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

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In the next section we will see friction

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Chapter 5.16 - Strings passing through Friction less Pulleys

In the previous section we completed a discussion about 'force in string'. In this section we will see 'strings through pulleys'.

We will write the steps:
1. In fig.5.61(a) below, a string is attached to a block of mass ‘m’ kg. 

Fig.5.61

• A person lifts the mass upwards by pulling at the free end of the string. 
• He then keeps it stationary at a height above the ground
2. Let us draw FBD's. The sub-systems are shown in fig.b
    ♦ The block is selected as a sub-system and is enclosed inside a red rectangle 
    ♦ The person 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
3. The FBD for the block is shown in fig.c
• We have seen this FBD in many problems in the previous sections. We can write 2 points:
(i) Magnitude of the tension $\small{\vec{T}}$ in the string is given by: $\small{|\vec{T}|=|\vec{W}|=mg}$    
(ii) It’s direction is vertical and upwards 
4. We know that, FBDs show the forces acting on the body
• So the $\small{\vec{T}}$ in fig.5.62(c) is acting on the block. It is provided by the person
• That is: The person has to provide an upward force of magnitude 'mg' in order to lift the block  
5. The FBD of the person is shown in fig.d
• We see that, the string is pulling his hands downwards. 
• The magnitude of this downward pulling force is the same 'mg'

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Now we introduce a pulley which is attached to the ceiling. This is shown in fig.5.62(a) below:

Fig.5.62

• The pulley is assumed to be friction less. We want to know what ‘friction less pulley’ implies.
Let us see it’s details:
(i) A pulley consists of a circular part rotating about a pivot.
• There will be a ‘contact surface’ between the circular part and the pivot
(ii) This contact surface is of particular interest to us
• If there is no friction in this contact surface, then the circular part can rotate freely
(iii) Let us see what happens if there is friction:
• Let the block in fig.5.62(a) be stationary. Then, the block will be trying to move downwards. 
• As a result, the pulley will be rotating in the clockwise direction
(iv) But if there is friction, the clockwise rotation will encounter resistance. 
• This is because, friction always acts in the direction 'opposite to the direction of motion’.
• Since the clockwise rotation is opposed, the person will need to apply a lesser force only to hold the block in position
• Now, if the person tries to lift the block continuously upwards  at a uniform speed, the pulley will be rotating in the anti-clockwise direction. 
• Since this rotation is opposed by the friction, the person will have to apply a greater force to lift it up

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■ If the pulley is friction less, the tension in both sides of the string will be the same. For our present discussion we consider only friction less pulleys

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Now we will continue our main discussion:
1. Let the block in fig.5.62(a) be stationary (or moving upwards/downwards with uniform velocity)
2. Let us draw FBD's. The sub-systems are shown in fig.5.62(b)
    ♦ The block is selected as a sub-system and is enclosed inside a red rectangle 
    ♦ The person 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. 'Same force' because, tension in a string is uniform along it's whole length even if a pulley (friction less) is present
3. The FBD for the block is shown in fig.5.62(c)
We can write 2 points:
(i) Magnitude of the tension $\small{\vec{T}}$ in the string is given by: $\small{|\vec{T}|=|\vec{W}|=mg}$    
(ii) It’s direction is vertical and upwards
4. We know that, FBDs show the forces acting on the body
• So the $\small{\vec{T}}$ in fig.5.62(c) is acting on the block. It is provided by the person
• That is: The person has to provide an upward force of magnitude 'mg' in order to lift the block
5. The FBD of the person is shown in fig5.62(d)
• We see that, the string is pulling his hands upwards
• We know that, since the pulley is friction less, this upward force is the same: $\small{|\vec{T}|=|\vec{W}|=mg}$
■ Thus we can conclude:
• If the right side of the string pulls the block with a tension $\small{\vec{T}}$, the left side will be pulling the hands of the person with the same tension $\small{\vec{T}}$ . 
• The magnitude of this upward pulling force is the same 'mg' 

6. We know that, FBDs show the forces acting on the bodies 

• So the $\small{\vec{T}}$ in fig.5.62(d) is acting on the person. It is trying to pull the hands upwards
7. So the person has to apply a pull in the opposite direction. That is., downwards
(In the FBDs, ‘forces applied by the bodies’ are not shown)
8. Thus he can conveniently lift the block by applying a downward force on the string
• Earlier, in fig.5.61(d), he lifts the block with the same force (but with difficulty) by applying the force in the upward direction
■ Thus we can write:
Pulleys help to ‘suitably change’ the direction of the force. The 'tension in a rope' between the end points of that rope will not change due to the introduction of a friction less pulley

