In the previous section we obtained the relation between force and 'time for which the force acts'.
• We obtained the relation: F∝ 1⁄t
In the sections before that, we obtained the following two relations:
• F ∝ m
• F ∝ (v2-v1)
Combining the three relations, we get:
5.1: $\mathbf\small{F\propto\frac{m(v_2-v_1)}{t}}$
Further steps can be written as follows:
1. Let us expand the expression in 5.1.
We get: $\mathbf\small{F\propto\frac{(mv_2-mv_1)}{t}}$
2. In the numerator, we have 'mv2' and 'mv1'
• The product of mass of a body and it's velocity is called momentum of that body.
♦ It is a vector quantity
♦ It is denoted as $\mathbf\small{\vec{p}}$
• So we can write:
♦ 'mv1' is the initial momentum
♦ 'mv2' is the final momentum
• The difference (mv2-mv1) is the 'change in momentum'
3. So $\mathbf\small{\frac{(mv_2-mv_1)}{t}}$ is the 'rate of change of momentum
Then the expression in 5.1 can be written as:
• F ∝ 'Rate of change of momentum'
■ Newton's second law states that:
The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts
• So, if we can measure the 'rate of change of momentum' of a body, we can obtain the force acting on it.
• There is an easy method for finding the 'rate of change of momentum'
• Consider again the expression in 5.1: $\mathbf\small{F\propto\frac{m(v_2-v_1)}{t}}$
• But $\mathbf\small{\frac{(v_2-v_1)}{t}}$ is acceleration 'a'.
■ So we get: $\mathbf\small{F\propto m\,a}$
• That is., Force is proportional to 'the product of mass and acceleration'
• We can write: $\mathbf\small{F=k\,m\,a}$
♦ Where 'k' is the proportionality constant
• We obtained the relation: F∝ 1⁄t
In the sections before that, we obtained the following two relations:
• F ∝ m
• F ∝ (v2-v1)
Combining the three relations, we get:
5.1: $\mathbf\small{F\propto\frac{m(v_2-v_1)}{t}}$
Further steps can be written as follows:
1. Let us expand the expression in 5.1.
We get: $\mathbf\small{F\propto\frac{(mv_2-mv_1)}{t}}$
2. In the numerator, we have 'mv2' and 'mv1'
• The product of mass of a body and it's velocity is called momentum of that body.
♦ It is a vector quantity
♦ It is denoted as $\mathbf\small{\vec{p}}$
• So we can write:
♦ 'mv1' is the initial momentum
♦ 'mv2' is the final momentum
• The difference (mv2-mv1) is the 'change in momentum'
3. So $\mathbf\small{\frac{(mv_2-mv_1)}{t}}$ is the 'rate of change of momentum
Then the expression in 5.1 can be written as:
• F ∝ 'Rate of change of momentum'
■ Newton's second law states that:
The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts
• So, if we can measure the 'rate of change of momentum' of a body, we can obtain the force acting on it.
• There is an easy method for finding the 'rate of change of momentum'
• Consider again the expression in 5.1: $\mathbf\small{F\propto\frac{m(v_2-v_1)}{t}}$
• But $\mathbf\small{\frac{(v_2-v_1)}{t}}$ is acceleration 'a'.
■ So we get: $\mathbf\small{F\propto m\,a}$
• That is., Force is proportional to 'the product of mass and acceleration'
• We can write: $\mathbf\small{F=k\,m\,a}$
♦ Where 'k' is the proportionality constant
So our next aim is to find 'k'. We will write the steps:
1. First we define a unit for measuring force
• Some examples for units are given below. We use:
♦ 'Meters' to measure length
♦ 'Kilograms' to measure mass
♦ 'Joules' to measure energy
2. We want an unit to measure force also
■ The SI unit for force is 'newton'. It's symbol is 'N'
■ We define '1 N' as: The force required to give an acceleration of '1 ms-2' to a body whose mass is 1 kg
This can be explained as follows:
(i) We have seen that, if a body moves with acceleration, a net force is acting on that body
• Conversely, if a net force acts on a body, it will experience acceleration
• In other words, we can accelerate a body by applying a net force on it
(ii) Now consider a body whose mass is 1 kg
• If we apply a small net force on on that 1 kg mass, it will move with a small acceleration
• If we apply a large net force on on that 1 kg mass, it will move with a large acceleration
(iii) We want that force 'F' which cause the 1 kg mass to move with an acceleration of exactly '1 ms-1'
■ Then we say F = 1 N
• That is., if we see a mass of 1 kg which is moving with an acceleration of 1 ms-2, we can be sure that, a net force of 1 N is acting on it
3. So now we know the definition of the unit 'newton'
• We can use this definition to find 'k'.
