Chapter 7.21 - Center of Gravity

In the previous section, we saw principle of moments. In this section we will see center of gravity

1. Consider the irregular flat object shown in fig.7.101(a) below:

Fig.7.101

• It has small but uniform thickness (for example, a cardboard)
• It's surface is parallel to the xy-plane
• It is somewhat transparent so that, we can see the support on which it is resting
• We see that, the support is a ‘pointed end’, like the tip of a pencil
2. Let the point (on the object) at which it is supported be 'G' 
• We see that there is no rotation or translation for the object
• It may seem amazing that an object (and that too, an irregular one) is able to rest at equilibrium on a pointed end
• But once we analyze the situation scientifically, it will no longer be a source of amazement
3. The analysis can be done as follows:
• The object contains infinite number of particles. Let us assume that there are n particles
• Three of them are shown as small red spheres in fig.7.101(b)
4. Let the mass of the particles be m₁, m₂, m₃, . . . so on . . .
• Then their weights will be $\mathbf\small{m_1\vec{g},\;m_2\vec{g},\;m_3\vec{g}\,\;.\;.\;.}$so on . . .
• These weights are, forces acting in the -ve z-direction
5. So total force in the -ve z–direction = $\mathbf\small{m_1\vec{g}+m_2\vec{g}+m_3\vec{g}\,+\;.\;.\;.\;+\;m_i\vec{g}\,+\;.\;.\;.\;+\;m_n\vec{g}}$
= $\mathbf\small{(m_1+m_2+m_3\,+\;.\;.\;.\;+\;m_i\,+\;.\;.\;.\;+\;m_n)\vec{g}=M\,\vec{g}}$
• Where M is the mass of the whole object
6. The direction of $\mathbf\small{M\,\vec{g}}$ will be towards the -ve z direction because $\mathbf\small{\vec{g}}$ acts in the -ve z-direction
• So we can write $\mathbf\small{M\,\vec{g}}$ as $\mathbf\small{-\left(M\,|\vec{g}|\right)\hat{k}}$  
7. For translational equilibrium:
• $\mathbf\small{\sum{\vec{F}_{z,ext}}=\vec{0}}$
• Let the reaction at the support point be $\mathbf\small{\vec{R}_G}$
• If this condition is to be satisfied, the vector sum ($\mathbf\small{M\,\vec{g}+\vec{R}_G}$) must be a zero vector

• This condition tells us that, if there is to be translational equilibrium, $\mathbf\small{\vec{R}_G}$ is indeed vertical