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Now we will see some solved examples
Solved example 5.27
Two masses 8 kg and 12 kg are connected at the two ends of a light inextensible string that goes over a friction less pulley. Find the acceleration of the masses, and the tension in the string when the masses are released. [Take g = 10 ms^-2]
Solution:
• Consider fig.5.63(a) below. It is clear that, when the masses are released, the block A will move up and B will move down

Fig.5.63

• Since they are connected together with string (inextensible and light), both will move with the same acceleration
• We are required to find this acceleration. The steps are given below:
1. First we draw the required FBDs. The sub-systems selected are shown in fig.b  
2. Consider the FBD of A. It is shown in fig.c
• The forces acting on A are: $\mathbf\small{\vec{T} \: \text {and}\: \vec{W_A}}$
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 $\mathbf\small{\vec{T}}$  
• Let $\mathbf\small{\vec{F_2}}$ be $\mathbf\small{\vec{W_A}}$
• Taking upward forces as positive and downward forces as negative, the equation in (3) becomes:
Net force = $\mathbf\small{\vec{T}-\vec{W_A}}$ 
5. According to Newton's second law, net force is equal to (mass ×acceleration). So we can write:
$\mathbf\small{\vec{T}-\vec{W_A}=m_A \times \vec{a}}$
$\mathbf\small{\Rightarrow (|\vec{T}|-|\vec{W_A}|)\hat{j}=(m_A \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow |\vec{T}|-|\vec{W_A}|=m_A \times |\vec{a}|}$
• Substituting the values we get:
$\mathbf\small{|\vec{T}|-80=8 \times |\vec{a}|}$
6. Consider the FBD of B shown in fig.d
• The forces acting on B are: $\mathbf\small{\vec{T} \: \text {and}\: \vec{W_B}}$

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Note that, $\mathbf\small{\vec{T}}$ will be same on both sides of the pulley. We know the reason. However, we will write it again:
• The tension in a light inextensible string will be the same through out it's whole length
• So the string will be pulling at both the ends with the same force
• This force will not change even if a pulley (which is friction less) is introduced between the ends of the string

---

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 equation in (7) becomes:
Net force = $\mathbf\small{\vec{T}-\vec{W_B}}$ 
9. 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:
$\mathbf\small{|\vec{T}|-120=-12 \times |\vec{a}|}$
10. Thus we get two equations:
• From (5), we have: $\mathbf\small{|\vec{T}|-80=8 \times |\vec{a}|}$
• From (9), we have: $\mathbf\small{|\vec{T}|-120=-12 \times |\vec{a}|}$
• Solving the two equations, we get:
$\mathbf\small{|\vec{a}|}$ = 2 ms^-2 and $\mathbf\small{|\vec{T}|}$ = 96 N

Solved example 5.28
Find the acceleration of the blocks and the tensions in the strings in the system shown in fig.5.64(a) below. Also find the reaction in the pulley. Assume the strings to be light and inextensible and the pulley to be light and friction less. [Take g = 10 ms^-2]

Fig.5.64

Solution:
• It is clear that, when the masses are released, block A will move up and blocks B and C will move down
• Since they are connected together with string (inextensible and light), both will move with the same acceleration
• We are required to find this acceleration. 
• The strings are marked as PQ and RS. We are required to find the tensions in those strings also
• The steps are given below:
1. First we draw the required FBDs. The sub-systems selected are shown in fig.b  
2. Consider the FBD of A shown in fig.c
The forces acting on A are: $\mathbf\small{\vec{T_{PQ}} \: \text {and}\: \vec{W_A}}$
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 $\mathbf\small{\vec{T_{PQ}}}$  
• Let $\mathbf\small{\vec{F_2}}$ be $\mathbf\small{\vec{W_A}}$
• Taking upward forces as positive and downward forces as negative, the equation in (3) becomes:
Net force = $\mathbf\small{\vec{T_{PQ}}-\vec{W_A}}$ 
5. According to Newton's second law, net force is equal to (mass ×acceleration). So we can write:
$\mathbf\small{\vec{T_{PQ}}-\vec{W_A}=m_A \times \vec{a}}$
$\mathbf\small{\Rightarrow (|\vec{T_{PQ}}|-|\vec{W_A}|)\hat{j}=(m_A \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow |\vec{T_{PQ}}|-|\vec{W_A}|=m_A \times |\vec{a}|}$
• Substituting the values we get:
$\mathbf\small{|\vec{T_{PQ}}|-20=2 \times |\vec{a}|}$
6. Next consider the FBD of B shown in fig.d
• The forces acting on B are: $\mathbf\small{\vec{T_{PQ}},\: \vec{T_{RS}} \: \text {and}\: \vec{W_B}}$

