In the previous section we saw coefficient of static friction. In this section we will see the kinetic friction.
• What happens if we apply a force greater than $\mathbf\small{|\vec{f_{s,max}}|}$?
• We already saw the answer in the previous section. We will write it again:
Behavior of the ridges and valleys:
• When the force exceeds $\mathbf\small{|\vec{f_{s,max}}|}$, the block will have to rise a little higher up so that, it’s inverted ridges gets freed from the valleys of the horizontal surface
• We do not notice this ‘rising of the block’ because, it is at a microscopic scale
• Once they are freed, motion can take place
• Also, when motion takes place, the tips of the ridges on both sides will be knocked off
Behavior of the adhesion
• When the force exceeds $\mathbf\small{|\vec{f_{s,max}}|}$, the ‘bonds of adhesion’ will break.
• Once those bonds are broken, motion can take place
• But the interlocking which takes place during motion will not be as effective as when the object is at rest
• Also new adhesive bonds will not be effectively formed when the object is in motion
■ In short, the object will be experiencing a lesser friction during motion. We can write:
$\mathbf\small{|\vec{f_s}|<|\vec{f_k}|}$
• We saw that, the static friction $\mathbf\small{|\vec{f_s}|}$ can reach a maximum value of $\mathbf\small{|\vec{f_{s,max}}|}$
• The kinetic friction (also known as sliding friction) does not have such a maximum value.
• Even if the velocity of an object changes to a greater value, the magnitude of the frictional force $\mathbf\small{\vec{f_k}}$ opposing that motion remains the same
• Similarly, even if the velocity of an object changes to a lesser value, the magnitude of the frictional force $\mathbf\small{\vec{f_k}}$ opposing that motion remains the same
1. Fig.5.79(a) below shows a block of mass ‘m’ kg
• Under the influence of an external force $\mathbf\small{\vec{F}}$, it is moving with an acceleration $\mathbf\small{\vec{a}}$
• It is moving on a horizontal surface which is neither too smooth or too rough
2. The block is chosen as a sub-system as shown in fig.b
• In the FBD shown in fig.c, we see two forces
(i) $\mathbf\small{\vec{F}}$ towards right
(ii) $\mathbf\small{\vec{f_k}}$ towards left
• The resultant of the two forces is given by the vector sum: $\mathbf\small{\vec{F}-\vec{f_k}}$
3. By the second law, this resultant must be equal to m×a
• So we get: $\mathbf\small{\vec{F}-\vec{f_k}=m \times \vec{a}}$
■ From this we get the relation:
5.4: $\mathbf\small{\vec{a}=\frac{\vec{F}-\vec{f_k}}{m}}$
4. If the body is moving with a constant velocity, the acceleration will be zero
• So from the above relation, we get:
$\mathbf\small{0=\frac{\vec{F}-\vec{f_k}}{m}}$
$\mathbf\small{\Rightarrow \vec{F}=\vec{f_k}}$
We see an interesting fact here:
• If we see an object in uniform motion on an ordinary surface, it does not mean that no external force is acting on it. On ordinary surface, there need to be an external force even if the motion is uniform. This external force will be cancelled by the kinetic frictional force
• Earlier, we saw objects in uniform motion on frictionless surfaces. There is no need for an external force in such cases
5. If we remove the external force, the relation becomes:
$\mathbf\small{\vec{a}=\frac{0-\vec{f_k}}{m}}$
$\mathbf\small{\Rightarrow \vec{a}=-\frac{\vec{f_k}}{m}}$
■That means, if the external force is removed, the object will begin to experience a negative acceleration
■All bodies which experience a negative acceleration will slow down and come to a stop
Solved example 5.35
What is the acceleration of the block and trolley system shown in fig.5.80(a) below, if the coefficient of kinetic friction between the trolley and the surface is 0.04? What is the tension in the string? Assume the string to be light and inextensible [g = 10 ms-2]
Solution:
1. The sub-systems are shown in fig.5.80.b
♦ Two arrows are shown at the ends of the string
♦ This is to help us remember that, a string in tension always pulls at it's ends (Details here)
