More Solved examples involving Torque and Center of gravity

This page shows two more solved examples in continuation of the examples that we saw here: Solved examples involving torque.

Example 1
From a uniform disk of radius R, a circular hole of radius R/2 is cut out. The centre of the hole is at R/2 from the centre of the original disc. Locate the centre of gravity of the resulting flat body
Solution:
1. The uniform circular disc can be supported at it's center as shown in fig.1(a) below:

Fig.1

• But when a hole is cut on the right side, the disc will tilt towards the left and fall off. This is shown in fig.1(b)
2. If we shift the support towards the left (along the diameter), the disc will balance again
• This is shown in fig.2(a) below:

Fig.2

• In fig.2(a), the disc is supported at the new CG
• We have to find the position of the new CG
3. In fig.2(b), the small circle is painted red and glued back to get the original disc
• The CG of this 'composite disc' will be same as the center of the original disc 
4. In such a situation, we apply the conditions of equilibrium
• Let the brown portion in fig.2(b) be indicated as 'Brown'
• Let the red portion in fig.2(b) be indicated as 'Red'
• Let 'mass per unit area' of the disc be M
• Since the disc is uniform, this 'M' will be same at all points on the disc
5. Fig.3(a) below shows the dimensions

Fig.3

• Area of Red = $\mathbf\small{\pi\,\left(\frac{R}{2}\right)^2=\frac{\pi R^2}{4}}$
    ♦ So $\mathbf\small{W_{Red}=\frac{\pi R^2Mg}{4}}$
• Area of Brown = $\mathbf\small{\pi\,R^2-\frac{\pi R^2}{4}=\frac{3\pi R^2}{4}}$
    ♦ So $\mathbf\small{W_{Brown}=\frac{3\pi R^2Mg}{4}}$ 
6. Fig.3(b) shows the forces
• The distance between the required CG and the support is denoted as 'x'
7. Since the composite disc is in translational equilibrium, we have:
R - W_Brown - W_Red = 0
8. Since the composite disc is in rotational equilibrium, we have: net torque = 0
• Let us take the torques about the support 
(i) Torque created by W_Brown about the support = W_Brown × x (anti clockwise)
(ii) Torque created by R about the support = 0
(iii) Torque created by W_Red about the support = W_Red × ^R⁄₂ (clockwise)
9. So applying the condition, we get:
-(W_Brown × x) + (W_Red × ^R⁄₂) = 0
$\mathbf\small{\Rightarrow x=\frac{R\,W_{Red}}{2\,W_{Brown} }}$
• Substituting the known values, we get: $\mathbf\small{x=\frac{\pi R^3Mg}{4}\div \frac{6\pi R^2Mg}{4}}$
$\mathbf\small{\Rightarrow x=\frac{\pi R^3Mg}{4}\times \frac{4}{6\pi R^2Mg}=\frac{R}{6}}$

Example 2
A metre stick is balanced on a knife edge at its centre. When two coins, each of mass 5 g are put one on top of the other at the 12.0 cm mark, the stick is found to be balanced at 45.0 cm. What is the mass of the metre stick?
Solution:
1. Given that, the metre stick is balanced on a knife edge at its centre 
• That means, if the support is given at the exact center, the metre stick will balance
• That means, it is a uniform scale
• This is shown in fig.2(a) below:

Fig.2

2. The condition when the two coins are placed at the 12 cm mark is shown in fig.b
• The forces and dimensions are also shown
• We see that, for equilibrium, the support has to be shifted by 5 cm towards the left
3. Since the system in fig.b is in translational equilibrium, we have:
R - W_Coin - W_Stick = 0
8. Since the system is in rotational equilibrium, we have: net torque = 0
• Let us take the torques about the support 
(i) Torque created by W_Coin about the support = W_Coin × 0.33 (anti clockwise)
(ii) Torque created by R about the support = 0
(iii) Torque created by W_Stick about the support = W_Stick × 0.05 (clockwise)
9. So applying the condition, we get:
-(W_Coin × 0.33) + (W_Stick × 0.05) = 0
W_Stick = (W_Coin × ^0.33⁄_0.05 ) = (10 ×  ¹⁄₁₀₀₀ ×  ^0.33⁄_0.05 ) = 0.066 kg = 66 g

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Chapter 7.22 - Solved examples involving Torque

In the previous section, we saw center of gravity. In this section we will see some solved examples

Solved example 7.21
The metal bar in fig.7.103(a) is 0.70 m long and has a mass of 4 kg. It is supported on two knife edges placed 0.10 m from each end. A 6 kg mass is suspended from a point P, which is 0.20 m from the left support

