Chapter 5.21 - Rolling friction

In the previous section we saw coefficient of kinetic friction. We also saw some solved examples. In this section we will see a few more solved examples. After the solved examples we will see rolling friction.

Solved example 5.36
Two bodies A and B of masses 5 kg and 10 kg in contact with each other rest on a table against a rigid wall (Fig. 5.82.a) The coefficient of friction between the bodies and the table is 0.15. A force of 200 N is applied horizontally to A. What are (a) the reaction of the partition (b) the action-reaction forces between A and B ? What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between μ_s and μ_k.

Fig.5.82

Solution:
Case 1: When the wall is present
• The blocks will not be able to move even the 'smallest possible distance' because of the wall
■ The friction between the blocks and the ground will begin to act only when the blocks try to move. 
• This is because, force will not develop in the interlocking ridges and valleys (and also the adhesion) if the blocks do not move
• So in this case, we need not consider friction at all
Part (a):
1. The FBD of (A+B) is shown in fig.b
(The vertical forces will cancel each other and hence are not shown)
• A force of 200 N acts on the blocks from left to right
• Clearly, a 200 N force must act in the opposite direction. That is., from right to left. Other wise there will not be equilibrium
2. This 200 N from right to left is provided by the wall
• That means., the reaction from the wall is 200 N
Part (b):
1. Fig.c shows the FBD of A
• A force of 200 N acts on A from left to right
• Clearly, a 200 N force must act from right to left. Other wise there will not be equilibrium
2. This 200 N from right to left is provided by the block B
• That means., the reaction from B is 200 N
3. By the third law, this reaction from B must be due to the action from A
• Action and reaction are equal in magnitude and opposite in direction
• So we can write:
    ♦ A applies an action of 200 N on B (from left to right) 
    ♦ B applies a reaction of 200 N on A (from right to left)
Case 2: When the wall is absent:
• Now the blocks are free to move as shown in fig.5.83(a) below:

Fig.5.83

• We must first check whether the friction is strong enough to prevent motion:
• The FBD of (A+B) is shown in fig.5.83(b)
(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)

