In the previous section we completed a discussion on 'strings passing through pulleys'. We also saw some solved examples. In this section we will see friction.
We will write the steps:
1. Consider a block of mass ‘m’ resting on a horizontal surface in fig.5.73(a) below:
• The horizontal surface in the fig. is an ordinary surface. It is not too smooth or too rough
2. Let us apply a small force $\mathbf\small{\vec{F_1}}$ on the block as shown in fig.b
• This force must satisfy both the conditions given below:
(i) $\mathbf\small{\vec{F_1}}$ should not be too small
(ii) $\mathbf\small{\vec{F_1}}$ should not cause the block to move
3. So we applied a force, and the block did not move
• But according to the second law, it should have moved
• Let us apply the second law:
• The block is at rest on a horizontal surface. We know that, the vertical forces acting on the block are:
(i) The self weight of the block $\mathbf\small{\vec{W}}$ (acting downwards)
(ii) The normal reaction from the surface $\mathbf\small{\vec{F_N}}$ (acting upwards)
• Those two will cancel each other. So we do not need to consider them.
• But what about the horizontal forces?
• We applied $\mathbf\small{\vec{F_1}}$ towards the right. We see no other force to counter it
• So the net force must be $\mathbf\small{\vec{F_1}}$
• According to the second law, if there is a net force, the block must move with acceleration.
• This acceleration is given by $\mathbf\small{\vec{a}=\frac{\vec{F_1}}{m}}$
• If the ‘m’ is very large, $\mathbf\small{\vec{a}}$ will be very small. This is because m is in the denominator
• Another possibility is that, $\mathbf\small{\vec{F_1}}$ is very small. Then also $\mathbf\small{\vec{a}}$ will be very small
■ But in our present case, we do not see even a smallest possible acceleration. That means, there is no net force
4. So what happened to the $\mathbf\small{\vec{F_1}}$ that we applied?
• Clearly, another force has played a role here
• It canceled out our $\mathbf\small{\vec{F_1}}$, thus causing the net force to be zero
5. This force is called the frictional force. Usually we simply call it friction
• More precisely, it is called the static friction. It is denoted as $\mathbf\small{\vec{f_s}}$
• The word 'static' is added because, it is the frictional force that acted when the block was at rest
[So is there a 'different frictional force' which will act when the block is in motion?
There sure is. We call it the kinetic friction. It is denoted as $\mathbf\small{\vec{f_k}}$. We will see it soon]
6. So $\mathbf\small{\vec{f_s}}$ is a force. It will be having magnitude as well as direction
• What is it’s direction?
The following steps (i) to (v) will give the direction:
(i) Consider the ‘surface of contact’ between the block and the surface
• $\mathbf\small{\vec{f_s}}$ is always parallel to that surface
(ii) In our present case, if the bottom surface of the block is level, the contact surface will be horizontal
• So the $\mathbf\small{\vec{f_s}}$ will also be horizontal
(iii) If the block and/or the surface on which the block rests, has an irregular or curved shape, we must carefully calculate the orientation of the ‘contact surface’
• The orientation may be horizontal, vertical or inclined
(iv) What ever be the orientation, the $\mathbf\small{\vec{f_s}}$ will always be parallel to that orientation
(v) Also, the direction of $\mathbf\small{\vec{f_s}}$ will always be opposite to the 'direction in which the object tries to move'.
• In our present case, we tried to move the object towards the right
♦ So $\mathbf\small{\vec{f_s}}$ was acting towards the left.
7. Next we will try to find the magnitude of $\mathbf\small{\vec{f_s}}$
The following steps (i) to (iv) will give the magnitude:
(i) We applied $\mathbf\small{\vec{F_1}}$ (towards right) and it was canceled by $\mathbf\small{\vec{f_s}}$ (which acted towards left)
(ii) If $\mathbf\small{\vec{f_s}}$ had a magnitude lesser than $\mathbf\small{\vec{F_1}}$, there would have been a net force and the block would have moved to the right
(iii) If $\mathbf\small{\vec{f_s}}$ had a magnitude greater than $\mathbf\small{\vec{F_1}}$, then also there would have been a net force and the block would have moved to the left
(iv) Since none of the above two happened, we can say:
$\mathbf\small{\vec{f_s}}$ had a magnitude exactly equal to $\mathbf\small{\vec{F_1}}$
8. So we obtained both magnitude and direction of $\mathbf\small{\vec{f_s}}$
• We can add it to the fig.5.73(b) above. The modified fig. is shown in 5.73(c)
• However, before making a conclusion, let us experiment a little more:
9. Consider the same block in fig.5.73(a) above
• Apply another force $\mathbf\small{\vec{F_2}}$. This is shown in fig.5.74(a) below:
• This $\mathbf\small{\vec{F_2}}$ must be such that both the two conditions below are satisfied:
(i) $\mathbf\small{\vec{F_2}}$ > $\mathbf\small{\vec{F_1}}$
(ii) $\mathbf\small{\vec{F_2}}$ should not cause the block to move
10. Here a surprising fact awaits us. Let us analyse:
• Earlier we saw that $\mathbf\small{\vec{F_1}}$ is canceled by $\mathbf\small{\vec{f_s}}$
• Now we see that, $\mathbf\small{\vec{F_2}}$ (which is greater than $\mathbf\small{\vec{F_1}}$) is also canceled
• How is that possible? Did any additional force (besides $\mathbf\small{\vec{f_s}}$) come into play?