Fig.7.103

Find the reactions at the supports. Assume the bar is of uniform cross section and homogeneous. Take g = 9.8 ms^-2
Solution:
1. Let us name the supports as A and B
■ The weight of the rod acts at it’s CG
• Given that, the metal bar is of uniform cross section and homogeneous.
■ So the CG of the rod will be at it’s geometric center
2. The detailed measurements are shown in fig.b
The CG is marked as G
3. Since the bar is in translational equilibrium, we have:
R_A + R_B - 6g - 4g = 0
⇒ R_A + R_B = 10g = 98.0 N.
4. Since the bar is in rotational equilibrium, we have: net torque = 0
• Let us take the torques about the support A
(i) Torque created by R_A about A = zero
(ii) Torque created by 6g about A = 6g × 0.2 = 1.2g Nm (clockwise)
(iii) Torque created by 4g about A = 4g × 0.25 = 1.0g Nm (clockwise)
(iv) Torque created by R_B about A = R_B × 0.5 Nm (anti clockwise)
5. So applying the condition, we get:
1.2g + g - 0.5R_B = 0
⇒ R_B = 43.12 N
• Substituting this value of R_B in (3), we get: R_A = (98-43.12) = 54.88 N

Solved example 7.22
A 3 m long ladder having a mass of 20 kg, leans on a frictionless wall. It’s feet rest on the ground 1 m from the wall as shown in fig.7.104(a) below:

Fig.7.104

Find the reaction forces on the wall and the floor. Take g = 9.8 ms^-2
Solution:
1. Let the end points of the ladder be A and B
• A is 1 m from the wall. This is shown in fig.b
• Let C be the foot of the wall
2. The reaction from the floor at A will be normal to the floor
• This reaction is denoted as R_A
3. The frictional force prevents the point A from moving away from C
• This frictional force is denoted as F. It pulls the ladder towards C. Other wise the ladder will slip
• Also recall that, the frictional force is always parallel to the surface
4. The reaction from the wall at B will be normal to the wall
• This reaction is denoted as R_B
5. Given that, the wall is frictionless
• If there was friction, a force F would have acted parallel to the wall in the upward direction
• This force would have made some contribution towards: 'preventing the movement of B towards C'   • In other words, this force would have made some contribution towards: 'preventing the ladder from slipping'
• In addition to that, this force would have made some contribution towards resisting the vertical force (20 g) of the ladder
    ♦ Where 'g' is the acceleration due to gravity
• But in this problem there is no such force
    ♦ The slipping is prevented entirely by the horizontal force F at A
    ♦ The vertical load 20 g is resisted entirely by the vertical force R_A at A
6. The weight of the ladder is 20 g
• It acts downwards at the CG of the ladder
• The CG is marked as G in fig.b
7. The above steps gives us all the 4 forces acting on the ladder
• Now we can apply the conditions of equilibrium
• Since the ladder is in translational equilibrium, we have:
(i) In the x direction: F - R_B = 0
(ii) In the y direction: R_A - 20 g = 0
⇒ R_A = 20 g = 196 N
8. Since the bar is in rotational equilibrium, we have: net torque = 0
Let us take the torques about A
(i) Torque created by R_A about A = zero
(ii) Torque created by F about A = zero
(iii) Torque created by R_B about A = R_B × BC
So we want the length of BC
• Applying Pythagoras theorem to the right triangle ABC, we get:
$\mathbf\small{BC=\sqrt{AB^2-AC^2}=\sqrt{3^2-1^2}=2\sqrt{2}\;\text{m}}$
Thus the required torque = $\mathbf\small{2\sqrt{2}R_B\;\text{N m}}$ (clockwise) 
(iv) Torque created by 20 g about A = 20 g × AD
• So we want the length of AD
• D is the foot of the perpendicular drawn from G
• Consider the similar triangles ABC and AGD
• We have: $\mathbf\small{\frac{AD}{AC}=\frac{AG}{AB}}$
$\mathbf\small{\Rightarrow \frac{AD}{1}=\frac{1.5}{3}}$
⇒ AD = 0.5 m
• Thus the torque = 20 g × 0.5 × 9.8 = 98 N m (anti clockwise)
9. Applying the condition, we get:
$\mathbf\small{2\sqrt{2}R_B-98=0}$
⇒ R_B = 34.65 N
10. Substituting this value of R_B in 7(i), we get: F = 34.65 N.
11. At the point A, two forces are acting on the ladder:
• R_A vertically and F horizontally. They are shown in fig.c
• The resultant of the two forces = $\mathbf\small{\sqrt{(R_A)^2+F^2}=\sqrt{196^2+34.65^2}=}$ 199.04 N
• Let this resultant make an angle α with the horizontal
• Then $\mathbf\small{\alpha=\tan^{-1}\frac{R_A}{F}=\tan^{-1}\frac{196}{34.65}=}$ 79.97^o

Solved example 7.23 
A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in fig.7.105(a) below:

Fig.7.105

The angles made by the strings with the vertical are 36.9^o and 53.1^o respectively. The bar is 2 m long. Calculate the distance d of the center of gravity of the bar from it’s left end
Solution:
1. The free body diagram is shown in fig.b
• Let T₁ and T₂ be the tensions in the strings
• In fig.a, we see that, the left string makes an angle of 36.9^o with the vertical wall
    ♦ In fig.b, we see that, this string makes the same angle with the vertical dotted line
    ♦ The angles are same because, they are alternate angles 
• In fig.a, we see that, the right string makes an angle of 53.1^o with the vertical wall
    ♦ In fig.b, we see that, this string makes the same angle with the vertical dotted line
    ♦ The angles are same because, they are alternate angles
• Now we can resolve the tensions into horizontal and vertical components. This is shown in fig.c
2. Since the bar is in translational equilibrium, we have:
(i) In the x direction: - T₁ sin θ₁ + T₂ sin θ₂  = 0
⇒ T₁ sin θ₁ = T₂ sin θ₂.
⇒ T₁ sin 36.9 = T₂ sin 53.1 = T₂ cos (90 - 36.9) = T₂ cos 36.9
⇒ $\mathbf\small{\frac{T_2}{T_1}=\tan36.9 =0.75}$
⇒ T₂ = 0.75 T₁
(ii) In the y direction: T₁ cos θ₁ + T₂ cos θ₂ - W = 0
⇒ T₁ cos 36.9 + T₂ cos 53.1 = W
⇒0.8 T₁ + 0.6 T₂ = W
3. Since the bar is in rotational equilibrium, we have: net torque = 0
• Let us take the torques about G
(i) Torque created by T₁ sin θ₁ about G = zero
(ii) Torque created by T₁ cos θ₁ about G = T₁ cos θ₁ × d (clockwise)
(iii) Torque created by W about G = zero
(iv) Torque created by T₂ cos θ₂ about G = T₂ cos θ₂ × (2-d) (anti clockwise)
(i) Torque created by T₂ sin θ₂ about G = zero
4. Applying the condition, we get:
[T₁ cos θ₁ × d] - [T₂ cos θ₂ × (2-d)] = 0
⇒ [T₁ × cos 36.9 × d] - [0.75 T₁ × cos 53.1 × (2-d)] = 0
(∵ from 2(i), we have: T₂ = 0.75 T₁)
⇒ [0.8 T₁ d] - [0.45 T₁ (2-d)] = 0
⇒ 0.8 T₁ d - 0.9 T₁ + 0.45 T₁ d = 0
⇒ 1.25 T₁ d = 0.9 T₁
⇒ 1.25 d = 0.9
⇒ d = 0.72 m

Solved example 7.24
A car weighs 1800 kg. The distance between it’s front and back axles is 1.8 m. It’s center of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel
Solution:
1. Fig.7.106 below shows the schematic diagram

Fig.106

• The small grey circles at the center of the wheels denote the axles
• The front and rear axles are named as A and B respectively
2. When the car is in translational equilibrium, we have:
R_A + R_B -1800g = 0
3. When the car is in rotational equilibrium, we have: net torque = 0
Let us take the torques about G
(i) Torque created by R_A about G = 1.05 R_A (clockwise) 
(ii) Torque created by 1800g about G = zero
(iii) Torque created by R_B about G = 0.75 R_B (anticlockwise)
4. Applying the condition, we get:
1.05 R_A - 0.75 R_B = 0
1.05 R_A = 0.75 R_B
R_B = 1.4 R_A
5. Substituting this in (2), we get:
R_A + 1.4R_A = 1800 × 9.8
⇒ R_A = 7350 N
• So R_B = 1.4 × 7350 = 10290 N
6. The reaction on the front axle A is 7350 N
• But the load from this axle is resisted by two front wheels
• So reaction on each of the front wheels = ⁷³⁵⁰⁄₂ = 3675 N
7. The reaction on the back axle B is 10290 N
• But the load from this axle is resisted by two back wheels
• So reaction on each of the back wheels = ¹⁰²⁹⁰⁄₂ = 5145 N

Solved example 7.25
As shown in fig.7.107(a), the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE 0.5 m long is tied half way up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. Take g = 9.8 ms^-2 [Hint: Consider the equilibrium of each side of the ladder separately]

Fig.7.107

Solution:
1. The detailed measurements are shown in fig.b
• Given:
    ♦ BD = 0.8 m
    ♦ DF = FA = 0.4 m
    ♦ DE = 0.5 m
• Let ∠ABC = θ
    ♦ Since DE is parallel to the ground, we get: ∠ADE = θ   
2. Let us calculate the other required dimensions:
• A perpendicular is dropped from A onto BC
    ♦ This is shown as red dashed line 
    ♦ The foot of the perpendicular on BC is O
    ♦ The foot of the perpendicular on DE is G
• Applying Pythagoras theorem to the right triangle ADG, we get:
$\mathbf\small{AG=\sqrt{AD^2-DG^2}=\sqrt{0.8^2-0.25^2}=0.76\;\text{m}}$
3. In the similar triangles ABO and ADG, we get:
• $\mathbf\small{\frac{AD}{AB}=\frac{AG}{AO}}$
$\mathbf\small{\Rightarrow \frac{0.8}{1.6}=\frac{0.76}{AO}}$
⇒ AO = 1.52 m
4. Again in the same similar triangles ABO and ADG, we get:
$\mathbf\small{\frac{AD}{AB}=\frac{DG}{BO}}$
$\mathbf\small{\Rightarrow \frac{0.8}{1.6}=\frac{0.25}{BO}}$
⇒ BO = 0.5 m
5. A perpendicular is dropped from F onto AG
• The foot of this perpendicular is H
• In the similar triangles AFH and ADG, we get:
$\mathbf\small{\frac{AF}{AD}=\frac{FH}{DG}}$
$\mathbf\small{\Rightarrow \frac{0.4}{0.8}=\frac{FH}{0.25}}$
⇒ FH = 0.125 m
6. Let us write the above distances together:
    ♦ BD = 0.8 m
    ♦ DF = FA = 0.4 m
    ♦ DE = 0.5 m
    ♦ AG =0.76 m
    ♦ AO = 1.52 m
    ♦ BO = 0.5 m
    ♦ FH = 0.125 m
• Thus we obtained all the distances that we will soon require for torque calculations
7. The forces are shown in fig.c below:

Fig.7.107 (c) & (d)

• R_B is the normal reaction at B
• R_C is the normal reaction at C
• Note that the tensions T in the string are internal forces and so will cancel each other
8. For translational equilibrium, we have:
• R_B + R_C - (40 × 9.8) = 0
⇒ R_B + R_C - 392 = 0
⇒ R_B + R_C = 392
9. For rotational equilibrium, we need to calculate the torques first:
• Let us find the torques about O
(i) Torque created by R_B about O = R_B × BO = R_B × 0.5 = 0.5R_B Nm (clockwise)
(ii) Torque created by the 40 kg load about O = (40 × 9.8) × FH = 392 0.125 = 49 Nm (anti clockwise)
(iii) Torque created by R_C about O = R_C × CO = R_C × 0.5 = 0.5R_C Nm (anti clockwise)
10. Applying the condition, we get:
⇒ 0.5R_B - 49 - 0.5R_C = 0
⇒ 0.5R_B - 0.5R_C = 49
⇒ R_B - R_C = 98
11. Solving the equations in (8) and (10), we get:
R_B = 245 N, R_C = 147 N
12. Next we want to find T
• In the fig.d above, the FBD of side AB is shown separately
• The forces acting are: R_B, T and the load of 40 kg
• Let us calculate the torques about A:
(i) Torque created by R_B about A = R_B × BO = R_B × 0.5 = 0.5R_B Nm (clockwise)
(ii) Torque created by T about A = T × AG = T × 0.76 = 0.76T Nm (anti clockwise)
(ii) Torque created by the 40 kg load about A = (40 × 9.8) × FH = 392 × 0.125 = 49 Nm (anti clockwise)
13. Applying the condition, we get:
0.5R_B - 0.76T - 49 = 0
0.5 × 245 - 0.76T - 49 = 0
T = 96.7 N

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Two more solved examples can be seen here

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In the next section, we will see moment of inertia

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Chapter 7.19 - Equilibrium of a Rigid body

In the previous section, we completed a discussion on torque and angular momentum. In this section we will see equilibrium of rigid body

1. We know that, if the net external force on a body is zero, that body will be having zero acceleration 
• That means
    ♦ If that body is at rest, it will continue to be at rest
    ♦ If that body is in uniform motion, it will continue to be in uniform motion
2. Another way of describing this situation is:
• There is no change in the linear momentum of the body 
3. In this situation, we are inclined to say that, the body is in equilibrium
• But that may not be true. Let us see why:
(i) We add all the external forces and find that $\mathbf\small{\sum{\vec{F}_{ext}}=\vec{0}}$
($\mathbf\small{\vec{0}}$ is the null vector which has zero magnitude and no direction)
(ii) But there may be a net torque. That is., $\mathbf\small{\sum{\vec{\tau}_{ext}}\neq \vec{0} }$
(iii) This torque will cause that body to rotate
(iv) So that body is not in equilibrium 
4. Another way of describing this situation is:
• Due to the presence of the net torque, the angular momentum of the body changes
5. So we can write:
• A body is in equilibrium only if both the conditions given below are satisfied:
(i) $\mathbf\small{\sum{\vec{F}_{ext}}= \vec{0} }$
    ♦ This condition corresponds to translational equilibrium
(ii) $\mathbf\small{\sum{\vec{\tau}_{ext}}= \vec{0}}$
    ♦ This condition corresponds to rotational equilibrium  
6. Now a question arises:
• We know that, to find the torque, we need a point of reference. We usually take the origin ‘O’ as the point of reference
• If we shift that origin to a suitable point,  can the net torque become zero?
7. This question can be elaborated as follows:
• A body is in translational equilibrium. 
• But it is not in rotational equilibrium due to the presence of a net torque about the selected origin. 
• Can we obtain rotational equilibrium by considering a new suitable point as the origin?
• The answer is ‘No’. The net torque does not depend on the position of 'O'. We will see the reason later in this section
8. The equation $\mathbf\small{\sum{\vec{F}_{ext}}=\vec{0} }$ is a vector equation
• So there will be 3 component equations:
(i) $\mathbf\small{\sum{\vec{F}_{x,ext}}=\vec{0} }$ (ii) $\mathbf\small{\sum{\vec{F}_{y,ext}}=\vec{0} }$ (iii) $\mathbf\small{\sum{\vec{F}_{z,ext}}=\vec{0} }$
    ♦ When all the forces in the x direction are added, the result must be a null vector ($\mathbf\small{\vec{0}}$)
    ♦ When all the forces in the y direction are added, the result must be ($\mathbf\small{\vec{0}}$)
    ♦ When all the forces in the z direction are added, the result must be ($\mathbf\small{\vec{0}}$) 
• If the null vector is not obtained in any one of the three directions, we will not get $\mathbf\small{\sum{\vec{F}_{ext}}=\vec{0}}$
9. Similarly, the equation $\mathbf\small{\sum{\vec{\tau}_{ext}}=\vec{0} }$ is a vector equation
• So there will be 3 component equations:
(i) $\mathbf\small{\sum{\vec{\tau}_{x,ext}}=\vec{0} }$ (ii) $\mathbf\small{\sum{\vec{\tau}_{y,ext}}=\vec{0} }$ (iii) $\mathbf\small{\sum{\vec{\tau}_{z,ext}}=\vec{0} }$
• When the torques of all the 'forces lying in the xy-plane' are added, the result must be $\mathbf\small{\vec{0}}$
• We know that, torques of 'forces lying in the xy-plane' will be in the z-direction
• Sum of all torques in the z-direction must be $\mathbf\small{\vec{0}}$     
    ♦ That is: $\mathbf\small{\sum{\vec{\tau}_{z,ext}}=\vec{0} }$
• When the torques of all the 'forces lying in the yz-plane' are added, the result must be $\mathbf\small{\vec{0}}$
• We know that, torques of 'forces lying in the yz-plane' will be in the x-direction
• Sum of all torques in the x-direction must be $\mathbf\small{\vec{0}}$     
    ♦ That is: $\mathbf\small{\sum{\vec{\tau}_{x,ext}}=\vec{0} }$
• When the torques of all the 'forces lying in the xz-plane' are added, the result must be $\mathbf\small{\vec{0}}$
• We know that, torques of 'forces lying in the xz-plane' will be in the y-direction
• Sum of all torques in the y-direction must be $\mathbf\small{\vec{0}}$     
    ♦ That is: $\mathbf\small{\sum{\vec{\tau}_{y,ext}}=\vec{0} }$
10. So there are 6 equations to be satisfied for equilibrium of a rigid body
Eq.7.23:
$\mathbf\small{\sum{\vec{F}_{ext}}=\vec{0} }$
$\mathbf\small{\Rightarrow \sum{\vec{F}_{x,ext}}=\vec{0},\;\;\sum{\vec{F}_{y,ext}}=\vec{0},\;\;\sum{\vec{F}_{z,ext}}=\vec{0}}$
Eq.7.24:
$\mathbf\small{\sum{\vec{\tau}_{ext}}=\vec{0} }$
$\mathbf\small{\Rightarrow \sum{\vec{\tau}_{x,ext}}=\vec{0},\;\;\sum{\vec{\tau}_{y,ext}}=\vec{0},\;\;\sum{\vec{\tau}_{z,ext}}=\vec{0}}$