• The FBD of ‘A’ is shown in fig.c
2. We see two forces acting on A.
(The vertical forces will cancel each other and hence are not shown. But remember that, the vertical reaction from the surface is required for calculating the frictional force)
• When two forces act, the net force is given by the vector sum:
$\mathbf\small{\vec{F_1}+\vec{F_2}}$
3. $\mathbf\small{\text{Let}\,\,\vec{F_1}=\vec{T}\,\,\text{and}\,\,\vec{F_2}=\vec{f_k}}$
• Considering forces towards right as positive and those towards left as negative, we get:
Net force on A = $\mathbf\small{\vec{T}-\vec{f_k}}$
4. By the second law, net force = mass × acceleration
• So we get: $\mathbf\small{\vec{T}-\vec{f_k}=m_A \times \vec{a}}$
$\mathbf\small{\Rightarrow (|\vec{T}|)\hat{i}-(\mu_k \times |\vec{F_N}|)\hat{i}=(m_A \times |\vec{a}|)\hat{i}}$
$\mathbf\small{\Rightarrow (|\vec{T}|)\hat{i}-(\mu_k \times m_A \times g)\hat{i}=(m_A \times |\vec{a}|)\hat{i}}$
• Substituting the values, we get:
$\mathbf\small{(|\vec{T}|)\hat{i}-(0.04 \times 20 \times 10)\hat{i}=(20 \times |\vec{a}|)\hat{i}}$
$\mathbf\small{\Rightarrow |\vec{T}|-8=(20 \times |\vec{a}|)}$
5. The FBD of ‘B’ is shown in fig.d
We see two forces acting on B. When two forces act, the net force is given by the vector sum:
$\mathbf\small{\vec{F_1}+\vec{F_2}}$
6. $\mathbf\small{\text{Let}\,\,\vec{F_1}=\vec{T}\,\,\text{and}\,\,\vec{F_2}=\vec{W_B}}$
• Considering upward forces as positive and downward forces as negative, we get:
Net force on B = $\mathbf\small{\vec{T}-\vec{W_B}}$
7. By the second law, net force = mass × acceleration
• So we get: $\mathbf\small{\vec{T}-\vec{W_B}=-m_B \times \vec{a}}$
$\mathbf\small{\Rightarrow (|\vec{T}|)\hat{j}-(m_B \times g)\hat{j}=-(m_B \times |\vec{a}|)\hat{j}}$
• Substituting the values, we get:
$\mathbf\small{(|\vec{T}|)\hat{j}-(3 \times 10)\hat{j}=-(3 \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow |\vec{T}|-30=-(3 \times |\vec{a}|)}$
8. Thus we get two equations:
• From (4) we have: $\mathbf\small{|\vec{T}|-8=(20 \times |\vec{a}|)}$
• From (7) we have: $\mathbf\small{|\vec{T}|-30=-(3 \times |\vec{a}|)}$
Solving them, we get:
• $\mathbf\small{|\vec{a}|=\frac{22}{23}=0.96\,\text{ms}^{-2}}$
• $\mathbf\small{|\vec{T}|=27.1\,\text{N}}$
• This 'state' is called the 'limiting state' because, if the external force is increased even by the smallest amount, the object will move.
• We will write the analysis in steps:
1. Consider the long 'platform on wheels' shown in fig.5.81 below:
• The block 'A' of mass 'm' kg is resting on it
2. Initially, the platform is at rest. So the block is also at rest.
• No horizontal forces are acting at this stage
3. Now, the platform starts to move. Some horizontal forces come into play. Let us see what they are:
• The platform starts from rest (velocity = 0) and attains a velocity v.
• So surely, there has to be an acceleration
• Let the acceleration be done gradually.