• The reason can be written in 5 steps:
(i) If $\mathbf\small{\vec{R}_G}$ is inclined, it will have a vertical as well as a horizontal component
(ii) But we see no other horizontal components to counter it
(iii) So, if there is to be translational equilibrium, $\mathbf\small{\vec{R}_G}$ must be truly vertical
(iv) Also, it must be directed towards the +ve side of the y-axis
(v) Only then we will get ($\mathbf\small{M\,\vec{g}+\vec{R}_G}$) equal to a zero vector
8. Since we know the directions, we can write the above equation in terms of unit vectors also:
$\mathbf\small{-\left(M\,|\vec{g}|\right)\hat{k}+|\vec{R}_G|\hat{k}=\vec{0}}$
$\mathbf\small{\Rightarrow \left(-M\,|\vec{g}|+|\vec{R}_G|\right)\hat{k}=\vec{0}}$
• In the above equation, we have vectors on both sides. So their magnitudes must be equal
• Thus the condition becomes:
$\mathbf\small{\left(-M\,|\vec{g}|+|\vec{R}_G|\right)=0}$
That means the two magnitudes are equal:
$\mathbf\small{M\,|\vec{g}|=|\vec{R}_G|}$
9. Now we can consider the rotational equilibrium:
• We want the net torque to be zero
• In this problem, the forces do not lie in a single plane. So we want all the 3 conditions (related to rotational equilibrium) to be satisfied
• That is, we want:
$\mathbf\small{\sum{\vec{\tau}_{x,ext}}=\vec{0},\;\;\sum{\vec{\tau}_{y,ext}}=\vec{0},\;\;\sum{\vec{\tau}_{z,ext}}=\vec{0}}$
• In general, we want: $\mathbf\small{\sum{\vec{\tau}_{ext}}=\vec{0} }$
10. Let 'G' be the origin 'O' from where position vectors are drawn
    ♦ Let $\mathbf\small{\vec{r}_1}$ be the position vector of the first particle
    ♦ Let $\mathbf\small{\vec{r}_2}$ be the position vector of the second particle
    ♦ Let $\mathbf\small{\vec{r}_3}$ be the position vector of the third particle . . . so on . . .
• Then we get: 
    ♦ Torque created by the first particle = $\mathbf\small{\vec{r}_1 \times m_1\vec{g}}$ 
    ♦ Torque created by the second particle = $\mathbf\small{\vec{r}_2 \times m_2\vec{g}}$
    ♦ Torque created by the third particle = $\mathbf\small{\vec{r}_3 \times m_3\vec{g}}$  . . . so on . . .
[These torques are called gravitational torques because, the 'forces causing the torques' are gravitational forces]
11. So total torque
= $\mathbf\small{(\vec{r}_1\times m_1\vec{g})+(\vec{r}_2\times m_2\vec{g})+(\vec{r}_3\times m_3\vec{g})\,+\;.\;.\;.\;+\;(\vec{r}_i\times m_i\vec{g})\,+\;.\;.\;.\;+\;(\vec{r}_n\times m_n\vec{g})}$
= $\mathbf\small{\sum{(\vec{r}_i\times m_i\vec{g}})}$
• The $\mathbf\small{\vec{g}}$ is the same for all particles. So it can be taken out of the brackets
• Thus we get:
Total torque = $\mathbf\small{\vec{g}\sum{\vec{r}_i\times m_i}}$
• We want this total torque to be equal to zero vector. So we can write:
$\mathbf\small{\vec{g}\sum{(\vec{r}_i\times m_i)}=\vec{0}}$
• $\mathbf\small{\vec{g}}$ cannot be zero. So we get:
• $\mathbf\small{\sum{(\vec{r}_i\times m_i)}=\vec{0}}$
12. This is the condition which is to be satisfied, if we want rotational equilibrium for the body
• We will write it in words:
(i) There is a unique point 'G' in the body
(ii) We consider that 'G' as the origin 'O' 
• From that 'O', we draw position vectors to every particle of the body
(iii) For each particle we calculate $\mathbf\small{(\vec{r}_i\times m_i)}$ 
(the above product is not a cross product. Because m_i is a scalar) 
(iv) Then we take their vector sum: $\mathbf\small{\sum{(\vec{r}_i\times m_i)}}$
(v) That vector sum should be a zero vector. Then only we get rotational equilibrium for the body
13. So our next aim is to find the location of 'G'
• In an earlier section of this chapter, we saw Eq.7.4: $\mathbf\small{\vec{r}_C=\frac{\sum{} \,m_i\,\vec{r}_i}{M}}$ 
• This equation is used to find the center of mass (C) of any object (Details here).
• For using this method, we first fix a convenient point as the origin 'O', and then take the position vectors from that 'O'
• The $\mathbf\small{\vec{r}_C}$ in Eq.7.4, is the position vector of 'C' from that 'O'
■ Now, what if we take the 'C' itself as the origin 'O'
• Then we will get: $\mathbf\small{\vec{r}_C=\vec{0}}$
• The Eq.7.4 will then become: $\mathbf\small{\vec{0}=\frac{\sum{\,m_i\,\vec{r}_i}}{M}}$
$\mathbf\small{\Rightarrow \vec{0} \times M={\sum{ \,m_i\,\vec{r}_i}}}$
$\mathbf\small{\Rightarrow \vec{0} ={\sum{m_i\,\vec{r}_i}}}$
14. We will write the above equation in words:
(i) We take the 'C' of the body
(ii) We consider that 'C' as the origin 'O' 
• From that 'O', we draw position vectors to every particle of the body
(iii) For each particle we calculate $\mathbf\small{(\vec{r}_i\times m_i)}$ 
(the above product is not a cross product. Because m_i is a scalar) 
(iv) Then we take their vector sum: $\mathbf\small{\sum{(\vec{r}_i\times m_i)}}$
(v) That vector sum will be a $\mathbf\small{\vec{0}}$
■ That means, if we take 'C' as 'O', then $\mathbf\small{\sum{(\vec{r}_i\times m_i)}}$ will definitely be $\mathbf\small{\vec{0}}$
15. Comparing (12) and (14), we find that:
• The 'G' in fig.7.101 is nothing else but 'C'
■ We can write: 'G' and 'C' are the same
16. We have seen the methods to find 'C' of various systems and bodies, regular or irregular
• So we can indeed find 'G' of any such system or body, regular or irregular
■ This 'G' is called the center of gravity of the body
• The 'center of gravity' is abbreviated as: CG 
17. We saw that, CG can be determined by the same methods that we saw for 'C'
• But there is a constraint
• The $\mathbf\small{\vec{g}}$ is not constant every where. For example, $\mathbf\small{\vec{g}}$ is slightly greater at the poles than at the equator
• So for very large bodies, we cannot take $\mathbf\small{\vec{g}}$ out of the brackets in step (11) above
■ If the body is so large that, the $\mathbf\small{\vec{g}}$ varies from part to part of that body, then the 'C' and 'CG' will not coincide
■ The 'C' has nothing to do with 'CG'. The 'C' depends only on the distribution of mass of the body

