Chapter 5.17 - Strings through Pulleys - Solved Examples

In the previous section we saw strings passing through pulleys. We also saw two solved examples. In this section we will see a few more solved examples.

Solved example 5.29
A block of mass 25 kg is raised by a 50 kg man in two different ways as shown in Fig. 5.65 below. What is the action on the floor by the man in the two cases ? If the floor yields to a normal force of 700 N, which mode should the man adopt to lift the block without the floor yielding ?

Fig.5.65

Solution:
Case 1: Lifting with out the use of pulley
1. Let us draw FBDs. The sub-systems are shown in fig.5.66(a) below:

Fig.5.66

    ♦ The block (B) is selected as a sub-system and is enclosed inside a red rectangle 
    ♦ The man (M) is selected as another sub-system and is enclosed inside a green polygon
• The two magenta arrows drawn on the string help us to remember two points:
(i) The string pulls the block upwards with a certain force
(ii) The string pulls the hands downwards with the same force
2. The FBD for the block is shown in fig.5.66(b)
• We know the following 2 points:
(i) Magnitude of the tension $\small{\vec{T}}$ in the string is given by: $\small{|\vec{T}|=|\vec{W_B}|=m_B \, g}$    
(ii) It’s direction is vertical and upwards 
3. We know that, FBDs show the forces acting on the body
• So the $\small{\vec{T}}$ in fig.5.66(b) is acting on the block. It is provided by the person
• That is: The person has to provide an upward force of magnitude 'm_Bg' in order to lift the block  
4. The FBD of the man is shown in fig.c
• We see that, the string is pulling his hands downwards with a force $\small{\vec{T}}$  
    ♦ The magnitude of this downward pulling force is the same 'm_Bg'
• In addition to $\small{\vec{T}}$, gravity exerts a downward pulling force on the man
    ♦ The magnitude of this downward pulling force is 'm_Mg'
• Also, the floor applies a reaction on the man. It is denoted as $\small{\vec{R_{MF}}}$ 
5. Net force on a 'body which is acted upon by three forces' is given by the vector sum: $\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}}$
6. Let $\mathbf\small{\vec{F_1}}$ be $\mathbf\small{\vec{T}}$  
• Let $\mathbf\small{\vec{F_2}}$ be $\mathbf\small{\vec{W_M}}$
• Let $\mathbf\small{\vec{F_3}}$ be $\mathbf\small{\vec{R_{MF}}}$
• Taking upward forces as positive and downward forces as negative, the equation in (5) becomes:
Net force = $\mathbf\small{\vec{R_{MF}}-\vec{T}-\vec{W_{M}}}$
7. But the man is stationary. That is., he is in equilibrium. 
• According to Newton's second law, there is no net force acting on the man. So we can write:
$\mathbf\small{\vec{R_{MF}}-\vec{T}-\vec{W_{M}}}$ = 0
$\mathbf\small{\Rightarrow \vec{R_{MF}}=\vec{T}+\vec{W_{M}}}$
$\mathbf\small{\Rightarrow \vec{R_{MF}}=(m_B \times g)\hat{j}+(m_M \times g)\hat{j}=[(m_B+m_M)g]\hat{j}}$
($\mathbf\small{\because \text{From (4), we know that}\: |\vec{T}|=m_B \times g}$)
$\mathbf\small{\Rightarrow |\vec{R_{MF}}|=(m_B+m_M)g}$
• Substituting the values, we get:
$\mathbf\small{|\vec{R_{MF}}|=(25+50)10=750 \,  \text{N}}$
8. So we can write:
• When the block is lifted with out using a pulley, the floor applies a normal reaction of 750 N on the man
• Applying Newton's third law, we get:
The man applies a normal reaction of 750 N on the floor
• But given that, the floor will yield under 700 N. So the man must not lift the block without using a pulley

Case 2: Lifting by using a pulley
1. Let us draw FBDs. The sub-systems are shown in fig.5.67(a) below:

Fig.5.67

    ♦ The block (B) is selected as a sub-system and is enclosed inside a red rectangle 
    ♦ The man (M) is selected as another sub-system and is enclosed inside a green polygon
• The two magenta arrows drawn on the string help us to remember two points:
(i) The string pulls the block upwards with a certain force
(ii) The string pulls the hands upwards with the same force
2. The FBD for the block is shown in fig.5.67(b)
• We know the following 2 points:
(i) Magnitude of the tension $\small{\vec{T}}$ in the string is given by: $\small{|\vec{T}|=|\vec{W_B}|=m_B \, g}$    
(ii) It’s direction is vertical and upwards 
3. We know that, FBDs show the forces acting on the body
• So the $\small{\vec{T}}$ in fig.5.66(b) is acting on the block. It is provided by the person
• That is: The person has to provide an upward force of magnitude 'm_Bg' in order to lift the block  
4. The FBD of the man is shown in fig.c
• We see that, the string is pulling his hands upwards with a force $\small{\vec{T}}$  
    ♦ The magnitude of this upward pulling force is the same 'm_Bg'
    ♦ (Note that, in the previous case, this was a downward pulling force)
• In addition to $\small{\vec{T}}$, gravity exerts a downward pulling force on the man
    ♦ The magnitude of this downward pulling force is 'm_Mg'
• Also, the floor applies a reaction on the man. It is denoted as $\small{\vec{R_{MF}}}$ 
5. Net force on a 'body which is acted upon by three forces' is given by the vector sum: $\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}}$
6. Let $\mathbf\small{\vec{F_1}}$ be $\mathbf\small{\vec{T}}$  
• Let $\mathbf\small{\vec{F_2}}$ be $\mathbf\small{\vec{W_M}}$
• Let $\mathbf\small{\vec{F_3}}$ be $\mathbf\small{\vec{R_{MF}}}$
• Taking upward forces as positive and downward forces as negative, the equation in (5) becomes:
Net force = $\mathbf\small{\vec{R_{MF}}+\vec{T}-\vec{W_{M}}}$
7. But the man is stationary. That is., he is in equilibrium. 
• According to Newton's second law, there is no net force acting on the man. So we can write:
$\mathbf\small{\vec{R_{MF}}+\vec{T}-\vec{W_{M}}}$ = 0
$\mathbf\small{\Rightarrow \vec{R_{MF}}=-\vec{T}+\vec{W_{M}}}$
$\mathbf\small{\Rightarrow \vec{R_{MF}}=-(m_B \times g)\hat{j}+(m_M \times g)\hat{j}=[(-m_B+m_M)g]\hat{j}}$
$\mathbf\small{\because \text{From (4), we know that}\: |\vec{T}|=m_B \times g}$
$\mathbf\small{\Rightarrow |\vec{R_{MF}}|=(-m_B+m_M)g}$
• Substituting the values, we get:
$\mathbf\small{|\vec{R_{MF}}|=(-25+50)10=250 \,  \text{N}}$
8. So we can write:
• When the block is lifted using a pulley, the floor applies a normal reaction of 250 N on the man
• Applying Newton's third law, we get:
The man applies a normal reaction of 250 N on the floor
• But given that, the floor will yield only under 700 N. So the man must lift the block using a pulley

Solved example 5.30
In fig.5.68(a) below, block A rests on a smooth horizontal platform P. 

Fig.5.68

What is the acceleration of the system when it is released?
Solution:
1. First we draw the FBDs. The sub-systems chosen are shown in fig.5.68(b)
• Two vectors are marked on the string
• They are marked to help us remember two points:
(i) The string pulls at block A
(ii) The string pulls at block B
• A string can exert only 'pulling force'. It cannot exert 'pushing force' 
■ We know that tension is uniform through out the length of a string. So the two 'pulls' are the same.
• This is true even if a pulley (light and friction less) is present between the ends of the string 
2. The FBD of 'A' is shown in fig.5.68(c)
• The vertical components cancel each other. There is no motion for 'A' in the vertical direction. So the vertical components are not taken into account
• In the FBD, the tension in the string is denoted as $\mathbf\small{\vec{T}}$
    ♦ It is the only unbalanced force
• So net force acting on 'A' = $\mathbf\small{\vec{T}}$
3. By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{{\vec{T}=(m_A \times |\vec{a}|})\hat{j}=(2 |\vec{a}|) \hat{j}}$