• We have: F = k×m×a
• Take a mass of 1 kg.
♦ Taking an exact 1 kg enables us to put m = 1 in the above equation
• Move it with an acceleration of 1 ms-2
♦ This enables us to put a = 1 in the above equation
• If that body of mass 1 kg is moving with an acceleration of 1 ms-2, then surely, 1 N is acting on it
♦ So we can put 1 on the left side
4. Thus we get: 1 = k × 1 × 1
■ So value of k = '1'
• Since the value of 'k' is '1', we no longer need to use it in the equation
• Because, multiplying by '1' will not change the value
• Thus we get the equation for force
Eq.5.2: F = ma
• In the above discussion, we defined 'newton' using a body in motion
• But earlier, we saw that a 'body at rest' can also be acted upon by forces. Let us see how the unit 'newton' applies to such cases.
1. Consider fig.5.3(b) that we saw in a previous section
• The book on the table is pulled down by 'gravitational force'
• Is there any chance that, 'the gravitational force which pulls it down' is 1 N?
2. Let us check:
• We know that, the gravitational force is the reason for the weight of a body.
• Weight is the product of two items: mass of the body and g'
• This weight is the force that pulls the book down
3. We want this force to be equal to 1 N
• So we can write: 1 N = m × 9.81
• Where m is the mass of the book
• Thus m = 1⁄9.81 = 0.102 kg = 102 grams
■ We can write:
If the mass of the book is 102 grams then, a force of 1 N is pulling it down towards the centre of the earth
■ But the book is observed to be at rest. So we can infer that, the table applies an equal force of 1 N in the upward direction
Another example:
• The mass of a medium sized apple is about 100 grams
• Consider a person holding a medium sized apple on his palm. See fig.5.10 below:
• He must hold it without any movement
■ If the mass of that apple is 102 grams, then he is applying an upward force of 1 N
The equation F = m a has a variety of applications. Let us see a few:
(i) We see a body moving with acceleration 'a'
• If we know the mass 'm' of that body, we can easily calculate the force F which is causing that acceleration
(ii) We want to move a body with an acceleration 'a'
• If we know the mass 'm' of that body. we can easily calculate the force F which is required for producing that acceleration
(iii) We want to move a heavy truck with an acceleration 'a'
• We want to move a car with the same acceleration 'a'
• Obviously the forces required will be different
♦ FC = mC × a
♦ FT = mT × a
• We see that FT will be greater than FC
• So we see the reason why a truck has 'a larger and more powerful engine' than that of a car
• Mass is a scalar quantity
• But acceleration is a vector quantity
■ So Force is a vector quantity and it's direction is same as the 'direction of acceleration'
Let us find the direction is some general cases:
Example 1:
1. Consider the wooden block shown in fig.5.11(a) below:
• It is moving with uniform velocity in a straight line
2. In fig.b, a force F acts on it.
• The direction of F is same as the direction of motion
3. As a result of the application of F, the 'uniform nature of motion' will change
• The block will experience acceleration
4. If, even after the application of F, the body continues to move in it's original direction, we can write the following two points:
(i) The body continues it's linear motion
(ii) The 'direction of acceleration' is same as the 'direction of motion'
■ So 'direction of force F' is same as the 'direction of motion'
5. The acceleration a experienced by the block is given by: a = F⁄m
• Where m is the mass of the block
Example 2:
1. Consider the wooden block in fig.5.12(a) below:
• It is moving with uniform velocity in a straight line
2. In fig.b, a force F acts on it.
• The direction of F is not same as the direction of motion.
