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)
• 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:
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 = 90o 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 = 90o 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
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.
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
• 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
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
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.
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 |
• The 'knife edge support' is shown clearly in the 3D view in fig.7.98 below:
 |
| Fig.7.98 |
• 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:
$\mathbf\small{-|\vec{F}_A|-|\vec{F}_B|+|\vec{R}_C|=0}$
• 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 = 90o 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 = 90o 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:
 |
| Fig.7.99 |
[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'
• 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 |
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
(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
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