$\mathbf\small{\Rightarrow |\vec{T}|=2 \times |\vec{a}|}$
4. The FBD of 'B' is shown in fig.5.68(d)
• The forces acting on B are: $\mathbf\small{\vec{T} \: \text {and}\: \vec{W_B}}$
5. Net force on a 'body which is acted upon by two forces' is given by the vector sum: $\mathbf\small{\vec{F_1}+\vec{F_2}}$
7. Let $\mathbf\small{\vec{F_1}}$ be $\mathbf\small{\vec{T}}$  
• Let $\mathbf\small{\vec{F_2}}$ be $\mathbf\small{\vec{W_B}}$
• Taking upward forces as positive and downward forces as negative, the equation in (7) becomes:
Net force = $\mathbf\small{\vec{T}-\vec{W_B}}$ 
8. According to Newton's second law, net force is equal to (mass ×acceleration). So we can write:
$\mathbf\small{\vec{T}-\vec{W_B}=-m_B \times \vec{a}}$
• The right side gets a netative sign because, the net force on B is in the downward direction
$\mathbf\small{\Rightarrow (|\vec{T}|-|\vec{W_B}|)\hat{j}=-(m_B \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow |\vec{T}|-|\vec{W_B}|=-m_B \times |\vec{a}|}$
• Substituting the values we get:
$\mathbf\small{|\vec{T}|-50=-5 \times |\vec{a}|}$
9. Thus we get two equations:
• From (3), we have: $\mathbf\small{|\vec{T}|=2 \times |\vec{a}|}$
• From (8), we have: $\mathbf\small{|\vec{T}|-50=-5 \times |\vec{a}|}$
• Solving the two equations, we get:
$\mathbf\small{|\vec{a}|=\frac{50}{7}\,\, \text{ms}^{-2}}$ and $\mathbf\small{|\vec{T}|=\frac{100}{7}\, \text{N}}$

Solved example 5.31
In fig.5.69(a) below, block A rests on a smooth inclined plane I

Fig.5.69

 What is the acceleration of the system when it is released?
Solution:
1. First we draw the FBDs. The sub-systems chosen are shown in fig.5.68(b)
• Two vectors are marked on the string
• They are marked to help us remember two points:
(i) The string pulls at block A
(ii) The string pulls at block B
• A string can exert only 'pulling force'. It cannot exert 'pushing force' 
■ We know that tension is uniform through out the length of a string. So the two 'pulls' are the same.
• This is true even if a pulley (light and friction less) is present between the ends of the string 
2. The FBD of 'A' is shown in fig.5.70(a) below:

Fig.5.70

• The components perpendicular to the inclined plane cancel each other. There is no motion in that direction. So we do not take those components into account. The modified FBD is shown in fig.b
• In the FBD, the tension in the string is denoted as $\mathbf\small{\vec{T}}$.
    ♦ It is parallel to the inclined surface
3. Net force on a 'body which is acted upon by two forces' is given by the vector sum: $\mathbf\small{\vec{F_1}+\vec{F_2}}$
4. Let $\mathbf\small{|\vec{F_1}|}$ be (m_Ag sin θ)
• Let $\mathbf\small{\vec{F_2}}$ be $\mathbf\small{|\vec{T}|}$
• Taking 'forces down the incline' as positive and 'forces up the incline' as negative, the magnitude of the net force is: $\mathbf\small{(m_A \times g \times \sin \theta )-|\vec{T}|}$
5. According to the second law, this net force is equal to (mass ×acceleration). So we get:
$\mathbf\small{(m_A \times g \times \sin \theta )-|\vec{T}|=m_A \times |\vec{a}|}$
(The right side is positive because, we assume that 'A' moves down the plane. And in this problem, 'down the plane ' is taken as positive) 
• Substituting the values, we get:
$\mathbf\small{(3 \times 10 \times \sin 30)-|\vec{T}|=3 \times |\vec{a}|}$
$\mathbf\small{\Rightarrow \text{15}-|\vec{T}|=3 \times |\vec{a}|}$
6. The FBD of 'B' is shown in fig.5.68(d)
• The forces acting on B are: $\mathbf\small{\vec{T} \: \text {and}\: \vec{W_B}}$
7. Net force on a 'body which is acted upon by two forces' is given by the vector sum: $\mathbf\small{\vec{F_1}+\vec{F_2}}$
8. Let $\mathbf\small{\vec{F_1}}$ be $\mathbf\small{\vec{T}}$  
• Let $\mathbf\small{\vec{F_2}}$ be $\mathbf\small{\vec{W_B}}$
• Taking upward forces as positive and downward forces as negative, the magnitude of the net force is:
$\mathbf\small{|\vec{T}|-|\vec{W_B}|}$ 
8. According to the second law, net force is equal to (mass ×acceleration). So we can write:
$\mathbf\small{|\vec{T}|-|\vec{W_B}|=m_B \times |\vec{a}|}$
• Substituting the values we get:
$\mathbf\small{|\vec{T}|-20=2 \times |\vec{a}|}$
9. Thus we get two equations:
• From (5), we have: $\mathbf\small{\text{15}-|\vec{T}|=3 \times |\vec{a}|}$
• From (8), we have: $\mathbf\small{|\vec{T}|-20=2 \times |\vec{a}|}$
• Solving the two equations, we get:
$\mathbf\small{|\vec{a}|=-1\, \text{ms}^{-2}}$ and $\mathbf\small{|\vec{T}|=18\, \text{N}}$
• We get a negative value for acceleration. That means, our 'assumed direcion' is wrong. The actual directions are:
    ♦ Block A moves up the plane
    ♦ Block B moves vertically downwards 

