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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Chapter 5.14 - Force in String

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

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

Fig.5.45

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

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

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

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

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

4. But how is that possible?

• We will write the answer in steps: 

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

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

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

• Other wise he will be pulled towards A

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

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

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

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

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

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

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

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

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

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

Fig.5.46

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

---

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

(1) The mass of string is negligible

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

(2) The string is 'inextensible'

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

Fig.5.47

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

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

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

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

---

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

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

---

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

There are 3 possible cases:

Case 1: The string is vertical

Case 2: The string is horizontal

Case 3: The string is inclined

We will now see each case in detail:

Case 1: The string is vertical

This 'case 1' has four sub cases 

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

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

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

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

We will see each case in detail:

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

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

Fig.5.48

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

4. So we can write:

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

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

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

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

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

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

• But the direction of acceleration must be reversed

2. We get:

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

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

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

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

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

---

Case 2: The string is horizontal

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

OR

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

(Body supported by other strings will be discussed later) 

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

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

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

Fig.5.49

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

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

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

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

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

Fig.5.50

• A string is attached to it's side

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

• Fig.(b) shows the FBD

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

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

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

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

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

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

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

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

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

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

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

3. We can write:

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

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

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

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

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

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

• The FBD is shown in fig.d

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

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

3. So we can write: 

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

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

---

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

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

Fig.5.51

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

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

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

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

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

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

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

Fig.5.52

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

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

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

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Chapter 5.13 - Normal reaction on Inclined surfaces

In the previous section we saw cases in which:
    ♦ Contact surfaces were all vertical
    ♦ So  the $\mathbf\small{\vec{F_N}}$ were all horizontal
In the section just before that, we saw cases in which:
    ♦ Contact surfaces were all horizontal. 
    ♦ So the $\mathbf\small{\vec{F_N}}$ were all vertical. 
In this section we will see cases in which:
    ♦ Contact surfaces are inclined

1. Consider Fig.5.41(a) below. A wooden block (B) is resting on a wedge shaped block (I). 
• The top surface of this wedge is a smooth inclined plane. 

Fig.5.41

• The plane is inclined at an angle θ to the horizontal
■ Before learning about the interaction between the block and the wedge, we must learn some geometrical features of the wedge. Steps (2) to (4) below enable us to learn those features:
2. The top surface of the wedge is not horizontal. It is inclined
• Draw a perpendicular to this inclined surface. It is the white dashed line in fig.5.41(b)
    ♦ This perpendicular intersects the inclined surface at D
    ♦ This perpendicular intersects the base of the wedge at C
• Draw a vertical through D. It is the green dashed line in fig.b
    ♦ This vertical intersects the base of the wedge at B
3. The base of the wedge is horizontal
• So now we have two right triangles:
    ♦ ⊿ACD right angled at D
    ♦ ⊿BCD right angled at B
4. In ⊿ACD, we have ∠ACD = (90-θ)
• From this we get: In ⊿BCD, ∠BDC = θ
■ This is an important result. It tells us that:
The following two angles are always the same:
(i) The angle between ‘the inclined surface’ and the 'horizontal'
    ♦ In the fig., it is the ∠DAB
(ii) The angle between ‘the normal to the inclined surface’ and the 'vertical'
    ♦ In the fig., it is the ∠CDB
5. Now we can learn about the interaction
• The $\mathbf\small{\vec{W_B}}$ in fig.a, denotes the weight of the block
• It will be always directed verticaly downwards (towards the center of the earth)   
• It's direction will not depend on the inclination of the plane
3. We can resolve $\mathbf\small{\vec{W_B}}$ into it's two perpendicular components
• For the convenience of resolution, let us choose the frame of reference (coordinate axes) in this way (see fig.5.41c):
    ♦ The x axis is parallel to the inclined surface
    ♦ The y axis is perpendicular to the inclined surface
4. So we have a new set of coordinate axes
• Consider fig.d. It shows a vector $\mathbf\small{\vec{F}}$ in the usual set of coordinate axes
• In this case, we know that:
    ♦ x component = $\mathbf\small{\vec{F_x}=(|\vec{F}|\cos \theta)\hat{i} }$ 
    ♦ y component = $\mathbf\small{\vec{F_y}=(|\vec{F}|\sin \theta)\hat{j} }$
(Details about the above results can be seen here)
• So we see that:
    ♦ The component which is adjacent to the angle gets the cosine
    ♦ The other component gets the sine
5. Now consider fig.e. It shows or present case
• Here θ is adjacent to the y axis 
Clearly:    
    ♦ x component = $\mathbf\small{\vec{W_{Bx}}=(|\vec{W_{B}}|\sin \theta)\hat{i} }$
    ♦ y component = $\mathbf\small{\vec{W_{By}}=(|\vec{W_{B}}|\cos \theta)\hat{j} }$
• The two components are shown in fig.5.42(a) below:

Fig.5.42

6. Next step is to draw the FBD of the block
• The component of $\mathbf\small{\vec{W_B}}$ parallel to the x axis is $\mathbf\small{\vec{W_{Bx}}}$ 
    ♦ This force is exerted by the earth. It is a force acting on the block. 
    ♦ So we show it in the FBD
• The component of $\mathbf\small{\vec{W_B}}$ parallel to the y axis is $\mathbf\small{\vec{W_{By}}}$ 
    ♦ This force is exerted by the earth. It is a force acting on the block. 
    ♦ So we show it also in the FBD
• Due to the action $\mathbf\small{\vec{W_{By}}}$ on the inclined surface, the inclined surface produces a reaction. 
• We can denote it as $\mathbf\small{\vec{F_{N(BI)}}}$
    ♦ This $\mathbf\small{\vec{F_{N}}}$ acts on the block. 
    ♦ So we show it in the FBD
• The FBD is now complete. It is shown in the fig.5.42(b)
7. We observe no movement in the y direction
• That is., there is equilibrium in the y direction
• So we can write: $\mathbf\small{\vec{W_{By}}=-\vec{F_{N(BI)}}}$
8. Now we consider the x direction
• There is no force to oppose $\mathbf\small{\vec{W_{Bx}}}$ 
(Remember that, the surface in this case is smooth. So there is no friction)
• So the block will move in the direction of $\mathbf\small{\vec{W_{Bx}}}$
• That means, the block will slide down along the inclined surface
• If there is friction between the ‘bottom surface of the block’ and the 'inclined surface', the block may be able to hold it’s position. This happens when the frictional force is equal in magnitude to $\mathbf\small{\vec{W_{Bx}}}$. We will learn more about  friction in later sections of this chapter

Solved example 5.21
A block of mass 3 kg slides down along a friction less inclined surface. What is the acceleration with which it slides down? Angle of inclination of the surface with the horizontal is 30^o. Take g = 10 ms^-2.
Solution:
1. Let the x axis be parallel to the inclined plane
• Let the y axis be perpendicular to the inclined plane
2. The force which causes the sliding is: 
$\mathbf\small{\vec{W_{Bx}}=(|\vec{W_{B}}|\sin \theta)\hat{i} }$
• Substituting the values, we get:
$\mathbf\small{\vec{W_{Bx}}=(mg \,\sin 30)\hat{i} }$
• So the magnitude of the force which causes the sliding is: mg sin30 = 0.5mg
3. Let 'a' be the acceleration with which the block slides
• Then we have: ma = 0.5mg
⟹ a = 0.5g = 0.5 × 10 = 5 ms^-2
■ Note that, this acceleration does not depend on the mass. But it certainly depends on the 'angle which the inclined surface makes with the horizontal'

Solved example 5.22
Three blocks A, B and C are placed on the smooth inclined surface of a wedge as shown in fig.5.43(a) below:

Fig.5.43

The masses of the blocks A, B and C are 1 kg, 2 kg and 3 kg respectively. External forces of 60 N and 18 N also act on the blocks as shown.  
(a) Find the acceleration of A, B and C
(b) Draw the following FBDs:
(i) FBD of (A+B+C)
(ii) FBD of A
(iii) FBD of B
(iv) FBD of C

Solution:
1. FBD of (A+B+C) is shown in fig.5.43(b)
• The modified frame of reference is also shown
• There is no motion in the y direction. So we need not consider the forces in the that direction. They will cancel each other
2. In the x direction, we have:
Net force = $\mathbf\small{60\, \hat{i}-\vec{W_{(A+B+C)x}}-18\, \hat{i}}$
⟹ Net force = $\mathbf\small{42\, \hat{i}-\vec{W_{(A+B+C)x}}}$
⟹ Net force = $\mathbf\small{42\, \hat{i}-(|\vec{W_{(A+B+C)}|\, \sin \theta )\, \hat{i}}}$
⟹ Net force = $\mathbf\small{42\, \hat{i}-[(m_{(A+B+C)} \times g) \, \sin \theta ]\, \hat{i}}$
• Substituting the values, we get:
• Net force = $\mathbf\small{\left[ 42\, \hat{i}-[(6 \times 10) \, \sin 30 ]\, \hat{i} \right ]=12\, \hat{i}}$
4. Since there is a net force, there will be acceleration
• The net force is the cause of the acceleration of (A+B+C). Applying the second law, we can write:
$\mathbf\small{(m_{(A+B+C)}\times|\vec{a_x}|)\, \hat{i}=12\, \hat{i}}$
$\mathbf\small{\Rightarrow (m_{(A+B+C)}\times|\vec{a_x}|)=12}$
• Substituting the values, we get:
$\mathbf\small{(6\times|\vec{a_x}|)=12}$
$\mathbf\small{\Rightarrow |\vec{a_x}|=2\, \text{ms}^{-1}}$
■ So each of the three blocks move down with an acceleration of 2 ms^-2

---

1. Now we draw the FBD of A. This is shown in the fig.5.44(a) below:

Fig.5.44

• There is no motion in the y direction. So we need not consider the forces in the that direction. They will cancel each other

2. In the x direction, we have:

Net force = $\mathbf\small{\vec{F_{N(AB)}}-\vec{W_{Ax}}-18\, \hat{i}}$

⟹ Net force = $\mathbf\small{\vec{F_{N(AB)}}-(|\vec{W_{A}}|\, \sin \theta )\, \hat{i}-18\, \hat{i}}$

⟹ Net force = $\mathbf\small{\vec{F_{N(AB)}}-(m_A \times g \times\sin \theta )\, \hat{i}-18\, \hat{i}}$

Substituting the values, we get:

Net force = $\mathbf\small{\vec{F_{N(AB)}}-(1 \times 10 \times\sin 30 )\, \hat{i}-18\, \hat{i}}$

⟹ Net force = $\mathbf\small{\vec{F_{N(AB)}}-5\, \hat{i}-18\, \hat{i}}$

⟹ Net force = $\mathbf\small{\vec{F_{N(AB)}}-23\, \hat{i}}$

3. This net force causes the acceleration of A. So we can write:

$\mathbf\small{\vec{F_{N(AB)}}-23\, \hat{i}=m_A \times \vec{a}}$

$\mathbf\small{\Rightarrow \vec{F_{N(AB)}}-23\, \hat{i}=(m_A \times |\vec{a}|)\, \hat{i}}$

$\mathbf\small{\Rightarrow \vec{F_{N(AB)}}-23\, \hat{i}=(1 \times 2)\, \hat{i}}$

$\mathbf\small{\Rightarrow \vec{F_{N(AB)}}= 25\, \hat{i}}$

$\mathbf\small{\Rightarrow |\vec{F_{N(AB)}}|= 25\, \text{N}}$

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1. Next we draw the FBD of B. This is shown in the fig.5.44(b) above

