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 →F1 on the block as shown in fig.b
• This force must satisfy both the conditions given below:
(i) →F1 should not be too small
(ii) →F1 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 →W (acting downwards)
(ii) The normal reaction from the surface →FN (acting upwards)
• Those two will cancel each other. So we do not need to consider them.
• But what about the horizontal forces?
• We applied →F1 towards the right. We see no other force to counter it
• So the net force must be →F1
• According to the second law, if there is a net force, the block must move with acceleration.
• This acceleration is given by →a=→F1m
• If the ‘m’ is very large, →a will be very small. This is because m is in the denominator
• Another possibility is that, →F1 is very small. Then also →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 →F1 that we applied?
• Clearly, another force has played a role here
• It canceled out our →F1, 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 →fs
• 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 →fk. We will see it soon]
6. So →fs 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
• →fs 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 →fs 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 →fs will always be parallel to that orientation
(v) Also, the direction of →fs 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 →fs was acting towards the left.
7. Next we will try to find the magnitude of →fs
The following steps (i) to (iv) will give the magnitude:
(i) We applied →F1 (towards right) and it was canceled by →fs (which acted towards left)
(ii) If →fs had a magnitude lesser than →F1, there would have been a net force and the block would have moved to the right
(iii) If →fs had a magnitude greater than →F1, 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:
→fs had a magnitude exactly equal to →F1
8. So we obtained both magnitude and direction of →fs
• 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 →F2. This is shown in fig.5.74(a) below:
• This →F2 must be such that both the two conditions below are satisfied:
(i) →F2 > →F1
(ii) →F2 should not cause the block to move
10. Here a surprising fact awaits us. Let us analyse:
• Earlier we saw that →F1 is canceled by →fs
• Now we see that, →F2 (which is greater than →F1) is also canceled
• How is that possible? Did any additional force (besides →fs) come into play?
The following steps (i) to (iii) will give the answer:
(i) There was no additional force. It was the same →fs. It simply adjusted itself to a higher magnitude
(ii) We can say: →fs is a self adjusting force. It has the ability to increase it’s magnitude.
(iii) Because of such an ability, the higher force →F2 was also canceled out. So the block did not move
11. So the →fs 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 →fs2 and the previous one →fs1. This is shown in figs.5.74(b) and (c)
• We can write →fs2 > →fs1
12. Now apply an even higher force →F3
• 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 →F3 was too much to handle
■ That is., there is a maximum value which →fs can take. It is denoted as →fs,max
• It is the 'maximum possible ability'. If a force higher than →fs,max is applied, the block will move.
14. We want to find this →fs,max
The following steps (i) to (v) should give us a method:
(i) Consider fig.5.75(b) above:
• A force →Fmax 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 →Fmax even by a minute amount, the block will move
• So →Fmax 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 (→fs) and it will be in it’s maximum possible value →fs,max
• Because, if →Fmax 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 →Fmax is completely canceled out by →fs,max
• So we get: |→Fmax| = |→fs,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, →Fmax is greater than →F2 but less than →F3
• That is., value of →Fmax lies some where between →F2 and →F3
• 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 →Fmax 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 →F1 on the block as shown in fig.b
• This force must satisfy both the conditions given below:
(i) →F1 should not be too small
(ii) →F1 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 →W (acting downwards)
(ii) The normal reaction from the surface →FN (acting upwards)
• Those two will cancel each other. So we do not need to consider them.
• But what about the horizontal forces?
• We applied →F1 towards the right. We see no other force to counter it
• So the net force must be →F1
• According to the second law, if there is a net force, the block must move with acceleration.
• This acceleration is given by →a=→F1m
• If the ‘m’ is very large, →a will be very small. This is because m is in the denominator
• Another possibility is that, →F1 is very small. Then also →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 →F1 that we applied?
• Clearly, another force has played a role here
• It canceled out our →F1, 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 →fs
• 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 →fk. We will see it soon]
6. So →fs 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
• →fs 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 →fs 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 →fs will always be parallel to that orientation
(v) Also, the direction of →fs 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 →fs was acting towards the left.
7. Next we will try to find the magnitude of →fs
The following steps (i) to (iv) will give the magnitude:
(i) We applied →F1 (towards right) and it was canceled by →fs (which acted towards left)
(ii) If →fs had a magnitude lesser than →F1, there would have been a net force and the block would have moved to the right
(iii) If →fs had a magnitude greater than →F1, 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:
→fs had a magnitude exactly equal to →F1
8. So we obtained both magnitude and direction of →fs
• 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 →F2. This is shown in fig.5.74(a) below:
 |
| Fig.5.74 |
(i) →F2 > →F1
(ii) →F2 should not cause the block to move
10. Here a surprising fact awaits us. Let us analyse:
• Earlier we saw that →F1 is canceled by →fs
• Now we see that, →F2 (which is greater than →F1) is also canceled
• How is that possible? Did any additional force (besides →fs) come into play?
The following steps (i) to (iii) will give the answer:
(i) There was no additional force. It was the same →fs. It simply adjusted itself to a higher magnitude
(ii) We can say: →fs is a self adjusting force. It has the ability to increase it’s magnitude.
(iii) Because of such an ability, the higher force →F2 was also canceled out. So the block did not move
11. So the →fs 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 →fs2 and the previous one →fs1. This is shown in figs.5.74(b) and (c)
• We can write →fs2 > →fs1
12. Now apply an even higher force →F3
• Let the block move this time. This is shown in fig.5.75(a) below:
 |
| Fig.5.75 |
13. It is clear that →F3 was too much to handle
■ That is., there is a maximum value which →fs can take. It is denoted as →fs,max
• It is the 'maximum possible ability'. If a force higher than →fs,max is applied, the block will move.
14. We want to find this →fs,max
The following steps (i) to (v) should give us a method:
(i) Consider fig.5.75(b) above:
• A force →Fmax 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 →Fmax even by a minute amount, the block will move
• So →Fmax 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 (→fs) and it will be in it’s maximum possible value →fs,max
• Because, if →Fmax 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 →Fmax is completely canceled out by →fs,max
• So we get: |→Fmax| = |→fs,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, →Fmax is greater than →F2 but less than →F3
• That is., value of →Fmax lies some where between →F2 and →F3
• 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 →Fmax by this method. Later, we will see a more accurate method
No comments:
Post a Comment