Chapter 7.28 - Dynamics of Fixed axis Rotation

In the previous section, we saw the kinematics of rotational motion about a fixed axis. In this section we will see dynamics of rotational motion about a fixed axis

• For our present discussion, we are considering ‘rotation about a fixed axis’
• A door is an example of this type of rotation. We will write the steps:
1. Consider fig.7.128(a) below:

Fig.7.128

• A force $\mathbf\small{\vec{F}}$ is applied at the door handle
2. If $\mathbf\small{\vec{F}}$ is perpendicular to the door, it is well and good
• But in the fig.a, the $\mathbf\small{\vec{F}}$ is inclined to the door
3. Let us resolve $\mathbf\small{\vec{F}}$ into it's rectangular components. This is shown in fig.b
• We know that, the handle will be moving in a circular path
    ♦ The red component in fig.b is tangential to that circular path
    ♦ The green component is perpendicular to that tangent
    ♦ The blue component is parallel to the hinge
4. What are the effects of the three components?
• The red component causes the door to rotate
• The green component tries to detach the door from the hinge
    ♦ And it tries to move the door (in a horizontal direction) away from the hinge
• The blue component also tries to detach the door from the hinge
    ♦ And it tries to move the door (in a vertical direction towards the roof) away from the hinge
5. But in or day to day life, we do not see doors getting detached
• So we conclude that, the green and blue components will not accomplish their objectives
• This is because, the hinge is firmly fixed
• The fixity of the hinge helps to develop forces of constraint which are required to cancel the effects of green and blue components
6. So we see that, green and blue components can be discarded. We will first discuss about the blue component
• We can discard any forces which are parallel to the blue component
• Because they have no effect on the rotation about the hinge
(If they are so large that, they do cause some effect, it means that, the hinge comes of from the wall, and thus it is not a case of ‘rotation about a fixed axis’ any more)
7. Discarding ‘forces parallel to blue component’ means that, discarding forces which are perpendicular to the ‘horizontal plane in which the door handle moves’
• That means, we discard all forces perpendicular to the red plane shown in fig.7.129 below:

Fig.7.129

• We see that the red circle is the path followed by the door handle
• The red plane is the plane of that red circle
8. So we are left with forces which lie exactly on the red plane in fig.7.129
• Those forces can be resolved into two components:
(i) Those which are tangential to the circle
(ii) Those which are perpendicular to the tangents (the green component in fig.7.128.b)
9. We can discard the item 8(ii) above
• But for that, we will have to resolve the forces at every point on the circle
■ So in general, we take into account all forces that lie exactly on that red plane in fig.7.129
9. Thus for the general case, in rotation about a fixed axis, we take all forces which lie on the planes perpendicular to the axis
• Forces which are perpendicular to those planes are discarded

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10. So now we know which forces are to be taken into account

• Next we will see which position vectors are to be taken into account:
• In fig.7.130 below, we see a force $\mathbf\small{\vec{F}}$ acting on the door handle