♦ It makes an angle 'θ' with the horizontal
3. As a result of the application of F, the 'uniform nature of motion' will change
• The block will experience acceleration
4. If, even after the application of F, the body continues to move in it's original direction, we can write the following two points:
(i) The body continues it's linear motion
(ii) The 'direction of acceleration' is same as the 'direction of motion'
5. But there is a difference from Example 1
• The acceleration experienced by the block will be due to the horizontal component (F cos θ) only
• We cannot say that acceleration is due to F
6. So what about the vertical component (F sin θ)
• Doesn't it have any effect?
Ans: There is no motion in the vertical direction because, the floor is firm
• So (F sin θ) will not have any effect on the motion
7. The acceleration 'a' experienced by the block is given by: a = (F cos θ)⁄m
• Where m is the mass of the block
Example 3:
1. Consider the projectile motion that we saw in the previous chapter. Details here
• While a body is in projectile motion, only one external force acts on it
• It is the gravitational force
2. Direction of this force is vertical
• So no force is available in the horizontal direction to affect the projectile
3. That means, there is no acceleration in the horizontal direction
• That is why, the horizontal component of the velocity of the projectile remains constant during it's entire flight
Example 4:
Circular motion
• The direction of force acting on a body which is in uniform circular motion has much significance.
• We will learn it in detail later in this chapter
Example 5:
■ Consider a body at rest
• If a second body moving with a certain velocity collide with it, obviously, it will experience a force
• How much force will it experience? Also what is the direction of the force?
■ Consider a body in motion
• If a second body moving with a certain velocity collide with it, obviously, it will experience a force
• How much force will it experience? Also what is the direction of the force?
• We will see the answers when we learn about 'conservation of momentum' in a later section of this chapter
Now we will see some solved examples
Solved example 5.1
Two forces F1 and F2 act on a body of mass 2 kg as shown in fig.5.13 below:
F1 = 8 N and F2 = 12 N. How much acceleration will the body experience? What is the direction of this acceleration?
Solution:
1. Net force acting on the body = 12-8 = 4 N
• Direction of this net force will be same as the direction of F2
2. We have: F = ma
• Substituting the values, we get: 4 = 2 × a
• So a = 2 ms-2
3. This acceleration will be in the direction of the net force
• But the net force is in the direction of F2
• So a is in the direction of F2
Solved example 5.2
A force of 40 N acts in the direction shown in fig.5.14 below:
The block has a mass of 5 kg and is resting on a smooth firm horizontal surface. What is the acceleration of the block?
Solution:
1. The horizontal component of the force will cause acceleration of the block in the horizontal direction
• The vertical component of the force has no effect because there is no movement in the vertical direction, as the surface is firm
2. The horizontal component is totally available for producing acceleration because there is no friction to resist the motion
• We have F = ma
• Substituting the values, we get:
Horizontal component of the force = m × acceleration in the horizontal direction
⟹ (40 × cos 30) = 5 × a
⟹ a = 6.93 ms-2
Solved example 5.3
A bullet of mass 0.04 kg moving with a speed of 90 ms-1 enters a heavy wooden block and is stopped after a distance of 60 cm. What is the average resistive force exerted by the block on the bullet?
