

Solved example 5.16
Two forces of equal magnitude are acting on an object. The angle between them is 60^o. If the magnitude of their resultant is 40√3 N, What is the magnitude of each force?
Solution:
■ We will use the usual sign convention:
• Forces towards right are positive
• Forces towards left are negative
• Forces in the upward direction are positive

• Forces in the downward direction are negative
1. Let the magnitude of each force be 'p' N
• Also let one of the forces be towards the positive side of the x axis 
• This is shown in fig.5.32(c) above
2. We have: 
$\mathbf\small{\vec{R_x}=p\: \hat{i}+(p \times \cos 60)\hat{i}=\frac{3p}{2}\: \hat{i}}$
$\mathbf\small{\vec{R_y}=0+-(p \times \sin 60)\hat{j}=-\frac{p\sqrt{3}}{2}\: \hat{j}}$ 
3. We have:
• Magnitude of the resultant = $\mathbf\small{|\vec{R}|=\sqrt{|\vec{R_x}|^2+|\vec{R_y}|^2}}$
• Substituting the values, we get: 
$\mathbf\small{|\vec{R}|=\sqrt{\left (\frac{3p}{2} \right )^2+\left (\frac{p\sqrt{3}}{2} \right )^2}=\sqrt{3p^2}=p\sqrt{3}}$
4. But given that, magnitude of the resultant is 40√3 N
• So we can write: 40√3 = p√3
• Thus we get: Magnitude of each force = p = 40 N    

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In the next section we will see some common forces in mechanics

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Chapter 4.17 - River crossing Problems

In our present chapter, we are discussing about motions in two dimensions. In the previous section we completed a discussion on relative motion in two dimensions. In this section, we will see problems in river crossing.

Case 1:
1. Consider a man swimming across a river
• The velocity of the man is $\mathbf\small{\vec v_M}$
• The velocity of the river current is $\mathbf\small{\vec v_R}$
2. The $\mathbf\small{\vec v_R}$ is in a direction perpendicular to $\mathbf\small{\vec v_M}$
• So the man will not be able to swim straight ahead. He will be deviated. 
• This is shown in fig.4.45(a) below:

Fig.4.45

3. The new direction will be the direction of $\mathbf\small{\vec v}$, which is the resultant of $\mathbf\small{\vec v_M}$ and $\mathbf\small{\vec v_R}$
• So the actual direction of travel is along $\mathbf\small{\vec v}$.
4. Not only the direction, but the 'magnitude of the velocity' with which the travel is made also changes.
• We expect the man to swim with a speed of $\mathbf\small{|\vec v_M|}$
• But the actual speed is $\mathbf\small{|\vec v|}$.
5. Let us apply the above results to an actual case. It is shown in fig.4.45(b)
• The yellow lines are the river banks. The width of the river is w
6. The man starts to swim from A. He wants to reach B, which is directely opposite A
• But due to the current, he will be travelling along $\mathbf\small{\vec v}$.
• Because of this change in direction, he will reach B'
■ The distance BB' is called drift. It is denoted by 'x'
7. If we know the details about $\mathbf\small{\vec v_M}$ and $\mathbf\small{\vec v_R}$, we can easily calculate the details of $\mathbf\small{\vec v}$
• That is., we can find the magnitude $\mathbf\small{|\vec v|}$ and direction θ. We have
(i) $\mathbf\small{|\vec v|=\sqrt{|\vec{v_M}|^2+|\vec{v_R}|^2}}$
(ii) $\mathbf\small{\theta =\tan^{-1}\frac{|\vec{v_M}|}{|\vec{v_R}|}}$
8. Once we find θ, we will be able to find some more useful details. Let us see what they are:
(a) The drift x
(i) Consider fig.b. A perpendicular B'C is dropped from B' to the other bank
(ii) In the right triangle ACB', we have:
$\mathbf\small{\tan \theta =\frac{B'C}{AC}=\frac{w}{x}}$
(iii) So we get Eq.4.34: $\mathbf\small{x=\frac{w}{\tan \theta}}$
(iv) But tan θ is equal to $\mathbf\small{\frac{|\vec{v_M}|}{|\vec{v_R}|}}$ also. So we can write:
$\mathbf\small{\tan \theta =\frac{B'C}{AC}=\frac{w}{x}=\frac{|\vec{v_M}|}{|\vec{v_R}|}}$
From this we get Eq.4.35: $\mathbf\small{x = \frac{w \times |\vec{v_R}|}{|\vec{v_M}|}}$
(b) The actual distance travelled:
(i) This is equal to AB'
(ii) Clearly, it is given by: $\mathbf\small{AB'=\sqrt{x^2+w^2}}$
(c) Time of travel T
(i) The travel is with a uniform speed of $\mathbf\small{|\vec v|}$ 
(ii) The distance covered is AB'
(iii) So time of travel $\mathbf\small{T=\frac{AB'}{|\vec{v}|}}$
(iv) Thus we get Eq.4.36: $\mathbf\small{T=\frac{\sqrt{x^2+w^2}}{\sqrt{|\vec{v_M}|^2+|\vec{v_R}|^2}}}$
(d) Another formula for 'x':
(i) Consider the horizontal travel alone
• The horizontal travel is with a uniform speed of $\mathbf\small{|\vec v_R|}$ 
• This travel has a duration of T
(ii) So we can write an equation for horizontal distance travelled as: 
Eq.4.37: x = $\mathbf\small{|\vec v_R| \times T}$