Fig.7.130

11. This $\mathbf\small{\vec{F}}$ is eligible to be taken into account because, it lies in a plane perpendicular to the axis
• To obtain the torque created about the hinge (the z-axis) by $\mathbf\small{\vec{F}}$, we take the cross product: $\mathbf\small{\vec{\tau}=\vec{r}\times \vec{F}}$
• But $\mathbf\small{\vec{r}=\vec{OC}+\vec{CP}}$
• So we get: $\mathbf\small{\vec{\tau}=\left(\vec{OC}+\vec{CP}\right)\times \vec{F}}$
• $\mathbf\small{\Rightarrow \vec{\tau}=\left(\vec{OC}\times \vec{F}\right)+\left(\vec{CP}\times \vec{F}\right)}$
12. That means, the torque created by $\mathbf\small{\vec{F}}$ is in fact, the vector sum of two torques. They are:
(i) $\mathbf\small{\vec{\tau}_1=\left(\vec{OC}\times \vec{F}\right)}$
(ii) $\mathbf\small{\vec{\tau}_2=\left(\vec{CP}\times \vec{F}\right)}$
13. Let us now closely examine each of those torques:
• First we consider $\mathbf\small{\vec{\tau}_1=\left(\vec{OC}\times \vec{F}\right)}$
• We know this:
The vector obtained as the result of cross product will be perpendicular to both the vectors which are being multiplied
• So in our present case, we can write this:
The torque $\mathbf\small{\vec{\tau}_1}$ will be perpendicular to both $\mathbf\small{\vec{OC}}$ and $\mathbf\small{\vec{F}}$
14. Next we consider $\mathbf\small{\vec{\tau}_2=\left(\vec{CP}\times \vec{F}\right)}$
• In our present case, we can write this:
The torque $\mathbf\small{\vec{\tau}_2}$ will be perpendicular to both $\mathbf\small{\vec{CP}}$ and $\mathbf\small{\vec{F}}$
• 'Perpendicular to $\mathbf\small{\vec{CP}}$' is same as 'parallel to $\mathbf\small{\vec{OC}}$
15. So we get two interesting results:
(i) $\mathbf\small{\vec{\tau}_1}$ is perpendicular to $\mathbf\small{\vec{OC}}$
(ii) $\mathbf\small{\vec{\tau}_2}$ is parallel to $\mathbf\small{\vec{OC}}$ 
16. The first result 15(i) is not desirable. Let us see the reason:
• $\mathbf\small{\vec{OC}}$ is aligned with the axis
• So the torque $\mathbf\small{\vec{\tau}_1}$ is perpendicular to the axis
• A torque which is perpendicular to the axis will be trying to tilt that axis
17. In our present case, $\mathbf\small{\vec{\tau}_1}$ will be trying to tilt the 'line of hinge' of the door
• As a result, the hinges will have a tendency to come off from the wall
• But in normal doors, we do not see the hinges coming off. This is because, the hinges have sufficient fixity
• So we can discard $\mathbf\small{\vec{\tau}_1}$
• In general, for rotation about a fixed axis, we can discard $\mathbf\small{\vec{\tau}_1}$
18. Now, $\mathbf\small{\vec{\tau}_1}$ is caused by the perpendicular component $\mathbf\small{\vec{OC}}$ of $\mathbf\small{\vec{r}}$
• So for the general case of rotation about a fixed axis, we can discard the perpendicular component of the position vectors
19. Can we discard $\mathbf\small{\vec{\tau}_2}$?
Let us find out:
• We have seen in 15(ii) that, $\mathbf\small{\vec{\tau}_2}$ is parallel to $\mathbf\small{\vec{OC}}$
• That means, $\mathbf\small{\vec{\tau}_2}$ is parallel to the axis
• So $\mathbf\small{\vec{\tau}_2}$ is in fact the torque which causes the door to rotate. We cannot discard it
• In general, for rotation about a fixed axis, we cannot discard $\mathbf\small{\vec{\tau}_2}$
20. Now, $\mathbf\small{\vec{\tau}_2}$ is caused by the parallel component $\mathbf\small{\vec{CP}}$ of $\mathbf\small{\vec{r}}$
• So for the general case of rotation about a fixed axis, we cannot discard the parallel component of the position vectors

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Now we can write a summary:

When a body rotates about a fixed axis:
1. Consider the planes perpendicular to the axis
• All forces lying in those planes should be taken into account
• Discard all forces which are perpendicular to those planes
2. Consider the position vectors of all particles in the body
• Components of the position vectors which are parallel to the planes in (1) should be taken into account
• Discard all components which are perpendicular to those planes

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This is a very interesting situation:

■ The red plane in fig.7.129 above is perpendicular to the axis of rotation
■ All the forces that we need, are parallel to that red plane
■ All the 'position vector components' that we need, are parallel to that red plane

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• So the 3D problem is now reduced to a 'problem in a plane'

• In other words, the 3D problem of 'rotation about a fixed axis' is now reduced to a 2D problem
• Let us see how to represent it as a 2D problem. We will write it in steps:
1. Consider the fig.7.130 that we saw above. The path of the door handle is shown as a pink circle
• Since it is now a 2D problem, the 'plane of that circle' is the plane which are are going to work on  
2. If we look at the 'plane of that circle' from above, we will get a 2D view
• In that view, the z-axis will be seen as a small blue circle. This is shown in fig.7.131 below:

Fig.7.131

• The entire path of the handle is not shown in fig.7.131. Rather, a portion of it is shown as the pink arc
    ♦ The center of this arc is C
• The door handle is shown as a small yellow sphere
x'-axis is parallel to the x-axis of the reference frame
    ♦ x'-axis passes through C   
• y'-axis is parallel to the y-axis of the reference frame
    ♦ y'-axis passes through C
3. A force $\mathbf\small{\vec{F}}$ acts on the door handle. This force lies on the x'y' plane. So we must take it into account
4. $\mathbf\small{\vec{r}_{\bot}}$ is the 'perpendicular component' of the 'position vector of the handle'. It lies on the x'y' plane. So we must take it into account

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Thus we have succeeded in representing the 'required items' in a 2D plane. Based on that, we can discuss the work done by a torque. We will see it in the next section

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Chapter 5.15 - Solved examples on Force in Strings

In the previous section we saw the basic details about 'force in string'. We saw the possible 3 cases and saw a solved example related to case 3. In this section we will see some solved examples related to cases 1 and 2.

