Chapter 7.22 - Solved examples involving Torque

In the previous section, we saw center of gravity. In this section we will see some solved examples

Solved example 7.21
The metal bar in fig.7.103(a) is 0.70 m long and has a mass of 4 kg. It is supported on two knife edges placed 0.10 m from each end. A 6 kg mass is suspended from a point P, which is 0.20 m from the left support

Fig.7.103

Find the reactions at the supports. Assume the bar is of uniform cross section and homogeneous. Take g = 9.8 ms^-2
Solution:
1. Let us name the supports as A and B
■ The weight of the rod acts at it’s CG
• Given that, the metal bar is of uniform cross section and homogeneous.
■ So the CG of the rod will be at it’s geometric center
2. The detailed measurements are shown in fig.b
The CG is marked as G
3. Since the bar is in translational equilibrium, we have:
R_A + R_B - 6g - 4g = 0
⇒ R_A + R_B = 10g = 98.0 N.
4. Since the bar is in rotational equilibrium, we have: net torque = 0
• Let us take the torques about the support A
(i) Torque created by R_A about A = zero
(ii) Torque created by 6g about A = 6g × 0.2 = 1.2g Nm (clockwise)
(iii) Torque created by 4g about A = 4g × 0.25 = 1.0g Nm (clockwise)
(iv) Torque created by R_B about A = R_B × 0.5 Nm (anti clockwise)
5. So applying the condition, we get:
1.2g + g - 0.5R_B = 0
⇒ R_B = 43.12 N
• Substituting this value of R_B in (3), we get: R_A = (98-43.12) = 54.88 N

Solved example 7.22
A 3 m long ladder having a mass of 20 kg, leans on a frictionless wall. It’s feet rest on the ground 1 m from the wall as shown in fig.7.104(a) below:

Fig.7.104

Find the reaction forces on the wall and the floor. Take g = 9.8 ms^-2
Solution:
1. Let the end points of the ladder be A and B
• A is 1 m from the wall. This is shown in fig.b
• Let C be the foot of the wall
2. The reaction from the floor at A will be normal to the floor
• This reaction is denoted as R_A
3. The frictional force prevents the point A from moving away from C
• This frictional force is denoted as F. It pulls the ladder towards C. Other wise the ladder will slip
• Also recall that, the frictional force is always parallel to the surface
4. The reaction from the wall at B will be normal to the wall
• This reaction is denoted as R_B
5. Given that, the wall is frictionless
• If there was friction, a force F would have acted parallel to the wall in the upward direction
• This force would have made some contribution towards: 'preventing the movement of B towards C'   • In other words, this force would have made some contribution towards: 'preventing the ladder from slipping'
• In addition to that, this force would have made some contribution towards resisting the vertical force (20 g) of the ladder
    ♦ Where 'g' is the acceleration due to gravity
• But in this problem there is no such force
    ♦ The slipping is prevented entirely by the horizontal force F at A
    ♦ The vertical load 20 g is resisted entirely by the vertical force R_A at A
6. The weight of the ladder is 20 g
• It acts downwards at the CG of the ladder
• The CG is marked as G in fig.b
7. The above steps gives us all the 4 forces acting on the ladder
• Now we can apply the conditions of equilibrium
• Since the ladder is in translational equilibrium, we have:
(i) In the x direction: F - R_B = 0
(ii) In the y direction: R_A - 20 g = 0
⇒ R_A = 20 g = 196 N
8. Since the bar is in rotational equilibrium, we have: net torque = 0
Let us take the torques about A
(i) Torque created by R_A about A = zero
(ii) Torque created by F about A = zero
(iii) Torque created by R_B about A = R_B × BC
So we want the length of BC
• Applying Pythagoras theorem to the right triangle ABC, we get:
$\mathbf\small{BC=\sqrt{AB^2-AC^2}=\sqrt{3^2-1^2}=2\sqrt{2}\;\text{m}}$
Thus the required torque = $\mathbf\small{2\sqrt{2}R_B\;\text{N m}}$ (clockwise) 
(iv) Torque created by 20 g about A = 20 g × AD
• So we want the length of AD
• D is the foot of the perpendicular drawn from G
• Consider the similar triangles ABC and AGD
• We have: $\mathbf\small{\frac{AD}{AC}=\frac{AG}{AB}}$
$\mathbf\small{\Rightarrow \frac{AD}{1}=\frac{1.5}{3}}$
⇒ AD = 0.5 m
• Thus the torque = 20 g × 0.5 × 9.8 = 98 N m (anti clockwise)
9. Applying the condition, we get:
$\mathbf\small{2\sqrt{2}R_B-98=0}$
⇒ R_B = 34.65 N
10. Substituting this value of R_B in 7(i), we get: F = 34.65 N.
11. At the point A, two forces are acting on the ladder:
• R_A vertically and F horizontally. They are shown in fig.c
• The resultant of the two forces = $\mathbf\small{\sqrt{(R_A)^2+F^2}=\sqrt{196^2+34.65^2}=}$ 199.04 N
• Let this resultant make an angle α with the horizontal
• Then $\mathbf\small{\alpha=\tan^{-1}\frac{R_A}{F}=\tan^{-1}\frac{196}{34.65}=}$ 79.97^o

