Chapter 7.8 - Force Acting on a System of Particles

In the previous section, we saw the velocity and acceleration of the 'C'. In this section we will see it's momentum and force

1. Let a system consist of two particles P and Q
• Let both P and Q be in motion
2. Consider any instant during that motion
• In earlier sections of this chapter, we have seen that, all particles in the system need not be having the same velocity at any instant (Details here)
• So let us assume that, at that instant, P and Q are having different velocities
3. Consider the instant at which the reading in the stop-watch is 't'
    ♦ Let at that instant, the velocity vector of P be $\mathbf\small{\vec{v}_{P(t)}}$
4. If we multiply this velocity by 'mass of P', we will get the momentum of P at that instant
So we can write: $\mathbf\small{\vec{p}_{P(t)}=m_P\,\vec{v}_{P(t)}}$  
5. Similarly, if the velocity of Q at that instant is $\mathbf\small{\vec{v}_{Q(t)}}$, we can write it's momentum:
$\mathbf\small{\vec{p}_{Q(t)}=m_Q\,\vec{v}_{Q(t)}}$
6. Sum of the two momenta will give the 'total momentum of the system'. So we can write:
$\mathbf\small{\vec{p}_{System(t)}=m_P\,\vec{v}_{P(t)}+m_Q\,\vec{v}_{Q(t)}}$
7. In general, if there are n particles, we can write:
$\mathbf\small{\vec{p}_{System(t)}=m_1 \,\, \vec{v}_{1(t)}+m_2 \,\, \vec{v}_{2(t)}+ \,\,.\,\,.\,\,.+m_i \,\, \vec{v}_{i(t)}+\,\,.\,\,.\,\,.+m_n \,\, \vec{v}_{n(t)}}$
8. Consider the right side of the above equation. We have seen it before
• In Eq.7.6 of the previous section, we wrote:
$\mathbf\small{M\,\,\vec{v}_{C(t)}=m_1 \,\, \vec{v}_{1(t)}+m_2 \,\, \vec{v}_{2(t)}+ \,\,.\,\,.\,\,.+m_i \,\, \vec{v}_{i(t)}+\,\,.\,\,.\,\,.+m_n \,\, \vec{v}_{n(t)}}$
9. Comparing (7) and (8), we can write:
Eq.7.11:
$\mathbf\small{\vec{p}_{System(t)}=M\,\,\vec{v}_{C(t)}}$
■ Based on this equation, we can write:
The total momentum of a system of particles is equal to the product of two items:
(i) The total mass of the system
(ii) Velocity of the 'C'

---

Force on a system

1. Let there be n particles in a system
• Let the system be in motion
2. Consider the instant at which the reading in the stop-watch is 't₁'
    ♦ Let at that instant, the total momentum of the system be $\mathbf\small{\vec{p}_{System(t1)}}$
• Consider the instant at which the reading in the stop-watch is 't₂'
    ♦ Let at that instant, the total momentum of the system be $\mathbf\small{\vec{p}_{System(t2)}}$
4. If we subtract the 'initial momentum vector' from the 'final momentum vector', we will get the change in momentum
• So we can write:
The 'change in momentum' suffered by the system during the time interval of (t₂-t₁) 
= $\mathbf\small{\vec{p}_{System(t2)}-\;\vec{p}_{System(t1)}}$
5. If we divide the 'change in momentum' by the 'time interval during which the change took place', we will get the average force (Details here)
• Time interval during which the displacement took place = (t₂-t₁) = Δt
• So we can write:
Average force experienced by the system during the time interval Δt =
$\mathbf\small{\vec{\bar{F}}_{System(\Delta t)}=\frac{\vec{\Delta p}_{System(\Delta t)}}{\Delta t}=\frac{\vec{p}_{System(t2)}-\vec{p}_{System(t1)}}{\Delta t}}$
6. Now we expand the numerator on the right side
Using Eq.7.11, we get:
$\mathbf\small{\vec{p}_{System(t2)}=M\,\,\vec{v}_{C(t2)}}$
$\mathbf\small{\vec{p}_{System(t1)}=M\,\,\vec{v}_{C(t1)}}$
• So the result in (5) becomes:
$\mathbf\small{\vec{\bar{F}}_{System(\Delta t)}=\frac{M\,\,\vec{v}_{C(t2)}-M\,\,\vec{v}_{C(t1)}}{\Delta t}}$
$\mathbf\small{\Rightarrow \vec{\bar{F}}_{System(\Delta t)}=\frac{M[\vec{v}_{C(t2)}-\,\,\vec{v}_{C(t1)}]}{\Delta t}}$
$\mathbf\small{\Rightarrow \vec{\bar{F}}_{System(\Delta t)}=M[\vec{a}_{C(\Delta t)}]}$
7. If we consider an interval of time Δt, we will be getting the average force $\mathbf\small{\vec{\bar{F}}_{System(\Delta t)}}$
• If we consider an instant 't', we will get the instantaneous force $\mathbf\small{\vec{F}_{System(t)}}$
• Thus we can write: 
Eq.7.12: $\mathbf\small{\Rightarrow \vec{F}_{System(t)}=M[\vec{a}_{C(t)}]}$
    ♦ Where $\mathbf\small{\vec{a}_{C(t)}}$ is the acceleration experienced by the 'C' at the instant 't'
• This is in a form familiar to us: Force = mass × acceleration 
8. Based on Eq.7.12, we can write:
• The force experienced by a system of particles is equal to the product of two items:
(i) The total mass of the system
(ii) Acceleration of the 'C'
• Thus we succeeded in extending Newton's second law to a 'system of particles'