Solved example 5.24
Three blocks A, B and C are connected vertically as shown in fig.5.53(a) below:

Fig.5.53

Their masses are 3 kg, 5 kg and 7 kg respectively. If the system of blocks is raised upwards by a force of 195 N, what will be the tensions in strings PQ and RS?
Solution:
1. First we draw the FBD of the whole system. This is shown in fig.5.53(b) above
• In the FBD,  there are 2 forces
• The weight W is the total weight of the 3 masses. So we can write:
• $\mathbf\small{|\vec{W}|=(m_A+m_B+m_C)g}$= 150 N
2. Net force acting on the system = $\mathbf\small{\vec{F}-\vec{W}=195\, \hat{j}-150\, \hat{j}=45\, \hat{j}}$
• By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\text{45}\, \hat{j}=\left[(m_A+m_B+m_C)\times |\vec{a}| \right ]\hat{j}}$
$\mathbf\small{\Rightarrow |\vec{a}|=\frac{45}{15}=3\, \text{ms}^{-2}}$
• So the whole system moves upwards with an acceleration of 3 ms^-2
3. Next we draw the FBD of 'C'. This is shown in fig.5.53(c) above
• In the segment RS, two vectors are marked on the rope
• They are marked to help us remember two points:
(i) The segment RS pulls at R
(ii) The segment RS pulls at S
• A string can exert only 'pulling force'. It cannot exert 'pushing force' 
■ We know that tension is uniform through out the length of a string. So the two 'pulls' are the same. 
4. In the FBD, the tension in segment RS is denoted as $\mathbf\small{\vec{T_{RS}}}$ 
• So net force acting on 'C' = $\mathbf\small{\vec{T_{RS}}-\vec{W_C}}$
5. By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{{\vec{T_{RS}}-\vec{W_C}=(m_C \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow {\vec{T_{RS}}=\vec{W_C}+(m_C \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow {\vec{T_{RS}}=(m_C \times g)\hat{j}+(m_C \times |\vec{a}|})\hat{j}=[m_C \times (|\vec{a}|+g)]\hat{j}}$
$\mathbf\small{\Rightarrow {|\vec{T_{RS}|}=[m_C \times (|\vec{a}|+g)]}}$
• Substituting the values, we get:
$\mathbf\small{{|\vec{T_{RS}|}=[7 \times (3+10)]}=91\, \text{N}}$
6. Next we draw the FBD of 'B'. This is shown in fig.5.54(a) below:

Fig.5.54

• In the segment PQ, two vectors are marked on the rope
• They are marked to help us remember two points:
(i) The segment PQ pulls at P
(ii) The segment PQ pulls at Q
• A string can exert only 'pulling force'. It cannot exert 'pushing force' 
■ We know that tension is uniform through out the length of a string. So the two 'pulls' are the same. 
7. In the FBD, the tension in segment PQ is denoted as $\mathbf\small{\vec{T_{PQ}}}$
• The tension in segment RS is denoted as $\mathbf\small{\vec{T_{RS}}}$
• In addition to the above two, the weight of 'B' will also be acting
• So net force acting on 'B' = $\mathbf\small{\vec{T_{PQ}}-\vec{T_{RS}}-\vec{W_B}}$
8. By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\vec{T_{PQ}}-{\vec{T_{RS}}-\vec{W_B}=(m_B \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow \vec{T_{PQ}}={\vec{T_{RS}}+\vec{W_B}+(m_B \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow \vec{T_{PQ}}={\vec{T_{RS}}+(m_B \times g)\hat{j}+(m_B \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow \vec{T_{PQ}}=\vec{T_{RS}}+\left[m_B \times (|\vec{a}|+g) \right ]\hat{j}}$
$\mathbf\small{\Rightarrow |\vec{T_{PQ}}|=|\vec{T_{RS}}|+\left[m_B \times (|\vec{a}|+g) \right ]}$
• Substituting the values, we get:
$\mathbf\small{|\vec{T_{PQ}}|=91+\left[5 \times (3+10) \right ]=156\, \text{N}}$
Thus we obtained the tensions in both the ropes
Check:
The FBD of 'A' is shown in fig.5.54(b). We have:
• So net force acting on 'A' = $\mathbf\small{\vec{F}-\vec{T_{PQ}}-\vec{W_A}}$
By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\vec{F}-\vec{T_{PQ}}-\vec{W_A}=(m_A \times |\vec{a}|)\hat{j}}$
Substituting the values, we get: $\mathbf\small{195 \hat{j}-156 \hat{j}-30 \hat{j}=(3 \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow|\vec{a}|= 3\, \text{ms}^{-2}}$
• We have already seen that each mass move upwards with an acceleration of 3 ms^-2. So our calculations are correct

