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 μs and μk.
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:
• 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)
• 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 μs and μk. So we took μk = 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)
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:
• 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?
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 μs,AB is the coefficient of static friction between A and B
• Substituting the values, we get: 12 × μs,AB = 40
⇒ μs,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: amax = μs×g (Details here)
• Substituting the values, we get: amax = 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 30o, 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 μk or μs
• But we can confirm that it is μk
• Because, μs 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 = μk × mg cosθ
(See details here)
2. So net force = (mg sinθ - μk × mg cosθ) = mg(sinθ - μk 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:
$\mathbf\small{v=0 +3.27 \times 5=16.35\, \text{ms}^{-1}}$
• 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
• 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
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 μs and μk.
 |
| Fig.5.82 |
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 |
• 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 μs and μk. So we took μk = 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 |
• 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 |
• 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 |
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 μs,AB is the coefficient of static friction between A and B
• Substituting the values, we get: 12 × μs,AB = 40
⇒ μs,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: amax = μs×g (Details here)
• Substituting the values, we get: amax = 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 30o, 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 μk or μs
• But we can confirm that it is μk
• Because, μs 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 = μk × mg cosθ
(See details here)
2. So net force = (mg sinθ - μk × mg cosθ) = mg(sinθ - μk 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:
$\mathbf\small{v=0 +3.27 \times 5=16.35\, \text{ms}^{-1}}$
Rolling Friction
In fig.5.87(a) below, a wheel is rolling over a horizontal plane. |
| Fig.5.87 |
• 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
• Thus we see that, friction comes into play even for the rolling motion.
• 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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