Chapter 7.6- Center of Mass in Three-Dimensional Problems

In the previous section, we saw the location of 'C' of some common objects. In this section we will see systems where particles are distributed in space

1. Consider a particle 'P' situated in space
• Let it be situated at a certain distance from the origin 'O' of the frame of reference
    ♦ x-axis is shown in red
    ♦ y-axis is shown in green
    ♦ z-axis is shown in blue
• This is shown in fig.7.42 (a) below:

Fig.7.42

2. The particle is shown as a yellow sphere
• It's mass is 0.5 kg
• We see that:
    ♦ It is not situated on the x-axis
    ♦ It is not situated on the y-axis
    ♦ It is not situated on the xy-plane
    ♦ It is situated in space
3. We know the method to specify the positions of such particles
• The method can be written in 5 steps:
(i) Drop a perpendicular from 'P' onto the xy-plane
• This is indicated by the blue line in fig.7.42(b) above
• Note that, the yellow sphere is a bit transparent. We see the blue line starting from the exact center of the sphere
(ii) Let P' be the foot of the perpendicular
• From P', draw a perpendicular onto the x-axis
• This is indicated by the green line
(iii) From P', draw a perpendicular onto the y-axis
• This is indicated by the red line
(iv) Let the length of the red line be x_P = 2 m
• Let the length of the green line be y_P = 3 m
• Let the length of the blue line be z_P = 2 m
(v) Then the coordinates of P are (x_P, y_P, z_P) = (2,3,2)
4. Consider another particle Q. It's mass is 0.75 kg. This is shown in fig.7.43(a) below:

Fig.7.43

• Q will have it's own red, green and blue lines
• Let the lengths of those lines be 1 m, 2 m, and 1 m respectively
• So the coordinates (x_Q, y_Q, z_Q) will be (1,2,1)
5. So we have a system consisting of two particles P and Q
• We want the location of the 'C' of this system
6. In such cases, the 'C' lies at a point whose coordinates are (X,Y,Z)
• We already know the method to find 'X' and 'Y'
7. We can apply the same method to find 'Z' 
• But to find 'Z', the method should be applied in the z-direction
• As usual, the method involves only 4 steps:
(i) Take the distances (from the xy-plane) for each particle
(ii) Apply the 'due weightage' 
(iii) Find the average of those 'weighted distances'
(iv) This average is the 'Z'
8. So we can write a new equation:
Eq.7.3: $\mathbf\small{Z=\frac{\sum{m_iz_i} }{\sum{m_i}}}$ 
9. Now we can form the table:

10. From the table, we get:
$\mathbf\small{\sum{m_i}=1.25}$
$\mathbf\small{\sum{m_ix_i}=1.75}$
$\mathbf\small{\sum{m_iy_i}=3}$
$\mathbf\small{\sum{m_iz_i}=1.75}$
• Thus we get: 
$\mathbf\small{X=\frac{\sum{m_ix_i} }{\sum{m_i}}=\frac{1.75}{1.25}=1.4\, \text{m}}$
$\mathbf\small{Y=\frac{\sum{m_iy_i} }{\sum{m_i}}=\frac{3}{1.25}=2.4\, \text{m}}$
$\mathbf\small{Z=\frac{\sum{m_iz_i} }{\sum{m_i}}=\frac{1.75}{1.25}=1.4\, \text{m}}$
11. Using these coordinates, we can mark 'C'. This is shown as a small white sphere in fig.7.43(b) above
• We see that, 'C' has it's own red, green and blue lines
    ♦ The length of it's red line = X = 1.4 m
    ♦ The length of it's green line = Y = 2.4 m
    ♦ The length of it's blue line = Z = 1.4 m
• Also note that, the 'C' lies on the line joining P and Q

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• Now we know the method to find 'C' of the particles distributed in 3 dimensional space

• In such 3-dimensional problems, another easier method can be used. Let us see the details of that method:

1. In fig.7.44(a) below, a vector is shown in magenta color

Fig.7.44

• It's tail end coincides with 'O'
• It's tip coincides with the center of 'P'
■ So it is the position vector of P
• We can denote it as $\mathbf\small{\vec{r}_P}$. (Details here)
2. Imagine that, a person wants to go from O to P
• He can take either one of the two paths given below:
Path 1:
• This path is along the $\mathbf\small{\vec{r}_P}$. It starts from O and ends at P
Path 2:
This path has 3 segments:
■ Segment 1:
• This segment starts from O
• The person travels a distance x_P
    ♦ x_P is the length of the red line
    ♦ So 'magnitude' of travel is x_P  
• Direction of travel for this segment is: 'along the x-axis'
• So the magnitude and direction of this travel can be represented in vector form as: $\mathbf\small{(x_P)\hat{i}}$
• At the end of this travel, the person is at the foot of the green line
■ Segment 2:
• This segment starts from the foot of the green line
• The person travels a distance y_P
    ♦ y_P is the length of the green line
    ♦ So 'magnitude' of travel is y_P  
• Direction of travel for this segment is: 'parallel to the y-axis'
• So the magnitude and direction of this travel can be represented in vector form as: $\mathbf\small{(y_P)\hat{j}}$
• At the end of this travel, the person is at the foot of the blue line
■ Segment 3:
• This segment starts from the foot of the blue line
• The person travels a distance z_P
    ♦ z_P is the length of the blue line
    ♦ So 'magnitude' of travel is z_P  
• Direction of travel for this segment is: 'parallel to the z-axis'
• So the magnitude and direction of this travel can be represented in vector form as: $\mathbf\small{(z_P)\hat{k}}$
• At the end of this travel, the person reaches the particle P
3. So we see 3 vectors in path 2. If we add those vectors, we will reach P
• That means:
Path 2 = $\mathbf\small{(x_P)\hat{i}+(y_P)\hat{j}+(z_P)\hat{k}}$
4. The initial and final points in both paths 1 and 2 are the same
• So we can write: Path 1 = Path 2
• That is: $\mathbf\small{\vec{r}_P=(x_P)\hat{i}+(y_P)\hat{j}+(z_P)\hat{k}}$
5. The above equality in (4) can be established based on fig.7.44(b) also:
(i) In fig.7.44(b), there are a total of 5 vectors:
Red, green, blue, yellow and magenta
(ii) We want to prove this:
Magenta = red + green + blue
(iii) Consider the following three vectors:
Red, green and yellow
• Applying triangle law of vector addition, we get:
yellow = red + green  
(iv) Consider the following three vectors:
Yellow, blue and magenta
• Applying triangle law of vector addition, we get:
Magenta = yellow + blue
(v) Now we expand 'yellow' using the result in (iii). We get:
Magenta = red + green + blue
• This is the required result that we mentioned in (ii)
6. So it is proved beyond doubt. We can confidently write:
$\mathbf\small{\vec{r}_P=(x_P)\hat{i}+(y_P)\hat{j}+(z_P)\hat{k}}$
• In this equation, 
    ♦ $\mathbf\small{\vec{r}_P}$ is the position vector  
    ♦ x_P, y_P and z_P are the coordinates of P
■ So we can write:
The following two items are closely related:
(i) position vector of a particle
(ii) The coordinates of that particle