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Now let us see another case:
Case 2:
1. Consider a man swimming across a river
• The velocity of the river current is $\mathbf\small{\vec v_R}$
• The velocity of the man is $\mathbf\small{\vec v_M}$
2. This time, the man aims to a point on the upstream.
• That is., $\mathbf\small{\vec v_R}$ is not directed towards the exact opposite point. It is directed towards a point on the upstream
• Due to the river current $\mathbf\small{\vec v_R}$, he will be deviated from his intended path
■ If the direction of $\mathbf\small{\vec v_M}$ is adjusted carefully, he can achieve an interesting result:
• The resulting $\mathbf\small{\vec v}$ will be directed towards the exact opposite point on the other bank
    ♦ Thus he will reach the exact opposite point on the other bank
This is shown in fig.4.46(a) below:

Fig.4.46

3. The new direction will be the direction of $\mathbf\small{\vec v}$, which is the resultant of $\mathbf\small{\vec v_M}$ and $\mathbf\small{\vec v_R}$
• So the actual direction of travel is along $\mathbf\small{\vec v}$.
4. Not only the direction, but the 'magnitude of the velocity' with which the travel is made also changes.
• We expect the man to swim with a speed of $\mathbf\small{|\vec v_M|}$
• But the actual speed is $\mathbf\small{|\vec v|}$
5. Let us apply the above results to an actual case. It is shown in fig.4.46(b)
• The yellow lines are the river banks. The width of the river is w.
6. The man starts to swim from A. He wants to reach B, which is directely opposite A
■ If he swim directely towards B, he will be carried away frm B due to the current
• So he aims for a point on the upstream
• We want to know the angle θ which will enable him to reach B
7. For obtaining θ, we can use the following steps:
$\mathbf\small{\vec v}$ is the resultant of $\mathbf\small{\vec v_M}$ and $\mathbf\small{\vec v_R}$
Let us write the steps for obtaining $\mathbf\small{\vec v}$:
(i) Considering horizontal components:
$\mathbf\small{\vec v_x=[\vec v_{Mx}+\vec v_{Rx}]=[-(|\vec v_{M}|\cos \theta )\hat{i}+|\vec v_{R}|\hat{i}]=[-(|\vec v_{M}|\cos \theta )+|\vec v_{R}|]\hat{i}}$
• The negative sign is due to the fact that, the horizontal component of $\mathbf\small{\vec v_M}$ is directed towards the left
(ii) Considering vertical components:
$\mathbf\small{\vec v_y=[\vec v_{My}+\vec v_{Ry}]=[(|\vec v_{M}|\sin \theta )\hat{j}+0]=[(|\vec v_{M}|\sin \theta )]\hat{j}}$
• The zero value comes in because, $\mathbf\small{\vec v_R}$ has no vertical component
8. Now, $\mathbf\small{\vec v=0}$ does not have horizontal component. That means: $\mathbf\small{\vec v_x=0}$
So we can equate 7(i) to zero. We get:
$\mathbf\small{[-(|\vec v_{M}|\cos \theta )+|\vec v_{R}|]\hat{i}=0}$
$\mathbf\small{\Rightarrow -(|\vec v_{M}|\cos \theta )+|\vec v_{R}|=0}$
$\mathbf\small{\Rightarrow |\vec v_{R}|=|\vec v_{M}|\cos \theta}$
$\mathbf\small{\Rightarrow \cos \theta=\frac{|\vec v_{R}|}{|\vec v_{M}|}}$
So we get Eq.4.38: $\mathbf\small{\theta=\cos^{-1}\frac{|\vec v_{R}|}{|\vec v_{M}|}}$
■ Thus we successfully calculated θ
9. Let us continue and find the magnitude of $\mathbf\small{\vec v}$ also:
(i) The [vertical component of $\mathbf\small{\vec v}$] is $\mathbf\small{\vec v}$ itself. That means: $\mathbf\small{\vec v_y=\vec v}$
(ii) So we can equate 7(ii) to $\mathbf\small{\vec v}$. We get:
$\mathbf\small{\vec v=[(|\vec v_{M}|\sin \theta )]\hat{j}}$
$\mathbf\small{\Rightarrow |\vec v|=|\vec v_{M}|\sin \theta}$
10. So, if we have 'sin θ', we can multiply it with $\mathbf\small{|\vec v_M|}$ to obtain $\mathbf\small{|\vec v|}$
(i) We have already obtained 'cos θ' in step 8. From that, we can easily calculate 'sin θ'
(ii) From math classes we know that $\mathbf\small{\sin \theta = \sqrt{1-\cos^2\theta }}$
So we get: $\mathbf\small{\sin \theta = \sqrt{1-\left( \frac{|\vec v_{R}|}{|\vec v_{M}|} \right )^2}}$
$\mathbf\small{\Rightarrow \sin \theta = \frac{\sqrt{(|\vec{v_M}|^2-|\vec{v_R}|^2})}{|\vec{v_M}|}}$
(iii) Thus, from the result in (9), we get:
$\mathbf\small{|\vec{v}| = |\vec{v_M}|\times \frac{\sqrt{|\vec{v_M}|^2-|\vec{v_R}|^2}}{|\vec{v_M}|}}$
■ So we can write Eq.4.39: $\mathbf\small{|\vec{v}| =\sqrt{|\vec{v_M}|^2-|\vec{v_R}|^2}}$
11. Once we obtain $\mathbf\small{|\vec v|}$, we can calculate the time of travel
(i) The travel is with a uniform speed of $\mathbf\small{|\vec v|}$ 
(ii) The distance covered is AB = w
(iii) So time of travel $\mathbf\small{T=\frac{AB}{|\vec{v}|}}$
(iv) Thus we get Eq.4.40: $\mathbf\small{T=\frac{w}{\sqrt{|\vec{v_M}|^2-|\vec{v_R}|^2}}}$