Solved example 5.25
Three blocks A, B and C connected horizontally rest on a friction less floor as shown in the fig.5.55(a) below:

Fig.5.55

Their masses are 7 kg, 5 kg and 2 kg respectively. If the system of blocks is moved towards the right by a force of 56 N applied on 'C', what will be the tensions in strings PQ and RS?
Solution:
1. First we draw the FBD of the whole system. This is shown in fig.5.55(b) above
• In the FBD,  there is only 1 force
(The vertical forces will all cancel each other, and hence are not shown) 
2. Net force acting on the system = $\mathbf\small{56\, \hat{j}}$
• By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\text{56}\, \hat{j}=\left[(m_A+m_B+m_C)\times |\vec{a}| \right ]\hat{j}}$
$\mathbf\small{\Rightarrow |\vec{a}|=\frac{56}{14}=4\, \text{ms}^{-2}}$
• So the whole system moves towards the right with an acceleration of 4 ms^-2
3. Next we draw the FBD of 'A'. This is shown in fig.5.56 below:

Fig.5.56

• In the segment PQ, two vectors are marked on the rope
• They are marked to help us remember two points:
(i) The segment PQ pulls at P
(ii) The segment PQ pulls at Q
• A string can exert only 'pulling force'. It cannot exert 'pushing force' 
■ We know that tension is uniform through out the length of a string. So the two 'pulls' are the same. 
4. In the FBD, the tension in segment PQ is denoted as $\mathbf\small{\vec{T_{PQ}}}$ 
• So net force acting on 'A' = $\mathbf\small{\vec{T_{PQ}}}$
5. By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{{\vec{T_{PQ}}=(m_A \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow {\vec{T_{PQ}}=(m_A \times |\vec{a}|})\hat{j}=28\, \hat{j}}$
$\mathbf\small{\Rightarrow {|\vec{T_{PQ}}|=28}\, \text{N}}$
6. Next we draw the FBD of 'B'. This is shown in fig.5.57 below:

Fig.5.57

• In the segment RS, two vectors are marked on the rope
• They are marked to help us remember two points:
(i) The segment RS pulls at R
(ii) The segment RS pulls at S
• A string can exert only 'pulling force'. It cannot exert 'pushing force' 
■ We know that tension is uniform through out the length of a string. So the two 'pulls' are the same. 
7. In the FBD, the tension in segment RS is denoted as $\mathbf\small{\vec{T_{RS}}}$ 
• So net force acting on 'B' = $\mathbf\small{\vec{T_{RS}}-\vec{T_{PQ}}}$
8. By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\vec{T_{RS}}-\vec{T_{PQ}}=(m_B \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow \vec{T_{RS}}=\vec{T_{PQ}}+(m_B \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow \vec{T_{RS}}=28\, \hat{j}+20\, \hat{j}}$
$\mathbf\small{\Rightarrow |\vec{T_{RS}}|=48\, \text{N}}$
Thus we obtained the tensions in both the ropes

Check:
The FBD of 'C' is shown in fig.5.58 below:

Fig.5.58

• So net force acting on 'C' = $\mathbf\small{\vec{F}-\vec{T_{RS}}}$
By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\vec{F}-\vec{T_{RS}}=(m_C \times |\vec{a}|)\hat{j}}$
Substituting the values, we get: $\mathbf\small{56 \hat{j}-48 \hat{j}=(2 \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow|\vec{a}|= 4\, \text{ms}^{-2}}$
• We have already seen that each mass move with an acceleration of 4 ms^-2. So our calculations are correct

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Now we will see a solved example based on case 3

Solved example 5.26
A block 'A' of mass 10 kg is suspended using 3 strings QP, QR and QS as shown in the fig.5.59(a) below. Find the tensions in the strings.