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The situation will be greatly simplified if all the forces lie in a plane. This can be explained with the help of an example:

1. In the fig.7.93(a) below, a light (ie. of negligible mass) rod AB lies on the xy-plane

Fig.7.93

• It’s length is ‘2a’
• It’s midpoint is ‘C’
• So we have AC = BC = a
2. Two forces are applied on the rod
• Those two forces have the following five peculiarities:
(i) Both forces have the same magnitude
(ii) Both forces have the same directions
    ♦ Let us assume the usual orientation: x-axis is horizontal and y-axis is vertical
    ♦ Then both forces act towards the negative side of the y-axis
(iii) One force is applied at A
(iv) The other force is applied at B
(v) Both forces are perpendicular to AB
3. Let us apply the conditions one by one:
(i) Check whether
$\mathbf\small{\sum{\vec{F}_{x,ext}}=\vec{0}}$
• There are no forces in the x-direction. So we do not have to check this condition
(ii) Check whether
$\mathbf\small{\sum{\vec{F}_{y,ext}}=\vec{0}}$
• If this condition is to be satisfied, ($\mathbf\small{\vec{F}_A+\vec{F}_B}$) must be zero vector
• Given that: $\mathbf\small{\vec{F}_B=\vec{F}_A}$
So ($\mathbf\small{\vec{F}_A+\vec{F}_B}$) is definitely not equal to a zero vector
(iii) Check whether
$\mathbf\small{\sum{\vec{F}_{z,ext}}=\vec{0}}$
• There are no forces in the z-direction. So we do not have to check this condition

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• Thus we applied the first 3 conditions for translational equilibrium

• We see that, the system does not satisfy the second condition
    ♦ Because $\mathbf\small{\sum{\vec{F}_{y,ext}} \neq \vec{0}}$ 
• So the rod AB is not in translational equilibrium

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4. For applying the next 3 conditions, we have to first find the magnitudes and directions of the torques