• We see that, the force from left to right is 200 N

• The force in the opposite direction (from right to left) is due to the frictional forces
• The total frictional force is: $\mathbf\small{\vec{f_{s,max(A)}}+\vec{f_{s,max(B)}}}$
• The magnitude of this total friction works out to:
$\mathbf\small{(\mu_s \times m_A \times g)+(\mu_s \times m_B \times g)=[\mu_s \times g \times (m_A+m_B)]=[0.15 \times 10 \times (5+10)]=22.5\,\text{N}}$
■ Since the 200 N is greater than this frictional force, the blocks will move
Now we can write the steps:
1. Since the two blocks move, the frictional force will be kinetic. So in the FBD, we must use $\mathbf\small{\vec{f_{k(A)}}\,\,\text{and}\,\,\vec{f_{k(B)}}}$
• This is shown in fig.5.83(c)
• When two forces act, the net force is given by the vector sum:
$\mathbf\small{\vec{F_1}+\vec{F_2}}$
2. $\mathbf\small{\text{Let}\,\,\vec{F_1}=200\,\text{N}}$
$\mathbf\small{\text{Let}\,\,\vec{F_2}=\vec{f_{k(A)}}+\vec{f_{k(B)}}}$
• The magnitude of this total friction works out to:
$\mathbf\small{(\mu_k \times m_A \times g)+(\mu_k \times m_B \times g)=[\mu_k \times g \times (m_A+m_B)]=[0.15 \times 10 \times (5+10)]=22.5\,\text{N}}$
• Note that, we are asked to ignore the difference between μ_s and μ_k. So we took μ_k = 0.15
• Thus we get: $\mathbf\small{\vec{F_2}=22.5\,\text{N}}$
• Considering forces towards right as positive and those towards left as negative, we get:
Net force on (A+B) = 200-22.5 = 177.5 N
3. By the second law, net force = mass × acceleration
• So we get: $\mathbf\small{177.5=m_{A+B} \times |\vec{a}|}$
$\mathbf\small{\Longrightarrow 177.5=15 \times |\vec{a}|}$
$\mathbf\small{\Longrightarrow |\vec{a}|=11.83\,\text{ms}^{-2}}$
• That means, the two blocks move together with an acceleration of 11.83 ms^-2.
4. The FBD of A is shown in fig.5.83(d) above
We see three forces acting on it. When three forces act, the net force is given by the vector sum:
$\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}}$
5. $\mathbf\small{\text{Let}\,\,\vec{F_1}=200\,\text{N}}$
$\mathbf\small{\text{Let}\,\,\vec{F_2}=\vec{F_{N(AB)}}}$
$\mathbf\small{\text{Let}\,\,\vec{F_3}=\vec{f_{k(A)}}}$
• The magnitude of the frictional force $\mathbf\small{\vec{F_3}}$ works out to:
$\mathbf\small{(\mu_k \times m_A \times g)=(0.15 \times 5 \times 10)=7.5\,\text{N}}$
• Considering forces towards right as positive and those towards left as negative, we get:  
Net force on A = $\mathbf\small{200 \hat{i}-\vec{F_{N(AB)}}-7.5 \hat{i}}$
6. By the second law, net force = mass × acceleration
• So we get: $\mathbf\small{192.5 \hat{i}-\vec{F_{N(AB)}}=(m_{A} \times |\vec{a}|)\hat{i}}$
$\mathbf\small{\Longrightarrow 192.5\hat{i}-\vec{F_{N(AB)}}=(5 \times 11.83)\hat{i}}$
$\mathbf\small{\Longrightarrow \vec{F_{N(AB)}}=133.35\hat{i}}$
$\mathbf\small{\Longrightarrow |\vec{F_{N(AB)}}|=133.35\,\,\text{N}}$
7. By the third law, this reaction from B must be due to the action from A
• Action and reaction are equal in magnitude and opposite in direction
• So we can write:
    ♦ A applies an action of 133.35 N on B (from left to right) 
    ♦ B applies a reaction of 133.35 N on A (from right to left)

Solved example 5.37
The rear side of a truck is open and a box A of 40 kg mass is placed 5 m away from the open end as shown in Fig. 5.84 below. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with 2 ms^-2. At what distance from the starting point does the box fall off the truck? (Ignore the size of the box)

Fig.5.84

Solution:
• First we have to determine whether the box will slip under the acceleration of 2 ms^-2
• We have: $\mathbf\small{|\vec{a_{max}}|=\mu_s \times g}$ (Details here)
• Substituting the values, we get: $\mathbf\small{|\vec{a_{max}}|=0.15 \times 10=1.5\,\text{ms}^{-2}}$
• But the truck accelerates at 2 ms^-2. So the box will definitely slip. That means, it will be sliding on the surface of the truck. 
Now we can write the steps:
1. If the surface of the truck is smooth, the box will not even move
• In that case, when the truck moves 5 m, the box will fall off
• Since such a situation does not arise, we can be sure that: Some force is dragging the box so that it moves with the truck
2. This dragging force is nothing but the frictional force between the truck and the box
• As the box is sliding, this frictional force is the 'kinetic friction'.
• We know how to calculate it's magnitude:
$\mathbf\small{|\vec{f_k}|=\mu_k \times |\vec{F_N}|=0.15 \times 400 = 60\, \text{N}}$
3. Now the FBD of the box A will be as shown in fig.5.85(a) below:

Fig.5.85

• We see that, 60 N is the only force acting on A
• So we can write: $\mathbf\small{m_A \times |\vec{a_A}|= 60\,\text{N}}$
$\mathbf\small{\Longrightarrow |\vec{a_A}|= \frac{60}{40}=1.5\,\text{ms}^{-2}}$
• That means, A moves towards left with an acceleration of 1.5 ms^-2
4. Remember that the truck is also moving towards the left. But it's acceleration is 2 ms^-2.
• So A will not be able to keep up with the truck. It will fall off after some time
• Before the motion begins, mark an arrow on the platform of the truck, right below the box. Let the tip of the arrow be 'P'
• This is shown in fig.5.85(b)
5. Let the truck start it's motion when the stop watch reading = 0
• Let the box fall off when stop watch reading = t seconds
• Let the box travel a distance of 'x' m during this 't' seconds
• Then, during this 't' seconds, point P must have traveled (5+x) meters
• This is shown in fig.c
6. So we have two sets of information:
(i) The box started from rest
• It moved with an acceleration of 1.5 ms^-2.
• It travelled for 't' s
• It travelled a distance of 'x' m during those 't' seconds 
(ii) The point 'P' started from rest
• It moved with an acceleration of 2.0 ms^-2.
• It travelled for 't' s
• It travelled a distance of '(5+x)' m during those 't' seconds 
7. We can use the equation: $\mathbf\small{s=ut+\frac{1}{2}at^2}$ 
• Applying the equation for the box, we get:
$\mathbf\small{x=0 \times t+\frac{1}{2}\times 1.5 \times t^2}$
$\mathbf\small{\Longrightarrow x=0.75 \times t^2}$
• Applying the equation for the point P, we get:
$\mathbf\small{5+x=0 \times t+\frac{1}{2}\times 2 \times t^2}$
$\mathbf\small{\Longrightarrow 5+x= t^2}$
• Solving the two equations, we get:
t = √20 s and x = 15 m
8. Thus we can write:
• When 'P' reaches a distance of (5+15) = 20 m from it's initial position, the box will fall off
• If 'P' travels 20 m, the truck also travels the same 20 m. 
■ So we can write:
When the truck reaches a distance of 20 m from it's initial position, the box will fall off

Solved example 5.38
In fig.5.86(a) below, a block 'A' of mass 4 kg is placed above another block 'B' of mass 5 kg. A force of 12 N (applied on 'A') is required to move 'A'. Then what is the maximum force that can be applied on 'B' so that 'A' and 'B' move together?

Fig.5.86

Solution:
1. Given that 12 N is the minimum required force. 
• That means, a force less than 12 N will not be sufficient to move A
• That means, 12 N is the limiting force $\mathbf\small{|\vec{f_{s,max}}|}$
2. We have:
$\mathbf\small{|\vec{f_{s,max}}|=\mu_{s,AB} \times |\vec{F_N}|}$
• Where μ_s,AB is the coefficient of static friction between A and B
• Substituting the values, we get: 12 × μ_s,AB = 40
⇒ μ_s,AB = 0.3
3. Next, we want to apply a force on B so that A and B 'move together'. 
• That means, A must not slip on B
• For small forces, A will not slip. But if the force is large, it will slip. 
• We want the maximum largest force which can be applied without causing the slip
4. We have: a_max = μ_s×g (Details here)
• Substituting the values, we get: a_max = 0.3 ×10 = 3 ms^-2
• That means., B can move with a maximum acceleration of 3 ms^-2
5. We have: Force = mass × acceleration
• But A and B are moving together. So mass is the total mass of (A+B) which is equal to 9 kg
• So we have a 9 kg mass moving at an acceleration of 3 ms^-2 
• So the required force = mass × acceleration = 9 × 3 = 27 N

Solved example 5.39
A block of mass 5 kg rests on an inclined plane as shown in fig.5.85(b) above. When the angle θ is 30^o, the block just starts sliding down. The coefficient of friction is 0.2. What is the velocity of the block 5 seconds after beginning the slide?
Solution:
• In this problem, the coefficient of friction is given as 0.2
• It is not specified whether 0.2 is μ_k or μ_s
• But we can confirm that it is μ_k 
• Because, μ_s will be tan 30, which is equal to 0.58
Now we can write the steps:
1. Resolving the forces parallel and perpendicular to the inclined plane, we get:
• Force causing the slide = mg sinθ
• Force resisting the slide = μ_k × mg cosθ
(See details here)
2. So net force = (mg sinθ - μ_k × mg cosθ) = mg(sinθ - μ_k cosθ) 
• Substituting the values, we get:
Net force = 5×10 (sin 30 - 0.2 × cos 30) = 16.34 N
3. Acceleration = ^{Net force}⁄_mass = ^16.34⁄₅  = 3.27 ms^-2.
4. We can use the equation: $\mathbf\small{v=u+at}$ 
• Substituting the values, we get:
$\mathbf\small{v=0 +3.27 \times 5=16.35\, \text{ms}^{-1}}$

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Rolling Friction

In fig.5.87(a) below, a wheel is rolling over a horizontal plane. 