♦ That is., the initial acceleration $\mathbf\small{\vec{a_1}}$ is small
• Then the block will move along with the platform with no slipping
4. However, the block will be trying to stay at it's position due to it's inertia
• But it cannot keep it's position
■ That means, some force is dragging the block
5. Obviously it is the frictional force between the block and the platform which causes the drag
• More precisely, it is the static friction between the block and the platform
• We can be sure that 'it is the static friction' and not 'the kinetic friction' because, there is no slipping between the block and the platform
• kinetic friction will come into play only when there is 'relative motion' between the block and the platform
• Here, the platform is accelerating gradually
♦ The block is in motion relative to the ground
♦ But the block is stationary relative to the platform
• That is why we say: 'at this stage, there is no relative motion between the block and the platform'.
• And so, the force of friction is that of 'static friction'
6. How does the static friction cause the drag?
Answer: It is through the interlocking and adhesion that we saw in fig.5.76 of the previous section
7. So the block is now moving as if it is clamped to the platform.
• The 'clamping force' is the 'force of static friction' $\mathbf\small{\vec{f_s}}$
• But this force has an upper limit. We denoted it as $\mathbf\small{\vec{f_{s,max}}}$
8. It is our duty to ensure that $\mathbf\small{\vec{f_s}}$ do not reach $\mathbf\small{\vec{f_{s,max}}}$
■ If it does, any further slightest increase will cause the block to slip
• If we want to prevent 'some thing', we must know the 'cause'
• So in this case, we must know 'cause of increase in $\mathbf\small{\vec{f_s}}$'
9. Imagine that, the platform is now moving with a greater acceleration $\mathbf\small{\vec{a_2}}$
• This higher acceleration $\mathbf\small{\vec{a_2}}$ should be provided for the block also.
• Then only it can keep up with the platform
10. So a force of $\mathbf\small{m_A \times \vec{a_2}}$ will be felt by the block
• The medium through which the acceleration is provided to the block is the same interlocking and adhesion that we saw earlier in fig.5.76 of the previous section
11. But due to inertia, the block does not want to receive this new acceleration. It wants to stay back
• However, due to the interlocking, the acceleration will be transfered
• So the ridges and valleys (and also the adhesive bonds) will begin to 'feel' the higher force $\mathbf\small{m_A \times \vec{a_2}}$
12. But there is a 'limiting force' which those ridges and valleys (and also the adhesive bonds) can take.
• We know that ' the limiting force' is the $\mathbf\small{\vec{f_{s,max}}}$
13. So if the $\mathbf\small{m_A \times \vec{a_2}}$ exceeds $\mathbf\small{\vec{f_{s,max}}}$, the ridges and valleys (and also the adhesive bonds) will break. The block will slip
• We must see that $\mathbf\small{m_A \times \vec{a}}$ is kept low. We cannot decrease mA. So the only option is to keep the acceleration low
14. The maximum acceleration possible can be obtained by equating the two forces. So we can write:
$\mathbf\small{m_A \times \vec{a_{max}}=\vec{f_{s,max}}}$
$\mathbf\small{\Longrightarrow m_A \times |\vec{a_{max}}|=|\vec{f_{s,max}}|}$
$\mathbf\small{\Longrightarrow m_A \times |\vec{a_{max}}|=\mu_s \times |\vec{F_N}|}$
$\mathbf\small{\Longrightarrow m_A \times |\vec{a_{max}}|=\mu_s \times m_A \times g}$
$\mathbf\small{\Longrightarrow |\vec{a_{max}}|=\mu_s \times g}$
• We see that, the 'maximum acceleration possible' is independent of the mass
An example:
Determine the maximum acceleration of a train in which a box lying on it's floor, will remain stationary, given that the coefficient of static friction between the box and the floor of the train is 0.15
Solution:
We have: $\mathbf\small{|\vec{a_{max}}|=\mu_s \times g}$
Substituting the values, we get: $\mathbf\small{|\vec{a_{max}}|=0.15 \times 10=1.5\,\text{ms}^{-2}}$
• What happens if we apply a force greater than $\mathbf\small{|\vec{f_{s,max}}|}$?