Solution:
1. The bullet is moving with a uniform velocity of 90 ms-1.
• Since the motion is uniform, no force is acting on it at this stage
2. But when it enters the wooden block, it experiences 'resistance'
• Resistance is a 'force acting in the direction opposite to the direction of motion'
• Because of this force, the bullet experiences acceleration/retardation
• Since the force is acting in the opposite direction, we can be sure that, the bullet experiences retardation
3. If we can find the 'magnitude of this retardation (a)', we can calculate the 'retarding force'
• To find the 'a', we can use equations of motion:
• We have: $\mathbf\small{{v_2}^2-{v_1}^2=2as}$
4. Substituting the values, we get: $\mathbf\small{{0}^2-{90}^2=2\times a\times 0.60}$
$\mathbf\small{\Rightarrow a=\frac{-{90}^2}{2\times 0.60}=-6750\: ms^{-2}}$
• The negative sign of the acceleration indicates that, it is opposite to the direction of motion, and hence a retardation
• The negative sign of the force indicates that, it is opposite to the direction of motion, and hence a retarding force
■ We successfully calculated the 'retarding force'. But the following points may be noted:
(i) When the bullet makes the first contact with the block, it begins to experience a retarding force
(ii) But as it continues to travel through the block, the retarding force goes on increasing
(iii) So we cannot say that the bullet experience a constant retarding force of 270 N
(iv) It is only an average force
Solved example 5.4
A constant force acting on a body of mass 3.0 kg changes it’s speed from 2.0 ms-1 to 3.5 ms-1 in 25 s. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force?
Solution:
1. Initial speed v1 = 2
• Final speed v2 = 3.5
• time t = 25 s
2. We have v2 = v1 + at
• Substituting the values, we get: a = 0.06 ms-2
3. Magnitude of the force = ma = 3 × 0.06 = 0.18 N
• Given that, the direction of motion remains unchanged. So the direction of force is same as the direction of motion
Solved example 5.5
A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body
Solution:
1. Given that, the forces are perpendicular to each other
• So the magnitude of the resultant force is given by: $\mathbf\small{|\vec{F}|=\sqrt{|\vec{F_1}|^2+|\vec{F_2}|^2}}$
• Substituting the values, we get: $\mathbf\small{|\vec{F}|=\sqrt{6^2+8^2}=\sqrt{100}=10\: N}$
2. The direction is given by $\mathbf\small{\theta =\tan^{-1}\left(\frac{|\vec{F_1}|}{|\vec{F_2}|}\right )}$
• Where θ is the angle made by the resultant with the force in the denominator
• Substituting the values, we get: $\mathbf\small{\theta =\tan^{-1}\left(\frac{6}{8}\right )=\tan^{-1}\left(\frac{3}{4}\right )=37.87\, ^o}$
3. So the resultant force has a magnitude of 10 N
• It makes an angle of 37.87o with the 8 N force
4. Given that, mass m = 5 kg
• So acceleration = F⁄m = 10⁄5 = 2 ms-2
• This acceleration makes an angle of 37.87o with the direction of the 8 N force
Solved example 5.6
A rocket with a lift-off mass of 20000 kg is blasted upwards with an initial acceleration of 5 ms-2. Calculate the initial force of the blast. [Take g = 10 ms-2]
Solution:
1. The ‘lift-off mass’ is the mass that the rocket will be having at the time of blast
• It moves upwards with an acceleration of 5
• Since there is acceleration, there must be a net force acting on the rocket
2. Let the ‘force of blast’ be F
• This F acts in the upward direction
3. Let the weight of the rocket be W.
• This W acts in the downward direction
♦ It is given by: W = mg = 20000 × 10 = 200000 N
4. Net force = F-W
5. We have acceleration a = Net force⁄Mass
• Substituting the values, we get: 5 = (F-W)⁄20000 = (F-200000)⁄20000
⇒ (F - 200000) = (5 × 20000) = 100000
⇒ F = (100000+200000) = 300000 N = 3 × 105 N
Solved example 5.6
A body of mass 0.4 kg moving initially with a constant speed of 10 ms-1 to the north is subjected to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0 s, the position of the body at that time to be x = 0, and predict it’s position at t = -5 s, 25 s, 100 s.
Solution:
• For convenience, let us consider the north to be towards the right and the south towards the left.
• Based on these directions, we can draw a 'position graph'. This is shown in fig.5.15(a) below:
Part (a):
• We want the position at t = -5 s
1. At the instant when the 8 N force is applied, a stop watch is turned on.
• So, the reading in the stop watch at that instant will be zero.