• Let ‘C’ be the origin. Then we can mark the position vectors as shown in fig.b
5. Torque created by $\mathbf\small{\vec{F}_A}$ about C 
= $\mathbf\small{(\vec{r}_A \times \vec{F}_A)}$
• Magnitude of this torque
= $\mathbf\small{|\vec{r}_A|\times |\vec{F}_A|\times \sin \theta}$
= $\mathbf\small{[a\times |\vec{F}_A|\times \sin 90]=a\times |\vec{F}_A|}$
(θ = 90^o because, forces are perpendicular to the rod)
6. For finding the direction of this torque:
(i) Assume that $\mathbf\small{\vec{F}_A}$ is shifted so that, it's tail end coincides with the tail end of $\mathbf\small{\vec{r}_A}$  
(ii) Imagine that a right handed screw is placed perpendicular to the computer screen
(iii) Turn it from $\mathbf\small{\vec{r}_A}$ to $\mathbf\small{\vec{F}_A}$
(iv) We see that, the screw moves away from the screen, towards us
(v) The rod is lying on the xy-plane. So the computer screen is the xy-plane
(vi) The z-axis is perpendicular to the computer screen
(vii) Let us assume that, the positive direction of the z-axis points towards us
(viii) So we can write:
The torque created by $\mathbf\small{\vec{F}_A}$ is perpendicular to the xy-plane and is directed towards the +ve side of the z axis
■ Combining the magnitude and direction, we can write:
Torque created by $\mathbf\small{\vec{F}_A}$ = $\mathbf\small{\left[a\times |\vec{F}_A|\right]\;\hat{k}}$ 
7. Next we find the torque created by $\mathbf\small{\vec{F}_B}$ about C 
= $\mathbf\small{(\vec{r}_B \times \vec{F}_B)}$
• Magnitude of this torque
= $\mathbf\small{|\vec{r}_B|\times |\vec{F}_B|\times \sin \theta}$
= $\mathbf\small{[a\times |\vec{F}_B|\times \sin 90]=a\times |\vec{F}_B|}$
(θ = 90^o because, forces are perpendicular to the rod)
8. For finding the direction of this torque:
(i) Assume that $\mathbf\small{\vec{F}_B}$ is shifted so that, it's tail end coincides with the tail end of $\mathbf\small{\vec{r}_B}$  
(ii) Imagine that a right handed screw is placed perpendicular to the computer screen
(iii) Turn it from $\mathbf\small{\vec{r}_B}$ to $\mathbf\small{\vec{F}_B}$
(iv) We see that, the screw moves into the screen, away from us
(v) The rod is lying on the xy-plane. So the computer screen is the xy-plane
(vi) The z-axis is perpendicular to the computer screen
(vii) We already assumed that, the positive direction of the z-axis points towards us
(viii) So we can write:
The torque created by $\mathbf\small{\vec{F}_B}$ is perpendicular to the xy-plane and is directed towards the -ve side of the z axis
■ Combining the magnitude and direction, we can write:
Torque created by $\mathbf\small{\vec{F}_B}$ = $\mathbf\small{\left[-a\times |\vec{F}_B|\right]\;\hat{k}}$
9. Let us write a summary about the torques:
• The torque created by $\mathbf\small{\vec{F}_A}$:
$\mathbf\small{\left[a\times |\vec{F}_A|\right]\;\hat{k}}$
• The torque created by $\mathbf\small{\vec{F}_B}$:
$\mathbf\small{\left[-a\times |\vec{F}_B|\right]\;\hat{k}}$
10. But $\mathbf\small{|\vec{F}_A|=|\vec{F}_B|}$
• So we can write:
The two torques have the same magnitude. But they are opposite in directions
• When we add such vectors, we get a null vector
11. So in this case, the net torque in the z-direction is zero
That is: $\mathbf\small{\sum{\vec{\tau}_{z,ext}}=\vec{0} }$
12. In this problem, all the forces lie in the xy-plane
• There will not be any torques in the x and y directions
• That is., we do not have to check these:
    ♦ Whether $\mathbf\small{\sum{\vec{\tau}_{x,ext}}=\vec{0} }$ 
    ♦ Whether $\mathbf\small{\sum{\vec{\tau}_{y,ext}}=\vec{0} }$

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So for this example, we can write:

• While applying the second set of 3 conditions, only one along the z direction is relevant
• The other two need not be checked
• Along the z direction, we get:  $\mathbf\small{\sum{\vec{\tau}_{z,ext}}=\vec{0} }$
• So the rod AB is in rotational equilibrium

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• In this example, out of the 6 conditions, we used only 2. They are:
    ♦ $\mathbf\small{\sum{\vec{F}_{y,ext}}=\vec{0}}$
    ♦ $\mathbf\small{\sum{\vec{\tau}_{z,ext}}=\vec{0} }$
• All the forces lie in the xy-plane
• Also all the forces are vertical
■ Some times we may get problems in which:
• Some of the forces in the xy-plane are horizontal
• Some of the forces in the xy-plane are inclined
    ♦ Then there will be horizontal components as well
• In such cases, we will have to check whether $\mathbf\small{\sum{\vec{F}_{x,ext}}=\vec{0}}$
■ In general, when all the forces lie in a plane, instead of 6, we need to check 3 conditions only

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The above example also helps us to conclude that, a rigid body can have rotational equilibrium without having translational equilibrium

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Let us see another example:
1. In the fig.7.94(a) below, the same rod AB lies on the same xy-plane