• We already saw the answer in the previous section. We will write it again:
Behavior of the ridges and valleys:
• When the force exceeds $\mathbf\small{|\vec{f_{s,max}}|}$, the block will have to rise a little higher up so that, it’s inverted ridges gets freed from the valleys of the horizontal surface
• We do not notice this ‘rising of the block’ because, it is at a microscopic scale
• Once they are freed, motion can take place
• Also, when motion takes place, the tips of the ridges on both sides will be knocked off
Behavior of the adhesion
• When the force exceeds $\mathbf\small{|\vec{f_{s,max}}|}$, the ‘bonds of adhesion’ will break.
• Once those bonds are broken, motion can take place
• So if the applied force is greater than $\mathbf\small{|\vec{f_{s,max}}|}$, the object will be in motion
• During motion also, the object will experience friction
The frictional force experienced during motion is called Kinetic friction. It is denoted as: $\mathbf\small{\vec{f_k}}$
• During motion also, the object will experience friction
The frictional force experienced during motion is called Kinetic friction. It is denoted as: $\mathbf\small{\vec{f_k}}$
• Also new adhesive bonds will not be effectively formed when the object is in motion
■ In short, the object will be experiencing a lesser friction during motion. We can write:
$\mathbf\small{|\vec{f_s}|<|\vec{f_k}|}$
• We saw that, the static friction $\mathbf\small{|\vec{f_s}|}$ can reach a maximum value of $\mathbf\small{|\vec{f_{s,max}}|}$
• The kinetic friction (also known as sliding friction) does not have such a maximum value.
• Even if the velocity of an object changes to a greater value, the magnitude of the frictional force $\mathbf\small{\vec{f_k}}$ opposing that motion remains the same
• Similarly, even if the velocity of an object changes to a lesser value, the magnitude of the frictional force $\mathbf\small{\vec{f_k}}$ opposing that motion remains the same
■Like static friction, kinetic friction also depends on the normal reaction $\mathbf\small{ |\vec{F_N}|}$. So, here also we have a similar relation:
5.3: $\mathbf\small{|\vec{f_k}|=\mu_s |\vec{F_N}|}$
Now let us analyse an object in motion. We will write it in steps:
 |
| Fig.5.79 |
• It is moving on a horizontal surface which is neither too smooth or too rough
2. The block is chosen as a sub-system as shown in fig.b
• In the FBD shown in fig.c, we see two forces
(i) $\mathbf\small{\vec{F}}$ towards right
(ii) $\mathbf\small{\vec{f_k}}$ towards left
• The resultant of the two forces is given by the vector sum: $\mathbf\small{\vec{F}-\vec{f_k}}$
3. By the second law, this resultant must be equal to m×a
• So we get: $\mathbf\small{\vec{F}-\vec{f_k}=m \times \vec{a}}$
■ From this we get the relation:
5.4: $\mathbf\small{\vec{a}=\frac{\vec{F}-\vec{f_k}}{m}}$
4. If the body is moving with a constant velocity, the acceleration will be zero
• So from the above relation, we get:
$\mathbf\small{0=\frac{\vec{F}-\vec{f_k}}{m}}$
$\mathbf\small{\Rightarrow \vec{F}=\vec{f_k}}$
We see an interesting fact here:
• If we see an object in uniform motion on an ordinary surface, it does not mean that no external force is acting on it. On ordinary surface, there need to be an external force even if the motion is uniform. This external force will be cancelled by the kinetic frictional force
• Earlier, we saw objects in uniform motion on frictionless surfaces. There is no need for an external force in such cases
5. If we remove the external force, the relation becomes:
$\mathbf\small{\vec{a}=\frac{0-\vec{f_k}}{m}}$
$\mathbf\small{\Rightarrow \vec{a}=-\frac{\vec{f_k}}{m}}$
■That means, if the external force is removed, the object will begin to experience a negative acceleration
■All bodies which experience a negative acceleration will slow down and come to a stop
Solved example 5.35
What is the acceleration of the block and trolley system shown in fig.5.80(a) below, if the coefficient of kinetic friction between the trolley and the surface is 0.04? What is the tension in the string? Assume the string to be light and inextensible [g = 10 ms-2]
 |
| Fig.5.80 |
1. The sub-systems are shown in fig.5.80.b
♦ Two arrows are shown at the ends of the string
♦ This is to help us remember that, a string in tension always pulls at it's ends (Details here)