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Note that, $\mathbf\small{\vec{T_{PQ}}}$ will be same on both sides of the pulley. We know the reason. However, we will write it again:
• The tension in a light inextensible string will be the same through out it's whole length
• So the string will be pulling at both the ends with the same force
• This force will not change even if a pulley (which is friction less) is introduced between the ends of the string

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7. Net force on a 'body which is acted upon by three forces' is given by the vector sum: $\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}}$
8. Let $\mathbf\small{\vec{F_1}}$ be $\mathbf\small{\vec{T_{PQ}}}$  
• Let $\mathbf\small{\vec{F_2}}$ be $\mathbf\small{\vec{W_B}}$
• Let $\mathbf\small{\vec{F_3}}$ be $\mathbf\small{\vec{T_{RS}}}$
• Taking upward forces as positive and downward forces as negative, the equation in (7) becomes:
Net force = $\mathbf\small{\vec{T_{PQ}}-\vec{W_B}-\vec{T_{RS}}}$
9. According to Newton's second law, net force is equal to (mass ×acceleration). So we can write:
$\mathbf\small{\vec{T_{PQ}}-\vec{W_B}-\vec{T_{RS}}=-m_B \times \vec{a}}$
(The right side gets a netative sign because, the net force on B is in the downward direction)
$\mathbf\small{\Rightarrow (|\vec{T_{PQ}}|-|\vec{W_B}|-|\vec{T_{RS}}|)\hat{j}=-(m_B \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow |\vec{T_{PQ}}|-|\vec{W_B}|-|\vec{T_{RS}}|=-m_B \times |\vec{a}|}$
• Substituting the values we get:
$\mathbf\small{|\vec{T_{PQ}}|-|\vec{T_{RS}}|-30=-3 \times |\vec{a}|}$
10. Next consider the FBD of C shown in fig.d
• The forces acting on C are: $\mathbf\small{\vec{T_{RS}} \: \text {and}\: \vec{W_C}}$
11. Net force on a 'body which is acted upon by two forces' is given by the vector sum: $\mathbf\small{\vec{F_1}+\vec{F_2}}$
12. Let $\mathbf\small{\vec{F_1}}$ be $\mathbf\small{\vec{T_{RS}}}$  
• Let $\mathbf\small{\vec{F_2}}$ be $\mathbf\small{\vec{W_C}}$
• Taking upward forces as positive and downward forces as negative, the equation in (11) becomes:
Net force = $\mathbf\small{\vec{T_{RS}}-\vec{W_C}}$ 
13. According to Newton's second law, net force is equal to (mass ×acceleration). So we can write:
$\mathbf\small{\vec{T_{RS}}-\vec{W_C}=-m_C \times \vec{a}}$
(The right side gets a netative sign because, the net force on C is in the downward direction)
$\mathbf\small{\Rightarrow (|\vec{T_{RS}}|-|\vec{W_C}|)\hat{j}=-(m_C \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow |\vec{T_{RS}}|-|\vec{W_C}|=-m_C \times |\vec{a}|}$
• Substituting the values we get:
$\mathbf\small{|\vec{T_{RS}}|-50=-5 \times |\vec{a}|}$
14. Thus we get three equations with three unknowns  :
• From (5), we have: $\mathbf\small{|\vec{T_{PQ}}|-20=2 \times |\vec{a}|}$
• From (9), we have: $\mathbf\small{|\vec{T_{PQ}}|-|\vec{T_{RS}}|-30=-3 \times |\vec{a}|}$
• From (13), we have: $\mathbf\small{|\vec{T_{RS}}|-50=-5 \times |\vec{a}|}$
• Solving the three equations, we get:
$\mathbf\small{|\vec{a}|}$ = 6 ms^-2 , $\mathbf\small{|\vec{T_{PQ}}|}$ = 32 N and $\mathbf\small{|\vec{T_{RS}}|}$ = 20 N
15. Finally, we have to calculate the reaction in the pulley
• The FBD is shown in fig.e
• Each rope will be pulling downwards with a force of $\mathbf\small{\vec{T_{PQ}}}$
• Given that, the pulley is of light weight. So we need not take it's weight into account  
16. Net force on a 'body which is acted upon by three forces' is given by the vector sum: $\mathbf\small{\vec{F_1}+\vec{F_2}+\vec{F_3}}$
17. Let $\mathbf\small{\vec{F_1}}$ be $\mathbf\small{\vec{R}}$ 
• Let $\mathbf\small{\vec{F_2}}$ be $\mathbf\small{\vec{T_{PQ}}}$   
• Let $\mathbf\small{\vec{F_3}}$ be $\mathbf\small{\vec{T_{PQ}}}$
• Taking upward forces as positive and downward forces as negative, the equation in (16) becomes:
Net force = $\mathbf\small{\vec{R}-\vec{T_{PQ}}-\vec{T_{PQ}}}$
18. But the pulley is at rest. That is., it is in equilibrium. 
• According to Newton's second law, there is no net force acting on the pulley. So we can write:
$\mathbf\small{\vec{R}-\vec{T_{PQ}}-\vec{T_{PQ}}}$ = 0
$\mathbf\small{\Rightarrow \vec{R}=2\, \vec{T_{PQ}}}$
$\mathbf\small{\Rightarrow |\vec{R}|=2 \times |\vec{T_{PQ}}|}$
$\mathbf\small{\Rightarrow |\vec{R}|=2 \times 32=64\, \text{N}}$