• That is., at the instant when 8 N is applied, t = 0
2. The position of the body at that instant is marked as the origin O
• At the origin, x = 0
♦ Distances to the right of O will be positive
♦ Distances to the left of O will be negative
3. Remember that, the body was already in motion towards the north.
• The force was applied while it was moving
4. So the body has already traveled some distance before t = 0
• How much distance did it move (for a duration of 5 s) just before reaching O?
Ans: It was moving with a uniform velocity of 10 ms-1 upto O
• So it would travel (10 × 5) = 50 m for those 5 seconds
• Thus, the position would be 50 m to the left of O
■ On the graph, it would be -50 m. It is shown as 'A' in fig.5.15(b) above
Part (b):
• We want the position at t = 25 s
1. The direction of the force is from north to south. It is opposite to the direction of motion
• So the body would not be able to travel towards the north
2. But, due to it's initial velocity which is directed towards the north, it would continue it's northward motion for 'some more time'
• We want this 'some more time' for which the body continues it's motion (beyond O) in the north direction
• We can use the equation: v2 = v1 + at
3. For using the above equation, we want to know the acceleration
• The body is experiences an acceleration due to the force of 8 N acting on it
• Using F= ma, we get: a = F⁄m = 8⁄0.4 = 20 ms-2 .
• Substituting the values in (2), we get: 0 = 10 -20t
⇒ t = 0.5 s
4. Now we want the distance traveled during this 0.5 s
• We can use the equation: s = v1t + 1⁄2 at2
• Substituting the values, we get: s = 10×0.5 - 1⁄2 × 20 ×0.52 = 2.5 m
• So at t = 0.5 s, the body is at a point 2.5 m to the right of O
■ This is shown as 'B' in the above fig.5.15(c) above
5. What we actually want is the position at t = 25 s
• That is., we want the distance traveled by the body for 25 s since it passed O
• That is., we want the distance traveled in 24.5 s, since it passed ‘B’
6. After passing ‘B’, the body would be traveling south
• ‘Traveling south’ means, traveling in the same direction as the direction of the force
• We can use the equation: s = v1t + 1⁄2 at2
• Substituting the values, we get: s = 0×24.5 + 1⁄2 × 20 ×24.52 = 6002.5 m.
[Note that, for the travel from B towards the south, the initial velocity is zero]
7. So the distance is 6002.5 m from B
• Thus the distance from O is (6002.5-2.5) = 6000 m
• The position on the graph is -6000 m
■ This is shown as 'C' in the fig.5.15(d)
Part (c):
We want the position at t = 100 s
1. But the force acts for only 30 s
• So we first want the position at t = 30 s
• That is., we want the distance traveled in 29.5 s, since it passed ‘B’
2. After passing ‘B’, the body would be traveling south
• ‘Traveling south’ means, traveling in the same direction as the direction of the force
• We can use the equation: s = v1t + 1⁄2 at2
• Substituting the values, we get: s = 0×29.5 + 1⁄2 × 20 ×29.52 = 8702.5 m
[Note that, for the travel from B towards the south, the initial velocity is zero]
3. So the distance is 8702.5 m from B
• Thus the distance from O is (8702.5-2.5) = 8700 m
• The position on the graph is -8700 m
■This is shown as 'D' in the fig.5.15(e)
4. So, after D, the body travels for 70 more seconds
• But during those 70 s, there is no force acting on the body
• That means, the body travels with uniform velocity
• That means, after 'D', the body travels with uniform velocity
5. So we want the velocity attained by the body when it reaches D
• For that, consider the travel from B to D
• The time for this travel is 29.5 s
• Initial velocity of this travel is 0
• The acceleration during this travel is 20 ms-2.