• The FBD of ‘A’ is shown in fig.c
2. We see two forces acting on A.
(The vertical forces will cancel each other and hence are not shown. But remember that, the vertical reaction from the surface is required for calculating the frictional force)
• When two forces act, the net force is given by the vector sum:
$\mathbf\small{\vec{F_1}+\vec{F_2}}$
3. $\mathbf\small{\text{Let}\,\,\vec{F_1}=\vec{T}\,\,\text{and}\,\,\vec{F_2}=\vec{f_k}}$
• Considering forces towards right as positive and those towards left as negative, we get:
Net force on A = $\mathbf\small{\vec{T}-\vec{f_k}}$
4. By the second law, net force = mass × acceleration
• So we get: $\mathbf\small{\vec{T}-\vec{f_k}=m_A \times \vec{a}}$
$\mathbf\small{\Rightarrow (|\vec{T}|)\hat{i}-(\mu_k \times |\vec{F_N}|)\hat{i}=(m_A \times |\vec{a}|)\hat{i}}$
$\mathbf\small{\Rightarrow (|\vec{T}|)\hat{i}-(\mu_k \times m_A \times g)\hat{i}=(m_A \times |\vec{a}|)\hat{i}}$
• Substituting the values, we get:
$\mathbf\small{(|\vec{T}|)\hat{i}-(0.04 \times 20 \times 10)\hat{i}=(20 \times |\vec{a}|)\hat{i}}$
$\mathbf\small{\Rightarrow |\vec{T}|-8=(20 \times |\vec{a}|)}$
5. The FBD of ‘B’ is shown in fig.d
We see two forces acting on B. When two forces act, the net force is given by the vector sum:
$\mathbf\small{\vec{F_1}+\vec{F_2}}$
6. $\mathbf\small{\text{Let}\,\,\vec{F_1}=\vec{T}\,\,\text{and}\,\,\vec{F_2}=\vec{W_B}}$
• Considering upward forces as positive and downward forces as negative, we get:
Net force on B = $\mathbf\small{\vec{T}-\vec{W_B}}$
7. By the second law, net force = mass × acceleration
• So we get: $\mathbf\small{\vec{T}-\vec{W_B}=-m_B \times \vec{a}}$
$\mathbf\small{\Rightarrow (|\vec{T}|)\hat{j}-(m_B \times g)\hat{j}=-(m_B \times |\vec{a}|)\hat{j}}$
• Substituting the values, we get:
$\mathbf\small{(|\vec{T}|)\hat{j}-(3 \times 10)\hat{j}=-(3 \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow |\vec{T}|-30=-(3 \times |\vec{a}|)}$
8. Thus we get two equations:
• From (4) we have: $\mathbf\small{|\vec{T}|-8=(20 \times |\vec{a}|)}$
• From (7) we have: $\mathbf\small{|\vec{T}|-30=-(3 \times |\vec{a}|)}$
Solving them, we get:
• $\mathbf\small{|\vec{a}|=\frac{22}{23}=0.96\,\text{ms}^{-2}}$
• $\mathbf\small{|\vec{T}|=27.1\,\text{N}}$
Problems in Limiting state of Static friction
• We have seen the basics of both static friction and kinetic friction. Now we can learn about the 'state' between the two types of frictions.• This 'state' is called the 'limiting state' because, if the external force is increased even by the smallest amount, the object will move.
• We will write the analysis in steps:
1. Consider the long 'platform on wheels' shown in fig.5.81 below:
 |
| Fig.5.81 |
2. Initially, the platform is at rest. So the block is also at rest.
• No horizontal forces are acting at this stage
3. Now, the platform starts to move. Some horizontal forces come into play. Let us see what they are:
• The platform starts from rest (velocity = 0) and attains a velocity v.
• So surely, there has to be an acceleration
• Let the acceleration be done gradually.