Fig.5.87

• At any instant, there is only a ‘point of contact’ between the wheel and the plane. 
• It is just like a tangent drawn to a circle. We know that, a tangent to a circle will touch it only at one point.
• Remember that, in all the cases that we saw so far in this chapter, there is an ‘area of contact’. 
    ♦ But here, for the wheel, there is only a ‘point of contact’.
• Consider any such point at an instant. This point has no relative motion with the plane. 
    ♦ This is because, the next instant, another point will be in contact.
• So there should not be any static or kinetic friction between the wheel and the surface. 
• That means, once the wheel is set rolling, it should continue to roll even without any external force. 
• But in practice, we do not see this happening. 
The reason can be written in the following 4 steps:
1. The ‘point of contact’ between the wheel and the surface is an ‘ideal situation’. But we do not get such a situation in practice. 
2..This is because, a small deformation happens to both the wheel and the surface at the point of contact. 
3. As a result, there will indeed be a ‘small area of contact’. This results in friction. 
• The deformation if enlarged, will look as in fig.c.
4. The deformation is however ‘momentary’. That means, when the next portion of the wheel and surface come into contact, the earlier deformations will recover their original shapes

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• Thus we see that, friction comes into play even for the rolling motion. 

• But when compared to the static and kinetic friction, this rolling frictional force is very small in magnitude. 
• That means, if we can introduce rollers at the contact surface between two sliding objects, the resistance to sliding will be very low. 
• So ‘Invention of the wheel’ was indeed a major milestone in the development of mankind

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In the next section we will see circular motion

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Chapter 5.20 - The Coefficient of Kinetic Friction

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

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• 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: $\mathbf\small{\vec{f_k}}$  

• 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

■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: $\mathbf\small{|\vec{f_k}|=\mu_s |\vec{F_N}|}$

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Now let us analyse an object in motion. We will write it in steps:

1. Fig.5.79(a) below shows a block of mass ‘m’ kg

Fig.5.79

• 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

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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

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: 
• $\mathbf\small{|\vec{a}|=\frac{22}{23}=0.96\,\text{ms}^{-2}}$
• $\mathbf\small{|\vec{T}|=27.1\,\text{N}}$

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

• 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 m_A. 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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In the next section we will see a few more solved examples

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Chapter 5.19 - The Coefficient of Static Friction

In the previous section, we saw the basics of static friction. In this section, we will see the causes of friction.
A detailed description of the ‘causes of friction’ is quite complex. We will see it in higher classes. At present, a general description based on the following steps (1) to (6) will be sufficient for our study of mechanics    
1. Fig.5.76 below shows the same block and the horizontal surface that we saw in the previous section

Fig.5.76 Irregularities at the contact surface

• A small magenta circle is marked at the surface of contact. The portion inside this circle is enlarged.
2. We see that the top side of the horizontal surface  is actually rough 
    ♦ It has many ridges and valleys
• The bottom side of the block is also rough.  
    ♦ It has many ‘inverted’ ridges and valleys. 
■ This is true for all surfaces that we come across in our day to day life. Even those surfaces which appear to be very smooth will have such irregularities.  
3. The ridges on the top side get locked between the valleys in the bottom side and vice versa. 
• So there is a sort of 'interlocking' between the two sides. 
• This interlocking will resist the motion of the block over the surface. 
4. Also, there will be adhesion between the molecules of the block and those of the horizontal surface
• This adhesion is predominant  at the contact surface between the two
5. Thus we see two causes for friction:
(i) The interlocking between the ridges and valleys
(ii) The adhesion between molecules on the two sides

6. In the previous section, we saw that a force greater than $\mathbf\small{\vec{F_{max}}}$ should not be applied if we want the block to remain stationary
• But if we do apply a force greater than $\mathbf\small{\vec{F_{max}}}$, what all changes will occur?