• There is no motion in the y direction. So we need not consider the forces in the that direction. They will cancel each other

2. In the x direction, we have:

Net force = $\mathbf\small{\vec{F_{N(BC)}}-\vec{W_{Bx}}-\vec{F_{N(BA)}}}$

⟹ Net force = $\mathbf\small{\vec{F_{N(BC)}}-(|\vec{W_{B}}|\, \sin \theta )\, \hat{i}-25\, \hat{i}}$

($\mathbf\small{\because\:  |\vec{F_{N(BA)}}|=|\vec{F_{N(AB)}}|= 25\, \text{N}}$)

⟹ Net force = $\mathbf\small{\vec{F_{N(BC)}}-(m_B \times g \times\sin \theta )\, \hat{i}-25\, \hat{i}}$

Substituting the values, we get:

Net force = $\mathbf\small{\vec{F_{N(BC)}}-(2 \times 10 \times\sin \text{30} )\, \hat{i}-25\, \hat{i}}$

⟹ Net force = $\mathbf\small{\vec{F_{N(BC)}}-10\, \hat{i}-25\, \hat{i}}$

⟹ Net force = $\mathbf\small{\vec{F_{N(BC)}}-35\, \hat{i}}$

3. This net force causes the acceleration of B. So we can write:

$\mathbf\small{\vec{F_{N(BC)}}-35\, \hat{i}=m_B \times \vec{a}}$

$\mathbf\small{\Rightarrow \vec{F_{N(BC)}}-35\, \hat{i}=(m_B \times |\vec{a}|)\, \hat{i}}$

$\mathbf\small{\Rightarrow \vec{F_{N(BC)}}-35\, \hat{i}=(2 \times 2)\, \hat{i}}$

$\mathbf\small{\Rightarrow \vec{F_{N(BC)}}= 39\, \hat{i}}$

$\mathbf\small{\Rightarrow |\vec{F_{N(BC)}}|= 39\, \text{N}}$

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1. Finally, we draw the FBD of C. This is shown in the fig.5.44(c) above

• There is no motion in the y direction. So we need not consider the forces in the that direction. They will cancel each other

2. In the x direction, we have:

Net force = $\mathbf\small{60 \, \hat{i}-\vec{W_{Cx}}-\vec{F_{N(CB)}}}$

⟹ Net force = $\mathbf\small{60 \, \hat{i}-(|\vec{W_{C}}|\, \sin \theta )\, \hat{i}-39\, \hat{i}}$

($\mathbf\small{\because\:  |\vec{F_{N(CB)}}|=|\vec{F_{N(BC)}}|= 39\, \text{N}}$)

⟹ Net force = $\mathbf\small{60 \, \hat{i}-(m_B \times g \times\sin \theta )\, \hat{i}-39\, \hat{i}}$ 

Substituting the values, we get:

Net force = $\mathbf\small{60 \, \hat{i}-(3 \times 10 \times\sin \text{30})\, \hat{i}-39\, \hat{i}}$  

⟹ Net force = $\mathbf\small{60 \, \hat{i}-15\, \hat{i}-39\, \hat{i}=6 \, \hat{i}}$

3. This net force causes the acceleration of C. So we can write:

$\mathbf\small{6 \hat{i}=m_C \times \vec{a}}$

$\mathbf\small{\Rightarrow 6 \hat{i}=(m_C \times |\vec{a}|)\, \hat{i}}$

$\mathbf\small{\Rightarrow 6 \hat{i}=(3 \times |\vec{a}|)\, \hat{i}}$
$\mathbf\small{\Rightarrow |\vec{a}|=2\, \text{ms}^{-1}}$
This is indeed true because, we saw that all the 3 blocks move with the same acceleration of 2 ms^-2.

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In the next section we will see force in strings.

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