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
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
• 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
• This FBD was but simple. For more complex cases, the FBD helps us to understand the situation and find the solutions easily
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 |
(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:
• 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
• Thus the FBD of the wooden block is complete
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 7th coin (counted from the bottom) due to all the coins on its top,
(b) the force on the 7th coin by the 8th coin,
(c) the reaction of the 6th coin on the 7th coin.
Solution:
1. Fig.5.34(a) below shows the ten coins put on top of each other
• Based on the discussion which we saw above, let us draw the FBD of the top most coin (10th 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 10th 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 10th coin applies a force on top of the 9th 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 9th coin by the 10th 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 9th coin. Because, it is acting on the 10th 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 10th coin by the 9th 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 9th coin
• First we put it inside a red rectangle as shown in fig.5.35(a) below:
(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 9th coin
• In the previous FBD of the 10th 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 9th coin
• We have to show it on the FBD of the 9th 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 9th coin tries to move downwards
• But the other coins obstruct this movement
• That means, the 9th coin applies a force on top of the 8th 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 8th coin by the 9th 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 8th coin. Because, it is acting on the 9th 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 9th coin by the 8th 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 10th 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 8th 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 7th coin (counted from the bottom) due to all the coins on its top
Solution:
1. We want a force on the 7th coin. So we must use the FBD of the 7th coin
2. From the result in (9), let us draw the FBD of the 7th 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:
3. From the fig., it is clear that, a force of magnitude '4mg' is pulling down on the 7th coin
• This force includes self weight of the 7th 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 7th coin by the 8th coin
Solution:
1. We want a force on the 7th coin. So we must use the FBD of the 7th coin
• From the fig.5.36, it is clear that, a force of magnitude '4mg' is pulling down on the 7th coin
2. But this force is a sum: (self weight + force from 8th coin)
• Magnitude of the self weight is 'mg'
3. So we can write: 4 mg = (mg + force from 8th coin)
• Thus force from 8th coin = 3 mg
Part (c): the reaction of the 6th coin on the 7th coin
Solution:
1. We want a force on the 7th coin. So we must use the FBD of the 7th coin
• $\mathbf\small{\vec{F_{N[7,6]}}}$ is indeed, the reaction from the 6th coin on the 7th 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]
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:
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)
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 7th coin (counted from the bottom) due to all the coins on its top,
(b) the force on the 7th coin by the 8th coin,
(c) the reaction of the 6th coin on the 7th coin.
Solution:
1. Fig.5.34(a) below shows the ten coins put on top of each other
 |
| Fig.5.34 |
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 10th 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 10th coin applies a force on top of the 9th 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 9th coin by the 10th 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 9th coin. Because, it is acting on the 10th 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 10th coin by the 9th 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 9th coin
• First we put it inside a red rectangle as shown in fig.5.35(a) below:
 |
| Fig.5.35 |
• 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 9th coin
• In the previous FBD of the 10th 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 9th coin
• We have to show it on the FBD of the 9th 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 9th coin tries to move downwards
• But the other coins obstruct this movement
• That means, the 9th coin applies a force on top of the 8th 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 8th coin by the 9th 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 8th coin. Because, it is acting on the 9th 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 9th coin by the 8th 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 10th 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 8th 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 7th coin (counted from the bottom) due to all the coins on its top
Solution:
1. We want a force on the 7th coin. So we must use the FBD of the 7th coin
2. From the result in (9), let us draw the FBD of the 7th 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 |
• This force includes self weight of the 7th 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 7th coin by the 8th coin
Solution:
1. We want a force on the 7th coin. So we must use the FBD of the 7th coin
• From the fig.5.36, it is clear that, a force of magnitude '4mg' is pulling down on the 7th coin
2. But this force is a sum: (self weight + force from 8th coin)
• Magnitude of the self weight is 'mg'
3. So we can write: 4 mg = (mg + force from 8th coin)
• Thus force from 8th coin = 3 mg
Part (c): the reaction of the 6th coin on the 7th coin
Solution:
1. We want a force on the 7th coin. So we must use the FBD of the 7th coin
• $\mathbf\small{\vec{F_{N[7,6]}}}$ is indeed, the reaction from the 6th coin on the 7th 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 |
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 |
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)
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