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Now we will see some solved examples
Solved example 4.19
A man can row a boat at a speed of 4 km/h in still water. He is crossing a river where the speed of current is 2 km/h
(a) In what direction should he be headed if he wants to reach a point directly opposite to his starting point?
(b) If the width of the river is 4 km, how long will it take to reach the opposite bank if he heads in the direction derived in (a)?
(c) In what direction should he be headed if he wants to reach the opposite bank in the shortest possible time? How much is this shortest time?
Solution:
Part (a)
1. This comes under case 2 that we saw above
• We can use Eq.4.38: $\mathbf\small{\theta=\cos^{-1}\frac{|\vec v_{R}|}{|\vec v_{M}|}}$
2. Substituting the values, we get: $\mathbf\small{\theta=\cos^{-1}\frac{2}{4}}$ 
Thus θ = cos^-1 (0.5) = 60°.
■ So the man should row the boat in such a way that, his direction makes an angle of 60° with the bank on his left side
Part (b)
1. We can use Eq.4.40: $\mathbf\small{T=\frac{w}{\sqrt{|\vec{v_M}|^2-|\vec{v_R}|^2}}}$
2. Substituting the values, we get:  $\mathbf\small{T=\frac{4}{\sqrt{4^2-2^2}}=\frac{4}{\sqrt{12}}=\frac{2}{\sqrt{3}}=1.155\: \text{h}}$
Part (c)
1. The shortest possible time is achieved when the direction is headed exactly to the opposite point   
• So this is case 1
2. We can use Eq.4.36: $\mathbf\small{T=\frac{\sqrt{x^2+w^2}}{\sqrt{|\vec{v_M}|^2+|\vec{v_R}|^2}}}$
3. But, first we have to calculate the drift 'x'. We can use Eq.4.35: $\mathbf\small{x = \frac{w \times |\vec{v_R}|}{|\vec{v_M}|}}$
• Substituting the values, we get: $\mathbf\small{x = \frac{4 \times 2}{4}=2\: \text{km}}$
4. Substituting in (2), we get: $\mathbf\small{T=\frac{\sqrt{2^2+4^2}}{\sqrt{4^2+2^2}}=\frac{\sqrt{20}}{\sqrt{20}}=1\: \text{h}}$
■ Note: To achieve the least possible time, the direction should be headed to the exact opposite point. Any other direction will take up a longer time.

Solved example 4.20
A man crosses a river in a boat. If he choose to cross within the least possible time, he can do so in 10 minutes. But there will be a drift of 120 m. If he choose to cross with the least possible distance, he can do so in 12.5 minutes. Find (a) Width of the river  (b) Speed of the boat (c) Speed of the river current
Solution:
1. Consider the 'crossing in least possible time'. This is case 1
• We can use Eq.4.37: drift = x = $\mathbf\small{|\vec v_R| \times T}$
2. Substituting the values, we get: 120 = $\mathbf\small{|\vec v_R| \times 10}$   
• So we get speed of the river current = $\mathbf\small{|\vec v_R|}$ = 12 m/min
3. Now we can use Eq.4.35: $\mathbf\small{x = \frac{w \times |\vec{v_R}|}{|\vec{v_M}|}}$
Substituting the values, we get: $\mathbf\small{120 = \frac{w \times 12}{|\vec{v_M}|}}$ 
So we get: $\mathbf\small{\frac{w}{|\vec{v_M}|}=10}$
4. Now we consider 'crossing in least possible distance'. This is case 2
• We can use Eq.4.40: $\mathbf\small{T=\frac{w}{\sqrt{|\vec{v_M}|^2-|\vec{v_R}|^2}}}$
• Substituting the values, we get: $\mathbf\small{12.5=\frac{10\times|\vec{v_M}|}{\sqrt{|\vec{v_M}|^2-12^2}}}$
• Squaring both sides: $\mathbf\small{12.5^2 \times\left ( |\vec{v_M}|^2-12^2 \right )=10^2 \times |\vec{v_M}|^2}$
$\mathbf\small{\Rightarrow (12.5^2-10^2)\times |\vec{v_M}|^2=(12.5 \times 12)^2}$
• Solving this, we get $\mathbf\small{|\vec{v_M}|}$ = 20 m/min
5. Substituting this value in (3), we get: w = 200 m

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With this we complete our present discussion on two dimensional motion. In the next chapter, we will see laws of motion.

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Chapter 4.5 - Addition and Subtraction of Vectors using Formula

In the previous section we saw an analytical method to find the resultant vector. In this section, we will see another analytical method which uses a formula.
The steps are given below:

1. Consider two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$ shown in fig.4.18(a) below:

Fig.4.18

2. Shift them so that, their tails coincide at a point O. This is shown in fig.(b) 
    ♦ Let the tip of $\small{\vec{A}}$ be P
    ♦ Let the tip of $\small{\vec{B}}$ be Q
• Draw a green line parallel to OP
• Draw a yellow line parallel to OQ
• The green and yellow lines meet at S
3. Place a vector between O and S. 
• We know that this vector between O and S is $\small{\vec{R}}$, which is the resultant of $\small{\vec{A}}$ and $\small{\vec{B}}$ (Details in a the previous section)
• The length of OS will give the magnitude of $\small{\vec{R}}$
4. So our first aim is to find the length OS
• For that, extend OP 
• Drop a perpendicular SN on to that extension
5. Now we have a new right triangle: ⊿ONS
• The base ON of this triangle = (OP + PN)
    ♦ Instead of OP, we can write $\left | \vec{A} \right |$

    ♦ Instead PN, we can write PS cos $\theta$

    ♦ But PS =  $\left | \vec{B} \right |$

    ♦ So instead of PN, we can write $\left | \vec{B} \right |$ cos $\theta$

• So the base ON = (OP+PN) = $\left | \vec{A} \right |$  +  $\left | \vec{B} \right |$ cos $\theta$

6. Now, altitude of⊿ONS = SN = PS sin $\theta$ =  $\left | \vec{B} \right |$ sin $\theta$

• Applying Pythagoras theorem, we get: 

