Chapter 5.13 - Normal reaction on Inclined surfaces

In the previous section we saw cases in which:
    ♦ Contact surfaces were all vertical
    ♦ So  the $\mathbf\small{\vec{F_N}}$ were all horizontal
In the section just before that, we saw cases in which:
    ♦ Contact surfaces were all horizontal. 
    ♦ So the $\mathbf\small{\vec{F_N}}$ were all vertical. 
In this section we will see cases in which:
    ♦ Contact surfaces are inclined

1. Consider Fig.5.41(a) below. A wooden block (B) is resting on a wedge shaped block (I). 
• The top surface of this wedge is a smooth inclined plane. 

Fig.5.41

• The plane is inclined at an angle θ to the horizontal
■ Before learning about the interaction between the block and the wedge, we must learn some geometrical features of the wedge. Steps (2) to (4) below enable us to learn those features:
2. The top surface of the wedge is not horizontal. It is inclined
• Draw a perpendicular to this inclined surface. It is the white dashed line in fig.5.41(b)
    ♦ This perpendicular intersects the inclined surface at D
    ♦ This perpendicular intersects the base of the wedge at C
• Draw a vertical through D. It is the green dashed line in fig.b
    ♦ This vertical intersects the base of the wedge at B
3. The base of the wedge is horizontal
• So now we have two right triangles:
    ♦ ⊿ACD right angled at D
    ♦ ⊿BCD right angled at B
4. In ⊿ACD, we have ∠ACD = (90-θ)
• From this we get: In ⊿BCD, ∠BDC = θ
■ This is an important result. It tells us that:
The following two angles are always the same:
(i) The angle between ‘the inclined surface’ and the 'horizontal'
    ♦ In the fig., it is the ∠DAB
(ii) The angle between ‘the normal to the inclined surface’ and the 'vertical'
    ♦ In the fig., it is the ∠CDB
5. Now we can learn about the interaction
• The $\mathbf\small{\vec{W_B}}$ in fig.a, denotes the weight of the block
• It will be always directed verticaly downwards (towards the center of the earth)   
• It's direction will not depend on the inclination of the plane
3. We can resolve $\mathbf\small{\vec{W_B}}$ into it's two perpendicular components
• For the convenience of resolution, let us choose the frame of reference (coordinate axes) in this way (see fig.5.41c):
    ♦ The x axis is parallel to the inclined surface
    ♦ The y axis is perpendicular to the inclined surface
4. So we have a new set of coordinate axes
• Consider fig.d. It shows a vector $\mathbf\small{\vec{F}}$ in the usual set of coordinate axes
• In this case, we know that:
    ♦ x component = $\mathbf\small{\vec{F_x}=(|\vec{F}|\cos \theta)\hat{i} }$ 
    ♦ y component = $\mathbf\small{\vec{F_y}=(|\vec{F}|\sin \theta)\hat{j} }$
(Details about the above results can be seen here)
• So we see that:
    ♦ The component which is adjacent to the angle gets the cosine
    ♦ The other component gets the sine
5. Now consider fig.e. It shows or present case
• Here θ is adjacent to the y axis 
Clearly:    
    ♦ x component = $\mathbf\small{\vec{W_{Bx}}=(|\vec{W_{B}}|\sin \theta)\hat{i} }$
    ♦ y component = $\mathbf\small{\vec{W_{By}}=(|\vec{W_{B}}|\cos \theta)\hat{j} }$
• The two components are shown in fig.5.42(a) below:

Fig.5.42

6. Next step is to draw the FBD of the block
• The component of $\mathbf\small{\vec{W_B}}$ parallel to the x axis is $\mathbf\small{\vec{W_{Bx}}}$ 
    ♦ This force is exerted by the earth. It is a force acting on the block. 
    ♦ So we show it in the FBD
• The component of $\mathbf\small{\vec{W_B}}$ parallel to the y axis is $\mathbf\small{\vec{W_{By}}}$ 
    ♦ This force is exerted by the earth. It is a force acting on the block. 
    ♦ So we show it also in the FBD
• Due to the action $\mathbf\small{\vec{W_{By}}}$ on the inclined surface, the inclined surface produces a reaction. 
• We can denote it as $\mathbf\small{\vec{F_{N(BI)}}}$
    ♦ This $\mathbf\small{\vec{F_{N}}}$ acts on the block. 
    ♦ So we show it in the FBD
• The FBD is now complete. It is shown in the fig.5.42(b)
7. We observe no movement in the y direction
• That is., there is equilibrium in the y direction
• So we can write: $\mathbf\small{\vec{W_{By}}=-\vec{F_{N(BI)}}}$
8. Now we consider the x direction
• There is no force to oppose $\mathbf\small{\vec{W_{Bx}}}$ 
(Remember that, the surface in this case is smooth. So there is no friction)
• So the block will move in the direction of $\mathbf\small{\vec{W_{Bx}}}$
• That means, the block will slide down along the inclined surface
• If there is friction between the ‘bottom surface of the block’ and the 'inclined surface', the block may be able to hold it’s position. This happens when the frictional force is equal in magnitude to $\mathbf\small{\vec{W_{Bx}}}$. We will learn more about  friction in later sections of this chapter

Solved example 5.21
A block of mass 3 kg slides down along a friction less inclined surface. What is the acceleration with which it slides down? Angle of inclination of the surface with the horizontal is 30^o. Take g = 10 ms^-2.
Solution:
1. Let the x axis be parallel to the inclined plane
• Let the y axis be perpendicular to the inclined plane
2. The force which causes the sliding is: 
$\mathbf\small{\vec{W_{Bx}}=(|\vec{W_{B}}|\sin \theta)\hat{i} }$
• Substituting the values, we get:
$\mathbf\small{\vec{W_{Bx}}=(mg \,\sin 30)\hat{i} }$
• So the magnitude of the force which causes the sliding is: mg sin30 = 0.5mg
3. Let 'a' be the acceleration with which the block slides
• Then we have: ma = 0.5mg
⟹ a = 0.5g = 0.5 × 10 = 5 ms^-2
■ Note that, this acceleration does not depend on the mass. But it certainly depends on the 'angle which the inclined surface makes with the horizontal'

Solved example 5.22
Three blocks A, B and C are placed on the smooth inclined surface of a wedge as shown in fig.5.43(a) below:

Fig.5.43

The masses of the blocks A, B and C are 1 kg, 2 kg and 3 kg respectively. External forces of 60 N and 18 N also act on the blocks as shown.  
(a) Find the acceleration of A, B and C
(b) Draw the following FBDs:
(i) FBD of (A+B+C)
(ii) FBD of A
(iii) FBD of B
(iv) FBD of C