---

Conservation of momentum

1. Consider the equation in (5) above. We will write it again:

$\mathbf\small{\vec{\bar{F}}_{System(\Delta t)}=\frac{\vec{p}_{System(t2)}-\vec{p}_{System(t1)}}{\Delta t}}$
• Suppose that, the system experiences no external force. Then $\mathbf\small{\vec{\bar{F}}_{System(\Delta t)}=0}$
• Then the numerator will become zero
• We get: $\mathbf\small{\vec{p}_{System(t2)}-\vec{p}_{System(t1)}=0}$
$\mathbf\small{\Rightarrow \vec{p}_{System(t2)}=\vec{p}_{System(t1)}}$
• That means: Final momentum = Initial momentum  
• That means: There is no change in momentum
• That means: Momentum is a constant
■ Thus we get the law of conservation of momentum for the system of particles. It can be stated as:
If the total external force acting on a system of particles is zero, the total linear momentum of that system is constant
2. But from Eq.7.11, we have: $\mathbf\small{\vec{p}_{System(t)}=M\,\,\vec{v}_{C(t)}}$
• If the left side is constant, the right side must also be constant
• In the right side, we have 'M' and '$\mathbf\small{\vec{v}_{C(t)}}$' 
• 'M' is already a constant because, we assume that, the mass of the system do not change
■ So we can write:
If the total external force acting on a system of particles is zero, the velocity of the 'C' of that system will be constant
• 'Constant velocity' implies that, both magnitude and direction of the velocity does not change
• So we can write:
If the total external force acting on a system of particles is zero, the 'C' of that system will travel at a constant speed along a straight line

---

In the next section, we will some examples where the 'C' moves with constant velocity

PREVIOUS           CONTENTS          NEXT

Copyright©2019 Higher Secondary Physics. blogspot.in - All Rights Reserved

Chapter 6.5 - Kinetic Energy and Work Done - Solved Examples

In the previous section, we saw the work-energy theorem for a variable force. 
• We saw that, even if the force acting on an object is not a steady constant force, the change in kinetic energy will be equal to the total work done by the force on the object. 
• In this section, we will see some solved examples.

Solved example 6.9
A woman pushes a trunk on a railway platform which has a rough surface. She applies a force of 100 N over a distance of 10 m. Thereafter, she gets progressively tired and her applied force reduces linearly with distance to 50 N. The total distance through which the trunk has been moved is 20 m. 
The frictional force is 50 N
(a) Plot the following two graphs:
(i) Force applied by the woman versus displacement
(ii) Frictional force versus displacement
(b) Calculate the work done by the two forces over 20 m
Solution:
Part (a):
1. Upto 10 m, the force is constant. So the graph will be a horizontal line for the first 10 m
2. The total distance is given as 20 m. So for the second 10 m, a variable force is applied
• Given that this force 'reduces linearly'. That means, the variation in the force is 'uniform'
• So the graph will be a sloping line 
3. The frictional force is constant. But it is negative because it acts in a direction opposite to that of the applied force
4. The 3 forces are shown in fig.6.15 below:

Fig.6.15

• The magenta horizontal line represent the force in (1)
• The cyan sloping line represent the force in (2)
• The yellow horizontal line represent the force in (3)
Part (b):
The work done by a force (constant or varying) is given by the enclosed area
The enclosed areas for each force is marked separately in fig.5.16 below:

Fig.6.16

• The work done by the magenta force (constant force applied by the woman) 
= Area of magenta rectangle = 100 × 10 = 1000 J  
• The work done by the cyan force (varying force applied by the woman) 
= Area of cyan trapezium = [^(100+50)⁄₂ × 10] = 75 × 10 =750 J
■ Total work done by the woman = 1000 + 750 = 1750 J
■ The work done by the yellow force (frictional force) 
= Area of yellow rectangle = -50 × 20 = -1000 J