Solved example 5.32
Explain how a system of two pulleys help us to 'lift a load with lesser force'.
Solution:
• Fig.5.71(a) below shows a man lifting a block 'A' mass 'm' kg using a two pulley system

Fig.5.71

• The system consists of two light friction less pulleys
    ♦ One of the pulleys is a fixed pulley. It is fixed to the ceiling
    ♦ The other is a movable pulley
• A single light string passes through both the pulleys
    ♦ One end of the string is attached firmly to a point in the ceiling
    ♦ The other end is pulled by the man
Now we will write the steps:
1. First we draw the FBDs. The sub-systems chosen are shown in fig.5.71(b)
• Two vectors are marked on the string
• They are marked to help us remember two points:
(i) The string pulls at the ceiling
(ii) The string pulls at the man's hands
• A string can exert only 'pulling force'. It cannot exert 'pushing force' 
■ We know that tension is uniform through out the length of a string. So the two 'pulls' are the same
• This is true even if two pulleys (light and friction less) are present between the ends of the string 
2. Consider the FBD of A shown in fig.5.72(a) below

Fig.5.72

• The forces acting on A are: $\mathbf\small{\vec{T},\: \vec{T} \: \text {and}\: \vec{W_A}}$

(Since the pulley is light, it's weight is not considered) 
3. Net force on a 'body which is acted upon by three forces' is given by the vector sum: $\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}}$
4. Let $\mathbf\small{\vec{F_1}}$ be $\mathbf\small{\vec{T}}$  
• Let $\mathbf\small{\vec{F_2}}$ be $\mathbf\small{\vec{T}}$
• Let $\mathbf\small{\vec{F_3}}$ be $\mathbf\small{\vec{W_A}}$
• Taking upward forces as positive and downward forces as negative, the equation in (3) becomes:
Net force = $\mathbf\small{\vec{T}+\vec{T}-\vec{W_A}}$
⟹ Net force = $\mathbf\small{2\,\vec{T}-\vec{W_A}}$
5. If the block A is being lifted with uniform velocity, the net force is zero
So we get: $\mathbf\small{2\,\vec{T}-\vec{W_A}}$ = 0
$\mathbf\small{\Rightarrow \vec{T}=0.5 \times \vec{W_A}}$
$\mathbf\small{\Rightarrow |\vec{T}|=0.5 \times |\vec{W_A}|=0.5 \times m_A \times \text{g}}$
6. Thus we obtained the magnitude of the $\mathbf\small{\vec{T}}$ in the string
• Since the tension in a light inextensible string is uniform through out it's length, the man will be experiencing this same $\mathbf\small{\vec{T}}$ in his hands
• This is shown in the FBD of the man in fig.5.72(b)
7. Thus we can write:
The 'pull that the man need to apply' is only half the weight of the block A

---

In the next section we will see friction

PREVIOUS           CONTENTS          NEXT

Copyright©2018 Higher Secondary Physics. blogspot.in - All Rights Reserved

Chapter 5.14 - Force in String

In the previous section we completed a discussion on 'Normal reaction forces' between objects. In this section we will see 'force in strings'. 
We will write the steps:

1. Let two men A and B, pull from each end of a string as shown in the fig.4.45 below:

Fig.5.45

• When the pull is made from both ends, the string is under tension. 