Solved example 7.23 
A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in fig.7.105(a) below:

Fig.7.105

The angles made by the strings with the vertical are 36.9^o and 53.1^o respectively. The bar is 2 m long. Calculate the distance d of the center of gravity of the bar from it’s left end
Solution:
1. The free body diagram is shown in fig.b
• Let T₁ and T₂ be the tensions in the strings
• In fig.a, we see that, the left string makes an angle of 36.9^o with the vertical wall
    ♦ In fig.b, we see that, this string makes the same angle with the vertical dotted line
    ♦ The angles are same because, they are alternate angles 
• In fig.a, we see that, the right string makes an angle of 53.1^o with the vertical wall
    ♦ In fig.b, we see that, this string makes the same angle with the vertical dotted line
    ♦ The angles are same because, they are alternate angles
• Now we can resolve the tensions into horizontal and vertical components. This is shown in fig.c
2. Since the bar is in translational equilibrium, we have:
(i) In the x direction: - T₁ sin θ₁ + T₂ sin θ₂  = 0
⇒ T₁ sin θ₁ = T₂ sin θ₂.
⇒ T₁ sin 36.9 = T₂ sin 53.1 = T₂ cos (90 - 36.9) = T₂ cos 36.9
⇒ $\mathbf\small{\frac{T_2}{T_1}=\tan36.9 =0.75}$
⇒ T₂ = 0.75 T₁
(ii) In the y direction: T₁ cos θ₁ + T₂ cos θ₂ - W = 0
⇒ T₁ cos 36.9 + T₂ cos 53.1 = W
⇒0.8 T₁ + 0.6 T₂ = W
3. Since the bar is in rotational equilibrium, we have: net torque = 0
• Let us take the torques about G
(i) Torque created by T₁ sin θ₁ about G = zero
(ii) Torque created by T₁ cos θ₁ about G = T₁ cos θ₁ × d (clockwise)
(iii) Torque created by W about G = zero
(iv) Torque created by T₂ cos θ₂ about G = T₂ cos θ₂ × (2-d) (anti clockwise)
(i) Torque created by T₂ sin θ₂ about G = zero
4. Applying the condition, we get:
[T₁ cos θ₁ × d] - [T₂ cos θ₂ × (2-d)] = 0
⇒ [T₁ × cos 36.9 × d] - [0.75 T₁ × cos 53.1 × (2-d)] = 0
(∵ from 2(i), we have: T₂ = 0.75 T₁)
⇒ [0.8 T₁ d] - [0.45 T₁ (2-d)] = 0
⇒ 0.8 T₁ d - 0.9 T₁ + 0.45 T₁ d = 0
⇒ 1.25 T₁ d = 0.9 T₁
⇒ 1.25 d = 0.9
⇒ d = 0.72 m

Solved example 7.24
A car weighs 1800 kg. The distance between it’s front and back axles is 1.8 m. It’s center of gravity is 1.05 m behind the front axle. Determine the force exerted by the level ground on each front wheel and each back wheel
Solution:
1. Fig.7.106 below shows the schematic diagram

Fig.106

• The small grey circles at the center of the wheels denote the axles
• The front and rear axles are named as A and B respectively
2. When the car is in translational equilibrium, we have:
R_A + R_B -1800g = 0
3. When the car is in rotational equilibrium, we have: net torque = 0
Let us take the torques about G
(i) Torque created by R_A about G = 1.05 R_A (clockwise) 
(ii) Torque created by 1800g about G = zero
(iii) Torque created by R_B about G = 0.75 R_B (anticlockwise)
4. Applying the condition, we get:
1.05 R_A - 0.75 R_B = 0
1.05 R_A = 0.75 R_B
R_B = 1.4 R_A
5. Substituting this in (2), we get:
R_A + 1.4R_A = 1800 × 9.8
⇒ R_A = 7350 N
• So R_B = 1.4 × 7350 = 10290 N
6. The reaction on the front axle A is 7350 N
• But the load from this axle is resisted by two front wheels
• So reaction on each of the front wheels = ⁷³⁵⁰⁄₂ = 3675 N
7. The reaction on the back axle B is 10290 N
• But the load from this axle is resisted by two back wheels
• So reaction on each of the back wheels = ¹⁰²⁹⁰⁄₂ = 5145 N

