Chapter 7.24 - Radius of Gyration

In the previous section, we saw the basics about moment of inertia. In this section we will see radius of gyration

■ In translational motion, we saw this:
• Inertia is a property of matter by which it continues in its existing state of rest or uniform motion in a straight line, unless that state is changed by an external force.
• Also, we saw that, mass is a measure of this inertia
■ Now, in rotational motion, we have not inertia, but moment of inertia 
• Rotational inertia is the a property of matter by which it continues in its existing state of rest or uniform rotational motion about an axis, unless that state is changed by an external force.
• 'moment of inertia' is the measure of ‘rotational inertia’

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• Different parts of a body are distributed at different distances from the axis

• We want to know how this distribution is present in a body 
• Moment of inertia helps us to acquire that information 
• But always remember these:
    ♦ Mass of a body is a constant
    ♦ But moment of inertia of a body depends upon the axis that we choose

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Let us see a practical application of moment of inertia. We will write it in steps:

1. We know the formula for finding I
$\mathbf\small{I=\sum\limits_{i=1}^{i=n}{\left(m_i\;r_{i(\bot)}^2 \right)} }$
• So, if mass is increased, I increases
• If the distances of the particles from the axis is increased, then also I increases
2. So large heavy discs will have a high I
• Because of this large I that they possess, 
    ♦ If they are initially at rest, it will be difficult to get them rotating
    ♦ If they are initially in rotational motion, it will be difficult to stop them
3. Such large heavy discs are used in steam engines and automobile engines
• They are called flywheels
    ♦ Because of their large I, we will not be able to put the vehicle (which is initially at rest) into a sudden motion
    ♦ Also we will not be able to bring a vehicle (which is initially in motion) into a sudden stop
• This helps to prevent jerky motion of the vehicle so that the passengers can experience a smoother ride

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We saw the I of 8 objects in the previous section. We will write them again here:

1. Body: Thin circular ring of radius R
Axis: Perpendicular to the plane of the ring, passing through the center
■ $\mathbf\small{I=MR^2}$ 
2. Body: Thin circular ring of radius R
Axis: Lies on the plane of the ring, passing through the center. Obviously, this axis is the extension of any one diameter of the ring
■ $\mathbf\small{I=\frac{MR^2}{2}}$
3. Body: Thin rod of length L
Axis: Perpendicular to the rod, passing through the midpoint
■ $\mathbf\small{I=\frac{ML^2}{12}}$
4. Body: Circular disc of radius R
Axis: Perpendicular to the disc, passing through the center
■ $\mathbf\small{I=\frac{MR^2}{2}}$

5. Body: Circular disc of radius R
Axis: Lies on the plane of the disc, passing through the center. Obviously, this axis is the extension of any one diameter of the disc
■ $\mathbf\small{I=\frac{MR^2}{4}}$
6. Body: Solid cylinder of radius R
Axis: Axis of the cylinder
■ $\mathbf\small{I=\frac{MR^2}{2}}$

7. Body: Hollow cylinder of radius R
Axis: Axis of the cylinder
■ $\mathbf\small{I=MR^2}$
8. Body: Solid sphere of radius R
Axis: Passes through the center. Obviously, this axis is the extension of any one diameter of the sphere
■ $\mathbf\small{I=\frac{2MR^2}{5}}$

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Based on the above results we can write an analysis. We will write it in steps: 