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In the next section we will see a few more solved examples

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Chapter 5.15 - Solved examples on Force in Strings

In the previous section we saw the basic details about 'force in string'. We saw the possible 3 cases and saw a solved example related to case 3. In this section we will see some solved examples related to cases 1 and 2.

Solved example 5.24
Three blocks A, B and C are connected vertically as shown in fig.5.53(a) below:

Fig.5.53

Their masses are 3 kg, 5 kg and 7 kg respectively. If the system of blocks is raised upwards by a force of 195 N, what will be the tensions in strings PQ and RS?
Solution:
1. First we draw the FBD of the whole system. This is shown in fig.5.53(b) above
• In the FBD,  there are 2 forces
• The weight W is the total weight of the 3 masses. So we can write:
• $\mathbf\small{|\vec{W}|=(m_A+m_B+m_C)g}$= 150 N
2. Net force acting on the system = $\mathbf\small{\vec{F}-\vec{W}=195\, \hat{j}-150\, \hat{j}=45\, \hat{j}}$
• By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\text{45}\, \hat{j}=\left[(m_A+m_B+m_C)\times |\vec{a}| \right ]\hat{j}}$
$\mathbf\small{\Rightarrow |\vec{a}|=\frac{45}{15}=3\, \text{ms}^{-2}}$
• So the whole system moves upwards with an acceleration of 3 ms^-2
3. Next we draw the FBD of 'C'. This is shown in fig.5.53(c) above
• In the segment RS, two vectors are marked on the rope
• They are marked to help us remember two points:
(i) The segment RS pulls at R
(ii) The segment RS pulls at S
• 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. 
4. In the FBD, the tension in segment RS is denoted as $\mathbf\small{\vec{T_{RS}}}$ 
• So net force acting on 'C' = $\mathbf\small{\vec{T_{RS}}-\vec{W_C}}$
5. By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{{\vec{T_{RS}}-\vec{W_C}=(m_C \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow {\vec{T_{RS}}=\vec{W_C}+(m_C \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow {\vec{T_{RS}}=(m_C \times g)\hat{j}+(m_C \times |\vec{a}|})\hat{j}=[m_C \times (|\vec{a}|+g)]\hat{j}}$
$\mathbf\small{\Rightarrow {|\vec{T_{RS}|}=[m_C \times (|\vec{a}|+g)]}}$
• Substituting the values, we get:
$\mathbf\small{{|\vec{T_{RS}|}=[7 \times (3+10)]}=91\, \text{N}}$
6. Next we draw the FBD of 'B'. This is shown in fig.5.54(a) below:

Fig.5.54

• 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
• 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. 
7. In the FBD, the tension in segment PQ is denoted as $\mathbf\small{\vec{T_{PQ}}}$
• The tension in segment RS is denoted as $\mathbf\small{\vec{T_{RS}}}$
• In addition to the above two, the weight of 'B' will also be acting
• So net force acting on 'B' = $\mathbf\small{\vec{T_{PQ}}-\vec{T_{RS}}-\vec{W_B}}$
8. By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\vec{T_{PQ}}-{\vec{T_{RS}}-\vec{W_B}=(m_B \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow \vec{T_{PQ}}={\vec{T_{RS}}+\vec{W_B}+(m_B \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow \vec{T_{PQ}}={\vec{T_{RS}}+(m_B \times g)\hat{j}+(m_B \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow \vec{T_{PQ}}=\vec{T_{RS}}+\left[m_B \times (|\vec{a}|+g) \right ]\hat{j}}$
$\mathbf\small{\Rightarrow |\vec{T_{PQ}}|=|\vec{T_{RS}}|+\left[m_B \times (|\vec{a}|+g) \right ]}$
• Substituting the values, we get:
$\mathbf\small{|\vec{T_{PQ}}|=91+\left[5 \times (3+10) \right ]=156\, \text{N}}$
Thus we obtained the tensions in both the ropes
Check:
The FBD of 'A' is shown in fig.5.54(b). We have:
• So net force acting on 'A' = $\mathbf\small{\vec{F}-\vec{T_{PQ}}-\vec{W_A}}$
By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\vec{F}-\vec{T_{PQ}}-\vec{W_A}=(m_A \times |\vec{a}|)\hat{j}}$
Substituting the values, we get: $\mathbf\small{195 \hat{j}-156 \hat{j}-30 \hat{j}=(3 \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow|\vec{a}|= 3\, \text{ms}^{-2}}$
• We have already seen that each mass move upwards with an acceleration of 3 ms^-2. So our calculations are correct

Solved example 5.25
Three blocks A, B and C connected horizontally rest on a friction less floor as shown in the fig.5.55(a) below:

Fig.5.55

Their masses are 7 kg, 5 kg and 2 kg respectively. If the system of blocks is moved towards the right by a force of 56 N applied on 'C', what will be the tensions in strings PQ and RS?
Solution:
1. First we draw the FBD of the whole system. This is shown in fig.5.55(b) above
• In the FBD,  there is only 1 force
(The vertical forces will all cancel each other, and hence are not shown) 
2. Net force acting on the system = $\mathbf\small{56\, \hat{j}}$
• By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\text{56}\, \hat{j}=\left[(m_A+m_B+m_C)\times |\vec{a}| \right ]\hat{j}}$
$\mathbf\small{\Rightarrow |\vec{a}|=\frac{56}{14}=4\, \text{ms}^{-2}}$
• So the whole system moves towards the right with an acceleration of 4 ms^-2
3. Next we draw the FBD of 'A'. This is shown in fig.5.56 below:

Fig.5.56

• 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
• 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. 
4. In the FBD, the tension in segment PQ is denoted as $\mathbf\small{\vec{T_{PQ}}}$ 
• So net force acting on 'A' = $\mathbf\small{\vec{T_{PQ}}}$
5. By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{{\vec{T_{PQ}}=(m_A \times |\vec{a}|})\hat{j}}$
$\mathbf\small{\Rightarrow {\vec{T_{PQ}}=(m_A \times |\vec{a}|})\hat{j}=28\, \hat{j}}$
$\mathbf\small{\Rightarrow {|\vec{T_{PQ}}|=28}\, \text{N}}$
6. Next we draw the FBD of 'B'. This is shown in fig.5.57 below:

Fig.5.57

• In the segment RS, two vectors are marked on the rope
• They are marked to help us remember two points:
(i) The segment RS pulls at R
(ii) The segment RS pulls at S
• 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. 
7. In the FBD, the tension in segment RS is denoted as $\mathbf\small{\vec{T_{RS}}}$ 
• So net force acting on 'B' = $\mathbf\small{\vec{T_{RS}}-\vec{T_{PQ}}}$
8. By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\vec{T_{RS}}-\vec{T_{PQ}}=(m_B \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow \vec{T_{RS}}=\vec{T_{PQ}}+(m_B \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow \vec{T_{RS}}=28\, \hat{j}+20\, \hat{j}}$
$\mathbf\small{\Rightarrow |\vec{T_{RS}}|=48\, \text{N}}$
Thus we obtained the tensions in both the ropes

Check:
The FBD of 'C' is shown in fig.5.58 below:

