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 ?
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:
♦ 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 'mBg' 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 'mBg'
• In addition to $\small{\vec{T}}$, gravity exerts a downward pulling force on the man
♦ The magnitude of this downward pulling force is 'mMg'
• 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:
♦ 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 'mBg' 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 'mBg'
♦ (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 'mMg'
• 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.
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
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:
• 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 (mAg 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
• 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
• 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
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 |
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 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 'mBg' 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 'mBg'
• In addition to $\small{\vec{T}}$, gravity exerts a downward pulling force on the man
♦ The magnitude of this downward pulling force is 'mMg'
• 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 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 'mBg' 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 'mBg'
♦ (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 'mMg'
• 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 |
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 |
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 |
• 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 (mAg 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
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| Fig.5.71 |
♦ 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
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| Fig.5.72 |
(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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