Solution:
1. FBD of (A+B+C) is shown in fig.5.43(b)
• The modified frame of reference is also shown
• There is no motion in the y direction. So we need not consider the forces in the that direction. They will cancel each other
2. In the x direction, we have:
Net force = $\mathbf\small{60\, \hat{i}-\vec{W_{(A+B+C)x}}-18\, \hat{i}}$
⟹ Net force = $\mathbf\small{42\, \hat{i}-\vec{W_{(A+B+C)x}}}$
⟹ Net force = $\mathbf\small{42\, \hat{i}-(|\vec{W_{(A+B+C)}|\, \sin \theta )\, \hat{i}}}$
⟹ Net force = $\mathbf\small{42\, \hat{i}-[(m_{(A+B+C)} \times g) \, \sin \theta ]\, \hat{i}}$
• Substituting the values, we get:
• Net force = $\mathbf\small{\left[ 42\, \hat{i}-[(6 \times 10) \, \sin 30 ]\, \hat{i} \right ]=12\, \hat{i}}$
4. Since there is a net force, there will be acceleration
• The net force is the cause of the acceleration of (A+B+C). Applying the second law, we can write:
$\mathbf\small{(m_{(A+B+C)}\times|\vec{a_x}|)\, \hat{i}=12\, \hat{i}}$
$\mathbf\small{\Rightarrow (m_{(A+B+C)}\times|\vec{a_x}|)=12}$
• Substituting the values, we get:
$\mathbf\small{(6\times|\vec{a_x}|)=12}$
$\mathbf\small{\Rightarrow |\vec{a_x}|=2\, \text{ms}^{-1}}$
■ So each of the three blocks move down with an acceleration of 2 ms^-2

---

1. Now we draw the FBD of A. This is shown in the fig.5.44(a) below:

Fig.5.44

• There is no motion in the y direction. So we need not consider the forces in the that direction. They will cancel each other

2. In the x direction, we have:

Net force = $\mathbf\small{\vec{F_{N(AB)}}-\vec{W_{Ax}}-18\, \hat{i}}$

⟹ Net force = $\mathbf\small{\vec{F_{N(AB)}}-(|\vec{W_{A}}|\, \sin \theta )\, \hat{i}-18\, \hat{i}}$

⟹ Net force = $\mathbf\small{\vec{F_{N(AB)}}-(m_A \times g \times\sin \theta )\, \hat{i}-18\, \hat{i}}$

Substituting the values, we get:

Net force = $\mathbf\small{\vec{F_{N(AB)}}-(1 \times 10 \times\sin 30 )\, \hat{i}-18\, \hat{i}}$

⟹ Net force = $\mathbf\small{\vec{F_{N(AB)}}-5\, \hat{i}-18\, \hat{i}}$

⟹ Net force = $\mathbf\small{\vec{F_{N(AB)}}-23\, \hat{i}}$

3. This net force causes the acceleration of A. So we can write:

$\mathbf\small{\vec{F_{N(AB)}}-23\, \hat{i}=m_A \times \vec{a}}$

$\mathbf\small{\Rightarrow \vec{F_{N(AB)}}-23\, \hat{i}=(m_A \times |\vec{a}|)\, \hat{i}}$

$\mathbf\small{\Rightarrow \vec{F_{N(AB)}}-23\, \hat{i}=(1 \times 2)\, \hat{i}}$

$\mathbf\small{\Rightarrow \vec{F_{N(AB)}}= 25\, \hat{i}}$

$\mathbf\small{\Rightarrow |\vec{F_{N(AB)}}|= 25\, \text{N}}$

---

1. Next we draw the FBD of B. This is shown in the fig.5.44(b) above

• There is no motion in the y direction. So we need not consider the forces in the that direction. They will cancel each other

2. In the x direction, we have:

Net force = $\mathbf\small{\vec{F_{N(BC)}}-\vec{W_{Bx}}-\vec{F_{N(BA)}}}$

⟹ Net force = $\mathbf\small{\vec{F_{N(BC)}}-(|\vec{W_{B}}|\, \sin \theta )\, \hat{i}-25\, \hat{i}}$

($\mathbf\small{\because\:  |\vec{F_{N(BA)}}|=|\vec{F_{N(AB)}}|= 25\, \text{N}}$)

⟹ Net force = $\mathbf\small{\vec{F_{N(BC)}}-(m_B \times g \times\sin \theta )\, \hat{i}-25\, \hat{i}}$

Substituting the values, we get:

Net force = $\mathbf\small{\vec{F_{N(BC)}}-(2 \times 10 \times\sin \text{30} )\, \hat{i}-25\, \hat{i}}$

⟹ Net force = $\mathbf\small{\vec{F_{N(BC)}}-10\, \hat{i}-25\, \hat{i}}$

⟹ Net force = $\mathbf\small{\vec{F_{N(BC)}}-35\, \hat{i}}$

3. This net force causes the acceleration of B. So we can write:

$\mathbf\small{\vec{F_{N(BC)}}-35\, \hat{i}=m_B \times \vec{a}}$

$\mathbf\small{\Rightarrow \vec{F_{N(BC)}}-35\, \hat{i}=(m_B \times |\vec{a}|)\, \hat{i}}$

$\mathbf\small{\Rightarrow \vec{F_{N(BC)}}-35\, \hat{i}=(2 \times 2)\, \hat{i}}$

$\mathbf\small{\Rightarrow \vec{F_{N(BC)}}= 39\, \hat{i}}$

$\mathbf\small{\Rightarrow |\vec{F_{N(BC)}}|= 39\, \text{N}}$

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1. Finally, we draw the FBD of C. This is shown in the fig.5.44(c) above

• There is no motion in the y direction. So we need not consider the forces in the that direction. They will cancel each other

2. In the x direction, we have:

Net force = $\mathbf\small{60 \, \hat{i}-\vec{W_{Cx}}-\vec{F_{N(CB)}}}$

⟹ Net force = $\mathbf\small{60 \, \hat{i}-(|\vec{W_{C}}|\, \sin \theta )\, \hat{i}-39\, \hat{i}}$

($\mathbf\small{\because\:  |\vec{F_{N(CB)}}|=|\vec{F_{N(BC)}}|= 39\, \text{N}}$)

⟹ Net force = $\mathbf\small{60 \, \hat{i}-(m_B \times g \times\sin \theta )\, \hat{i}-39\, \hat{i}}$ 

Substituting the values, we get:

Net force = $\mathbf\small{60 \, \hat{i}-(3 \times 10 \times\sin \text{30})\, \hat{i}-39\, \hat{i}}$  

⟹ Net force = $\mathbf\small{60 \, \hat{i}-15\, \hat{i}-39\, \hat{i}=6 \, \hat{i}}$

3. This net force causes the acceleration of C. So we can write:

$\mathbf\small{6 \hat{i}=m_C \times \vec{a}}$

$\mathbf\small{\Rightarrow 6 \hat{i}=(m_C \times |\vec{a}|)\, \hat{i}}$

$\mathbf\small{\Rightarrow 6 \hat{i}=(3 \times |\vec{a}|)\, \hat{i}}$
$\mathbf\small{\Rightarrow |\vec{a}|=2\, \text{ms}^{-1}}$
This is indeed true because, we saw that all the 3 blocks move with the same acceleration of 2 ms^-2.