• To find the answer, we have to see two items: 
(i) Behavior of the ridges and valleys when a force greater than $\mathbf\small{\vec{F_{max}}}$ is applied
(ii) Behavior of the adhesion when a force greater than $\mathbf\small{\vec{F_{max}}}$ is applied 

Behavior of the ridges and valleys:
• When the force exceeds $\mathbf\small{\vec{F_{max}}}$, 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 $\mathbf\small{\vec{F_{max}}}$, the ‘bonds of adhesion’ will break.
• Once those bonds are broken, motion can take place

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• So we now know the basic causes of friction. 
• It is clear that, if the block is pressed  harder (as shown in fig.5.77.a below) from the top, the interlocking will increase

Fig.5.77

• The adhesion will also increase. 
• Thus the friction will increase. 
• That means, the magnitude of $\mathbf\small{\vec{F_{max}}}$ required to start motion will increase

■ Can we think of any practical situations where such ‘extra pressing’ occurs?
• Of course we can. When a second block is placed above the original block, the original block will be pressed more towards the horizontal surface. This is shown in fig.5.77(b) above 
• When such a second block is placed, it will be more difficult to start motion. 
■ So it is clear that static friction is proportional to the ‘total amount of pressing’ from the top. 
• The ‘total amount of pressing’ can be calculated just by adding the weights of the individual blocks
• There can be any number of blocks stacked above one another. We will need to calculate the total weight of all the blocks which are stacked