Solved example 6.10
The force F acting on an object moving in a straight line varies with distance as shown in the graph in fig.6.17(a) below. Find the work done during the displacement of 16 m

Fig.6.17

Solution:
1. Work done = Area enclosed by the Force graph
• This area is shown in fig.b
2. So we can write:
• Total work done = Area of magenta rectangle + Area of yellow triangle
    ♦ Area of magenta rectangle = 5 × (9-4) = 5 × 5 = 25 J 
    ♦ Area of yellow triangle = ¹⁄₂ × 5 × (16-9) = ¹⁄₂ × 5 × (7) = 17.5 J
• Thus we get: Total work = 25 + 17.5 = 42.5 J

Solved example 6.11
The force F acting on an object moving in a straight line varies with distance as shown in the graph in fig.6.18(a) below. 

Fig.6.18

Find the work done in the following two cases:
(a) Work done when the object is displaced from x = -3 m to x = 0
(b) Work done when the object is displaced from x = 0 to x = 2 m
Solution:
1. Work done = Area enclosed by the Force graph
• This area is shown in fig.b
2. So we can write:
• Total work done = Area of magenta triangle + Area of yellow triangle + Area of brown trapezium
3. To find those areas, we need their sides
• We can use analytic geometry:
• From the given data in fig.a, the coordinates of the ends of the force graph (cyan line) are:
P(-3,10) and Q(2,10)
• Slope 'm' of the cyan line PQ = $\mathbf\small{\frac{y_2-y_1}{x_2-x_1}=\frac{10-(-10)}{2-(-3)}=\frac{20}{5}=4}$
• Equation of line PQ is given by: $\mathbf\small{y-y_1=m(x-x_1)\Rightarrow y-(-10)=4(x-(-3))}$
$\mathbf\small{\Rightarrow y-(-10)=4(x-(-3))}$
$\mathbf\small{\Rightarrow y+10=4x+12}$
$\mathbf\small{\Rightarrow y=4x+2}$
• PQ intersects the x axis at R
    ♦ Coordinates of R can be obtained by putting y = 0 in the equation of PQ
    ♦ Thus we get: R(-0.5,0)
• PQ intersects the y axis at S
    ♦ Coordinates of S can be obtained by putting x = 0 in the equation of PQ
    ♦ Thus we get: S(0,2)
• So we have all the required points
4. For the magenta triangle:
• Base = (3-0.5) = 2.5
• Height = 10
• So area = ¹⁄₂ × 2.5 × 10 = 12.5 J 
■ This is negative energy because the force corresponding to this area lies below the x axis and so is negative
5. For the yellow triangle:
• Base = 0.5
• Height = 2
• So area = ¹⁄₂ × 0.5 × 2 = 0.5 J
6. For the brown trapezium:
• Height on left side = 2
• Height on right side = 10
• Base = 2
• So area = [⁽²⁺¹⁰⁾⁄₂ × 2] = 6 × 2 =12 J

---

Now we can calculate the total works
Part (a): Work done when the object is displaced from x = -3 m to x = 0
• This is given by:
Area of magenta triangle + Area of yellow triangle
= -12.5+0.5 = -12 J
Part (b): Work done when the object is displaced from x = 0 m to x = 2 m
• This is given by:
Area of brown trapezium = 12 J

Solved example 6.12
The force F acting on an object moving in a straight line varies with distance as shown in the graph in fig.6.17(a) below. Find the work done during the displacement from x = 1 to x = 5 m

Fig.6.19

Solution:
1. The area enclosed by the given graph (upto x = 5 m) can be considered to be made up of rectangles 
2. The various areas are:
Area 1: 20 × 1 = 20 J (+ve) 
Area 2: 40 × 1 = 40 J (+ve)
Area 3: 40 × 1 = 40 J (-ve)
Area 4: 20 × 1 = 20 J (+ve)
3. So work done = 20+40-40+20 = 40 J

Solved example 6.13
A 15 kg block is placed on a rough horizontal surface and is pulled by a horizontal force of 90 N. Coefficient of friction between the block and the floor is 0.4. Find the work done by each individual force acting on the block over a displacement of 8 m [Take g = 10 ms^-2]
Solution:
• The FBD of the block will show all the forces acting on it. This is shown in fig.6.20(a) below:

Fig.6.20

• There are four forces. We will consider each separately
• Note that, $\mathbf\small{\vec{W}}$ and $\mathbf\small{\vec{F_N}}$ will cancel each other. 
    ♦ That means, each of them becomes zero. Work done by a zero force is zero. 
    ♦ However, we will apply the usual procedure (for finding work done) to them also 
1. Force $\mathbf\small{\vec{W}}$
• This is the weight of the block. It acts vertically downwards
• But there is no displacement in the vertical direction. So work done by this force is zero
2. Force $\mathbf\small{\vec{F_N}}$
• This is the normal reaction from the floor. It acts vertically upwards

• But there is no displacement in the vertical direction. So work done by this force is zero
3. Force 90 N
• This force acts towards the right. This force is pulling the block
• The displacement is in the same direction as the force
• So work done = Force × displacement = 90 × 8 = 720 J 
4. The frictional force $\mathbf\small{\vec{f_k}}$ 
• The magnitude of this force is given by μF_N
• So magnitude = 0.4 × 15 × 10 = 60 N
• This force acts in a direction opposite to the displacement
• So we take frictional force as -60 N
• Then work done = Force × displacement = -60 × 8 = -480 J
■ Kinetic energy attained by the block during the displacement of 8 m 
= Net work done = 720-480 = 240 J

Solved example 6.14
A body of mass 2 kg, initially at rest, moves under the action of an applied horizontal force of 7 N on a table with coefficient of kinetic friction 0.1 Compute the
(a) Work done by applied force in 10 s
(b) Work done by friction in 10 s
(c) Work done by the net force on the body in 10 s
(d) Change in kinetic energy of the body in 10 s
[Take g = 9.8 ms^-2]
Solution:
The FBD is shown in fig.6.20(b) above
1. First we want the displacement 'x' that takes place during the 10 s
• The initial velocity v₀ is 0 ms^-1
• The final velocity v has to be calculated
2. We can use the equation: $\mathbf\small{v=v_0 +at}$
3. a = ^{Net force}⁄_mass = ^{Applied force - frictional force}⁄_mass
Frictional force = μF_N = 0.1 × 2 × 9.8 = 1.96 N
So a = ^{7 - 1.96}⁄₂ = ^5.04⁄₂ = 2.52 ms^-2   
4. Entering known values in (2), we get:
v = 0 + 2.52 × 10 = 25.2 ms^-1 
5. Now we can use the equation: $\mathbf\small{v^2-v_0^2=2ax}$
• Entering known values, we get: $\mathbf\small{25.2^2-0^2=2 \times 2.52 \times x}$  
• From this we get: x = 126 m
6. So work done by the applied force = Force × displacement = 7 × 126 = 882 J  
• This is the answer for part (a)
7. Work done by friction = Frictional force × displacement = 1.96 × 126 = 246.96 J
• This is the answer for part (b)
8. Net force = Applied force - frictional force = 7 - 1.96 = 5.04
• Work done by net force = Net force × displacement = 5.04 × 126 = 635.04 J
• This is the answer for part (c)
9. According to work-energy theorem, we have:
• Change in kinetic energy = K_f - K_i = work done by net force
• So change in kinetic energy = work done by net force = 635.04 J (from part (c))
• This is the answer for part (d)

---

In the next section, we will see potential energy

PREVIOUS           CONTENTS          NEXT

Copyright©2019 Higher Secondary Physics. blogspot.in - All Rights Reserved

Chapter 5.14 - Force in String

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

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

Fig.5.45

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

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

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

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

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

4. But how is that possible?

• We will write the answer in steps: 

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

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

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

• Other wise he will be pulled towards A

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

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

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

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

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

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

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

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

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

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

Fig.5.46

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

---

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

(1) The mass of string is negligible

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

(2) The string is 'inextensible'

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

Fig.5.47

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

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

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

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

---

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

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

---

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

There are 3 possible cases:

Case 1: The string is vertical

Case 2: The string is horizontal

Case 3: The string is inclined

We will now see each case in detail:

Case 1: The string is vertical

This 'case 1' has four sub cases 

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

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

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

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

We will see each case in detail:

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

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

Fig.5.48

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

4. So we can write:

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

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

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

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

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

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

• But the direction of acceleration must be reversed

2. We get:

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

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

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

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

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

---

Case 2: The string is horizontal

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

OR

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

(Body supported by other strings will be discussed later) 

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

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

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

Fig.5.49

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

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

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

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

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

Fig.5.50

• A string is attached to it's side

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

• Fig.(b) shows the FBD

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

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

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

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

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

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

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

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

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

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

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

3. We can write:

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

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

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

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

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

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

• The FBD is shown in fig.d

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

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

3. So we can write: 

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

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

---

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

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

Fig.5.51

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

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

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

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

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

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

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

Fig.5.52

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

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

---

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

PREVIOUS           CONTENTS          NEXT

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