♦ That is., the initial acceleration $\mathbf\small{\vec{a_1}}$ is small
• Then the block will move along with the platform with no slipping
4. However, the block will be trying to stay at it's position due to it's inertia
• But it cannot keep it's position
■ That means, some force is dragging the block
5. Obviously it is the frictional force between the block and the platform which causes the drag
• More precisely, it is the static friction between the block and the platform
• We can be sure that 'it is the static friction' and not 'the kinetic friction' because, there is no slipping between the block and the platform
• kinetic friction will come into play only when there is 'relative motion' between the block and the platform
• Here, the platform is accelerating gradually
♦ The block is in motion relative to the ground
♦ But the block is stationary relative to the platform
• That is why we say: 'at this stage, there is no relative motion between the block and the platform'.
• And so, the force of friction is that of 'static friction'
6. How does the static friction cause the drag?
Answer: It is through the interlocking and adhesion that we saw in fig.5.76 of the previous section
7. So the block is now moving as if it is clamped to the platform.
• The 'clamping force' is the 'force of static friction' $\mathbf\small{\vec{f_s}}$
• But this force has an upper limit. We denoted it as $\mathbf\small{\vec{f_{s,max}}}$
8. It is our duty to ensure that $\mathbf\small{\vec{f_s}}$ do not reach $\mathbf\small{\vec{f_{s,max}}}$
■ If it does, any further slightest increase will cause the block to slip
• If we want to prevent 'some thing', we must know the 'cause'
• So in this case, we must know 'cause of increase in $\mathbf\small{\vec{f_s}}$'
9. Imagine that, the platform is now moving with a greater acceleration $\mathbf\small{\vec{a_2}}$
• This higher acceleration $\mathbf\small{\vec{a_2}}$ should be provided for the block also.
• Then only it can keep up with the platform
10. So a force of $\mathbf\small{m_A \times \vec{a_2}}$ will be felt by the block
• The medium through which the acceleration is provided to the block is the same interlocking and adhesion that we saw earlier in fig.5.76 of the previous section
11. But due to inertia, the block does not want to receive this new acceleration. It wants to stay back
• However, due to the interlocking, the acceleration will be transfered
• So the ridges and valleys (and also the adhesive bonds) will begin to 'feel' the higher force $\mathbf\small{m_A \times \vec{a_2}}$
12. But there is a 'limiting force' which those ridges and valleys (and also the adhesive bonds) can take.
• We know that ' the limiting force' is the $\mathbf\small{\vec{f_{s,max}}}$
13. So if the $\mathbf\small{m_A \times \vec{a_2}}$ exceeds $\mathbf\small{\vec{f_{s,max}}}$, the ridges and valleys (and also the adhesive bonds) will break. The block will slip
• We must see that $\mathbf\small{m_A \times \vec{a}}$ is kept low. We cannot decrease mA. So the only option is to keep the acceleration low
14. The maximum acceleration possible can be obtained by equating the two forces. So we can write:
$\mathbf\small{m_A \times \vec{a_{max}}=\vec{f_{s,max}}}$
$\mathbf\small{\Longrightarrow m_A \times |\vec{a_{max}}|=|\vec{f_{s,max}}|}$
$\mathbf\small{\Longrightarrow m_A \times |\vec{a_{max}}|=\mu_s \times |\vec{F_N}|}$
$\mathbf\small{\Longrightarrow m_A \times |\vec{a_{max}}|=\mu_s \times m_A \times g}$
$\mathbf\small{\Longrightarrow |\vec{a_{max}}|=\mu_s \times g}$
• We see that, the 'maximum acceleration possible' is independent of the mass
An example:
Determine the maximum acceleration of a train in which a box lying on it's floor, will remain stationary, given that the coefficient of static friction between the box and the floor of the train is 0.15
Solution:
We have: $\mathbf\small{|\vec{a_{max}}|=\mu_s \times g}$
Substituting the values, we get: $\mathbf\small{|\vec{a_{max}}|=0.15 \times 10=1.5\,\text{ms}^{-2}}$
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