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In the next section we will see force in strings.

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Chapter 5.12 - Normal reaction on Vertical surfaces

In the previous section we saw that the 'normal reaction force' will be always perpendicular to the surface of contact. 
• We saw cases in which:
    ♦ Contact surfaces were all horizontal. 
    ♦ So the $\mathbf\small{\vec{F_N}}$ were all vertical. 
• In this section we will see cases in which:
    ♦ Contact surfaces are vertical
    ♦ So  the $\mathbf\small{\vec{F_N}}$ will be horizontal
We will learn them through some solved examples: 

Solved example 5.19
Two blocks A and B of masses 5 kg and 3 kg respectively are placed side by side on a smooth horizontal surface. A horizontal force of 12 N is applied on A as shown in fig.5.37(a) below:

Fig.5.37

(a) Find the acceleration of A and B
(b)Draw the following free body diagrams:
(i) FBD of (A+B)
(ii) FBD of A
(iii) FBD of B
Solution:
Part (a):
1. Under the action of the 12 N force, both A and B will move together
2. Applying Newton's second law, we have: F = ma
12 = (5+3)×a 
⟹ a = 1.5 ms^-2 
Part (b)(i):
1. Let us choose (A+B) as the sub-system. This is shown in fig.5.37(b)
The following forces are acting on (A+B)
(i) Self weight of A: $\mathbf\small{\vec{W_A}}$ downwards  
(ii) Self weight of B: $\mathbf\small{\vec{W_B}}$ downwards
(iii) Reaction from the horizontal surface on A: $\mathbf\small{\vec{F_{N(AS)}}}$ upwards
• The subscript '(AS)' indicates that, the normal reaction $\mathbf\small{\vec{F_N}}$ is applied on A by the surface (S)
(iv) Reaction from the horizontal surface on B: $\mathbf\small{\vec{F_{N(BS)}}}$ upwards
• The subscript '(BS)' indicates that, the normal reaction $\mathbf\small{\vec{F_N}}$ is applied on B by the surface (S)
(v) The external horizontal force of 12 N
• The FBD showing the above 5 forces is given in fig.5.37(c)
2. Since there is no motion in the vertical direction, we can say that, there is equilibrium in the vertical direction
Thus we get:
$\mathbf\small{\vec{W_A}}$ = $\mathbf\small{-\vec{F_{N(AS)}}}$ 
    ♦ The two forces cancel each other
$\mathbf\small{\vec{W_B}}$ = $\mathbf\small{-\vec{F_{N(BS)}}}$ 
    ♦ The two forces cancel each other
3. Since the 12 N is the only horizontal force, (A+B) will move with an acceleration of 1.5 ms^-2 as calculated in part (a)

Part (b)(ii):
1. Let us choose A as the sub-system. This is shown in fig.5.38(b) below:

Fig.5.38

The following forces are acting on A
(i) Self weight of A: $\mathbf\small{\vec{W_A}}$ downwards  
(ii) Reaction from the horizontal surface on A: $\mathbf\small{\vec{F_{N(AS)}}}$ upwards
• The subscript '(AS)' indicates that, the normal reaction $\mathbf\small{\vec{F_N}}$ is applied on A by the surface (S)
(iii) The external horizontal force of 12 N
(iv) The normal reaction $\mathbf\small{\vec{F_{N(AB)}}}$ horizontally towards A
• The subscript '(AB)' indicates that, the normal reaction $\mathbf\small{\vec{F_N}}$ is applied on A by B
• Since the contact surface is vertical, this $\mathbf\small{\vec{F_N}}$ will be horizontal

---

■ Let us see the reason for existence of this $\mathbf\small{\vec{F_N}}$:
• Under the influence of the 12 N force, 'A' begins to move towards the right. 
• But it is obstructed by 'B'
• B will not want to move because of inertia
• So A will apply a force on B
    ♦ This force can be denoted as: $\mathbf\small{\vec{F_{N(BA)}}}$
    ♦ This force cannot be shown in the 'FBD of A' because, it is applied by A
• But B will apply a normal reaction $\mathbf\small{\vec{F_{N(AB)}}}$ in accordance with Newton's third law
    ♦ This force is the reaction to $\mathbf\small{\vec{F_{N(BA)}}}$ 
    ♦ This force must be shown in the 'FBD of A' because, it is applied on A

---

The 4 forces are shown in the 'FBD of A' in fig.5.38(c) above
2. Since there is vertical equilibrium, we can say: The vertical forces will cancel each other
3. In the horizontal direction, there is no equilibrium because, A is moving with acceleration
• We can say: Whatever horizontal acceleration 'A' is experiencing, will be due to the net horizontal force on it
• So we have:
$\mathbf\small{12 \, \, \hat{i}+\vec{F_{N(AB)}}=m_A \times \vec{a_A}}$
• Applying the usual sign convention, we have:
$\mathbf\small{12 \, \, \hat{i}-\vec{F_{N(AB)}}=m_A \times \vec{a_A}}$
$\mathbf\small{\Rightarrow \vec{F_{N(AB)}}=12 \, \, \hat{i}\: \: -\: \: m_A \times \vec{a_A}}$
• Substituting the values, we get:
$\mathbf\small{\vec{F_{N(AB)}}=(12 \, \, \hat{i}\: \: -\: \: 5 \times 1.5 \, \, \hat{i})=4.5 \, \, \hat{i}}$

Part (b)(iii):
1. Let us choose B as the sub-system. This is shown in fig.5.39(b) below:

Fig.5.39

The following forces are acting on B
(i) Self weight of B: $\mathbf\small{\vec{W_B}}$ downwards  
(ii) Reaction from the horizontal surface on B: $\mathbf\small{\vec{F_{N(BS)}}}$ upwards
• The subscript '(BS)' indicates that, the normal reaction $\mathbf\small{\vec{F_N}}$ is applied on B by the surface (S)
(iii) The normal reaction $\mathbf\small{\vec{F_{N(BA)}}}$ horizontally towards B
• The subscript '(BA)' indicates that, the normal reaction $\mathbf\small{\vec{F_N}}$ is applied on B by A
• Since the contact surface is vertical, this $\mathbf\small{\vec{F_N}}$ will be horizontal

---

■ Let us see the reason for existence of this $\mathbf\small{\vec{F_{N(BA)}}}$:
• Under the influence of the 12 N force, A begins to move towards the right. 
• But it is obstructed by B
• B will not want to move because of inertia
• So A will apply a force on B. This is the $\mathbf\small{\vec{F_{N(BA)}}}$
    ♦ This force must be shown in the 'FBD of B' because, it is applied on B 
• B will apply a normal reaction $\mathbf\small{\vec{F_{N(AB)}}}$ in accordance with Newton's third law
    ♦ This force cannot be shown in the 'FBD of B' because, it is applied by A
    ♦ This force was shown in the 'FBD of A'

---

The 3 forces are shown in the 'FBD of B' in fig.5.39(c) above
2. Since there is vertical equilibrium, we can say: The vertical forces will cancel each other
3. In the horizontal direction, $\mathbf\small{\vec{F_{N(BA)}}}$ is the net force
• So whatever horizontal acceleration 'B' is experiencing, will be due to $\mathbf\small{\vec{F_{N(BA)}}}$
• We can write: $\mathbf\small{\vec{F_{N(BA)}}=m_B \times \vec{a_B}}$
4. Now, $\mathbf\small{\vec{F_{N(BA)}}}$ is equal and opposite to $\mathbf\small{\vec{F_{N(AB)}}}$ 
• We have already found out that $\mathbf\small{\vec{F_{N(AB)}}=4.5\, \hat{i}}$
    ♦ $\mathbf\small{\vec{F_{N(AB)}}}$ is acting towards left
• So $\mathbf\small{\vec{F_{N(BA)}}}$ will be having the same magnitude and will be acting towards the right 
5. Substituting the values in (3), we get: $\mathbf\small{4.5\, \hat{i}=(3 \times |\vec{a_B}|)\hat{i}}$
$\mathbf\small{\Rightarrow |\vec{a_B}|=1.5\: ms^{-2}}$
■ In part (a), we have already seen that, both A and B move with an acceleration of 1.5 ms^-2
• So our calculations are correct

---

• We see the existence of both $\mathbf\small{\vec{F_{N(AB)}}}$ and $\mathbf\small{\vec{F_{N(BA)}}}$  

• We used $\mathbf\small{\vec{F_{N(AB)}}}$ in the FBD of  'A'
• We used $\mathbf\small{\vec{F_{N(BA)}}}$ in the FBD of 'B'
■ But we did not use any one of them in the FBD of (A+B). Why is that so?
Answer: In the individual FBDs of 'A" and 'B', those forces are external forces. They will not cancel
• But in the FBD of (A+B), they are 'equal and opposite internal forces'. So they cancel each other and so will not be having any effect in the state of (A+B)
• It is just like trying to move a stationary car by pushing it while sitting inside
    ♦ The force of that push will be cancelled by the reaction from the car

---

Solved example 5.20

Three blocks A, B and C of masses 9 kg, 7 kg and 4 kg respectively are placed side by side on a smooth horizontal surface. A horizontal force of 45 N is applied on A as shown in fig.5.40(a) below. Find the normal reaction on B due to C

Fig.5.40

Solution:
In this problem, we will write only the minimum required steps:
1. Under the action of the 45 N force, A, B and C will move together
2. Applying Newton's second law, we have: F = ma
45 = (9+7+4)×a 
⟹ a = 2.25 ms^-2. 
3. Fig.5.40(b) shows the free body diagrams
The vertical forces will cancel each other. So they are not shown
4. For A, we have:
$\mathbf\small{45 \, \, \hat{i}-\vec{F_{N(AB)}}=m_A \times \vec{a_A}}$
$\mathbf\small{\Rightarrow \vec{F_{N(AB)}}=12 \, \, \hat{i}\: \: -\: \: m_A \times \vec{a_A}}$
• Substituting the values, we get:
$\mathbf\small{\vec{F_{N(AB)}}=(45 \, \, \hat{i}\: \: -\: \: 9 \times 2.25 \, \, \hat{i})=24.75 \, \, \hat{i}}$
5. For B, we have:
$\mathbf\small{24.75 \, \, \hat{i}-\vec{F_{N(BC)}}=m_B \times \vec{a_B}}$
$\mathbf\small{\Rightarrow \vec{F_{N(BC)}}=24.75 \, \, \hat{i}\: \: -\: \: m_B \times \vec{a_B}}$
• Substituting the values, we get:
$\mathbf\small{\vec{F_{N(BC)}}=(24.75 \, \, \hat{i}\: \: -\: \: 7 \times 2.25 \, \, \hat{i})=9 \, \, \hat{i}}$
• We are asked to find $\mathbf\small{\vec{F_{N(BC)}}}$
• So '$\mathbf\small{9 \, \, \hat{i}}$ (towards left)' is the required answer 
Check:
For C, we have:
$\mathbf\small{\vec{F_{N(CB)}}=m_C \times \vec{a_C}}$
$\mathbf\small{\Rightarrow 9\, \hat{i}=(4 \times |\vec{a_C}|)\hat{i}}$
$\mathbf\small{\Rightarrow |\vec{a_C}|=2.25\: ms^{-2}}$
■ We have already seen that, A, B and C move with an acceleration of 2.25 ms^-2
• So our calculations are correct

---

• In the previous section, we saw cases in which:
    ♦ Contact surfaces were all horizontal. 
    ♦ So the $\mathbf\small{\vec{F_N}}$ were all vertical. 
• In this section we saw cases in which:
    ♦ Contact surfaces were all vertical
    ♦ So  the $\mathbf\small{\vec{F_N}}$ were all horizontal
In the next section we will see cases in which:
    ♦ Contact surfaces are all inclined

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Chapter 5.11 - The Normal Reaction Force

In the previous section we saw equilibrium of a particle. Now we will see some common forces in mechanics. In the present section we will see 'normal reaction force'. In this discussion, we will learn about 'free body diagrams' also

We can learn about the 'normal reaction force' through some examples:
Example 1:
1. In fig.5.33(a) below, a wooden block rests on a level ground