---

Experimental method to find CG of irregular laminar bodies       

1. Fig.7.102(a) below shows an irregular shaped cardboard

Fig.7.102

• A hole is made at any convenient point near the edge. That hole is marked as 'A'
2. A thin pin is passed through the hole and the cardboard is suspended from the point S
• Now the cardboard is perpendicular to the floor of the room
• The hole A should be just large enough so that, the cardboard is able to rotate freely about the pin
• If the hole is too small, friction will arise between the pin and the cardboard, thereby obstructing free rotation
• Also, when suspended from S, the cardboard should not touch side walls or other objects 
3. When the suspended cardboard becomes stable, draw a vertical dashed line through A. This is shown in red color in fig.a
■ The ‘CG’ of the cardboard lies somewhere on the red dashed line
Let us see the reason. It can be written in 5 steps:
(i) The particles on the left side of the red dashed line create anti clockwise 'gravitational torques' about G
(ii) The particles on the right side of the red dashed line create clockwise 'gravitational torques' about G
(iii) The particles on the red dashed line create no torques at all about G
(iv) The anti clockwise torques must balance out the clockwise torques. Other wise, the cardboard will rotate
(v) So we can be sure that, when the cardboard is stable, the G lies somewhere on the red vertical dashed line
4. Next, make a similar hole at another point B
• Repeat the experiment
• We get the green vertical dashed line shown in fig.b
• The CG lies somewhere on the green dashed line
5. So we have two information:
(i) The CG lies on the red dashed line
(ii) The CG lies on the green dashed line
■ So obviously, the CG lies at the point of intersection of the red and green lines
6. To make a check:
• Make a similar hole at a third point C
• Repeat the experiment
• We get the yellow vertical dashed line shown in fig.c
■ This yellow line will pass through the CG obtained in fig.b
7. Note that, since the cardboard is small, all particles of the cardboard will be experiencing the same $\mathbf\small{\vec{g}}$ 
■ So the CG that we determined above, is also the center of mass ('C') of the cardboard

---

In the next section, we will see some solved examples

PREVIOUS           CONTENTS          NEXT

Copyright©2019 Higher Secondary Physics. blogspot.in - All Rights Reserved

Chapter 7.20 - Principle of Moments

In the previous section, we saw conditions of equilibrium of rigid bodies. In this section we will see Principle of Moments

1. In the fig.7.97(a) below, a light (ie. of negligible mass) rod AB lies on the xy-plane
• It’s length is (d₁+d₂)

Fig.7.97

• It is supported at point 'C' by a 'knife edge support'
• The 'knife edge support' is shown clearly in the 3D view in fig.7.98 below:

Fig.7.98

2. Two forces are applied on the rod
• Those two forces have the following five peculiarities:
(i) The forces do not 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 -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. As a result of the forces $\mathbf\small{\vec{F}_A}$ and $\mathbf\small{\vec{F}_B}$, a reaction force will develop at C
• This reaction force is denoted as $\mathbf\small{\vec{R}_C}$
• $\mathbf\small{\vec{R}_C}$ is shown to be vertical and directed towards the +ve side of the y-axis
• We will soon prove mathematically that, it is indeed vertical and directed towards the +ve side of y-axis
4. We want the rod AB to remain in equilibrium (both translational and rotational)
• How can we achieve that?
Ans: By ensuring that, the conditions of equilibrium are satisfied
• So let us apply the conditions:
• 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:
$\mathbf\small{\sum{\vec{F}_{y,ext}}=\vec{0}}$
$\mathbf\small{\sum{\vec{\tau}_{z,ext}}=\vec{0}}$
5. First we will apply:
$\mathbf\small{\sum{\vec{F}_{y,ext}}=\vec{0}}$
• If this condition is to be satisfied, the vector sum ($\mathbf\small{\vec{F}_A+\vec{F}_B+\vec{R}_C}$) must be a zero vector
• This condition tells us that, if there is to be translational equilibrium, $\mathbf\small{\vec{R}_C}$ is indeed vertical
• The reason can be written in 5 steps:
(i) If $\mathbf\small{\vec{R}_C}$ is inclined, it will have a vertical as well as a horizontal component
(ii) But we see no other horizontal components to counter it
(iii) So, if there is to be translational equilibrium, $\mathbf\small{\vec{R}_C}$ must be truly vertical
(iv) Also, it must be directed towards the +ve side of the y-axis
(v) Only then we will get ($\mathbf\small{\vec{F}_A+\vec{F}_B+\vec{R}_C}$) equal to a zero vector
6. Since we know the directions, we can write the above equation in terms of unit vectors also:
$\mathbf\small{-|\vec{F}_A|\hat{j}-|\vec{F}_B|\hat{j}+|\vec{R}_C|\hat{j}=\vec{0}}$
$\mathbf\small{\Rightarrow \left(-|\vec{F}_A|-|\vec{F}_B|+|\vec{R}_C|\right)\hat{j}=\vec{0}}$
• In the above equation, we have vectors on both sides. So their magnitudes must be equal
• Thus the condition becomes:
$\mathbf\small{-|\vec{F}_A|-|\vec{F}_B|+|\vec{R}_C|=0}$

