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.
• 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'
Now we introduce a pulley which is attached to the ceiling. This is shown in fig.5.62(a) below:
• 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
• 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
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 |
• 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'
 |
| Fig.5.62 |
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
■ 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
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'
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
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
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
• 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: $\mathbf\small{\vec{T} \: \text {and}\: \vec{W_B}}$
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]
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: $\mathbf\small{\vec{T_{PQ}},\: \vec{T_{RS}} \: \text {and}\: \vec{W_B}}$
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
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
$\mathbf\small{\Rightarrow \vec{R}=2\, \vec{T_{PQ}}}$
$\mathbf\small{\Rightarrow |\vec{R}|=2 \times |\vec{T_{PQ}}|}$
$\mathbf\small{\Rightarrow |\vec{R}|=2 \times 32=64\, \text{N}}$
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 |
• 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: $\mathbf\small{\vec{T} \: \text {and}\: \vec{W_B}}$
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 |
• 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: $\mathbf\small{\vec{T_{PQ}},\: \vec{T_{RS}} \: \text {and}\: \vec{W_B}}$
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
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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