2. This tension is a force. We can call it 'tensile force'. 

    ♦ It is denoted by the vector $\small{\vec{T}}$.
• It's magnitude can take any value like 5 N, 7 N, 20 N etc.,

3. What ever be the value of magnitude of the tensile force, it will be the same at all points (between the two men) along the string 

• What if one of the men exert a greater force?
• Then will the tension in the string become 'non-uniform'? 
• The answer is 'No'. Even if one of the men increase or decrease his pull, the tension will be uniform through out the string.

4. But how is that possible?

• We will write the answer in steps: 

(i) Let the initial $\small{|\vec{T}|}$ be $\small{|\vec{T_1}|}$ = 20 N 

(ii) Now, 'A' increases his pull to 25 N

(iii) Just then, B must increase his own pull to 25 N

• Other wise he will be pulled towards A

(iv) Thus the $\small{|\vec{T}|}$ will attain a new value $\small{|\vec{T_2}|}$ = 25 N

• This $\small{|\vec{T_2}|}$ will be uniform through out the length of the string

5. Also note that, a string can never take a compressive force. That is: 

    ♦ We can pull from two ends of a string. 
    ♦ But we cannot push from the two ends
■ So we can write a conclusion:

If we see a string which is pulled tight from it's ends, we can be sure that, the string is experiencing a tension $\small{|\vec{T}|}$ and this $\small{|\vec{T}|}$ is uniform through out it's length

■ Based on the above conclusion, we can write another one:

If we see a 'string in tension $\small{|\vec{T}|}$' between two objects P and Q, we can be sure about two points:

(i) The string is pulling the object P towards the 'string itself' with a force of $\small{|\vec{T}|}$ newtons

(ii) The string is pulling the object Q towards the 'string itself' with the same force of $\small{|\vec{T}|}$ newtons
• It is as if, the string is trying to decrease in length

• This will be clear if we draw the FBD of each men in fig.5.45. This is shown in fig.5.46(b) below:

Fig.5.46

• In the FBDs, the force exerted by the string is indicated by the magenta arrows.
• If those magenta arrows are absent, obviously, both the men will fall backwards
    ♦ So the string is indeed pulling 'A'
    ♦ It is also pulling 'B' 

---

For our present discussion on 'force in strings', we make two assumptions:

(1) The mass of string is negligible

• That is., we assume that, the string has zero mass. 
• While doing problems in our present discussion, we do not take the self weight of the string in to account.

(2) The string is 'inextensible'

This can be explained by comparing the actions of two strings. One which is extensible and the other inextensible. This is shown in fig.5.47 below:

Fig.5.47

(i) In fig.a, one end of a string is tied firmly to the ceiling

• When a mass is attached to the free end at bottom, the length of the string do not increase
• We say: the string is inextensible

(ii) In fig.b, one end of another string is tied firmly to the ceiling         

• When a mass is attached to the free end at bottom, the length of the string increases
• We say: the string is extensible
■ If a string is extensible, $\small{|\vec{T}|}$ will not be uniform along it's length

---

• Both extensible and inextensible strings/cables find applications in many scientific and engineering fields

• It is interesting to note that in some cases, it is prohibited to use inextensible strings
• We will learn it’s details in higher classes
• In this present discussion however, we deal with inextensible strings only

---

Calculation of the magnitude and direction of $\small{\vec{T}}$  

There are 3 possible cases:

Case 1: The string is vertical

Case 2: The string is horizontal

Case 3: The string is inclined

We will now see each case in detail:

Case 1: The string is vertical

This 'case 1' has four sub cases 

Case 1(a): The body to which the string is attached is at rest  

Case 1(b): The body to which the string is attached is in uniform motion

Case 1(c): The body to which the string is attached is in accelerated motion upwards