Fig.5.59

Solution:
1. First we draw the FBD of 'A'
• So 'A' is taken as the sub-system. A red rectangle is drawn around 'A' in fig.5.59(b)
• What ever is enclosed in the red rectangle, should be used in the FBD 
• 'A' is in equilibrium under the action of two forces. So we apply the condition for equilibrium as follows:

$\mathbf\small{\vec{F_1}+\vec{F_2}=0}$
    ♦ Let $\small{\vec{F_1}}$ indicate $\small{\vec{T_{QS}}}$ 
    ♦ Let $\small{\vec{F_2}}$ indicate $\small{\vec{W_A}}$ 

• Considering upward forces as positive and downward forces as negative, we get:

$\mathbf\small{(|\vec{T_{QS}}|)\hat{j}-(|\vec{W_A}|)\hat{j}=0}$

$\mathbf\small{\Rightarrow (|\vec{T_{QS}}|)\hat{j}-(m_A \times g)\hat{j}=0}$

$\mathbf\small{\Rightarrow (|\vec{T_{QS}}|)\hat{j}=(m_A \times g)\hat{j}}$

$\mathbf\small{\Rightarrow |\vec{T_{QS}}|=m_A \times g= 10 \times 10 = 100\,N}$
• Direction of $\small{\vec{T_{QS}}}$ is obviously, 'vertical and upwards' 
2. Now we draw the FBD of a sub-system around the point 'Q'
This is shown in fig.5.60 below:

Fig.5.60

• What ever is enclosed in the red rectangle, should be used in the FBD.
3. In fig.b, 'Q' is in equilibrium under the action of 3 forces. So we apply the condition for equilibrium as follows:

$\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}=0}$
    ♦ Let $\small{\vec{F_1}}$ indicate $\small{\vec{T_{PQ}}}$ 
    ♦ Let $\small{\vec{F_2}}$ indicate $\small{\vec{T_{QR}}}$  
    ♦ Let $\small{\vec{F_3}}$ indicate $\small{\vec{T_{QS}}}$
■ Here two forces are inclined. So we must take vertical and horizontal components separately
8. Considering horizontal components, we have:
$\mathbf\small{\vec{F_{1x}}+\vec{F_{2x}}+\vec{F_{3x}}=0}$
• Considering forces to the right as positive and forces to the left as negative, we get:
$\mathbf\small{-(|\vec{T_{PQ}}|\cos 60 )\hat{i}+(|\vec{T_{QR}}|\cos 30 )\hat{i}+\,0=0}$
(The last term is zero because, $\small{\vec{T_{QS}}}$ does not have a horizontal component)
$\mathbf\small{\Rightarrow |\vec{T_{PQ}}|\cos 60=|\vec{T_{QR}}|\cos 30}$
$\mathbf\small{\Rightarrow |\vec{T_{PQ}}|=\sqrt{3}\, |\vec{T_{QR}}|}$
9. Considering the vertical components, we have:
$\mathbf\small{\vec{F_{1y}}+\vec{F_{2y}}+\vec{F_{3y}}=0}$
• Considering upward forces as positive and downward forces as negative, we get:
$\mathbf\small{(|\vec{T_{PQ}}|\sin 60 )\hat{j}+(|\vec{T_{QR}}|\sin 30 )\hat{j}-(|\vec{T_{QS}}|)\hat{j}=0}$
$\mathbf\small{\Rightarrow (|\vec{T_{PQ}}|\sin 60 +|\vec{T_{QR}}|\sin 30 )\hat{j}=(100)\hat{j}}$
$\mathbf\small{\Rightarrow (|\vec{T_{PQ}}|\sin 60 +|\vec{T_{QR}}|\sin 30 )=100}$
$\mathbf\small{\Rightarrow \sqrt{3}\, |\vec{T_{PQ}}|+|\vec{T_{QR}}|=200}$
10. But from (8) we have: $\mathbf\small{|\vec{T_{PQ}}|=\sqrt{3}\, |\vec{T_{QR}}|}$
• So the result in (9) becomes: $\mathbf\small{ \sqrt{3}\, (\sqrt{3}\,|\vec{T_{QR}}|)+|\vec{T_{QR}}|=200}$
$\mathbf\small{\Rightarrow 4\,|\vec{T_{QR}}|=200}$
$\mathbf\small{\Rightarrow |\vec{T_{QR}}|=50 \,\text{N}}$
11. So from (8) we get: $\mathbf\small{|\vec{T_{PQ}}|=50\sqrt{3}\,\text{N}}$

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In the next section we will see strings through pulleys

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Chapter 4.3 - Resolution of Vector

In the previous section we saw addition and subtraction of vectors. In this section we will see resolution of vectors.
1. Consider any three vectors $\small{\vec{A}}$, $\small{\vec{B}}$ and $\small{\vec{C}}$ as shown in fig.4.10 below:

Fig.4.10

• They are 'any three' vectors. 
• If we add $\small{\vec{B}}$ and $\small{\vec{C}}$, we will not get $\small{\vec{A}}$. This is shown in fig.4.10(b)
2. Now isolate $\small{\vec{A}}$. Let it's tail be P and tip be Q
• Draw a line parallel to $\small{\vec{B}}$ through P. This is shown in green colour in fig.c  
• Draw a line parallel to $\small{\vec{C}}$ through Q. This is shown in yellow colour in fig.d
• Let the green and yellow lines intersect at R
3. Consider the two points: P and R
• A vector can fit between those two points. This is shown in fig.e
• So we have a $\small{\vec{PR}}$. It is drawn over the green line.
• But the green line is parallel to $\small{\vec{B}}$  
• So $\small{\vec{PR}}$ can be obtained by multiplying $\small{\vec{B}}$ by a scalar λ
■ Thus we get: $\small{\vec{PR}}$ = λ $\small{\vec{B}}$
4. Consider the two points: Q and R
• A vector can fit between those two points. This is shown in fig.e
• So we have a $\small{\vec{QR}}$. It is drawn over the yellow line.
• But the yellow line is parallel to $\small{\vec{C}}$  
• So $\small{\vec{QR}}$ can be obtained by multiplying $\small{\vec{C}}$ by a scalar μ 
■ Thus we get: $\small{\vec{QR}}$ = μ $\small{\vec{C}}$.
5. In the fig.e, we see that: $\small{\vec{A}}$ = λ $\small{\vec{B}}$ + μ $\small{\vec{C}}$
■ Even if A, B and C are 'any three' vectors, we are able to express A in terms of B and C 
6. Can we express $\small{\vec{B}}$ in terms of $\small{\vec{A}}$ and $\small{\vec{C}}$?
Let us try. We will write only the minimum required steps:
(i) Isolate B and draw the green and yellow lines through it's ends. See fig.4.11 below:
    ♦ Green line through P is parallel to A
    ♦ Yellow line through Q is parallel to C

Fig.4.11

7. In the final fig.d, we see that: $\small{\vec{B}}$ = λ $\small{\vec{A}}$ + μ $\small{\vec{C}}$
• But direction of μ $\small{\vec{C}}$ is opposite to that of $\small{\vec{C}}$
• This indicates that, the scalar μ in this case is a negative real number
8. Anyway, we are able to express B in terms of A and C
• Following a similar procedure, we can express C in terms of A and B also
• Reader may draw different sets of 'any three' vectors and try all the three combinations for each set

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The above discussion leads us to write this:
■ Any vector lying in a plane can be expressed in terms of 'any other two vectors' lying in the same plane
• Concentrate on the words: 'any other two vectors'
• Those words tell us that, the two vectors can have any magnitude and any direction
■ So why not make them like this:
(i) One of them have a magnitude of one unit
• It is directed towards the positive side of the x axis
• We will call it unit vector i. In vector notation, it is written as: $\small{\hat{i}}$ 
• It is read as 'i cap' 
[A cap sign ('^') is given above all unit vectors. This is to distinguish them other vectors]
See fig.4.12(a) below:

Fig.4.12

(ii) The other also have a magnitude of one unit
• But this one is directed towards the positive side of the y axis
• We will call it unit vector j. In vector notation, it is written as: $\small{\hat{j}}$

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Let us see some properties of unit vectors:
1. A unit vector is a vector having a magnitude of one unit. It points in a particular direction.
2. It has no dimension (like mass, length, time etc.,)
• So it has no unit (like kg, m, s etc.,)
3. It is used to specify a direction only
4. Unit vectors along the x-, y- and z-axes are denoted by $\small{\hat{i},\hat{j}}$ and $\small{\hat{k}}$ respectively
5. Since the magnitude of unit vectors is one, we can write:
$\small{\left | \hat{i} \right |}$ = $\small{\left | \hat{j} \right |}$ = $\small{\left | \hat{k} \right |}$ = 1
■ In our present chapter, we are discussing motion in two dimensions. So in this chapter, we will need $\small{\hat{i}}$ and $\small{\hat{j}}$ only.
6. We can specify unit vectors in any required direction we want. 
• If we multiply a unit vector $\small{\hat{n}}$ by a scalar λ, we will get a new vector: λ$\small{\hat{n}}$
• The magnitude of this new vector will be λ
• The direction of this new vector will be same as that of $\small{\hat{n}}$
7. In general, any $\small{\vec{A}}$ can be written as:
$\small{\vec{A}}$ = $\small{\left | \vec{A} \right |}$ $\small{\hat{n}}$  
• Where n is a unit vector which has the same direction as A