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• Instead of ‘total weight of all the blocks’, we can use a simpler item: ‘normal reaction’, Which we saw in a previous section. 
• We represented it as: $\mathbf\small{\vec{F_N}}$ 
• When we use $\mathbf\small{\vec{F_N}}$, all individual weights will be accounted for. This is clear from fig.5.77(c) above
• Also, when the contact surface is inclined, we must use only 'that pressing force' which is perpendicular to the surface.
■ Thus it is clear that we must use $\mathbf\small{\vec{F_N}}$ instead of 'total weight' 
• The following steps (1) to (10) will help us to understand the relation between $\mathbf\small{\vec{F_N}}$ and static friction:
1. The total pressing is equal to the total weight
2. But according to third law, total weight is equal to the $\mathbf\small{\vec{F_N}}$
That means: 
Magnitude of the total pressing = Magnitude of the total weight = $\mathbf\small{|\vec{F_N}|}$
3. We saw that the magnitude of $\mathbf\small{\vec{F_{max}}}$ required to start motion depends on the total pressing
• That is., $\mathbf\small{|\vec{F_{max}}|}$ depends on the total pressing
• Now, we can write '$\mathbf\small{|\vec{F_N}|}$' in place of 'total pressing'
• Thus we get: 
$\mathbf\small{|\vec{F_{max}}|}$ depends on $\mathbf\small{|\vec{F_N}|}$
4. We know that $\mathbf\small{|\vec{F_{max}}|=|\vec{f_{s,max}}|}$ 
• So in (3), we can write $\mathbf\small{|\vec{f_{s,max}}|}$ in the place of $\mathbf\small{|\vec{F_{max}}|}$    
• Thus we get: $\mathbf\small{|\vec{f_{s,max}}|}$ depends on $\mathbf\small{|\vec{F_N}|}$ 
5. That is., greater the $\mathbf\small{|\vec{F_N}|}$, greater the static friction
• In other words, static friction is directly proportional to $\mathbf\small{|\vec{F_N}|}$
• We can write: $\mathbf\small{|\vec{f_{s,max}}|\propto |\vec{F_N}|}$
6. Now we introduce a constant of proportionality. We get:
• $\mathbf\small{|\vec{f_{s,max}}|=\mu_s |\vec{F_N}|}$
• Where $\mathbf\small{\mu_s }$ is the constant of proportionality
■ It is called the coefficient of static friction
7. This coefficient depends on the type of surfaces in contact. 
• Scientists and Engineers have done numerous trials and experiments and have determined it's value for a large number of surfaces. 
• Those values are published in the form of standard tables. We can obtain the required values from those tables. One such table can be seen here.
8. The $\mathbf\small{\mu_s }$ depends on the type of surfaces
• So when we use the equation $\mathbf\small{|\vec{f_{s,max}}|=\mu_s |\vec{F_N}|}$, we do not have to worry about the ridges and valleys in the surface of contact. 
• Also, we do not have to worry about the adhesion. 
• All those are take account of in $\mathbf\small{\mu_s }$        
9. Thus the equation $\mathbf\small{|\vec{f_{s,max}}|=\mu_s |\vec{F_N}|}$ helps us to find the minimum force required to start motion
• Alternatively, if that minimum required force is known, we can find the normal reaction by rearranging the equation as: $\mathbf\small{|\vec{F_N}|=\frac{|\vec{f_{s,max}}|}{\mu_s}}$
10. Consider the following scenario:
• A force is being applied to an object
• But the object is not moving
• If friction is the only force which resists the motion, we can write the following two points:
(i) The external force $\mathbf\small{|\vec{F}|}$ which is trying to move the object is less than or equal to $\mathbf\small{|\vec{F_{max}}|}$ 
• That is why the object is not moving
• Why do we use the words: 'less than or equal to'?
• Note that, the object will not move even if $\mathbf\small{|\vec{F}|}$ is equal to $\mathbf\small{|\vec{F_{max}}|}$
• This is because, if they are equal, it is the limiting state. Motion will start only if $\mathbf\small{|\vec{F}|}$ exceeds $\mathbf\small{|\vec{F_{max}}|}$     
(ii) The frictional force $\mathbf\small{|\vec{f_s}|}$ developed at the contact surface is always less than or equal to $\mathbf\small{|\vec{f_{s,max}}|}$
• That is., $\mathbf\small{|\vec{f_s}|\leq |\vec{f_{s,max}}|}$
$\mathbf\small{\Rightarrow |\vec{f_s}|\leq \mu_s |\vec{F_N}|}$
11. Thus we get two important relations:
• From step (6) above we have:
5.1:
$\mathbf\small{|\vec{f_{s,max}}|=\mu_s |\vec{F_N}|}$
• From step (10) above, we have:
5.2:
$\mathbf\small{|\vec{f_s}|\leq \mu_s |\vec{F_N}|}$

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Now we will see an experiment to find μ_s. We will write it in steps:
1. In fig.5.78(a) below, a block 'A' of mass 'm' kg rests on an inclined plane 'I'