Case 1(d): The body to which the string is attached is in accelerated motion downwards

We will see each case in detail:

Case 1(a): The body to which the string is attached is at rest 

1. In fig.5.48(a) below, a block 'A' of mass m is suspended from the ceiling using a string

Fig.5.48

• If the ceiling is at rest, the block will also be at rest

2. There will be a tension $\small{\vec{T}}$ in the string. We want to find the magnitude and direction of that $\small{\vec{T}}$ 

• The magnitude and direction can be calculated using FBD. This is shown in fig.5.48(b)
    ♦ The earth pulls the block with a force $\small{\vec{W}}$  
    ♦ It is a force experienced by the block. So we show it in the FBD
    ♦ The string pulls the block by the force $\small{\vec{T}}$ 
    ♦ It is a force experienced by the block. So we show it also in the FBD
■ Direction of $\small{\vec{T}}$ is obvious: vertical and upwards. 
■ 'Upwards' because, as seen from fig.5.46 above, a string in tension will be pulling both the 'objects at it's ends' towards itself.

3. To find the magnitude, we apply the condition for equilibrium:

$\mathbf\small{\vec{F_1}+\vec{F_2}=0}$

Considering upward forces as positive and downward forces as negative, we get:

$\mathbf\small{(|\vec{T}|)\hat{j}-(|\vec{W}|)\hat{j}=0}$

$\mathbf\small{\Rightarrow (|\vec{T}|)\hat{j}-(mg)\hat{j}=0}$

$\mathbf\small{\Rightarrow (|\vec{T}|)\hat{j}=(mg)\hat{j}}$

$\mathbf\small{\Rightarrow |\vec{T}|=mg}$

Case 1(b): The body to which the string is attached is in uniform motion

1. In fig.5.45(a) above, if the ceiling moves up or down with uniform speed, there will be no net force on the block

• So the FBD in fig.b is valid in this case also

2. We will get: $\mathbf\small{|\vec{T}|=mg}$ 

Case 1(c): The body to which the string is attached is in accelerated motion upwards

1. In fig.5.48(c) above, the block 'A' of mass m is suspended from the ceiling using a string

• If the ceiling is in accelerated motion, the block will also experience acceleration
    ♦ In that case, we will have to do some calculations to find $\small{|\vec{T}|}$

2. There will be a tension $\small{\vec{T}}$ in the string. We want to find the magnitude and direction of that $\small{\vec{T}}$ 

• The magnitude and direction can be calculated using FBD. This is shown in fig.5.48(d)
■ Direction of $\small{\vec{T}}$ will obviously be: vertical and upwards. 
■ 'upwards' because, as seen from fig.5.46 above, a string in tension will be pulling both the 'objects at it's ends' towards itself.

3. To find the magnitude, we calculate the net force on the block. The net force is:

$\mathbf\small{(|\vec{T}|)\hat{j}-(|\vec{W}|)\hat{j}}$

• This net force will be equal to $\mathbf\small{m \times \vec{a}}$ 

4. So we can write:

$\mathbf\small{(|\vec{T}|)\hat{j}-(|\vec{W}|)\hat{j}=(m \times |\vec{a}|)\hat{j}}$

$\mathbf\small{\Rightarrow (|\vec{T}|)\hat{j}-(mg)\hat{j}=(m \times |\vec{a}|)\hat{j}}$

$\mathbf\small{\Rightarrow |\vec{T}|=m(g+|\vec{a}|)}$

• So we find that the tension in the string will be greater than in case 1(a)

Case 1(d): The body to which the string is attached is in accelerated motion downwards

1. In this case the same FBD in fig.d is valid

• But the direction of acceleration must be reversed

2. We get:

$\mathbf\small{(|\vec{T}|)\hat{j}-(|\vec{W}|)\hat{j}=-(m \times |\vec{a}|)\hat{j}}$

(∵ acceleration vector in the downward direction is taken as negative)

$\mathbf\small{\Rightarrow (|\vec{T}|)\hat{j}-(mg)\hat{j}=-(m \times |\vec{a}|)\hat{j}}$

$\mathbf\small{\Rightarrow |\vec{T}|=m(g-|\vec{a}|)}$

• So we find that the tension in the string will be lesser than in case 1(a)