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Now we will see how unit vectors can be helpful to find the components of a given vector
1. In fig.4.12(a) above, a vector $\small{\vec{A}}$ is shown
2. We draw a green line (parallel to $\small{\hat{i}}$ ) through P. This is shown in fig.b
• But $\small{\hat{i}}$ is parallel to the x axis.
• So the green line is parallel to the x axis
3. We draw a yellow line (parallel to $\small{\hat{j}}$) through Q
• But $\small{\hat{j}}$ is parallel to the y axis.
• So the yellow line is parallel to the y axis
4. The green and yellow lines meet at R
• $\small{\vec{PR}}$ is parallel to $\small{\hat{i}}$. So we can write:
$\small{\vec{PR}}$ = λ$\small{\hat{i}}$
• Where λ is a real number
6. $\small{\vec{QR}}$ is parallel to $\small{\hat{j}}$. So we can write:
$\small{\vec{QR}}$ = μ$\small{\hat{j}}$
• Where μ is a real number 
7. From fig.4.12(b), we can see that  [λ$\small{\hat{i}}$ + μ$\small{\hat{j}}$] = $\small{\vec{A}}$
■ That is., λ$\small{\hat{i}}$ and μ$\small{\hat{j}}$ are the components of $\small{\vec{A}}$
8. Since $\small{\hat{i}}$ is parallel to x axis, λ$\small{\hat{i}}$ is called the horizontal component of $\small{\vec{A}}$
• Another name for horizontal component is x component
• The x component of $\small{\vec{A}}$ is denoted as $\small{\vec{A_x}}$
■ So we can write: $\small{\vec{A_x}}$ = λ$\small{\hat{i}}$
This is shown in fig.4.12(c) above.
9. Since $\small{\hat{j}}$ is parallel to y axis, μ$\small{\hat{y}}$ is called the vertical component of $\small{\vec{A}}$
• Another name for vertical component is y component
• The y component of $\small{\vec{A}}$ is denoted as $\small{\vec{A_y}}$
[The x and y components together are called rectangular components of a vector] 
■ So we can write: $\small{\vec{A_y}}$ = μ$\small{\hat{j}}$ 
10. If we can find the values of λ and μ, we can calculate the components A_x and A_y
• So our next aim is to find λ and μ. 
11. For that, consider PQR as a right angled triangle. This is shown in fig.d
Length of PQ will be equal to the magnitude of A. 
That is., PQ = $\small{\left | \vec{A} \right |}$  
12. Length of the base PR will be equal to the magnitude of λ$\small{\hat{i}}$. 
That is., PR = λ 
13. Length of the altitude QR will be equal to the magnitude of μ$\small{\hat{i}}$. 
That is., QR = μ.
14, Note the angle 'θ' shown in fig.c
It is the angle which $\small{\vec{A}}$ makes with the x axis
So it is the direction of the $\small{\vec{A}}$. It will be given to us.
15. Now we can apply the trigonometric ratios
• We have cos $\theta$ = $\frac{\lambda}{magnitude\:of\: \vec{A}}$
⟹ $\lambda$ = ${\left | \vec{A} \right |}$ cos $\theta$
• We have sin $\theta$ = $\frac{\mu}{magnitude\:of\: \vec{A}}$
⟹ $\mu$ = ${\left | \vec{A} \right |}$ sin $\theta$
16. We can write the final results:
Eq.4.1:
(i) Magnitude of the x component of $\small{\vec{A}}$ = $\small{\left | \vec{A_x} \right |}$ = Length of PR = $\lambda$ = ${\left | \vec{A} \right |}$ cos $\theta$
(ii) Magnitude of the y component of $\small{\vec{A}}$ = $\small{\left | \vec{A_y} \right |}$ = Length of QR = $\mu$ = ${\left | \vec{A} \right |}$ sin $\theta$
An example:
• Fig.4.13(a) below shows $\small{\vec{A}}$
    ♦ It has a magnitude of 5 units
    ♦ It makes 30^o with the x axis

