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
• 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
• 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
■ 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
• 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:
$\mathbf\small{|\vec{f_s}|\leq \mu_s |\vec{F_N}|}$
Now we will see an experiment to find μs. 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'
• 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 90o' because, '90o' is 'vertical'. The block will begin to slide long before reaching 90o
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 θmax 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 = θmax, 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'
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 15o
• 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 25o 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 θmax= 0.18
• Thus θmax = 10.2o
3. That means, the mass will slide if the angle is greater than 10.2o
• The angle given in the problem is 25o. 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 = 10.2o
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 |
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?
• But if we do apply a force greater than $\mathbf\small{\vec{F_{max}}}$, what all changes will occur?
(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
• 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
• 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
• 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 |
• Thus the friction will increase.
• That means, the magnitude of $\mathbf\small{\vec{F_{max}}}$ required to start motion will increase
• 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
• 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:
$\mathbf\small{|\vec{f_s}|\leq \mu_s |\vec{F_N}|}$
Now we will see an experiment to find μs. 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 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 90o' because, '90o' is 'vertical'. The block will begin to slide long before reaching 90o
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 θmax 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 = θmax, 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'
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 15o
• 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 25o 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 θmax= 0.18
• Thus θmax = 10.2o
3. That means, the mass will slide if the angle is greater than 10.2o
• The angle given in the problem is 25o. 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 = 10.2o
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