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7. Now, if 'Q' is another particle with coordinates (x_Q,y_Q,z_Q), we can write:
$\mathbf\small{\vec{r}_Q=(x_Q)\hat{i}+(y_Q)\hat{j}+(z_Q)\hat{k}}$
8. Consider the system consisting of the two particles P and Q
• Let 'C' be the center of mass of the system
• The coordinates of 'C' are (X,Y,Z)
9. If we know those coordinates, we can easily write the position vector of 'C'
• Based on (6) above, we get: $\mathbf\small{\vec{r}_C=(X)\hat{i}+(Y)\hat{j}+(Z)\hat{k}}$
10. Now we expand X, Y and Z. We have:
• $\mathbf\small{X=\frac{m_p\,x_P+m_Q\,x_Q}{m_P+m_Q}=\frac{m_p\,x_P+m_Q\,x_Q}{M}}$
• $\mathbf\small{Y=\frac{m_p\,y_P+m_Q\,y_Q}{M}}$
• $\mathbf\small{Z=\frac{m_p\,z_P+m_Q\,z_Q}{M}}$
11. So the equation in (9) becomes:
$\mathbf\small{\vec{r}_C=\left[\frac{m_p\,x_P+m_Q\,x_Q}{M} \right]\hat{i}+\left[\frac{m_p\,y_P+m_Q\,y_Q}{M} \right]\hat{j}+\left[\frac{m_p\,z_P+m_Q\,z_Q}{M} \right]\hat{k}}$
$\mathbf\small{\Rightarrow \vec{r}_C=\frac{(m_P\,x_P)\hat{i}+(m_Q\,x_Q)\hat{i}+(m_P\,y_P)\hat{j}+(m_Q\,y_Q)\hat{j}+(m_P\,z_P)\hat{k}+(m_Q\,z_Q)\hat{k}}{M}}$
12. Now we rearrange the above equation by the two steps:
(i) Bringing the terms with m_P together
(ii) Bringing the terms with m_Q together
• We get:
$\mathbf\small{\vec{r}_C=\frac{[(m_P\,x_P)\hat{i}+(m_P\,y_P)\hat{j}+(m_P\,z_P)\hat{k}]+[(m_Q\,x_Q)\hat{i}+(m_Q\,y_Q)\hat{j}+(m_Q\,z_Q)\hat{k}]}{M}}$
• Taking m_P and m_Q outside, we get:
$\mathbf\small{\vec{r}_C=\frac{m_P[(x_P)\hat{i}+(y_P)\hat{j}+(z_P)\hat{k}]+m_Q[(x_Q)\hat{i}+(y_Q)\hat{j}+(z_Q)\hat{k}]}{M}}$
13. But:
    ♦ $\mathbf\small{(x_P)\hat{i}+(y_P)\hat{j}+(z_P)\hat{k}=\vec{r}_P}$ 
    ♦ $\mathbf\small{(x_Q)\hat{i}+(y_Q)\hat{j}+(z_Q)\hat{k}=\vec{r}_Q}$
• Thus (12) becomes:
$\mathbf\small{\vec{r}_C=\frac{m_P[\vec{r}_P]+m_Q[\vec{r}_Q]}{M}}$
14. This gives us an easy method to write the 'position vector of C'
• Let us check and see if we will get the same answer as before:
(i) The coordinates of P are (2,3,2)
• So $\mathbf\small{\vec{r}_P=2\hat{i}+3\hat{j}+2\hat{k}}$
• So $\mathbf\small{m_P\,\vec{r}_P=0.5(2\hat{i}+3\hat{j}+2\hat{k})=\hat{i}+1.5\hat{j}+\hat{k}}$
(ii) The coordinates of Q are (1,2,1)
• So $\mathbf\small{\vec{r}_Q=\hat{i}+2\hat{j}+\hat{k}}$
• So $\mathbf\small{m_Q\,\vec{r}_P=0.75(\hat{i}+2\hat{j}+\hat{k})=0.75\hat{i}+1.5\hat{j}+0.75\hat{k}}$
(iii) Thus the numerator in (13) becomes:
$\mathbf\small{(\hat{i}+1.5\hat{j}+\hat{k})+(0.75\hat{i}+1.5\hat{j}+0.75\hat{k})}$
$\mathbf\small{=(1.75\hat{i}+3.0\hat{j}+1.75\hat{k})}$
• The denominator M is the total mass = (m_P + m_Q) = (0.5+0.75) = 1.25 kg
(iv) Thus we get:
$\mathbf\small{\vec{r}_C=\frac{1.75\hat{i}+3.0\hat{j}+1.75\hat{k}}{1.25}=1.4\hat{i}+2.4\hat{j}+1.4\hat{k}}$
(v) So the coordinates of C are: (1.4,2.4,1.4)
• This is the same result that we obtained before

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15. We will generalize this method:

• P, Q, R, S, . . . are various particles in a system
• There are a total of 'n' particles in the system
    ♦ P is the first particle
    ♦ Q is the second particle
    ♦ R is the third particle . . . so on . . .
• Then:
    ♦ m_P is the first mass
    ♦ m_Q is the second mass
    ♦ m_R is the third mass . . . so on . . .
16. We can write:
    ♦ m₁ which is equal to m_P is the first mass
    ♦ m₂ which is equal to m_Q is the second mass
    ♦ m₃ which is equal to m_R is the third mass . . . so on . . .
• Then:
    ♦ m_i is the i^th mass
    ♦ m_n is the last mass
17. Also we can write:
    ♦ $\mathbf\small{\vec{r}_1}$ which is equal to $\mathbf\small{\vec{r}_P}$ is the first position vector
    ♦ $\mathbf\small{\vec{r}_2}$ which is equal to $\mathbf\small{\vec{r}_Q}$ is the second position vector
    ♦ $\mathbf\small{\vec{r}_3}$ which is equal to $\mathbf\small{\vec{r}_R}$ is the third position vector . . . so on . . .
• Then:
    ♦ $\mathbf\small{\vec{r}_i}$ is the i^th position vector
    ♦ $\mathbf\small{\vec{r}_n}$ is the last position vector
18. So the equation in (13) becomes:
$\mathbf\small{\vec{r}_C=\frac{m_1\,\vec{r}_1+m_2\,\vec{r}_2\,+\,.\,.\,.\,+\,m_i\,\vec{r}_i+\,.\,.\,.\,+\,m_n\,\vec{r}_n}{m_1+m_2\,+\,.\,.\,.\,+\,m_i+\,.\,.\,.\,+\,m_n}}$
$\mathbf\small{\Rightarrow \vec{r}_C=\frac{m_1\,\vec{r}_1+m_2\,\vec{r}_2\,+\,.\,.\,.\,+\,m_i\,\vec{r}_i+\,.\,.\,.\,+\,m_n\,\vec{r}_n}{M}}$
19. Thus we get 
Eq.7.4: $\mathbf\small{\vec{r}_C=\frac{\sum{} \,m_i\,\vec{r}_i}{M}}$
• This equation is applicable to 2- dimensional and 1-dimensional problems as well

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• Let us apply it to a two dimensional problem that we solved in the previous section

• We will consider solved example 7.3. We will do it as a new solved example:

Solved example 7.7
Three particles P, Q and R are situated at the vertices of an equilateral triangle of side 0.5 m. Their masses are 100 g, 150 g and 200 g respectively. Find the location of the 'C' of this system of 3 particles
Solution:
1. We have three particles so situated in space that, they are at the vertices of an equilateral triangle
• For ease in calculations, we arrange the frame of reference in the following way:
    ♦ The plane of the triangle lies in the xy-plane
    ♦ One of the sides (say PQ), lies on the x-axis
    ♦ The left vertex (P) of that side coincides with O
• This arrangement is shown in fig.7.35 below:

Fig.7.35

2. Once this arrangement is fixed, we can easily write the coordinates of the bottom vertices
• The coordinates of P will be (0,0)
• The coordinates of Q will be (0.5,0)
• The x-coordinate of R will be 0.25
3. To find the y-coordinate of R, we use the following steps:
• Drop a perpendicular from R. This is shown in fig.b
• The foot of this perpendicular is R'
• In the right triangle PR'R, we have:
$\mathbf\small{\sin 60=\frac{RR'}{PR}=\frac{RR'}{0.5}}$
$\mathbf\small{\Rightarrow RR'=\sin 60 \times 0.5=0.433}$
So the y-coordinate of R is 0.433
4. Once we obtain the coordinates, we can write the position vectors:
(i) $\mathbf\small{\vec{r}_P=0\hat{i}+0\hat{j}}$
• This is a null vector
(ii) $\mathbf\small{\vec{r}_Q=0.5\hat{i}+0\hat{j}}$
(iii) $\mathbf\small{\vec{r}_R=0.25\hat{i}+0.433\hat{j}}$
5. We have: $\mathbf\small{\vec{r}_C=\frac{\sum{} \,m_i\,\vec{r}_i}{M}}$
• The numerator becomes: $\mathbf\small{0.1 \times 0+0.15 \times 0.5\hat{i}+0.2 \times 0.25\hat{i}+0.2 \times 0.433\hat{j}}$
= $\mathbf\small{0.125\hat{i}+0.0866\hat{j}}$ 
• The denominator is the total mass M = 0.450 kg
6. Thus we get: $\mathbf\small{\vec{r}_C=\frac{0.125\hat{i}+0.0866\hat{j}}{0.45}=0.278 \hat{i}+0.192 \hat{j}}$
7. So the coordinates of 'C' are: (0.278,0.192)
• This is the same result that we obtained earlier
• Note that, the unit vector $\mathbf\small{\hat{k}}$ does not come in the calculations because, this is a 2-dimensional problem

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So now we know how to find the location of the 'C' of any given system. In the next section, we will see the significance of 'C'

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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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