---

7. 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
= $\mathbf\small{|\vec{r}_A|\times |\vec{F}_A|\times \sin \theta}$
= $\mathbf\small{[d_1\times |\vec{F}_A|\times \sin 90]=d_1\times |\vec{F}_A|}$
(θ = 90^o because, forces are perpendicular to the rod)
8. For finding the direction of this torque:
(i) Assume that $\mathbf\small{\vec{F}_A}$ 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 the direction, we can write:
The torque created by $\mathbf\small{\vec{F}_A}$ is: $\mathbf\small{[d_1\times |\vec{F}_A|]\; \hat{k}}$ 
9. 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
= $\mathbf\small{|\vec{r}_B|\times |\vec{F}_B|\times \sin \theta}$
= $\mathbf\small{[d_2\times |\vec{F}_B|\times \sin 90]=d_2\times |\vec{F}_B|}$
(θ = 90^o because, forces are perpendicular to the rod)
10. 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 the direction, we can write:
The torque created by $\mathbf\small{\vec{F}_B}$ is: $\mathbf\small{[-d_2\times |\vec{F}_B|]\; \hat{k}}$ 
11. Next we find the torque created by $\mathbf\small{\vec{R}_C}$ about C
• But $\mathbf\small{\vec{R}_C}$ passes through C
• So the torque is $\mathbf\small{\vec{0}}$  
12. Now we can write a summary about the torques:
• The torque created by $\mathbf\small{\vec{F}_A}$:
$\mathbf\small{[d_1\times |\vec{F}_A|]\; \hat{k}}$
• The torque created by $\mathbf\small{\vec{F}_B}$:
$\mathbf\small{[-d_2\times |\vec{F}_B|]\; \hat{k}}$
• The torque created by $\mathbf\small{\vec{R}_C}$ is $\mathbf\small{\vec{0}}$
13. We want $\mathbf\small{\sum{\vec{\tau}_{z,ext}}=0}$
• Let us input the values:
$\mathbf\small{[(d_1\times |\vec{F}_A|)\hat{k}]-[(d_2\times |\vec{F}_B|)\hat{k}]+\vec{0}}$ must be equal to $\mathbf\small{\vec{0}}$
• That means, the following equation must be satisfied:
$\mathbf\small{[(d_1\times |\vec{F}_A|)-(d_2\times |\vec{F}_B|)]\hat{k}=\vec{0}}$
14. In the above equation, we have vectors on both sides. So their magnitudes must be equal
• Thus the condition becomes:
$\mathbf\small{[(d_1\times |\vec{F}_A|)-(d_2\times |\vec{F}_B|)]=0}$
15. The above equation can be rearranged as: $\mathbf\small{|\vec{F}_B|=\frac{|\vec{F}_A|d_1}{d_2}}$
• From the fig.7.97, we see that d₁ is less than d₂. So $\mathbf\small{\frac{d_1}{d_2}}$ will be less than '1'
• Thus we find that: $\mathbf\small{\vec{F}_B}$ is less than $\mathbf\small{\vec{F}_A}$
16. Also we can write:
• The magnitude of the force applied at B should be equal to $\mathbf\small{\frac{|\vec{F}_A|d_1}{d_2}}$
• Then only we can achieve equilibrium for the rod AB
17. We can write the other way also:
• The magnitude of the force applied at A should be equal to $\mathbf\small{\frac{|\vec{F}_B|d_2}{d_1}}$
• Then only we can achieve equilibrium for the rod AB

---

Let us see a practical application of the equation:

In fig.7.99 below, a block of 10 kg hangs from the end A of the rod AB. When the rod is exactly horizontal, the block rests completely on the floor and the string is straight.

Fig.7.99

What force must be applied at B so that, the block just loses contact with the floor? 
[Take g = 10 ms^-2]
Solution:
1. We have: $\mathbf\small{\vec{F}_A}$ = mg = 10 × 10 = 100 N
• Using equation in (16) above, we get:
$\mathbf\small{|\vec{F}_B|=\frac{|\vec{F}_A|d_1}{d_2}=\frac{100 \times 0.2}{0.8}=25\, \text{N}}$
2. 25 N corresponds to a mass of 2.5 kg
• So we are able to lift a large mass of 10 kg by a small mass of 2.5 kg

---

Note that, the 25 N corresponds to 'just losing contact between the block and the floor'

• If we lift the block further, the rod AB will no longer be horizontal
• As a result, the forces $\mathbf\small{\vec{F}_A}$ and $\mathbf\small{\vec{F}_B}$ will no longer be perpendicular to AB
• This situation is shown in fig.7.99(b)
■ If we want to keep the block stationary at a height much above the floor, what force must be applied at B?
Solution:
1. Fig.7.100(a) below shows a schematic diagram