Fig.5.78

• The inclination of the plane with the horizontal can be adjusted using a screw mechanism
• The inclination in fig.a is θ₁
2. We have seen this situation before in a previous section. But we will write the steps again:
The three forces acting on the block are:
(i) $\mathbf\small{\vec{W_A}}$ vertically downwards
• This can be resolved into two components:
    ♦ Component parallel to the inclined plane:
        ♦ $\mathbf\small{|\vec{W_A}|\sin\theta_1 =m_A \times g \times \sin\theta_1 }$
    ♦ Component perpendicular to the inclined plane:
        ♦ $\mathbf\small{|\vec{W_A}|\cos\theta_1 =m_A \times g \times \cos\theta_1 }$
        ♦ (This perpendicular component is not shown in the fig.a)
(ii) The normal reaction $\mathbf\small{\vec{F_{N(AI)}}}$ which is perpendicular to the inclined surface
• We know the value of this normal reaction:
$\mathbf\small{\vec{F_{N(AI)}}=|\vec{W_A}|\cos\theta_1 =m_A \times g \times \cos\theta_1 }$
(iii) The force of friction $\mathbf\small{\vec{f_s}}$ along the contact surface 
3. The components perpendicular to the inclined surface cancel each other. There is no motion in that direction. So we need not consider those components
■ But when there is friction, it becomes necessary to take $\mathbf\small{\vec{F_{N(AI)}}}$ into account. So we cannot ignore it
4. In fig.a, the block is stationary. 
• The force trying to slide the block down is $\mathbf\small{m_A \times g \times \sin\theta_1 }$ 
    ♦ In the fig.a, it is denoted as $\mathbf\small{\vec{F_1}}$  
• Clearly, this $\mathbf\small{\vec{F_1}}$ is cancelled out by the frictional force $\mathbf\small{\vec{f_{s1}}}$    
5. We know that friction acts parallel to the contact surface. 
■ That means, $\mathbf\small{\vec{F_1}}$ and $\mathbf\small{\vec{f_{s1}}}$ acts along the same line
• So the 'cancelling process' can be expressed mathematically in terms of magnitudes as:
$\mathbf\small{|\vec{F_1}|=|\vec{f_{s1}}|}$  
• The block does not slide down because the above magnitudes are equal
6. Next we gradually increase the angle of inclination. Let the new angle be θ₂. This is shown in fig.b
• The block still does not move. So we can write: 
$\mathbf\small{|\vec{F_2}|=|\vec{f_{s2}}|}$
7. Note that:
$\mathbf\small{|\vec{F_2}|}$ [which is equal to ($\mathbf\small{m_A \times g \times \sin\theta_2}$)] 
Is a larger force than 
$\mathbf\small{|\vec{F_1}|}$ [which is equal to ($\mathbf\small{m_A \times g \times \sin\theta_1}$)]
• This is because, when θ increases from zero to 90, the 'sine value' increases 
• We need not consider the range beyond '0 to 90^o' because, '90^o' is 'vertical'. The block will begin to slide long before reaching 90^o
8. Since $\mathbf\small{|\vec{F_2}|}$ is greater than $\mathbf\small{|\vec{F_1}|}$, we can write:
• Frictional force in fig.b is more than that in fig.a
• That is., $\mathbf\small{|\vec{f_{s2}}|> |\vec{f_{s1}}|}$
• That means., when θ increases, the resisting frictional force also increases
9. As we gradually increase θ, it takes on values θ₃, θ₄, θ₅, . . .  
• As a result, the $\mathbf\small{|\vec{f_s}|}$ will also increase
• It will take on values $\mathbf\small{|\vec{f_{s3}}|,\,|\vec{f_{s4}}|,\,|\vec{f_{s5}}|,\,\text{. . . }}$
• Each one of the above frictional forces is greater than the preceeding ones 
10. We want the maximum possible value. We denoted it as: $\mathbf\small{|\vec{f_{s,max}}|}$
• For finding that maximum value, we find than inclination θ_max at which the block just begins to slip
• At that instant, we have: $\mathbf\small{|\vec{F_{max}}|=|\vec{f_{s,max}}|}$
Where $\mathbf\small{|\vec{F_{max}}|}$ is the maximum force which causes the sliding at that instant. 
• This is shown in fig.c
[• Note that, when θ increases from zero to 90, the 'cosine value' decreases 
• So when θ = θ_max, The normal reaction $\mathbf\small{\vec{F_{N(AI)}}}$ will be having the minimum value
• But this will not affect our steps to find the coefficient] 
11. We have the value of $\mathbf\small{|\vec{F_{max}}|}$
• It is given by: $\mathbf\small{|\vec{F_{max}}|=(m_A \times g \times \sin \theta _{max})}$
12. We have the value of $\mathbf\small{|\vec{f_{s,max}}|}$ also  
• It is given by: $\mathbf\small{|\vec{f_{s,max}}|=\left[ \mu _s \times \vec{F_{N(AI)}}\right ]}$
$\mathbf\small{\Rightarrow |\vec{f_{s,max}}|=\left[\mu_s \times (m_A \times g \times \cos \theta _{max}) \right ] }$
13. We take the results from (11) and (12) and put them in (10).  We get:
 $\mathbf\small{(m_A \times g \times \sin \theta _{max})=\left[\mu_s \times (m_A \times g \times \cos \theta _{max}) \right ] }$
$\mathbf\small{\Rightarrow \mu _s=\tan \theta _{max}}$
14. So we can write:
■ Coefficient of static friction is equal to the 'tangent of the maximum possible angle'