1. We see that, in all case,
    ♦ Mass occurs once
    ♦ Length occurs twice
2. The dimensions of I is ML²
• The SI unit of I is: kg m²
3. So we can write I in another form:
I = Mk².
    ♦ Where k is called the radius of gyration 
    ♦ It has the dimension: L
4. Let us see an example:
(i) In case 3 above, we have: $\mathbf\small{I=\frac{ML^2}{12}}$
(ii) But I = Mk²
(iii) Equating the two, we get: $\mathbf\small{I=\frac{ML^2}{12}=Mk^2}$
$\mathbf\small{\Rightarrow k^2=\frac{L^2}{12}}$
$\mathbf\small{\Rightarrow k=\frac{L}{\sqrt{12}}}$
(iv) So for case 3, we can write:
Body: Thin rod of length L
Axis: Perpendicular to the rod, passing through the midpoint
■ $\mathbf\small{I=\frac{ML^2}{12}}$
■ $\mathbf\small{k=\frac{L}{\sqrt{12}}}$
Note that, just like I, k also depends on the axis
5. Another example:
(i) In case 5 above, we have: $\mathbf\small{I=\frac{MR^2}{4}}$
(ii) But I = Mk²
(iii) Equating the two, we get: $\mathbf\small{I=\frac{MR^2}{4}=Mk^2}$
$\mathbf\small{\Rightarrow \frac{R^2}{4}=k^2}$
$\mathbf\small{\Rightarrow k=\frac{R}{2}}$
(iv) So for case 5, we can write:
Body: Circular disc of radius R
Axis: Lies on the plane of the disc, passing through the center. Obviously, this axis is the extension of any one diameter of the disc
■ $\mathbf\small{I=\frac{MR^2}{4}}$
■ $\mathbf\small{k=\frac{R}{2}}$
6. We know that, I of a body is given by: $\mathbf\small{I=\sum\limits_{i=1}^{i=n}{\left(m_i\;r_{i(\bot)}^2 \right)} }$
• It is clear that, we have to consider each one of the n particles in the body
    ♦ We want the mass of each particle
    ♦ We want the perpendicular distance of each particle from the axis
7. But if the body consists of only one particle, then the I of that body is given by:
$\mathbf\small{I={m\;r_{(\bot)}^2}}$
• I of this body can be obtained using it's k also:
$\mathbf\small{I={m\;k^2}}$
• So, for this body, we get:
$\mathbf\small{I={m\;r_{(\bot)}^2}={m\;k^2}}$
8. Based on the above analysis, we can write the following:
(i) Let a body 'A' of mass M, contain n particles
(ii) Let it's moment of inertia about an axis 'L' be I
(iii) Let it's radius of gyration about 'L' be k
• Then we get: $\mathbf\small{I={M\;k^2}}$
(iv) Now consider a point mass 'B'. See fig.7.116 below
• Let the mass of 'B' be the same 'M' possessed by A
(v) Place 'B' at a perpendicular distance of 'k' from the axis 'L'
• Then moment of inertia of B about 'L' = $\mathbf\small{{M\;k^2}}$
• This is the same I possessed by the body 'A'
9. So we can write the definition of k of a body A about an axis L:
The definition can be written in 4 steps:
(i) Take a point mass B such that:
• It's mass is equal to the mass of A
(ii) We want B to have the same I (about L) possessed by A
(iii) For that, we must place B at a 'particular perpendicular distance' from L
(iv) This 'particular perpendicular distance' is k
This is shown in fig.7.116 below:

Fig.7.116

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In the next section, we will see theorem of perpendicular axes

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Chapter 7.23 - Moment of Inertia

In the previous section, we completed a discussion on torque and center of gravity. We saw some solved examples also. In this section we will see moment of inertia

In our discussions so far, we saw these:
• An important role is played by 'force’ in translational motion
    ♦ In rotational motion, that role is played by ‘moment of force’ or 'torque' ($\mathbf\small{\vec{\tau}}$)
• An important role is played by 'linear momentum’ in translational motion
    ♦ In rotational motion, that role is played by 'angular momentum' ($\mathbf\small{\vec{l}}$)
    ♦ Angular momentum can be considered as the ‘moment of linear momentum’
• Moment of a force is given by: $\mathbf\small{\vec{\tau}=\vec{r}\times \vec{F}}$ 
    ♦ In a similar way, moment of linear momentum ($\mathbf\small{\vec{p}}$) is given by: $\mathbf\small{\vec{l}=\vec{r}\times \vec{p}}$
■ We know that, 'mass' plays an important role in translational motion
• So next, we want to see who plays the role of 'mass' in rotational motion