The following steps (i) to (iii) will give the answer:
(i) There was no additional force. It was the same $\mathbf\small{\vec{f_s}}$. It simply adjusted itself to a higher magnitude
(ii) We can say: $\mathbf\small{\vec{f_s}}$ is a self adjusting force. It has the ability to increase it’s magnitude.
(iii) Because of such an ability, the higher force $\mathbf\small{\vec{F_2}}$ was also canceled out. So the block did not move
11. So the $\mathbf\small{\vec{f_s}}$ that is experienced in the present experiment is clearly greater than the one experienced in the previous experiment
• We will denote the present one as $\mathbf\small{\vec{f_{s2}}}$ and the previous one $\mathbf\small{\vec{f_{s1}}}$. This is shown in figs.5.74(b) and (c)
• We can write $\mathbf\small{\vec{f_{s2}}}$ > $\mathbf\small{\vec{f_{s1}}}$
12. Now apply an even higher force $\mathbf\small{\vec{F_3}}$
• Let the block move this time. This is shown in fig.5.75(a) below:
• But that is puzzling. What happened to the 'ability'?
13. It is clear that $\mathbf\small{\vec{F_3}}$ was too much to handle
■ That is., there is a maximum value which $\mathbf\small{\vec{f_s}}$ can take. It is denoted as $\mathbf\small{\vec{f_{s,max}}}$
• It is the 'maximum possible ability'. If a force higher than $\mathbf\small{\vec{f_{s,max}}}$ is applied, the block will move.
14. We want to find this $\mathbf\small{\vec{f_{s,max}}}$
The following steps (i) to (v) should give us a method:
(i) Consider fig.5.75(b) above:
• A force $\mathbf\small{\vec{F_{max}}}$ is applied.
• The block is in a state of ‘impending motion’.
• That is., the block is about to start moving. But it has not moved yet
(ii) If we increase $\mathbf\small{\vec{F_{max}}}$ even by a minute amount, the block will move
• So $\mathbf\small{\vec{F_{max}}}$ is the 'maximum possible force' which can be applied with out causing motion
(iii) Clearly the friction which is playing now, will be the static friction ($\mathbf\small{\vec{f_s}}$) and it will be in it’s maximum possible value $\mathbf\small{\vec{f_{s,max}}}$
• Because, if $\mathbf\small{\vec{F_{max}}}$ is increased further, the block will start ‘movement’ and so it is no longer ‘static’ and consequently, the friction can no longer be called ‘static friction’
♦ Also, we can no longer use the subscript 's'
(iv) So in fig.5.75(b), the block is in the limiting state of equilibrium
• In this state, the $\mathbf\small{\vec{F_{max}}}$ is completely canceled out by $\mathbf\small{\vec{f_{s,max}}}$
• So we get: |$\mathbf\small{\vec{F_{max}}}$| = |$\mathbf\small{\vec{f_{s,max}}}$|
■ That means, if we can find the external force at the limiting state, we will have found the ‘maximum value of the ability’ to remain static
(v) Clearly, $\mathbf\small{\vec{F_{max}}}$ is greater than $\mathbf\small{\vec{F_2}}$ but less than $\mathbf\small{\vec{F_3}}$
• That is., value of $\mathbf\small{\vec{F_{max}}}$ lies some where between $\mathbf\small{\vec{F_2}}$ and $\mathbf\small{\vec{F_3}}$
• To find it, we will have to use trial and error method by trying different forces.