Fig.5.33

2. We already know the following 5 facts:
(i) On any object which is on/near the surface of the earth, gravitational force will be acting
(ii) Due to this gravitational force, the block will experience a pull towards the center of the earth.
(iii) This pulling force is denoted as the weight (W)
(iv) Since it is a force, we can use vector notation and write:
• The block is pulled down by the force $\mathbf\small{\vec{W}}$
• This is shown in fig.5.33(b) 
(v) We have: magnitude of $\mathbf\small{\vec{W}}$ = $\mathbf\small{|\vec{W}|=mg}$
• Where: 
    ♦ m is the mass of the block  
    ♦ g is the acceleration due to gravity
3. So, the block is trying to reach the center of the earth
• But the floor is getting in the way
• That means, the block will be exerting a force on the ground
4. By Newton's third law, the floor will exert an equal and opposite reaction force
• This reaction force will be always perpendicular to the 'surface of contact' (Details here) 
5. Another word for 'perpendicular' is 'normal'
• Instead of 'perpendicular', we usually use the word 'normal' in physics
• So the reaction force is called 'normal reaction force'
    ♦ It is denoted as $\mathbf\small{\vec{F_N}}$
6. In this particular example, we find that the block is at rest
• So no net force is acting on the block
    ♦ That is., the vector sum of the forces acting on the block is zero  
    ♦ Note that, it is a 'vector addition'. When we do a 'vector addition', we get a resultant
    ♦ All rules for such an addition (analytical method OR graphical methods like triangle law, parallelogram law, polygon law etc.,) should be strictly followed to obtain the resultant
• In our present case, both the vectors are along the same line. So it is essentially a one-dimensional problem and so the addition is easy
• However, we will use vector notations so as to get a general picture
• Thus we have:
$\mathbf\small{\vec{W}+\vec{F_N}=0}$
$\mathbf\small{\Rightarrow \vec{W}=-\vec{F_N}}$
7. So in this example, the magnitude of $\mathbf\small{\vec{F_N}}$ is equal to the magnitude of $\mathbf\small{\vec{W}}$
8. We can now draw the 'free body diagram (FBD)' for the wooden block. 
The procedure is given below:
(i) Isolate the 'sub-system' for which we want to draw the FBD
• Here, we want to draw the FBD for the 'wooden block'. So it is our sub-system. We must isolate it
• We must isolate it from the rest of the main-system 
• The main-system consists of the block and the floor
• So we want to isolate the block from the floor
(ii) For that, draw a polygon of convenient shape around the block
• For simple sub-systems, the polygon can be a simple rectangle or a square
• Here we draw a rectangle of convenient size around the block     
• This rectangle is shown in red color in fig.5.33(c)
(iii) Draw a schematic diagram of what ever is present inside the polygon 
• In our present case, we have a wooden block inside the red rectangle
• So we draw a small rectangle filled with yellow color to represent the wooden block
• This is shown in fig.d
(iv) Now we draw the vectors representing those forces which are experienced by the sub-system
■ Remember that we must draw only those forces which are experienced by the sub-system
■ We must not draw any of those forces which the sub-system exerts on the surroundings
• Let us see the various forces in our present case:
• In our present case, the sub-system is the wooden block
A. It experiences a downward pull from the earth. We have denoted it as $\mathbf\small{\vec{W}}$ 
     ♦ So we draw the vector $\mathbf\small{\vec{W}}$ in fig.d
B. It experiences an upward push from the ground. We have denoted it as $\mathbf\small{\vec{F_N}}$
    ♦ So we draw the vector $\mathbf\small{\vec{F_N}}$ in fig.d

---

Note:

• The block exerts a force on the floor because, the floor is obstructing the block from travelling to the center of the earth
• But we do not show this force in the FBD because, it is not the force experienced by the block
    ♦ It is the force exerted by the block
• $\mathbf\small{\vec{W}}$ appears to be a force exerted by the block. But in reality, it is not
    ♦ It is a force applied on the block by the earth
    ♦ In other words, it is a force experienced by the block. 
    ♦ So we can show it in the FBD
• $\mathbf\small{\vec{F_N}}$ is the 'reaction' produced from the floor due to the 'action' $\mathbf\small{\vec{W}}$
    ♦ The block suffers the effect of $\mathbf\small{\vec{F_N}}$
    ♦ In other words, $\mathbf\small{\vec{F_N}}$ is a force experienced by the block. 
    ♦ So we show it in the FBD

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• Thus the FBD of the wooden block is complete

• This FBD was but simple. For more complex cases, the FBD helps us to understand the situation and find the solutions easily

---

Let us see some solved examples which demonstrates the application of FBD
Solved example 5.17
Ten identical coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of
(a) the force on the 7^th coin (counted from the bottom) due to all the coins on its top,
(b) the force on the 7^th coin by the 8^th coin,
(c) the reaction of the 6^th coin on the 7^th coin.
Solution:
1. Fig.5.34(a) below shows the ten coins put on top of each other

Fig.5.34

• Based on the discussion which we saw above, let us draw the FBD of the top most coin (10^th coin counted from the bottom):
2. The FBD is shown in fig.5.34c
It can be explained as follows:
• First we put the concerned coin inside a red rectangle as shown in fig.5.34(b)
    ♦ This is to indicate that, whatever is within the rectangle is selected as a sub-system
    ♦ And the FBD is drawn for that sub-system
(i) The earth pulls down on it
• This pulling force is given by: $\mathbf\small{\vec{W_{10}}}$
• We have: magnitude of $\mathbf\small{\vec{W_{10}}}$ = $\mathbf\small{|\vec{W_{10}}|=mg}$
Where: 
    ♦ m is the mass of one coin
    ♦ g is the acceleration due to gravity
• The subscript '10' denotes the 10^th coin
• Since $\mathbf\small{\vec{W_{10}}}$ is a force which is experienced by the coin, we show it in the FBD
(ii) As a result of this pull by the earth, the coin tries to move downwards
• But the other coins obstruct this movement
• That means, the 10^th coin applies a force on top of the 9^th coin
    ♦ We can denote this force as: $\mathbf\small{\vec{F_{N(9,10)}}}$
    ♦ The subscript (9,10) indicates that, this $\mathbf\small{\vec{F_N}}$ is applied on the 9^th coin by the 10^th coin   
• Since this force is 'applied by' the coin, we do not show it in the FBD
(iii) But we do show the reaction from the 9^th coin. Because, it is acting on the 10^th coin
• This reaction is indicated as $\mathbf\small{\vec{F_{N(10,9)}}}$
    ♦ The subscript (10,9) indicates that, this $\mathbf\small{\vec{F_N}}$ is applied on the 10^th coin by the 9^th coin   