OS² = ON² + SN²

$\small{\Rightarrow {OS}^2\, =\, \left ( \left | \vec{A} \right |\, +\, \left | \vec{B} \right |\, cos\, \theta  \right )^2\: +\: \left ( \left | \vec{B} \right |\, sin\, \theta  \right )^2}$
• The right side can be expanded and simplified:
$\small{\Rightarrow {OS}^2\, =\, \left ( \left | \vec{A} \right |^2\, +\,2\left | \vec{A} \right |\, \left | \vec{B} \right |\, cos\,\theta \, +\,   \left | \vec{B} \right |^2 \, {cos}^2\, \theta  \right )\: +\: \left ( \left | \vec{B} \right |^2\, {sin}^2\, \theta  \right )}$
• Group the last two terms together:
$\small{\Rightarrow {OS}^2\, =\, \left | \vec{A} \right |^2\, +\,2\left | \vec{A} \right |\, \left | \vec{B} \right |\, cos\,\theta \, +\,   \left ( \left | \vec{B} \right |^2 \, {cos}^2\, \theta  \: +\:  \left | \vec{B} \right |^2\, {sin}^2\, \theta  \right )}$
$\small{\Rightarrow\, {OS}^2\: =\: \left | \vec{A} \right |^2 \, +\, 2\left | \vec{A} \right | \left | \vec{B} \right |cos\, \theta \, +\, \left | \vec{B} \right |^2\, \left ( {cos}^2\, \theta \, +\, {sin}^2\, \theta  \right )}$
• But $\small{ \left ( {cos}^2\, \theta \, +\, {sin}^2\, \theta  \right ) = 1}$
• So we get:
$\small{{OS}^2\: =\: \left | \vec{A} \right |^2 \, +\, 2\left | \vec{A} \right | \left | \vec{B} \right |cos\, \theta \, +\, \left | \vec{B} \right |^2}$
Thus we can write:
Eq.4.4:
OS = $\small{\left | \vec{R} \right |=\left ( \left | \vec{A}+\vec{B} \right | \right )=\sqrt{\left | \vec{A} \right |^2 + \left | \vec{B} \right |^2 + 2\left | \vec{A} \right |\left | \vec{B} \right |cos \theta}}$
■ Thus we get the magnitude of $\small{\vec{R}}$
■ Eq.4.4 is known as the Law of cosines
7. Next we want the direction of $\small{\vec{R}}$
• For that, we consider 𝜟ONS
• We have: $tan\,  \alpha \, =\, \frac{SN}{ON}$ = $\frac{SN}{OP\, +\, PN}$
• We have already obtained the expressions for SN, OP and PN 
• Thus we get:
Eq.4.5:
$tan\,  \alpha \, =\, \frac{\left | \vec{B} \right |\, sin\, \theta }{\left | \vec{A} \right |\, +\,\left | \vec{B} \right |\, cos\, \theta }$
• From this expression, we get 'α', which is the angle made by $\small{\vec{R}}$ with one of the vectors $\small{\vec{A}}$
■ We may not be always able to denote the vectors as $\small{\vec{A}}$ and $\small{\vec{B}}$. So remember that, α in Eq.4.5 is the angle between $\small{\vec{R}}$ and the first term (the term without the cosine ratio) in the denominator 
8. Another method to find 'α':
We have two right triangles: ⊿ONS and ⊿PNS
• Consider ⊿ONS. We have:
SN = OS sin α = |$\small{\vec{R}}$| sin α
• Consider ⊿PNS. We have:
SN = SP sin θ = |$\small{\vec{B}}$| sin θ.
• Equating the two, we get: |$\small{\vec{R}}$| sin α = |$\small{\vec{B}}$| sin θ. 
So we can write: 
Eq.4.6:
$\large{\frac{\left | \vec{R} \right |}{sin\, \theta}}$ = $\large{\frac{\left | \vec{B} \right |}{sin\, \alpha}}$
• So, after calculating the magnitude of $\small{\vec{R}}$, we can use Eq.4.6 to calculate 'α'
9. Now we want the angle 'β', which $\small{\vec{R}}$ makes with the other vector $\small{\vec{B}}$
• For that, we drop a perpendicular from P onto OS. This is shown in fig.4.18(c)
• Now we have two new right triangles: ⊿OPM and ⊿SPM