• That is., by trial and error, we have to find 'that force which when just increased' will cause motion
• But it will be difficult to obtain the precise $\mathbf\small{\vec{F_{max}}}$ by this method. Later, we will see a more accurate method
We will write the steps:
1. Consider a block of mass ‘m’ resting on a horizontal surface in fig.5.73(a) below:
 |
| Fig.5.73 |
2. Let us apply a small force $\mathbf\small{\vec{F_1}}$ on the block as shown in fig.b
• This force must satisfy both the conditions given below:
(i) $\mathbf\small{\vec{F_1}}$ should not be too small
(ii) $\mathbf\small{\vec{F_1}}$ should not cause the block to move
3. So we applied a force, and the block did not move
• But according to the second law, it should have moved
• Let us apply the second law:
• The block is at rest on a horizontal surface. We know that, the vertical forces acting on the block are:
(i) The self weight of the block $\mathbf\small{\vec{W}}$ (acting downwards)
(ii) The normal reaction from the surface $\mathbf\small{\vec{F_N}}$ (acting upwards)
• Those two will cancel each other. So we do not need to consider them.
• But what about the horizontal forces?
• We applied $\mathbf\small{\vec{F_1}}$ towards the right. We see no other force to counter it
• So the net force must be $\mathbf\small{\vec{F_1}}$
• According to the second law, if there is a net force, the block must move with acceleration.
• This acceleration is given by $\mathbf\small{\vec{a}=\frac{\vec{F_1}}{m}}$
• If the ‘m’ is very large, $\mathbf\small{\vec{a}}$ will be very small. This is because m is in the denominator
• Another possibility is that, $\mathbf\small{\vec{F_1}}$ is very small. Then also $\mathbf\small{\vec{a}}$ will be very small
■ But in our present case, we do not see even a smallest possible acceleration. That means, there is no net force
4. So what happened to the $\mathbf\small{\vec{F_1}}$ that we applied?
• Clearly, another force has played a role here
• It canceled out our $\mathbf\small{\vec{F_1}}$, thus causing the net force to be zero
5. This force is called the frictional force. Usually we simply call it friction
• More precisely, it is called the static friction. It is denoted as $\mathbf\small{\vec{f_s}}$
• The word 'static' is added because, it is the frictional force that acted when the block was at rest
[So is there a 'different frictional force' which will act when the block is in motion?
There sure is. We call it the kinetic friction. It is denoted as $\mathbf\small{\vec{f_k}}$. We will see it soon]
6. So $\mathbf\small{\vec{f_s}}$ is a force. It will be having magnitude as well as direction
• What is it’s direction?
The following steps (i) to (v) will give the direction:
(i) Consider the ‘surface of contact’ between the block and the surface
• $\mathbf\small{\vec{f_s}}$ is always parallel to that surface
(ii) In our present case, if the bottom surface of the block is level, the contact surface will be horizontal
• So the $\mathbf\small{\vec{f_s}}$ will also be horizontal
(iii) If the block and/or the surface on which the block rests, has an irregular or curved shape, we must carefully calculate the orientation of the ‘contact surface’
• The orientation may be horizontal, vertical or inclined
(iv) What ever be the orientation, the $\mathbf\small{\vec{f_s}}$ will always be parallel to that orientation
(v) Also, the direction of $\mathbf\small{\vec{f_s}}$ will always be opposite to the 'direction in which the object tries to move'.
• In our present case, we tried to move the object towards the right
♦ So $\mathbf\small{\vec{f_s}}$ was acting towards the left.
7. Next we will try to find the magnitude of $\mathbf\small{\vec{f_s}}$
The following steps (i) to (iv) will give the magnitude:
(i) We applied $\mathbf\small{\vec{F_1}}$ (towards right) and it was canceled by $\mathbf\small{\vec{f_s}}$ (which acted towards left)
(ii) If $\mathbf\small{\vec{f_s}}$ had a magnitude lesser than $\mathbf\small{\vec{F_1}}$, there would have been a net force and the block would have moved to the right
(iii) If $\mathbf\small{\vec{f_s}}$ had a magnitude greater than $\mathbf\small{\vec{F_1}}$, then also there would have been a net force and the block would have moved to the left
(iv) Since none of the above two happened, we can say:
$\mathbf\small{\vec{f_s}}$ had a magnitude exactly equal to $\mathbf\small{\vec{F_1}}$
8. So we obtained both magnitude and direction of $\mathbf\small{\vec{f_s}}$
• We can add it to the fig.5.73(b) above. The modified fig. is shown in 5.73(c)
• However, before making a conclusion, let us experiment a little more:
9. Consider the same block in fig.5.73(a) above
• Apply another force $\mathbf\small{\vec{F_2}}$. This is shown in fig.5.74(a) below:
 |
| Fig.5.74 |
(i) $\mathbf\small{\vec{F_2}}$ > $\mathbf\small{\vec{F_1}}$
(ii) $\mathbf\small{\vec{F_2}}$ should not cause the block to move
10. Here a surprising fact awaits us. Let us analyse:
• Earlier we saw that $\mathbf\small{\vec{F_1}}$ is canceled by $\mathbf\small{\vec{f_s}}$
• Now we see that, $\mathbf\small{\vec{F_2}}$ (which is greater than $\mathbf\small{\vec{F_1}}$) is also canceled
• How is that possible? Did any additional force (besides $\mathbf\small{\vec{f_s}}$) come into play?