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Solved example 5.33
A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined until at an angle θ = 15° with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface ?
Solution:
1. The coefficient of static friction between two surfaces do not depend on the mass
• It is the tangent of the maximum possible angle at which the mass can hold it's the position with out sliding
2. In this problem, that angle is given as 15^o
• So the required  coefficient of static friction is tan 15 = 0.268

Solved example 5.34
A block of mass 3 kg is intended to be placed on an inclined surface. Check whether it is safe to do so if the angle of inclination is 25^o and coefficient of static friction is 0.18
Solution:
• We will do this problem by two methods
    ♦ First, the easy method using formula
    ♦ Second, the method using basics
Easy method:
1. The coefficient of static friction is the tangent of the maximum angle possible
• In this problem, the coefficient is given as 0.18
2. So we have: tan θ_max= 0.18
• Thus θ_max = 10.2^o
3. That means, the mass will slide if the angle is greater than 10.2^o
• The angle given in the problem is 25^o. So it is not safe

Method using basics:
1. Assume that the block is resting on the inclined plane. Also, assume that it is in the limiting state
■ If we draw the FBD of the block, we will see three forces acting on the block. They are:
(i) $\mathbf\small{\vec{W}}$ vertically downwards
• This can be resolved into two components:
    ♦ Component parallel to the inclined plane:
        ♦ $\mathbf\small{|\vec{W}|\sin\theta =m \times g \times \sin\theta }$
    ♦ Component perpendicular to the inclined plane:
        ♦ $\mathbf\small{|\vec{W}|\cos\theta =m \times g \times \cos\theta }$
(ii) The normal reaction $\mathbf\small{\vec{F_N}}$ which is perpendicular to the inclined surface
• We know the value of this normal reaction:
$\mathbf\small{\vec{F_N}=|\vec{W}|\cos\theta =m \times g \times \cos\theta }$
(iii) The force of friction $\mathbf\small{\vec{f_s}}$ along the contact surface. This force tries to prevent sliding
• Since we assume the block to be at the limiting state, we can use the maximum possible friction whose magnitude is denoted as: $\mathbf\small{|\vec{f_{s,max}}|}$
2. The actual magnitude of item (iii) above can be calculated using the relation 5.1 that we saw above:
$\mathbf\small{|\vec{f_{s,max}}|=\left[ \mu _s \times \vec{F_N}\right ]}$
$\mathbf\small{\Rightarrow |\vec{f_{s,max}}|=\left[\mu_s \times (m \times g \times \cos \theta ) \right ] }$
• Substituting the values we get: $\mathbf\small{|\vec{f_{s,max}}|=\left[0.18 \times (3 \times 10 \times \sin 25 ) \right ]=2.28\,\text{N}}$
3. The actual magnitude of item (i) above is: $\mathbf\small{3 \times 10 \times \sin25 = 12.68\,\text{N}}$
4. The result in (3) is greater than that in (2). That means, the mass cannot be held in position by friction. Hence it is not safe
5. It would have been safe if the angle was a lesser value. 
• When the angle is lowered, the normal force $\mathbf\small{\vec{F_N}}$ increases. So the friction increases
• When the angle is lowered, the sliding force $\mathbf\small{m \times g \times \sin\theta }$ decreases
■ Thus lowering the angle has double advantage
6. Let us find that angle below which it is safe to keep the block:
• The sliding force should be equal to the frictional force. That is:
$\mathbf\small{m \times g \times \sin \theta = \mu_s \times m \times g \times \cos \theta }$
$\mathbf\small{\Rightarrow \mu _s=\tan \theta }$
• Substituting the values, we get: 0.18 = tan θ
• So θ = 10.2^o

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In the next section we will see kinetic friction

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