Solved example 7.25
As shown in fig.7.107(a), the two sides of a step ladder BA and CA are 1.6 m long and hinged at A. A rope DE 0.5 m long is tied half way up. A weight 40 kg is suspended from a point F, 1.2 m from B along the ladder BA. Assuming the floor to be frictionless and neglecting the weight of the ladder, find the tension in the rope and forces exerted by the floor on the ladder. Take g = 9.8 ms^-2 [Hint: Consider the equilibrium of each side of the ladder separately]

Fig.7.107

Solution:
1. The detailed measurements are shown in fig.b
• Given:
    ♦ BD = 0.8 m
    ♦ DF = FA = 0.4 m
    ♦ DE = 0.5 m
• Let ∠ABC = θ
    ♦ Since DE is parallel to the ground, we get: ∠ADE = θ   
2. Let us calculate the other required dimensions:
• A perpendicular is dropped from A onto BC
    ♦ This is shown as red dashed line 
    ♦ The foot of the perpendicular on BC is O
    ♦ The foot of the perpendicular on DE is G
• Applying Pythagoras theorem to the right triangle ADG, we get:
$\mathbf\small{AG=\sqrt{AD^2-DG^2}=\sqrt{0.8^2-0.25^2}=0.76\;\text{m}}$
3. In the similar triangles ABO and ADG, we get:
• $\mathbf\small{\frac{AD}{AB}=\frac{AG}{AO}}$
$\mathbf\small{\Rightarrow \frac{0.8}{1.6}=\frac{0.76}{AO}}$
⇒ AO = 1.52 m
4. Again in the same similar triangles ABO and ADG, we get:
$\mathbf\small{\frac{AD}{AB}=\frac{DG}{BO}}$
$\mathbf\small{\Rightarrow \frac{0.8}{1.6}=\frac{0.25}{BO}}$
⇒ BO = 0.5 m
5. A perpendicular is dropped from F onto AG
• The foot of this perpendicular is H
• In the similar triangles AFH and ADG, we get:
$\mathbf\small{\frac{AF}{AD}=\frac{FH}{DG}}$
$\mathbf\small{\Rightarrow \frac{0.4}{0.8}=\frac{FH}{0.25}}$
⇒ FH = 0.125 m
6. Let us write the above distances together:
    ♦ BD = 0.8 m
    ♦ DF = FA = 0.4 m
    ♦ DE = 0.5 m
    ♦ AG =0.76 m
    ♦ AO = 1.52 m
    ♦ BO = 0.5 m
    ♦ FH = 0.125 m
• Thus we obtained all the distances that we will soon require for torque calculations
7. The forces are shown in fig.c below:

Fig.7.107 (c) & (d)

• R_B is the normal reaction at B
• R_C is the normal reaction at C
• Note that the tensions T in the string are internal forces and so will cancel each other
8. For translational equilibrium, we have:
• R_B + R_C - (40 × 9.8) = 0
⇒ R_B + R_C - 392 = 0
⇒ R_B + R_C = 392
9. For rotational equilibrium, we need to calculate the torques first:
• Let us find the torques about O
(i) Torque created by R_B about O = R_B × BO = R_B × 0.5 = 0.5R_B Nm (clockwise)
(ii) Torque created by the 40 kg load about O = (40 × 9.8) × FH = 392 0.125 = 49 Nm (anti clockwise)
(iii) Torque created by R_C about O = R_C × CO = R_C × 0.5 = 0.5R_C Nm (anti clockwise)
10. Applying the condition, we get:
⇒ 0.5R_B - 49 - 0.5R_C = 0
⇒ 0.5R_B - 0.5R_C = 49
⇒ R_B - R_C = 98
11. Solving the equations in (8) and (10), we get:
R_B = 245 N, R_C = 147 N
12. Next we want to find T
• In the fig.d above, the FBD of side AB is shown separately
• The forces acting are: R_B, T and the load of 40 kg
• Let us calculate the torques about A:
(i) Torque created by R_B about A = R_B × BO = R_B × 0.5 = 0.5R_B Nm (clockwise)
(ii) Torque created by T about A = T × AG = T × 0.76 = 0.76T Nm (anti clockwise)
(ii) Torque created by the 40 kg load about A = (40 × 9.8) × FH = 392 × 0.125 = 49 Nm (anti clockwise)
13. Applying the condition, we get:
0.5R_B - 0.76T - 49 = 0
0.5 × 245 - 0.76T - 49 = 0
T = 96.7 N

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Two more solved examples can be seen here

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In the next section, we will see moment of inertia

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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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