Fig.4.13

• Find the horizontal component $\small{\vec{A_x}}$ and the vertical component $\small{\vec{A_y}}$ of $\small{\vec{A}}$
Solution:
1. Using Eq.4.1, we have:
(i) $\small{\left | \vec{A_x} \right |}$ = ${\left | \vec{A} \right |}$ cos $\theta$
• Substituting the values, we get: 
$\small{\left | \vec{A_x} \right |}$ = 5 cos30 = 5 × 0.8660 = 4.33 units
• Direction of $\small{\vec{A_x}}$ is towards the positive side of the x axis 
• So we can write: $\small{\vec{A_x}}$ = 4.33$\small{\hat{i}}$. This is shown in fig.4.13(b)
(ii) $\small{\left | \vec{A_y} \right |}$ = ${\left | \vec{A} \right |}$ sin $\theta$
• Substituting the values, we get: 
$\small{\left | \vec{A_y} \right |}$ = 5 sin30 = 5 × 0.5 = 2.5 units  
• Direction of $\small{\vec{A_y}}$ is towards the positive side of the y axis 
• So we can write: $\small{\vec{A_y}}$ = 2.5$\small{\hat{j}}$
2. Let u see a practical application:
• Since we are dealing with free vectors in this chapter, we will shift $\small{\vec{A_y}}$
• We will shift it so that it's tail coincides with the tails of $\small{\vec{A}}$ and $\small{\vec{A_x}}$
• This is shown in fig.c
3. If a force of 5 N acts at an angle of 30^o at a point, the effect will be the combination of the following two:
(i) A force of 4.33 N pushing the point towards the positive side of x axis 
(i) A force of 2.5 N pushing the point towards the positive side of y axis 
• The combined action is the vector sum (4.33$\small{\hat{i}}$ + 2.5$\small{\hat{j}}$) 
• This sum produces the same effect of 5 N acting at an angle 
■ So there are two methods to represent the vector.
Method 1:
• A force vector acts at a point
• It has a magnitude 5 N
• It has a direction which makes 30^o with the x axis
Method 2:
A force vector 4.33$\small{\hat{i}}$ + 2.5$\small{\hat{j}}$ acts at a point

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17. Thus we successfully calculated the x and y components of the given $\small{\vec{A}}$
• Can we do the reverse?
• That is., if we are given the x and y components of a vector, we must be able to find the original vector. Let us try:
(i) First we will find the magnitude.
• From the right triangle PQR in fig.4.12(d), we have:
|$\small{\vec{A}}$|² = λ² + μ² .
⟹ |$\small{\vec{A}}$|² = |$\small{\vec{A_x}}$|² + |$\small{\vec{A_y}}$|²
From this we get:
Eq.4.2:
|$\small{\vec{A}}$| = ⎷[|$\small{\vec{A_x}}$|² + |$\small{\vec{A_y}}$|²]
(ii) Now we want the direction of $\small{\vec{A}}$
• From the right triangle PQR in fig.4.12(d), we have:
tan $\theta$ = $\frac{\mu}{\lambda}$
• From this we get:
Eq.4.3:
 tan $\theta$ = $\frac{|\vec{A_y}|}{|\vec{A_x}|}$
• We can write: $\theta$ = tan^-1$\frac{|\vec{A_y}|}{|\vec{A_x}|}$
An example:
Given that:
• x component $\small{\vec{A_x}}$ of a vector $\small{\vec{A}}$ is 9$\small{\hat{i}}$
• y component $\small{\vec{A_y}}$ of a vector $\small{\vec{A}}$ is 11$\small{\hat{i}}$
■ Find $\small{\vec{A}}$
Solution:
1. $\small{\vec{A_x}}$ and $\small{\vec{A_y}}$ are shown in fig.4.14(a) below:

Fig.4.14

2. Shift $\small{\vec{A_y}}$ until it's tail coicide with the tip of $\small{\vec{A_x}}$
This is shown in fig.b
3. Join the tail of $\small{\vec{A_x}}$ and tip of $\small{\vec{A_y}}$. This gives $\small{\vec{A}}$. It is shown in fig.c
4. In the resulting right angled triangle, 
Base = |$\small{\vec{A_x}}$| = 9 units   
Altitude = |$\small{\vec{A_y}}$| = 11 units   
5. So hypotenuse = |$\small{\vec{A}}$| = √[9² + 11²] = 14.213 units   
6. Now we want the direction. 
• We have: tan $\theta$ = $\frac{|\vec{A_y}|}{|\vec{A_x}|}$ = $\frac{11}{9}$ = 1.222
• So $\theta$ = tan^-1 1.222 = 50.71^o.
7. So $\small{\vec{A}}$ has a magnitude of 14.213 units and it makes an angle of 50.71^o with the x axis

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In the next section, we will see the analytical method of vector addition.

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