■ Note that, in ⊿SPM, the angle at S is the same $\small{\beta}$ that we are trying to find. This is because, ∠QOS and ∠OSP are alternate angles 
10. Consider ⊿OPM. We have:
PM = OP sin α = |$\small{\vec{A}}$| sin α
• Consider ⊿SPM. We have:
PM = SP sin β = |$\small{\vec{B}}$| sin β.
• Equating the two, we get: |$\small{\vec{A}}$| sin α = |$\small{\vec{B}}$| sin β.
$\large{\Rightarrow \, \frac{\left | \vec{A} \right |}{sin\, \beta}}$ = $\large{\frac{\left | \vec{B} \right |}{sin\, \alpha}}$
• But Eq.4.6 above gives another expression for $\large{\frac{\left | \vec{B} \right |}{sin\, \alpha}}$
• So we can equate the three items. We get:
Eq.4.7:
$\large{\frac{\left | \vec{R} \right |}{sin\, \theta}}$ = $\large{\frac{\left | \vec{A} \right |}{sin\, \beta}}$ = $\large{\frac{\left | \vec{B} \right |}{sin\, \alpha}}$
■ Eq.4.7 gives the relation between the three vectors and the angles. It is known as the Law of sines.
• We can see a pattern:
(i) When $\small{\left | \vec{A} \right |}$ is in the numerator
    ♦ Angle 'β', which is related to $\small{\vec{B}}$ is in the denominator
(ii) When $\small{\left | \vec{B} \right |}$ is in the numerator
    ♦ Angle 'α', which is related to $\small{\vec{A}}$ is in the denominator
(iii) When $\small{\left | \vec{R} \right |}$ is in the numerator
    ♦ Angle 'θ', which is related to both $\small{\vec{A}}$ and $\small{\vec{B}}$ is in the denominator
• If we select the first two ratios from Eq.4.7, we will be using $\small{\left | \vec{R} \right |}$, $\small{\left | \vec{A} \right |}$ and θ to calculate β
• If we select the last two ratios from Eq.4.7, we will be using $\small{\left | \vec{A} \right |}$, $\small{\left | \vec{B} \right |}$ and α to calculate β
11. Special cases:
(i) Maximum magnitude for resultant
• Consider Eq.4.4 that we saw above. It gives the magnitude of the resultant. We want to know the 'condition for obtaining the maximum magnitude' 
• From our math classes, we know that, the range of cosine function varies from -1 to +1 both inclusive
• In most cases, the cosine value that we will be using in Eq.4.4 will be a fraction
• We know that, when a quantity is multiplied by a fraction, it's value will decrease
• The only 'non-fraction values' that cos θ can take are: -1 and +1
• So, maximum value of $\small{\left | \vec{R} \right |}$ will be obtained when cos θ = 1
• For cos θ to be 1, θ should be equal to zero
• That is., for maximum magnitude, the two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$ should have the same direction  
(ii) Minimum magnitude for resultant
• We want to know the 'condition for obtaining the least possible magnitude'
• When we add two vectors, we may get a resultant which has a negative direction
• But even if the direction is negative, the resultant will have a magnitude.
• So the least possible magnitude is zero. 
• That is., for the least possible magnitude, the two vectors should have the same magnitude but opposite directions
• Resultant of such a pair will have a zero magnitude
• Resultant of all other pairs will have a certain magnitude

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Now we will see a solved example.