• Thus we get the FBD in fig.c

■ In the FBD, we see all the forces experienced by the coin
• Also, we see that the coin is at rest
• So no net force is acting on the coin
    ♦ That is., the vector sum of the forces acting on the coin is zero  
    ♦ Note that, it is a 'vector addition'. When we do a 'vector addition', we get a resultant
    ♦ All rules for such an addition (analytical method OR graphical methods like triangle law, parallelogram law, polygon law etc.,) should be strictly followed to obtain the resultant
• In our present case, both the vectors are along the same line. So it is essentially a one-dimensional problem and so the addition is easy
• However, we will use vector notations so as to get a general picture
• Thus we have:
$\mathbf\small{\vec{W_{10}}+\vec{F_{N(10,9)}}=0}$
$\mathbf\small{\Rightarrow \vec{F_{N(10,9)}}=-\vec{W_{10}}}$
• So in this example, the magnitude of $\mathbf\small{\vec{F_{N(10,9)}}}$ is equal to the magnitude of $\mathbf\small{\vec{W_{10}}}$
3. In a similar way, we will draw the FBD of the 9^th coin 
• First we put it inside a red rectangle as shown in fig.5.35(a) below:

Fig.5.35

(i) The earth pulls down on it
• This pulling force is given by: $\mathbf\small{\vec{W_9}}$
• We have: magnitude of $\mathbf\small{\vec{W_9}}$ = $\mathbf\small{|\vec{W_9}|=mg}$
Where: 
    ♦ m is the mass of one coin
    ♦ g is the acceleration due to gravity
(ii) But this is not the only downward force experienced by the 9^th coin
• In the previous FBD of the 10^th coin, we saw a force $\mathbf\small{\vec{F_{N(10,9)}}}$
• This was a 'reaction' to the action $\mathbf\small{\vec{F_{N(9,10)}}}$
• So $\mathbf\small{\vec{F_{N(9,10)}}}$ is indeed acting on the 9^th coin
• We have to show it on the FBD of the 9^th coin
• So in the FBD, we will have two downward force vectors
• This is shown in fig.5.35(b)
4. under the influence of the two downward vectors, the 9^th coin tries to move downwards
• But the other coins obstruct this movement
• That means, the 9^th coin applies a force on top of the 8^th coin
    ♦ We can denote this force as: $\mathbf\small{\vec{F_{N(8,9)}}}$
    ♦ The subscript (8,9) indicates that, this $\mathbf\small{\vec{F_N}}$ is applied on the 8^th coin by the 9^th coin   
    ♦ Since this force is 'applied by' the coin, we do not show it in the FBD
(iii) But we do show the reaction from the 8^th coin. Because, it is acting on the 9^th coin
• This reaction is indicated as $\mathbf\small{\vec{F_{N(9,8)}}}$
    ♦ The subscript (9,8) indicates that, this $\mathbf\small{\vec{F_N}}$ is applied on the 9^th coin by the 8^th coin   

• Thus we get the FBD in fig.c

5. In the FBD, we see all the forces experienced by the coin
• Also, we see that the coin is at rest
• So no net force is acting on the coin
    ♦ That is., the vector sum of the forces acting on the coin is zero  
• Thus we have:
$\mathbf\small{\vec{W_{9}}+\vec{F_{N(9,8)}}+\vec{F_{N(9,10)}}=0}$
    ♦ But $\mathbf\small{\vec{F_{N(9,10)}}=-\vec{F_{N(10,9)}}}$
    ♦ But $\mathbf\small{\vec{F_{N(10,9)}}=-\vec{W_{10}}}$ [From FBD of the 10^th coin]
    ♦ Thus $\mathbf\small{\vec{F_{N(9,10)}}=-(-\vec{W_{10}})=\vec{W_{10}}}$ 
• So we get: $\mathbf\small{\vec{W_{9}}+\vec{F_{N(9,8)}}+\vec{W_{10}}=0}$
6. The weights of all the coins are the same. Let us denote it as $\mathbf\small{\vec{W}}$
So we get: $\mathbf\small{\vec{W}+\vec{F_{N(9,8)}}+\vec{W}=0}$
$\mathbf\small{\Rightarrow 2\vec{W}+\vec{F_{N(9,8)}}=0}$
$\mathbf\small{\Rightarrow \vec{F_{N(9,8)}}=-2\vec{W}}$
7. In the same way, the FBD of the 8^th coin is shown in fig.5.35(d)
Since there is equilibrium, we can write:
$\mathbf\small{\vec{W_{8}}+\vec{F_{N(8,7)}}+\vec{F_{N(8,9)}}=0}$
    ♦ But $\mathbf\small{\vec{F_{N(8,9)}}=-\vec{F_{N(9,8)}}}$
    ♦ But $\mathbf\small{\vec{F_{N(9,8)}}=-2\vec{W}}$ [From step (6)]
    ♦ Thus $\mathbf\small{\vec{F_{N(8,9)}}=-(-2\vec{W})=2\vec{W}}$ 
• So we get: $\mathbf\small{\vec{W_{8}}+\vec{F_{N(8,7)}}+2\vec{W}=0}$
$\mathbf\small{\Rightarrow \vec{F_{N(8,7)}}+3\vec{W}=0}$
$\mathbf\small{\Rightarrow \vec{F_{N(8,7)}}=-3\vec{W}}$
8. Let us write a summary:
• From step (2), we have: $\mathbf\small{\vec{F_{N(10,9)}}=-1\vec{W}}$
• From step (6), we have: $\mathbf\small{\vec{F_{N(9,8)}}=-2\vec{W}}$
• From step (7), we have: $\mathbf\small{\vec{F_{N(8,7)}}=-3\vec{W}}$
9. We see a pattern. Based on that, we can write:
$\mathbf\small{\vec{F_{N[n,(n-1)]}}=-1 \times [10-(n-1)] \times \vec{W}}$
$\mathbf\small{\Rightarrow \vec{F_{N[n,(n-1)]}}=-1 \times [11-n] \times \vec{W}}$
• From this we can find the reaction on any coin from the 'coin just below it'