---

Case 2: The string is horizontal

• A horizontal string can not support the weight of a body
• So, if the string is to be horizontal, the body to which the string is attached, must be resting on a platform or floor

OR

• The body must be supported by other strings
■ We will consider the cases when the body rests on a floor

(Body supported by other strings will be discussed later) 

• The floor may be with or without friction
■ We will consider the cases when the body rests on smooth floors

(Body resting on floors with friction will be discussed later) 

• Further, the bodies attached to the string may be at rest or in motion
• Further still, the body in motion may be in 'uniform motion' or in 'accelerated motion'   
• So we will prepare a flow chart like presentation to make the above possibilities clear. It is shown in fig.5.49 below:

Fig.5.49

So for our present discussion, we will be considering the following three sub cases:

Case 2(a): Body is at rest on a smooth floor 

Case 2(b): Body is in uniform motion on a smooth floor

Case 2(c): Body is in accelerated motion on a smooth floor

Case 2(a): Body is at rest on a smooth floor
1. In fig.5.50(a) below, a block of mass m rests on a smooth horizontal floor

Fig.5.50

• A string is attached to it's side

2. Let us apply a tension $\small{\vec{T}}$ in the string

• Fig.(b) shows the FBD

(The vertical forces will cancel each other. So we do not take them into account)

• We see that, since the floor is friction less, there is no force to oppose $\small{\vec{T}}$

• When $\small{\vec{T}}$ is applied, the block will certainly begin to move

3. So, on an object resting on a friction less horizontal floor, we cannot apply a force through a horizontal string, without causing it to move

• In other words, on an object resting on a friction less horizontal floor, the magnitude of $\small{\vec{T}}$ applied through a horizontal string is zero

Case 2(b): Body is in uniform motion on a smooth floor     

1. When a body is in uniform motion, there is no net force acting on that body

• In our present case, since the floor is smooth, there is no force to oppose motion

2. Once the body attains a uniform velocity, it will continue to move in that velocity

• There is no need to apply tension through the horizontal string

• The FBD in fig.5.50(b) is valid in this case also

3. We can write:

On an object in uniform motion on a friction less horizontal floor, the magnitude of $\small{\vec{T}}$ applied through a horizontal string is zero

4. If a need arise to change the speed, then the string will become useful

• A $\small{\vec{T}}$ applied through the string will cause an acceleration and thus a change in speed

• We will want to know the magnitude of $\small{\vec{T}}$ when a certain $\small{\vec{a}}$ is applied. This exactly is our next case

Case 2(c): Body is in accelerated motion on a smooth floor

1. This is shown in fig.5.50(c) above

• The FBD is shown in fig.d

2. Since there is no friction, the net force acting on A is $\small{\vec{T}}$

• This net force $\small{\vec{T}}$ is the cause for acceleration

3. So we can write: 

$\mathbf\small{(|\vec{T}|)\hat{i}=(m \times |\vec{a}|)\hat{j}}$

$\mathbf\small{\Rightarrow |\vec{T}|=m(|\vec{a}|)}$

---

Case 3: The string is inclined
• In this case, we resolve $\small{\vec{T}}$ into horizontal and vertical components and then apply the conditions for equilibrium
• The solved example given below will demonstrate the procedure

Solved example 5.23
In fig.5.51(a) below, a mass 'A' of 6 kg is suspended by a rope of length 2 m from the ceiling. 

Fig.5.51

A force of 50 N in the horizontal direction is applied at the midpoint P of the rope, as shown. What is the angle which the rope makes with the vertical in equilibrium ? (Take g = 10 ms^-2). Neglect the mass of the rope
Solution:
1. Let the point 'on the ceiling' at which the rope is attached be 'O'
• Let the point 'on the rope' at which the mass 'A' is attached be 'Q'
• These are shown in fig.b
2. In the segment OP, two vectors are marked on the rope
• They are marked to help us remember two points:
(i) The segment OP pulls at O
(ii) The segment OP pulls at P
■ We know that tension is uniform through out the length of a string. So the two 'pulls' are the same. It is denoted as $\small{\vec{T_1}}$  
3. Similarly, in the segment PQ, two vectors are marked on the rope
• They are marked to help us remember two points:
(i) The segment PQ pulls at P
(ii) The segment PQ pulls at Q
■  We know that tension is uniform through out the length of a string. So the two 'pulls' are the same. It is denoted as $\small{\vec{T_2}}$  
4. Next step is to find $\small{\vec{T_1}}$ and $\small{\vec{T_2}}$
For that, we need to draw FBDs
5. First we draw the FBD of 'A'
• So 'A' is taken as the sub-system. A red rectangle is drawn around 'A' in fig.b
• What ever is enclosed in the red rectangle, should be used in the FBD. Thus we get fig.c
• In fig.c, 'A' is in equilibrium under the action of two forces. So we apply the condition for equilibrium as follows:

$\mathbf\small{\vec{F_1}+\vec{F_2}=0}$
    ♦ Let $\small{\vec{F_1}}$ indicate $\small{\vec{T_2}}$ 
    ♦ Let $\small{\vec{F_2}}$ indicate $\small{\vec{W}}$ 

• Considering upward forces as positive and downward forces as negative, we get:

$\mathbf\small{(|\vec{T_2}|)\hat{j}-(|\vec{W}|)\hat{j}=0}$

$\mathbf\small{\Rightarrow (|\vec{T_2}|)\hat{j}-(mg)\hat{j}=0}$

$\mathbf\small{\Rightarrow (|\vec{T_2}|)\hat{j}=(mg)\hat{j}}$

$\mathbf\small{\Rightarrow |\vec{T_2}|=mg= 6 \times 10 = 60\,N}$
• Direction of $\small{\vec{T_2}}$ is obviously, 'vertical and upwards' 
6. Now we draw the FBD of a sub-system around the point 'P'
This is shown in fig.5.52(a) below:

Fig.5.52

• What ever is enclosed in the red rectangle, should be used in the FBD. Thus we get fig.5.52(b)
7. In fig.b, 'P' is in equilibrium under the action of 3 forces. So we apply the condition for equilibrium as follows:

$\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}=0}$
    ♦ Let $\small{\vec{F_1}}$ indicate $\small{\vec{T_1}}$ 
    ♦ Let $\small{\vec{F_2}}$ indicate $\small{50 \, \hat{i}}$ 
    ♦ Let $\small{\vec{F_3}}$ indicate $\small{\vec{T_2}}$ 
■ Here one of the forces is inclined. So we must take vertical and horizontal components separately
8. Considering horizontal components, we have:
$\mathbf\small{\vec{F_{1x}}+\vec{F_{2x}}+\vec{F_{3x}}=0}$
• Considering forces to the right as positive and forces to the left as negative, we get:
$\mathbf\small{(|\vec{T_1}|\sin \theta )\hat{i}-50\,\hat{i}+\,0=0}$
(The last term is zero because, $\small{\vec{T_2}}$ does not have a horizontal component)
$\mathbf\small{\Rightarrow |\vec{T_1}|\sin \theta=50}$
9. Considering the vertical components, we have:
$\mathbf\small{\vec{F_{1y}}+\vec{F_{2y}}+\vec{F_{3y}}=0}$
• Considering upward forces as positive and downward forces as negative, we get:
$\mathbf\small{(|\vec{T_1}|\cos \theta )\hat{j}+0-(|\vec{T_2}|)\hat{j}=0}$
(The second term is zero because, $\small{50 \, \hat{i}}$ does not have a vertical component)
$\mathbf\small{\Rightarrow (|\vec{T_1}|\cos \theta )=|\vec{T_2}|}$
10. But from (5) we have: $\mathbf\small{|\vec{T_2}|=60}$
(i) So the result in (9) becomes: $\mathbf\small{\Rightarrow (|\vec{T_1}|\cos \theta )=60}$
(ii) From (8) we have: $\mathbf\small{|\vec{T_1}|\sin \theta=50}$  
• Dividing (ii) by (i), we get: $\mathbf\small{\tan \theta =\frac{5}{6}}$
• So θ = 39.806^o.

---

In the next section we will see solved examples based on cases 1 and 2

PREVIOUS           CONTENTS          NEXT

Copyright©2018 Higher Secondary Physics. blogspot.in - All Rights Reserved