Fig.7.94

2. Two forces are applied on the rod
• Those two forces have the following five peculiarities:
(i) Both forces have the same magnitude
(ii) But they have opposite directions
    ♦ One force acts towards the +ve side of the y-axis
    ♦ The other force acts towards the -ve side of the y-axis
(iii) One force is applied at A
(iv) The other force is applied at B
(v) Both forces are perpendicular to AB
3. We have already seen that, if all the forces lie in the xy-plane, we need to check 3 conditions only:
$\mathbf\small{\sum{\vec{F}_{x,ext}}=\vec{0}}$
$\mathbf\small{\sum{\vec{F}_{y,ext}}=\vec{0}}$
$\mathbf\small{\sum{\vec{\tau}_{z,ext}}=\vec{0}}$
Further, there are no forces in the x-direction. So we need to check 2 conditions only:
$\mathbf\small{\sum{\vec{F}_{y,ext}}=\vec{0}}$
$\mathbf\small{\sum{\vec{\tau}_{z,ext}}=\vec{0}}$
4. First we will check whether
$\mathbf\small{\sum{\vec{F}_{y,ext}}=\vec{0}}$
• If this condition is to be satisfied, ($\mathbf\small{\vec{F}_A+\vec{F}_B}$) must be zero
• Given that: $\mathbf\small{\vec{F}_B=-\vec{F}_A}$
So ($\mathbf\small{\vec{F}_A+\vec{F}_B}$) is indeed equal to a zero vector

---

• Thus we applied the required condition for translational equilibrium

• We see that, the system satisfies the condition
    ♦ Because $\mathbf\small{\sum{\vec{F}_{y,ext}}}$ is indeed equal to zero vector 
• So the rod AB is in translational equilibrium

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5. For applying the next condition, we have to first find the magnitudes and directions of the torques created by the forces in the xy-plane
• Let ‘C’ be the origin. Then we can mark the position vectors as shown in fig.b
• Torque created by $\mathbf\small{\vec{F}_A}$ about C 
= $\mathbf\small{(\vec{r}_A \times \vec{F}_A)}$
• Magnitude of this torque
= $\mathbf\small{|\vec{r}_A|\times |\vec{F}_A|\times \sin \theta}$
= $\mathbf\small{[a\times |\vec{F}_A|\times \sin 90]=a\times |\vec{F}_A|}$
(θ = 90^o because, forces are perpendicular to the rod)
6. For finding the direction of this torque:
(i) Assume that $\mathbf\small{\vec{F}_A}$ is shifted so that, it's tail end coincides with the tail end of $\mathbf\small{\vec{r}_A}$  
(ii) Imagine that a right handed screw is placed perpendicular to the computer screen
(iii) Turn it from $\mathbf\small{\vec{r}_A}$ to $\mathbf\small{\vec{F}_A}$
(iv) We see that, the screw moves away from the screen, towards us
(v) The rod is lying on the xy-plane. So the computer screen is the xy-plane
(vi) The z-axis is perpendicular to the computer screen
(vii) Let us assume that, the positive direction of the z-axis points towards us
(viii) So we can write:
The torque created by $\mathbf\small{\vec{F}_A}$ is perpendicular to the xy-plane and is directed towards the +ve side of the z axis
■ Combining the magnitude and direction, we can write:
Torque created by $\mathbf\small{\vec{F}_A}$ = $\mathbf\small{\left[a\times |\vec{F}_A|\right]\;\hat{k}}$
7. Next we find the torque created by $\mathbf\small{\vec{F}_B}$ about C 
= $\mathbf\small{(\vec{r}_B \times \vec{F}_B)}$
• Magnitude of this torque
= $\mathbf\small{|\vec{r}_B|\times |\vec{F}_B|\times \sin \theta}$
= $\mathbf\small{[a\times |\vec{F}_B|\times \sin 90]=a\times |\vec{F}_B|}$
(θ = 90^o because, forces are perpendicular to the rod)
8. For finding the direction of this torque:
(i) Assume that $\mathbf\small{\vec{F}_B}$ is shifted so that, it's tail end coincides with the tail end of $\mathbf\small{\vec{r}_B}$  
(ii) Imagine that a right handed screw is placed perpendicular to the computer screen
(iii) Turn it from $\mathbf\small{\vec{r}_B}$ to $\mathbf\small{\vec{F}_B}$
(iv) We see that, the screw moves away from the screen, towards us
(v) The rod is lying on the xy-plane. So the computer screen is the xy-plane
(vi) The z-axis is perpendicular to the computer screen
(vii) We already assumed that, the positive direction of the z-axis points towards us
(viii) So we can write:
The torque created by $\mathbf\small{\vec{F}_B}$ is perpendicular to the xy-plane and is directed towards the +ve side of the z axis
■ Combining the magnitude and direction, we can write:
Torque created by $\mathbf\small{\vec{F}_B}$ = $\mathbf\small{\left[a\times |\vec{F}_B|\right]\;\hat{k}}$
9. Let us write a summary about the torques:
• The torque created by $\mathbf\small{\vec{F}_A}$:
$\mathbf\small{\left[a\times |\vec{F}_A|\right]\;\hat{k}}$
• The torque created by $\mathbf\small{\vec{F}_B}$:
$\mathbf\small{\left[a\times |\vec{F}_B|\right]\;\hat{k}}$
10. We have: $\mathbf\small{|\vec{F}_A|=|\vec{F}_B|}$
• So we can write:
The two torques have the same magnitude. Also they have the same directions
• When we add such vectors, we will not get a null vector
11. So in this case, the net torque in the z-direction is not a zero vector
• That is: $\mathbf\small{\sum{\vec{\tau}_{z,ext}} \neq \vec{0} }$
• So the rod AB does not have rotational equilibrium
• It will continue to rotate. Note that, it is a 'rotation with out translation'
• So the AB will appear to be rotating with 'C' as pivot, even though there is no such pivot at 'C'