Fig.5.58

• So net force acting on 'C' = $\mathbf\small{\vec{F}-\vec{T_{RS}}}$
By Newton's second law, we have: Net force = ma
• So we have: $\mathbf\small{\vec{F}-\vec{T_{RS}}=(m_C \times |\vec{a}|)\hat{j}}$
Substituting the values, we get: $\mathbf\small{56 \hat{j}-48 \hat{j}=(2 \times |\vec{a}|)\hat{j}}$
$\mathbf\small{\Rightarrow|\vec{a}|= 4\, \text{ms}^{-2}}$
• We have already seen that each mass move with an acceleration of 4 ms^-2. So our calculations are correct

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Now we will see a solved example based on case 3

Solved example 5.26
A block 'A' of mass 10 kg is suspended using 3 strings QP, QR and QS as shown in the fig.5.59(a) below. Find the tensions in the strings.

Fig.5.59

Solution:
1. First we draw the FBD of 'A'
• So 'A' is taken as the sub-system. A red rectangle is drawn around 'A' in fig.5.59(b)
• What ever is enclosed in the red rectangle, should be used in the FBD 
• '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 $\small{\vec{T_{QS}}}$ 
    ♦ Let $\small{\vec{F_2}}$ indicate $\small{\vec{W_A}}$ 

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

$\mathbf\small{(|\vec{T_{QS}}|)\hat{j}-(|\vec{W_A}|)\hat{j}=0}$

$\mathbf\small{\Rightarrow (|\vec{T_{QS}}|)\hat{j}-(m_A \times g)\hat{j}=0}$

$\mathbf\small{\Rightarrow (|\vec{T_{QS}}|)\hat{j}=(m_A \times g)\hat{j}}$

$\mathbf\small{\Rightarrow |\vec{T_{QS}}|=m_A \times g= 10 \times 10 = 100\,N}$
• Direction of $\small{\vec{T_{QS}}}$ is obviously, 'vertical and upwards' 
2. Now we draw the FBD of a sub-system around the point 'Q'
This is shown in fig.5.60 below:

Fig.5.60

• What ever is enclosed in the red rectangle, should be used in the FBD.
3. In fig.b, 'Q' 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 $\small{\vec{T_{PQ}}}$ 
    ♦ Let $\small{\vec{F_2}}$ indicate $\small{\vec{T_{QR}}}$  
    ♦ Let $\small{\vec{F_3}}$ indicate $\small{\vec{T_{QS}}}$
■ Here two forces are 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:
$\mathbf\small{-(|\vec{T_{PQ}}|\cos 60 )\hat{i}+(|\vec{T_{QR}}|\cos 30 )\hat{i}+\,0=0}$
(The last term is zero because, $\small{\vec{T_{QS}}}$ does not have a horizontal component)
$\mathbf\small{\Rightarrow |\vec{T_{PQ}}|\cos 60=|\vec{T_{QR}}|\cos 30}$
$\mathbf\small{\Rightarrow |\vec{T_{PQ}}|=\sqrt{3}\, |\vec{T_{QR}}|}$
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:
$\mathbf\small{(|\vec{T_{PQ}}|\sin 60 )\hat{j}+(|\vec{T_{QR}}|\sin 30 )\hat{j}-(|\vec{T_{QS}}|)\hat{j}=0}$
$\mathbf\small{\Rightarrow (|\vec{T_{PQ}}|\sin 60 +|\vec{T_{QR}}|\sin 30 )\hat{j}=(100)\hat{j}}$
$\mathbf\small{\Rightarrow (|\vec{T_{PQ}}|\sin 60 +|\vec{T_{QR}}|\sin 30 )=100}$
$\mathbf\small{\Rightarrow \sqrt{3}\, |\vec{T_{PQ}}|+|\vec{T_{QR}}|=200}$
10. But from (8) we have: $\mathbf\small{|\vec{T_{PQ}}|=\sqrt{3}\, |\vec{T_{QR}}|}$
• So the result in (9) becomes: $\mathbf\small{ \sqrt{3}\, (\sqrt{3}\,|\vec{T_{QR}}|)+|\vec{T_{QR}}|=200}$
$\mathbf\small{\Rightarrow 4\,|\vec{T_{QR}}|=200}$
$\mathbf\small{\Rightarrow |\vec{T_{QR}}|=50 \,\text{N}}$
11. So from (8) we get: $\mathbf\small{|\vec{T_{PQ}}|=50\sqrt{3}\,\text{N}}$

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In the next section we will see strings through pulleys

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