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1. Consider a body rotating about a fixed axis
• Consider any one particle in that body
• In a previous section, we have seen Eq.7.16 which gives the linear velocity of 'a particle in a rotating body' (Details here)
• Eq.7.16: $\mathbf\small{\vec{v}=\vec{r}\times \vec{\omega}}$
• So we can write:
Linear velocity (v_i) of the i^th particle at any instant is given by: $\mathbf\small{\vec{v}_i=\vec{r}_i\times \vec{\omega}}$
• Where:
    ♦ $\mathbf\small{\vec{r}_i}$ is the position vector of that particle
       (with origin 'O' on the axis of rotation)
    ♦ $\mathbf\small{\vec{\omega}}$ is the angular velocity of the whole body at that instant
2. This is a vector cross product. The result will be the velocity vector
• We want the magnitude of that resulting velocity vector. This is because, kinetic energy is calculated using the 'magnitude of velocity'
3. We have: $\mathbf\small{|\vec{v}_i|=|\vec{r}_i|\times |\vec{\omega}| \times \sin \theta}$
• This can be rearranged as:
$\mathbf\small{|\vec{v}_i|=\left(|\vec{r}_i|\times \sin \theta \right)\times |\vec{\omega}|}$
$\mathbf\small{\Rightarrow |\vec{v}_i|=r_{i(\bot)}\times |\vec{\omega}|}$
• Where $\mathbf\small{r_{i(\bot)}}$ is the perpendicular distance of that particle from the axis of rotation
4. Once we get the above magnitude, we can calculate the kinetic energy possessed by that particle at that instant
• We have the familiar equation:
Kinetic energy (K) = $\mathbf\small{\frac{1}{2}m|\vec{v}|^2}$
• So, for the i^th particle, we have: $\mathbf\small{K_i=\frac{1}{2}m_i|\vec{v}_i|^2}$
• Substituting for v_i from (3), we get:
$\mathbf\small{K_i=\frac{1}{2}m_i\left(r_{i(\bot)}\;|\vec{\omega}|\right)^2}$
$\mathbf\small{\Rightarrow K_i=\frac{1}{2}m_i\;r_{i(\bot)}^2\;|\vec{\omega}|^2}$
5. So we obtained the kinetic energy of one particle of the body
• If we add the kinetic energies of all the particles, we will get the 'kinetic energy possessed by the whole body'
• Thus we get: $\mathbf\small{K=\sum\limits_{i=1}^{i=n}{K_i} }$
$\mathbf\small{\Rightarrow K=\sum\limits_{i=1}^{i=n}{\left(\frac{1}{2}m_i\;r_{i(\bot)}^2\;|\vec{\omega}|^2 \right)} }$
6. $\mathbf\small{\frac{1}{2}}$ and $\mathbf\small{|\vec{\omega}|^2}$ are same for all the particles. So they can be taken outside the summation
• Thus we get:
$\mathbf\small{K=\frac{1}{2}\;|\vec{\omega}|^2\sum\limits_{i=1}^{i=n}{\left(m_i\;r_{i(\bot)}^2 \right)} }$
7. In the above result, consider the quantity inside the brackets: $\mathbf\small{m_i\;r_{i(\bot)}^2}$
• We can elaborate it as follows:
Take any particle in the rigid body. Note down three items related to that particle:
(i) It's mass. Let it be $\mathbf\small{m_i}$
(ii) It's perpendicular distance from the axis. Let it be $\mathbf\small{r_{i(\bot)}}$ 
(iii) Calculate $\mathbf\small{m_i\;r_{i(\bot)}^2}$ for that particle
• In this way calculate $\mathbf\small{m_i\;r_{i(\bot)}^2}$ for all the n particles
• Then take their sum: $\mathbf\small{\sum\limits_{i=1}^{i=n}{\left(m_i\;r_{i(\bot)}^2 \right)} }$
■ This sum has much significance. It is given a special name: Moment of inertia. It's symbol is: I
• So we can write: 
Eq.7.25: $\mathbf\small{I=\sum\limits_{i=1}^{i=n}{\left(m_i\;r_{i(\bot)}^2 \right)} }$
8. Let us see the important features of I:
(i) We see that, for calculating I, we do not require $\mathbf\small{\vec{\omega}}$
• That means, what ever be the angular velocity of the body. The I of that body will not change 
(ii) We know that 'masses of the particles' will not change
• Also the 'perpendicular distances of the particles from the axis' will not change
■ So I of a body is a unique value
■ In other words, every rigid body has it's own unique I
9. Is there any situation in which the I of a body changes?
■ There are indeed such situations. Let us analyse it:
• We know that, 'perpendicular distances of the particles from the axis' will not change
• But if the 'axis' is changed, then those 'perpendicular distances' will change
• Consequently, the I will change
10. So we will modify the final statement that we wrote in (8) above
• We can write it in 3 steps:
(i) A rigid body can have infinite number of lines as it's 'axis of rotation'
(ii) That rigid body may rotate about any one of those possible axes
(iii) That rigid body will have a unique I for each of those axes
• Another way of writing it in 3 steps:
(i) A rigid body can have infinite number of lines as it's 'axis of rotation'
(ii) Select any one axis from among them
(iii) That rigid body will have a unique I with respect to that selected axis
■ It is given the status 'unique' because, once the axis is selected, the I will not change unless the shape of the body itself changes
11. Now that we have decided to represent $\mathbf\small{\sum\limits_{i=1}^{i=n}{\left(m_i\;r_{i(\bot)}^2 \right)} }$ by I, the result in (6) becomes :
Eq.7.26: $\mathbf\small{K=\frac{1}{2}\;I\;|\vec{\omega}|^2}$
• This equation gives us the kinetic energy of a body in pure rotation
12. We know that, the kinetic energy of a body in pure translation is given by:
$\mathbf\small{K=\frac{1}{2}\;m\;|\vec{v}|^2}$
13. Let us compare (11) and (12):
• We already know that $\mathbf\small{\vec{\omega}}$ in rotational motion is analogous to $\mathbf\small{\vec{v}}$ in translational motion
■ So we can write:   
• 'Mass' plays an important role in translational motion 
• In rotational motion, that role is played by 'moment of inertia (I)'