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The above example helps us to conclude that, a rigid body can have translational equilibrium without having rotational equilibrium

---

• In the second example above, note the following points:
(i) There are a total of 2 forces. We can call them a pair
(ii) Both the forces have the same magnitude
(iii) The forces are opposite in direction
(iv) Line of action of one force is different from that of the other

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■ A pair of equal and opposite forces with different lines of action is known as a couple

• Note that, the lines of action must be parallel. Only then will the forces become truly ‘opposite’

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Let us see an example:

1. Fig.7.95(a) below, shows a magnetic needle

Fig.7.95

• The 'magnetic south pole of the earth' will exert a pulling force on the 'north pole of the magnetic needle'
• This force is shown by the magenta vector
2. The 'magnetic north pole of the earth' will exert a pulling force on the 'south pole of the magnetic needle'
• This force is shown by the yellow vector
3. The yellow vector has the same magnitude as the magenta vector
• Also, the yellow vector is parallel to the magenta vector. They act in opposite directions
4. So the magnetic needle is acted upon by a couple
• This couple will bring the needle into alignment with the south and north magnetic poles of the earth
5. Once that alignment is achieved, the situation will be as shown in fig.b
• We see that, the magenta and yellow vectors now have the same line of action. There is no couple any more
'Opening of a lid' is also an example of ‘application of couple’
We will write it in steps:
1. We apply two forces:
(i) A force towards the left using the tip of the forefinger
(ii) A force towards the right using the tip of the thump
2. Note that, the following two points are diametrically opposite:
(i) The point of contact between the forefinger and the lid
(ii) The point of contact between the thump and the lid
• So the two forces have different lines of action
3. Thus a couple is formed and the lid will open

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Now we will see a solved example:

Solved example 7.21
Show that the moment of a couple does not depend on the point about which you take the moments
Solution:
1. In the fig.7.96 below, A and B are two points on a rigid body

Fig.7.96

• A force $\mathbf\small{\vec{F}_A}$ acts at A
• An equal and opposite force $\mathbf\small{\vec{F}_B}$ acts at B
• A suitable point 'O' is selected as the origin
• $\mathbf\small{\vec{r}_A}$ is the position vector of A
• $\mathbf\small{\vec{r}_B}$ is the position vector of B
2. First we will calculate the 'torque created by $\mathbf\small{\vec{F}_A}$ about O'
['torque created by $\mathbf\small{\vec{F}_A}$ about O'
Can also be called:
'moment of $\mathbf\small{\vec{F}_A}$ about O']
• So we have:
moment of $\mathbf\small{\vec{F}_A}$ about O = $\mathbf\small{\vec{r}_A \times \vec{F}_A}$
• Applying right hand screw rule, we get the direction of the vector obtained by the cross product:
It is perpendicular to the computer screen and towards us
3. Similarly, moment of $\mathbf\small{\vec{F}_B}$ about O = $\mathbf\small{\vec{r}_B \times \vec{F}_B}$
• Applying right hand screw rule, we get the direction of the vector obtained by the cross product:
It is perpendicular to the computer screen and away from us
4. Net moment = Vector sum = $\mathbf\small{(\vec{r}_A \times \vec{F}_A)+(\vec{r}_B \times \vec{F}_B)}$
• But $\mathbf\small{\vec{F}_A=-\vec{F}_B}$
• So we get:
Net moment = $\mathbf\small{[\vec{r}_A \times (-\vec{F}_B)]+(\vec{r}_B \times \vec{F}_B)}$
= $\mathbf\small{(\vec{r}_B \times \vec{F}_B)-(\vec{r}_A \times \vec{F}_B)}$
= $\mathbf\small{(\vec{r}_B - \vec{r}_A)\times \vec{F}_B}$
5. But $\mathbf\small{\vec{r}_A+\vec{AB}=\vec{r}_B}$
So $\mathbf\small{\vec{r}_B-\vec{r}_A=\vec{AB}}$
6. Thus the result in (4) becomes:
Net moment = $\mathbf\small{\vec{AB}\times \vec{F}_B}$   
7. This 'net moment' is the vector sum of two items:
(i) Moment of $\mathbf\small{\vec{F}_A}$
(ii) Moment of $\mathbf\small{\vec{F}_B}$ 
• But $\mathbf\small{\vec{F}_A}$ and $\mathbf\small{\vec{F}_B}$ have the same magnitude, and they form the couple
• So the 'net moment' that we calculated above is called 'moment of the couple'
8. So we can write:
Moment of the couple in fig.7.96 above = $\mathbf\small{\vec{AB}\times \vec{F}_B}$ 
• So to find the moment of the couple, we need 2 items:
(i) The vector from A to B
(ii) One of the force vectors
• Whatever be the position of 'O', the above two items will not change
■ So we can write:
Moment of a couple does not depend on the point about which you take the moments

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In the next section, we will see Principle of Moments

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