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:
$\mathbf\small{\sum{\vec{\tau}_{ext}}=\vec{0} }$
$\mathbf\small{\Rightarrow \sum{\vec{\tau}_{x,ext}}=\vec{0},\;\;\sum{\vec{\tau}_{y,ext}}=\vec{0},\;\;\sum{\vec{\tau}_{z,ext}}=\vec{0}}$
1. In the fig.7.93(a) below, a light (ie. of negligible mass) rod AB lies on the xy-plane
• 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
• 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
• 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 = 90o 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 = 90o 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 $\mathbf\small{\sum{\vec{\tau}_{y,ext}}=\vec{0} }$
• 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
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:
$\mathbf\small{\sum{\vec{\tau}_{ext}}=\vec{0} }$
$\mathbf\small{\Rightarrow \sum{\vec{\tau}_{x,ext}}=\vec{0},\;\;\sum{\vec{\tau}_{y,ext}}=\vec{0},\;\;\sum{\vec{\tau}_{z,ext}}=\vec{0}}$
The situation will be greatly simplified if all the forces lie in a plane. This can be explained with the help of an example:
 |
| Fig.7.93 |
• 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
• Thus we applied the first 3 conditions for translational equilibrium
♦ Because $\mathbf\small{\sum{\vec{F}_{y,ext}} \neq \vec{0}}$
• So the rod AB is not in translational equilibrium
4. For applying the next 3 conditions, we have to first find the magnitudes and directions of the torques
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 = 90o 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 = 90o 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 $\mathbf\small{\sum{\vec{\tau}_{y,ext}}=\vec{0} }$
So for this example, we can write:
• 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
• 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
♦ $\mathbf\small{\sum{\vec{F}_{y,ext}}=\vec{0}}$
♦ $\mathbf\small{\sum{\vec{\tau}_{z,ext}}=\vec{0} }$
• 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
The above example also helps us to conclude that, a rigid body can have rotational equilibrium without having translational equilibrium
Let us see another example:
1. In the fig.7.94(a) below, the same rod AB lies on the same xy-plane
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
• 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
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 = 90o 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 = 90o 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'
1. In the fig.7.94(a) below, the same rod AB lies on the same xy-plane
 |
| Fig.7.94 |
• 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
♦ Because $\mathbf\small{\sum{\vec{F}_{y,ext}}}$ is indeed equal to zero vector
• So the rod AB is in translational equilibrium
• 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 = 90o 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 = 90o 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'
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
• Note that, the lines of action must be parallel. Only then will the forces become truly ‘opposite’
1. Fig.7.95(a) below, shows a magnetic needle
• 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
(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
■ A pair of equal and opposite forces with different lines of action is known as a couple
Let us see an example:
 |
| 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
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
• 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
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 |
• 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
In the next section, we will see Principle of Moments
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