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• Moment of inertia (I) is an important parameter. It is essential that, we know it’s value for commonly occurring bodies like rectangles, rings, cylinders, spheres, etc.,

• Let us now calculate the I of a thin circular ring. We will write it in steps:   
1. We know that, the I depends on the axis of rotation
• In our present case:
    ♦ The ring is rotating on it’s own plane
    ♦ Also, it is rotating about it’s center
• So it is clear that:
    ♦ The axis is perpendicular to the ring 
    ♦ Also, the axis passes through the center of the ring
• This is shown in the animation in fig.7.108 below:

Fig.7.108

• The magenta line is the axis of rotation
2. In fig.7.109 below, a particle in the ring is isolated for analysis

Fig.7.109

• It is shown as a red sphere
• A point on the axis is chosen as the origin 'O'
• The position vector of the particle is $\mathbf\small{\vec{r}}$
• The perpendicular distance of the particle from the axis is $\mathbf\small{r_{\bot}}$
• The direction of rotation is indicated by the blue curved arrow
3. Let there be n particles in the ring
• Each of them will have it's own position vector
• But all those position vectors will be having the same magnitude $\mathbf\small{|\vec{r}|}$
• And all of them will be making the same angle θ with the axis
• So the perpendicular distance ($\mathbf\small{r_{\bot}}$) of the particle (given by $\mathbf\small{|\vec{r}|\sin \theta}$) will be same for all particles
4. We want to find I
• We have: $\mathbf\small{I=\sum\limits_{i=1}^{i=n}{\left(m_i\;r_{i(\bot)}^2 \right)} }$
• Since the perpendicular distances are the same, it can be taken outside the brackets. So we get:
$\mathbf\small{I=r_{\bot}^2\sum\limits_{i=1}^{i=n}{\left(m_i \right)}}$
5. But $\mathbf\small{\sum\limits_{i=1}^{i=n}{\left(m_i \right)}}$ is the total mass M of the ring
• Also, $\mathbf\small{r_{\bot}}$ is the radius R of the ring
6. So the result in (4) becomes: I = MR².
• This is the moment of inertia of a thin ring, about the 'perpendicular axis passing through it's center'
• In this example, we used the position vector in the analysis
    ♦ This is for obtaining a basic understanding only
    ♦ What we actually need is the perpendicular distances $\mathbf\small{r_{i(\bot)}}$  