We saw Solved example 4.2 in the previous section. We will do the same problem by the new method. The link to the new file is given below:

Solved example 4.3
We find that, the results obtained for $\small{\left | \vec{R} \right |}$, α, β, and $\small{\theta_R}$ are the same in both methods
• One more solved example is given below:
Solved example 4.4

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Now we will see the analytical method for vector subtraction
1. Consider two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$ shown in fig.4.19(a) below:

Fig.4.19

• We want to find ($\small{\vec{A}}$ - $\small{\vec{B}}$)
• That is., we want to find [$\small{\vec{A}}$ + (-$\small{\vec{B}}$)]
2. Draw a vector with the same magnitude of $\small{\vec{B}}$, but opposite in direction.
• This new vector is $\small{\vec{-B}}$. It is also shown in fig.a
3. Shift $\small{\vec{-B}}$ until it's tail coincide with the tail of $\small{\vec{A}}$. This is shown in fig.b
Clearly:
IF angle between $\small{\vec{A}}$ and $\small{\vec{B}}$ is θ,
THEN angle between $\small{\vec{A}}$ and $\small{\vec{-B}}$ is (180-θ).
4. Draw green line parallel to $\small{\vec{A}}$
• Draw yellow line parallel to $\small{\vec{B}}$  
• Thus we get the parallelogram
• Based on the parallelogram, we can derive expressions for the following quantities:
$\small{\left | \vec{R} \right |}$, α, β, and $\small{\theta_R}$
(i) Magnitude of resultant:

$\small{\left | \vec{R} \right |=\left ( \left | \vec{A}-\vec{B} \right | \right )=\sqrt{\left | \vec{A} \right |^2 + \left | \vec{B} \right |^2 + 2\left | \vec{A} \right |\left | \vec{B} \right |cos(180-\theta)}}$
• But cos (180-θ) = -cos θ. So we get:
Eq.4.4(a):
$\small{\left | \vec{R} \right |=\left ( \left | \vec{A}-\vec{B} \right | \right )=\sqrt{\left | \vec{A} \right |^2 + \left | \vec{B} \right |^2 - 2\left | \vec{A} \right |\left | \vec{B} \right |cos \theta}}$
(ii) Direction of resultant:
$\tan\alpha =\frac{\left | \vec{B} \right |\sin(180-\theta )}{\left | \vec{A} \right |+\left | \vec{B} \right |\cos(180-\theta )}$
• But sin (180-θ) = sin θ and 
• cos (180-θ) = -cos θ. So we get:
Eq.4.5(a):
$\tan\alpha =\frac{\left | \vec{B} \right |\sin \theta}{\left | \vec{A} \right |-\left | \vec{B} \right |\cos \theta}$
■ In the same way, the other equations can also be written. Readers are advised to write all steps in their own notebooks and derive Eqs.4.4(a), 4.5(a), 4.6(a) and 4.7(a) independently.

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■ A  note about 'angle between two vectors':

1. Consider fig.4.19(c) above.
• Two vectors $\small{\vec{A}}$ and $\small{\vec{B}}$ are placed in such a way that, the head of $\small{\vec{B}}$ coincide with the tail of $\small{\vec{A}}$
2. In this position, the angle between the two vectors is $\theta_1$.
• But this is not the correct way for finding the angle between two vectors
3. Both the tails must coincide at the same point. This is shown in fig.d
• In this position, the angle between the two vectors is $\theta_2$.
■ So we can write:
• The angle $\theta$ between the two vectors is $\theta_2$.

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In the next section, we will see vectors applied to motion in a plane.

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