The following steps (i) to (iii) will give the answer:
(i) There was no additional force. It was the same $\mathbf\small{\vec{f_s}}$. It simply adjusted itself to a higher magnitude
(ii) We can say: $\mathbf\small{\vec{f_s}}$ is a self adjusting force. It has the ability to increase it’s magnitude.
(iii) Because of such an ability, the higher force $\mathbf\small{\vec{F_2}}$ was also canceled out. So the block did not move
11. So the $\mathbf\small{\vec{f_s}}$ that is experienced in the present experiment is clearly greater than the one experienced in the previous experiment
• We will denote the present one as $\mathbf\small{\vec{f_{s2}}}$ and the previous one $\mathbf\small{\vec{f_{s1}}}$. This is shown in figs.5.74(b) and (c)
• We can write $\mathbf\small{\vec{f_{s2}}}$ > $\mathbf\small{\vec{f_{s1}}}$
12. Now apply an even higher force $\mathbf\small{\vec{F_3}}$
• Let the block move this time. This is shown in fig.5.75(a) below:
 |
| Fig.5.75 |
13. It is clear that $\mathbf\small{\vec{F_3}}$ was too much to handle
■ That is., there is a maximum value which $\mathbf\small{\vec{f_s}}$ can take. It is denoted as $\mathbf\small{\vec{f_{s,max}}}$
• It is the 'maximum possible ability'. If a force higher than $\mathbf\small{\vec{f_{s,max}}}$ is applied, the block will move.
14. We want to find this $\mathbf\small{\vec{f_{s,max}}}$
The following steps (i) to (v) should give us a method:
(i) Consider fig.5.75(b) above:
• A force $\mathbf\small{\vec{F_{max}}}$ is applied.
• The block is in a state of ‘impending motion’.
• That is., the block is about to start moving. But it has not moved yet
(ii) If we increase $\mathbf\small{\vec{F_{max}}}$ even by a minute amount, the block will move
• So $\mathbf\small{\vec{F_{max}}}$ is the 'maximum possible force' which can be applied with out causing motion
(iii) Clearly the friction which is playing now, will be the static friction ($\mathbf\small{\vec{f_s}}$) and it will be in it’s maximum possible value $\mathbf\small{\vec{f_{s,max}}}$
• Because, if $\mathbf\small{\vec{F_{max}}}$ is increased further, the block will start ‘movement’ and so it is no longer ‘static’ and consequently, the friction can no longer be called ‘static friction’
♦ Also, we can no longer use the subscript 's'
(iv) So in fig.5.75(b), the block is in the limiting state of equilibrium
• In this state, the $\mathbf\small{\vec{F_{max}}}$ is completely canceled out by $\mathbf\small{\vec{f_{s,max}}}$
• So we get: |$\mathbf\small{\vec{F_{max}}}$| = |$\mathbf\small{\vec{f_{s,max}}}$|
■ That means, if we can find the external force at the limiting state, we will have found the ‘maximum value of the ability’ to remain static
(v) Clearly, $\mathbf\small{\vec{F_{max}}}$ is greater than $\mathbf\small{\vec{F_2}}$ but less than $\mathbf\small{\vec{F_3}}$
• That is., value of $\mathbf\small{\vec{F_{max}}}$ lies some where between $\mathbf\small{\vec{F_2}}$ and $\mathbf\small{\vec{F_3}}$
• To find it, we will have to use trial and error method by trying different forces.
• That is., by trial and error, we have to find 'that force which when just increased' will cause motion
• But it will be difficult to obtain the precise $\mathbf\small{\vec{F_{max}}}$ by this method. Later, we will see a more accurate method
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