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Another example:

1. Take a rigid massless rod of length l. 
• Attach two equal masses at the ends. Let each mass be $\mathbf\small{\frac{M}{2}}$
• The axis of rotation passes through the center of the rod
• Also, the axis is perpendicular to the rod
• This is shown in the animation in fig.7.110 below:

Fig.7.110

• The magenta line is the axis of rotation
2. We want to find I
We have: $\mathbf\small{I=\sum\limits_{i=1}^{i=n}{\left(m_i\;r_{i(\bot)}^2 \right)} }$
• The distances are shown in fig.7.111 below:

Fig.7.111

• Since the perpendicular distances are the same, it can be taken outside the brackets. So we get:
$\mathbf\small{I=r_{\bot}^2\sum\limits_{i=1}^{i=n}{\left(m_i \right)}}$
3. But $\mathbf\small{\sum\limits_{i=1}^{i=n}{\left(m_i \right)}}$ is the total mass of the system
• This is equal to $\mathbf\small{\frac{M}{2}+\frac{M}{2}=M}$ 
• Also, $\mathbf\small{r_{\bot}=\frac{l}{2}}$
4. So the result in (2) becomes: $\mathbf\small{I=M\left(\frac{l}{2} \right)^2=\frac{Ml^2}{4}}$

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Thus we saw the I of two simple systems. For more complex bodies and systems, we can find I with the help of calculus. The I of some common bodies are given below:

1. Body: Thin circular ring of radius R
Axis: Perpendicular to the plane of the ring, passing through the center
• We already saw this case. We obtained $\mathbf\small{I=MR^2}$ 
2. Body: Thin circular ring of radius R
Axis: Lies on the plane of the ring, passing through the center. Obviously, this axis is the extension of any one diameter of the ring
• This is shown in fig.7.112(a) below:

Fig.7.112

■ In this case, $\mathbf\small{I=\frac{MR^2}{2}}$
3. Body: Thin rod of length L
Axis: Perpendicular to the rod, passing through the midpoint
• This is shown in fig.7.112(b) above
■ In this case, $\mathbf\small{I=\frac{ML^2}{12}}$
4. Body: Circular disc of radius R
Axis: Perpendicular to the disc, passing through the center
• This is shown in fig.7.113(a) below
■ In this case, $\mathbf\small{I=\frac{MR^2}{2}}$

Fig.7.113

5. Body: Circular disc of radius R
Axis: Lies on the plane of the disc, passing through the center. Obviously, this axis is the extension of any one diameter of the disc
• This is shown in fig.7.113(b) above
■ In this case, $\mathbf\small{I=\frac{MR^2}{4}}$
6. Body: Solid cylinder of radius R
Axis: Axis of the cylinder
• This is shown in fig.7.114(a) below
■ In this case, $\mathbf\small{I=\frac{MR^2}{2}}$

Fig.7.114

7. Body: Hollow cylinder of radius R
Axis: Axis of the cylinder
• This is shown in fig.7.114(b) above
■ In this case, $\mathbf\small{I=MR^2}$
8. Body: Solid sphere of radius R
Axis: Passes through the center. Obviously, this axis is the extension of any one diameter of the sphere
• This is shown in fig.7.115 below:

Fig.7.115

■ In this case, $\mathbf\small{I=\frac{2MR^2}{5}}$

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In the next section, we will see radius of gyration

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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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