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 →FN were all vertical.
• In this section we will see cases in which:
♦ Contact surfaces are vertical
♦ So the →FN 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: →WA downwards
(ii) Self weight of B: →WB downwards
(iii) Reaction from the horizontal surface on A: →FN(AS) upwards
• The subscript '(AS)' indicates that, the normal reaction →FN is applied on A by the surface (S)
(iv) Reaction from the horizontal surface on B: →FN(BS) upwards
• The subscript '(BS)' indicates that, the normal reaction →FN 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:
→WA = −→FN(AS)
♦ The two forces cancel each other
→WB = −→FN(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: →WA downwards
(ii) Reaction from the horizontal surface on A: →FN(AS) upwards
• The subscript '(AS)' indicates that, the normal reaction →FN is applied on A by the surface (S)
(iii) The external horizontal force of 12 N
(iv) The normal reaction →FN(AB) horizontally towards A
• The subscript '(AB)' indicates that, the normal reaction →FN is applied on A by B
• Since the contact surface is vertical, this →FN will be horizontal
■ Let us see the reason for existence of this →FN:
• 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: →FN(BA)
♦ This force cannot be shown in the 'FBD of A' because, it is applied by A
• But B will apply a normal reaction →FN(AB) in accordance with Newton's third law
♦ This force is the reaction to →FN(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:
12ˆi+→FN(AB)=mA×→aA
• Applying the usual sign convention, we have:
12ˆi−→FN(AB)=mA×→aA
⇒→FN(AB)=12ˆi−mA×→aA
• Substituting the values, we get:
→FN(AB)=(12ˆi−5×1.5ˆi)=4.5ˆi
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: →WB downwards
(ii) Reaction from the horizontal surface on B: →FN(BS) upwards
• The subscript '(BS)' indicates that, the normal reaction →FN is applied on B by the surface (S)
(iii) The normal reaction →FN(BA) horizontally towards B
• The subscript '(BA)' indicates that, the normal reaction →FN is applied on B by A
• Since the contact surface is vertical, this →FN will be horizontal
■ Let us see the reason for existence of this →FN(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 →FN(BA)
♦ This force must be shown in the 'FBD of B' because, it is applied on B
• B will apply a normal reaction →FN(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, →FN(BA) is the net force
• So whatever horizontal acceleration 'B' is experiencing, will be due to →FN(BA)
• We can write: →FN(BA)=mB×→aB
4. Now, →FN(BA) is equal and opposite to →FN(AB)
• We have already found out that →FN(AB)=4.5ˆi
♦ →FN(AB) is acting towards left
• So →FN(BA) will be having the same magnitude and will be acting towards the right
5. Substituting the values in (3), we get: 4.5ˆi=(3×|→aB|)ˆi
⇒|→aB|=1.5ms−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 →FN(AB) in the FBD of 'A'
• We used →FN(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:
45ˆi−→FN(AB)=mA×→aA
⇒→FN(AB)=12ˆi−mA×→aA
• Substituting the values, we get:
→FN(AB)=(45ˆi−9×2.25ˆi)=24.75ˆi
5. For B, we have:
24.75ˆi−→FN(BC)=mB×→aB
⇒→FN(BC)=24.75ˆi−mB×→aB
• Substituting the values, we get:
→FN(BC)=(24.75ˆi−7×2.25ˆi)=9ˆi
• We are asked to find →FN(BC)
• So '9ˆi (towards left)' is the required answer
Check:
For C, we have:
→FN(CB)=mC×→aC
⇒9ˆi=(4×|→aC|)ˆi
⇒|→aC|=2.25ms−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 →FN were all vertical.
• In this section we will see cases in which:
♦ Contact surfaces are vertical
♦ So the →FN 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: →WA downwards
(ii) Self weight of B: →WB downwards
(iii) Reaction from the horizontal surface on A: →FN(AS) upwards
• The subscript '(AS)' indicates that, the normal reaction →FN is applied on A by the surface (S)
(iv) Reaction from the horizontal surface on B: →FN(BS) upwards
• The subscript '(BS)' indicates that, the normal reaction →FN 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:
→WA = −→FN(AS)
♦ The two forces cancel each other
→WB = −→FN(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: →WA downwards
(ii) Reaction from the horizontal surface on A: →FN(AS) upwards
• The subscript '(AS)' indicates that, the normal reaction →FN is applied on A by the surface (S)
(iii) The external horizontal force of 12 N
(iv) The normal reaction →FN(AB) horizontally towards A
• The subscript '(AB)' indicates that, the normal reaction →FN is applied on A by B
• Since the contact surface is vertical, this →FN will be horizontal
■ Let us see the reason for existence of this →FN:
• 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: →FN(BA)
♦ This force cannot be shown in the 'FBD of A' because, it is applied by A
• But B will apply a normal reaction →FN(AB) in accordance with Newton's third law
♦ This force is the reaction to →FN(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:
12ˆi+→FN(AB)=mA×→aA
• Applying the usual sign convention, we have:
12ˆi−→FN(AB)=mA×→aA
⇒→FN(AB)=12ˆi−mA×→aA
• Substituting the values, we get:
→FN(AB)=(12ˆi−5×1.5ˆi)=4.5ˆi
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: →WB downwards
(ii) Reaction from the horizontal surface on B: →FN(BS) upwards
• The subscript '(BS)' indicates that, the normal reaction →FN is applied on B by the surface (S)
(iii) The normal reaction →FN(BA) horizontally towards B
• The subscript '(BA)' indicates that, the normal reaction →FN is applied on B by A
• Since the contact surface is vertical, this →FN will be horizontal
■ Let us see the reason for existence of this →FN(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 →FN(BA)
♦ This force must be shown in the 'FBD of B' because, it is applied on B
• B will apply a normal reaction →FN(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, →FN(BA) is the net force
• So whatever horizontal acceleration 'B' is experiencing, will be due to →FN(BA)
• We can write: →FN(BA)=mB×→aB
4. Now, →FN(BA) is equal and opposite to →FN(AB)
• We have already found out that →FN(AB)=4.5ˆi
♦ →FN(AB) is acting towards left
• So →FN(BA) will be having the same magnitude and will be acting towards the right
5. Substituting the values in (3), we get: 4.5ˆi=(3×|→aB|)ˆi
⇒|→aB|=1.5ms−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 →FN(AB) and →FN(BA)
• We used →FN(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:
45ˆi−→FN(AB)=mA×→aA
⇒→FN(AB)=12ˆi−mA×→aA
• Substituting the values, we get:
→FN(AB)=(45ˆi−9×2.25ˆi)=24.75ˆi
5. For B, we have:
24.75ˆi−→FN(BC)=mB×→aB
⇒→FN(BC)=24.75ˆi−mB×→aB
• Substituting the values, we get:
→FN(BC)=(24.75ˆi−7×2.25ˆi)=9ˆi
• We are asked to find →FN(BC)
• So '9ˆi (towards left)' is the required answer
Check:
For C, we have:
→FN(CB)=mC×→aC
⇒9ˆi=(4×|→aC|)ˆi
⇒|→aC|=2.25ms−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 →FN were all vertical.
• In this section we saw cases in which:
♦ Contact surfaces were all vertical
♦ So the →FN 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 →FN were all vertical.
• In this section we saw cases in which:
♦ Contact surfaces were all vertical
♦ So the →FN were all horizontal
In the next section we will see cases in which:
♦ Contact surfaces are all inclined
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