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:
(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:
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:
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 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
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
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
• 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 |
(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 |
(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 |
(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 $\mathbf\small{\vec{F_{N(BA)}}}$ 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
 |
| Fig.5.40 |
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
♦ 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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