---

Now we can answer the given questions
Part (a): Force on the 7^th coin (counted from the bottom) due to all the coins on its top
Solution:
1. We want a force on the 7^th coin. So we must use the FBD of the 7^th coin
2. From the result in (9), let us draw the FBD of the 7^th coin:
• n = 7. So we get: $\mathbf\small{\Rightarrow \vec{F_{N[7,(7-1)]}}=-1 \times [11-7] \times \vec{W}}$
$\mathbf\small{\Rightarrow \vec{F_{N[7,6]}}=-1 \times [4] \times \vec{W}}$
• The FBD is shown in fig.5.36 below:

Fig.5.36

3. From the fig., it is clear that, a force of magnitude '4mg' is pulling down on the 7^th coin
• This force includes self weight of the 7^th coin
• Magnitude of the self weight is 'mg'
• So the force due to the coins on top is (4mg-mg) = 3mg

Part (b): Force on the 7^th coin by the 8^th coin
Solution:
1. We want a force on the 7^th coin. So we must use the FBD of the 7^th coin
• From the fig.5.36, it is clear that, a force of magnitude '4mg' is pulling down on the 7^th coin
2. But this force is a sum: (self weight + force from 8^th coin)
• Magnitude of the self weight is 'mg'
3. So we can write: 4 mg = (mg + force from 8^th coin)
• Thus force from 8^th coin = 3 mg

Part (c): the reaction of the 6^th coin on the 7^th coin
Solution:
1. We want a force on the 7^th coin. So we must use the FBD of the 7^th coin
• $\mathbf\small{\vec{F_{N[7,6]}}}$ is indeed, the reaction from the 6th coin on the 7^th coin
2. We get: $\mathbf\small{\vec{F_{N[7,6]}}=-1 \times [4] \times \vec{W}}$ 
• So magnitude of this force is 4mg
■ Note:
For this problem, we have written detailed steps. They are for easy understanding only. Once the basics are understood, the reader should do a large number of practice problems and thus must be able to draw rough sketches and solve problems with minimum steps.

Solved example 5.18
Two masses A and B of 3 kg and 5 kg respectively rests on a platform. See fig.5.37(a) below. The platform moves upwards with an acceleration of 4 ms^-2. What is the normal reaction exerted by B on A? [Take g = 10 ms^-2]

Fig.5.36

Solution:
1. We want a force (due to B) experienced by A. So let us choose A as a sub-system. This is shown in fig.5.36(b)
2. A will experience the following 3 forces:
(i) Self weight: $\mathbf\small{\vec{W}}$ downwards  
(ii) Reaction from B: $\mathbf\small{\vec{F_{N(AB)}}}$ upwards
• The subscript '(AB)' indicates that, the normal reaction $\mathbf\small{\vec{F_N}}$ is applied on A by B
(iii) Since A is under an acceleration, it will be experiencing a force of $\mathbf\small{m_A \times \vec{a}}$
• The 3 forces are shown in the FBD in fig.c
3. Forces (i) and (ii) are not equal
• We can be sure about it because, the mass A is not in equilibrium. It is experiencing an acceleration
• The resultant of (i) and (ii) is the force which causes the acceleration of A
4. The vector sum (resultant) of (i) and (ii) is the cause for acceleration
• So we can write: $\mathbf\small{\vec{F_{N(AB)}}+\vec{W}=m_A \times \vec{a}}$
• We will use the usual sign convention:
    ♦ Upward forces are positive
    ♦ Downward forces are negative
• So we can write: $\mathbf\small{\vec{F_{N(AB)}}-\vec{W}=m_A \times \vec{a}}$
$\mathbf\small{\Rightarrow \vec{F_{N(AB)}}=\vec{W}+m_A \times \vec{a}}$
• Substituting the values, we get:
$\mathbf\small{\vec{F_{N(AB)}}=(3 \times 10)\hat{j}+(3 \times 4)\hat{j}=42\: \hat{j}}$
• So the magnitude of $\mathbf\small{\vec{F_{N(AB)}}}$ is 42 N
Check:
We will check the above answer by considering B as a sub-system. This is shown in fig.5.36(d)
5. B will experience the following 4 forces:
(i) Self weight: $\mathbf\small{\vec{W}}$ downwards  
(ii) Reaction from A: $\mathbf\small{\vec{F_{N(BA)}}}$ 
• The subscript '(BA)' indicates that, the normal reaction $\mathbf\small{\vec{F_N}}$ is applied on B by A
• By Newton's third law, this force will be equal and opposite to $\mathbf\small{\vec{F_{N(AB)}}}$ 
• We have already calculated the magnitude of $\mathbf\small{\vec{F_{N(AB)}}}$ as 42 N
(iii) Since B is under an acceleration, it will be experiencing a force of $\mathbf\small{m_B \times \vec{a}}$
(iv)  Reaction from platform: $\mathbf\small{\vec{F_{N(BP)}}}$ 
• The subscript '(BP)' indicates that, the normal reaction $\mathbf\small{\vec{F_N}}$ is applied on B by the platform
• The 4 forces are shown in the FBD in fig.e
6. The resultant of (i), (ii) and (iv) is the force which causes the acceleration of B
• So we can write: $\mathbf\small{\vec{F_{N(BP)}}-\vec{W}-\vec{F_{N(BA)}}=m_B \times \vec{a}}$
• $\mathbf\small{\Rightarrow \vec{F_{N(BP)}}=\vec{W}+\vec{F_{N(BA)}}+m_B \times \vec{a}}$
• Substituting the values, we get:
$\mathbf\small{\vec{F_{N(BP)}}=(5 \times 10)\hat{j}+42\: \hat{j}+(5 \times 4)\hat{j}=112\: \hat{j}}$
• So the magnitude of $\mathbf\small{\vec{F_{N(BP)}}}$ is 112 N
7. That means, the mass B will experience a force of 112 N due to the reaction from the platform
• That means, the platform is providing a reaction force of 112 N
• If we place a weighing scale below B, then that scale will be providing this reaction of 112 N
• This reaction will appear as the reading on the scale
• That means, the reading on the scale will be 112 N
8. In a previous section, we have seen how to predict the apparent weight in a lift
• Here we have a total mass of 8 kg moving up with an acceleration of 4 ms^-2
• If we put a scale below B, the reading will be: m(a+g) = 8×(4+10) = 112 N
9. In step (6), we obtained 112 N by using the result of 42 N obtained in (4)
• So the result in (4) is correct

Solved example 5.19
A wooden block of mass 2 kg rests on a soft horizontal ground. See Fig.5.38(a) below:

Fig.5.38

When an iron cylinder of mass 25 kg is placed on top of the block, the ground yields steadily and the block and the cylinder together go down with an acceleration of 0.1 ms^-2. See Fig.(b). What is the action of the block on the ground (a) before and (b) after the ground yields ? (c) Identify the action-reaction pairs in the problem. Take g = 10 ms^-2
Solution:
1. Consider the situation before the yielding of the ground takes place
Let us choose the wooden block as the sub-system. It's FBD is shown in fig.5.37(c)
2. The normal reaction force from the ground is denoted as $\mathbf\small{\vec{F_{N(BG)}}}$
• The subscript '(BG)' indicates that, the normal reaction $\mathbf\small{\vec{F_N}}$ is applied on Block by Ground
3. The ground exerts this $\mathbf\small{\vec{F_{N(BG)}}}$ as a 'reaction' to the 'action' $\mathbf\small{\vec{F_{N(GB)}}}$ applied on it by the block 
• By Newtons third law, this $\mathbf\small{\vec{F_{N(BG)}}}$ will be equal and opposite to $\mathbf\small{\vec{F_{N(GB)}}}$
4. So, if we can find the magnitude and direction of $\mathbf\small{\vec{F_{N(BG)}}}$, we will get the magnitude and direction of $\mathbf\small{\vec{F_{N(GB)}}}$  
• For that, consider the block in the FBD in fig.5.37(c):
• Since the block is in equilibrium, we have:
$\mathbf\small{\vec{W_B}+\vec{F_{N(BG)}}=0}$
$\mathbf\small{\Rightarrow \vec{W_B}=-\vec{F_{N(BG)}}}$
5. We have: magnitude of $\mathbf\small{\vec{W_B}}$ = $\mathbf\small{|\vec{W_B}|=m_B g}$
= 2 × 10 = 20 N
• So magnitude of $\mathbf\small{\vec{F_{N(BG)}}}$ is 20 N
6. Thus, using the result in (3), we get:
• Magnitude of the 'action of block on the ground' is 20 N
• It's direction is 'downwards'  
■ This is the answer for part (a)
7. Consider the situation after yielding
• C will experience the following 3 forces:
(i) Self weight: $\mathbf\small{\vec{W_C}}$ downwards  
(ii) Reaction from B: $\mathbf\small{\vec{F_{N(CB)}}}$ upwards
• The subscript '(CB)' indicates that, the normal reaction $\mathbf\small{\vec{F_N}}$ is applied on C by B
(iii) Since C is under an acceleration, it will be experiencing a force of $\mathbf\small{m_C \times \vec{a}}$
• The 3 forces are shown in the FBD in fig.d
8. Forces (i) and (ii) are not equal
• We can be sure about it because, the mass C is not in equilibrium. It is experiencing an acceleration
• The resultant of (i) and (ii) is the force which causes the acceleration of A
9. The vector sum (resultant) of (i) and (ii) is the cause for acceleration
• So we can write: $\mathbf\small{\vec{F_{N(CB)}}+\vec{W_C}=m_C \times \vec{a}}$
• We will use the usual sign convention:
    ♦ Upward forces are positive
    ♦ Downward forces are negative
10. So we can write: $\mathbf\small{\vec{F_{N(CB)}}-\vec{W_A}=-m_A \times \vec{a}}$
• $\mathbf\small{\Rightarrow \vec{F_{N(CB)}}=\vec{W_A}-m_A \times \vec{a}}$
• Substituting the values, we get:
$\mathbf\small{\vec{F_{N(CB)}}=(25 \times 10)\hat{j}-(25 \times 0.1)\hat{j}=247.5\: \hat{j}}$
• So the magnitude of $\mathbf\small{\vec{F_{N(CB)}}}$ is 247.5 N 
• It acts in the upward direction
11. Now we will draw the FBD of B after yielding:
• B will experience the following 3 forces:
(i) Self weight: $\mathbf\small{\vec{W_B}}$ downwards  
(ii) Reaction from Ground G: $\mathbf\small{\vec{F_{N(BG)}}}$ upwards
(iii) Since B is under an acceleration, it will be experiencing a force of $\mathbf\small{m_B \times \vec{a}}$
(iv) There is one more force
It is $\mathbf\small{\vec{F_{N(BC)}}}$ upwards, the force exerted by C on B. 
It will be equal and opposite to $\mathbf\small{\vec{F_{N(CB)}}}$ 
• But we have already calculated $\mathbf\small{\vec{F_{N(CB)}}}$ in step (10)
    ♦ It has a magnitude of 247.5 N
    ♦ It acts in the upward direction 
• So our present $\mathbf\small{\vec{F_{N(BC)}}}$ will have:
    ♦ A magnitude of 247.5 N
    ♦ A downward direction
 • The 4 forces are shown in the FBD in fig.e
12. The vector sum (resultant) of (i), (ii) and (iv) is the cause for acceleration of B
• So we can write: $\mathbf\small{\vec{W_B}+\vec{F_{N(BG)}}+\vec{F_{N(BC)}}=m_B \times \vec{a}}$
• We will use the usual sign convention:
    ♦ Upward forces are positive
    ♦ Downward forces are negative
13. So we can write: $\mathbf\small{-\vec{W_B}+\vec{F_{N(BG)}}-\vec{F_{N(BC)}}=-m_B \times \vec{a}}$
$\mathbf\small{\Rightarrow \vec{F_{N(BG)}}=\vec{W_B}+\vec{F_{N(BC)}}-m_B \times \vec{a}}$
• Substituting the values, we get:
$\mathbf\small{\vec{F_{N(BG)}}=(2 \times 10)\hat{j}+247.5 \hat{j}-(2 \times 0.1)\hat{j}=267.3\: \hat{j}}$
• So the magnitude of $\mathbf\small{\vec{F_{N(BG)}}}$ is 267.3 N 
• It acts in the upward direction
■ This is the answer for part (b)
14. The action reaction pair before yielding is:
    ♦ $\mathbf\small{\vec{F_{N(BG)}}}$ and $\mathbf\small{\vec{F_{N(GB)}}}$
• The action reaction pairs after yielding are:
    ♦ Pair 1: $\mathbf\small{\vec{F_{N(CB)}}}$ and $\mathbf\small{\vec{F_{N(BC)}}}$
    ♦ Pair 2: $\mathbf\small{\vec{F_{N(BG)}}}$ and $\mathbf\small{\vec{F_{N(GB)}}}$
■ This is the answer for part (c)

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We saw that the 'normal reaction force' will be always perpendicular to the surface of contact. In the above discussion we saw cases in which contact surfaces are all horizontal. So the $\mathbf\small{\vec{F_N}}$ are all vertical. If the contact surfaces are vertical, then $\mathbf\small{\vec{F_N}}$ will be horizontal. We will see such cases in the next section

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