Fig.7.100

AB makes an angle θ with the horizontal
2. We see that:
• $\mathbf\small{\vec{F}_A}$ and $\mathbf\small{\vec{F}_B}$ are vertical
    ♦ But $\mathbf\small{\vec{F}_A}$ is not perpendicular to AB
    ♦ $\mathbf\small{\vec{F}_B}$ is also not perpendicular to AB
3. So we resolve the two forces into components:
(i) $\mathbf\small{\vec{F}_A}$ has two components:
• One component is parallel to AB. It has magnitude: $\mathbf\small{|\vec{F}_A|\sin \theta}$ 
• The other component is perpendicular to AB. It has magnitude: $\mathbf\small{|\vec{F}_A|\cos \theta}$
(ii) $\mathbf\small{\vec{F}_B}$ has two components:
• One component is parallel to AB. It has magnitude: $\mathbf\small{|\vec{F}_B|\sin \theta}$ 
• The other component is perpendicular to AB. It has magnitude: $\mathbf\small{|\vec{F}_B|\cos \theta}$
• These components are shown in blue color in fig.b
4. The parallel components pass through C
• So they do not create any torque about C
• We need to consider the perpendicular components only
5. We can easily show the following two items:
(i) The torque created by the perpendicular component of $\mathbf\small{\vec{F}_A}$ is:
$\mathbf\small{[d_1\times |\vec{F}_A|\times \cos \theta]\; \hat{k}}$
(ii) The torque created by the perpendicular component of $\mathbf\small{\vec{F}_B}$ is:
$\mathbf\small{[-d_2\times |\vec{F}_B|\times \cos \theta]\; \hat{k}}$
(The reader may write the proofs in his/her own note books)
6. We want $\mathbf\small{\sum{\vec{\tau}_{z,ext}}=0}$
• Let us input the values:
$\mathbf\small{[(d_1\times |\vec{F}_A|\times \cos \theta)\hat{k}]-[(d_2\times |\vec{F}_B|\times \cos \theta)\hat{k}]}$ must be equal to $\mathbf\small{\vec{0}}$
• That means, the following equation must be satisfied:
$\mathbf\small{\left[(d_1\times |\vec{F}_A|\times \cos \theta)-(d_2\times |\vec{F}_B|\times \cos \theta)\right]\hat{k}=\vec{0}}$
14. In the above equation, we have vectors on both sides. So their magnitudes must be equal
• Thus the condition becomes:
$\mathbf\small{\left[(d_1\times |\vec{F}_A|\times \cos \theta)-(d_2\times |\vec{F}_B|\times \cos \theta)\right]=0}$
$\mathbf\small{\Rightarrow (d_1\times |\vec{F}_A|)-(d_2\times |\vec{F}_B|)=0}$
• This is the same result that we obtained earlier
15. The above equation can be rearranged as: $\mathbf\small{|\vec{F}_B|=\frac{|\vec{F}_A|d_1}{d_2}}$
16. Let us input the values:
$\mathbf\small{|\vec{F}_B|=\frac{100 \times 0.2}{0.8}=25\, \text{N}}$
• This is the same result that we obtained before

---

■ The rod AB in fig.7.99 is a lever

Let us see the important features of a lever:
(i) An ideal lever is a light (ie. of negligible mass) rod pivoted at a point along its length
(ii) The pivot point is called the fulcrum
(iii) The force $\mathbf\small{\vec{F}_A}$ is usually some weight to be lifted
• This $\mathbf\small{\vec{F}_A}$ is called the load
• Distance (d₁) of load from the fulcrum is called load arm
(iv) The force $\mathbf\small{\vec{F}_B}$ applied to lift the load
• This $\mathbf\small{\vec{F}_B}$ is called the effort
• Distance (d₂) of effort from the fulcrum is called effort arm
(v) Consider the expression '$\mathbf\small{(d_1\times |\vec{F}_A|)-(d_2\times |\vec{F}_B|)=0}$' that we derived in (14) above
• It can be rearranged as: $\mathbf\small{d_1\times |\vec{F}_A|=d_2\times |\vec{F}_B|}$
• That is: Load arm × Load = Effort arm × Effort
• This is the mathematical expression of the Principle of moments for a lever
(vi) The ratio $\mathbf\small{\frac{Load}{Effort}}$ is called Mechanical advantage
• From (v) above, we get: mechanical advantage = $\mathbf\small{\frac{\text{Effort arm}}{\text{Load arm}}}$
(vii) If $\mathbf\small{\frac{\text{Effort arm}}{\text{Load arm}}}$ is greater than '1', we will be able to lift a larger load by a smaller effort
(viii) A see saw at the children’s park is an example of a lever
Some other examples are:
Cutting pliers
Wheel barrow
Nut cracker etc.

---

In the next section, we will see Center of gravity

PREVIOUS           CONTENTS          NEXT

Copyright©2019 Higher Secondary Physics. blogspot.in - All Rights Reserved