• We can use the equation: v2 = v1 + at
• Substituting the values we get: v2 = 0 + 20×29.5 = 590 ms-1
• So, at D, the velocity is 590 ms-1
6. For the next 70 s, the body will be travelling with a uniform velocity of 590 ms-1.
• So distance traveled = 590 ×70 = 41300 m
7. So the distance from D to the last point E is 41300 m
• Thus the position of E on the graph = -(DE+OD) = -(41300+8700) = -50000 m
■ This is shown in fig.5.15(f)
■ A, C and E are the required points. They are shown in red color
1. First we define a unit for measuring force
• Some examples for units are given below. We use:
♦ 'Meters' to measure length
♦ 'Kilograms' to measure mass
♦ 'Joules' to measure energy
2. We want an unit to measure force also
■ The SI unit for force is 'newton'. It's symbol is 'N'
■ We define '1 N' as: The force required to give an acceleration of '1 ms-2' to a body whose mass is 1 kg
This can be explained as follows:
(i) We have seen that, if a body moves with acceleration, a net force is acting on that body
• Conversely, if a net force acts on a body, it will experience acceleration
• In other words, we can accelerate a body by applying a net force on it
(ii) Now consider a body whose mass is 1 kg
• If we apply a small net force on on that 1 kg mass, it will move with a small acceleration
• If we apply a large net force on on that 1 kg mass, it will move with a large acceleration
(iii) We want that force 'F' which cause the 1 kg mass to move with an acceleration of exactly '1 ms-1'
■ Then we say F = 1 N
• That is., if we see a mass of 1 kg which is moving with an acceleration of 1 ms-2, we can be sure that, a net force of 1 N is acting on it
3. So now we know the definition of the unit 'newton'
• We can use this definition to find 'k'.
• We have: F = k×m×a
• Take a mass of 1 kg.
♦ Taking an exact 1 kg enables us to put m = 1 in the above equation
• Move it with an acceleration of 1 ms-2
♦ This enables us to put a = 1 in the above equation
• If that body of mass 1 kg is moving with an acceleration of 1 ms-2, then surely, 1 N is acting on it
♦ So we can put 1 on the left side
4. Thus we get: 1 = k × 1 × 1
■ So value of k = '1'
• Since the value of 'k' is '1', we no longer need to use it in the equation
• Because, multiplying by '1' will not change the value
• Thus we get the equation for force
Eq.5.2: F = ma
• In the above discussion, we defined 'newton' using a body in motion
• But earlier, we saw that a 'body at rest' can also be acted upon by forces. Let us see how the unit 'newton' applies to such cases.
1. Consider fig.5.3(b) that we saw in a previous section
• The book on the table is pulled down by 'gravitational force'
• Is there any chance that, 'the gravitational force which pulls it down' is 1 N?
2. Let us check:
• We know that, the gravitational force is the reason for the weight of a body.
• Weight is the product of two items: mass of the body and g'
• This weight is the force that pulls the book down
3. We want this force to be equal to 1 N
• So we can write: 1 N = m × 9.81
• Where m is the mass of the book
• Thus m = 1⁄9.81 = 0.102 kg = 102 grams
■ We can write:
If the mass of the book is 102 grams then, a force of 1 N is pulling it down towards the centre of the earth
■ But the book is observed to be at rest. So we can infer that, the table applies an equal force of 1 N in the upward direction
Another example:
• The mass of a medium sized apple is about 100 grams
• Consider a person holding a medium sized apple on his palm. See fig.5.10 below:
 |
| Fig.5.10 |
■ If the mass of that apple is 102 grams, then he is applying an upward force of 1 N
The equation F = m a has a variety of applications. Let us see a few:
(i) We see a body moving with acceleration 'a'
• If we know the mass 'm' of that body, we can easily calculate the force F which is causing that acceleration
(ii) We want to move a body with an acceleration 'a'
• If we know the mass 'm' of that body. we can easily calculate the force F which is required for producing that acceleration
(iii) We want to move a heavy truck with an acceleration 'a'
• We want to move a car with the same acceleration 'a'
• Obviously the forces required will be different
♦ FC = mC × a
♦ FT = mT × a
• We see that FT will be greater than FC
• So we see the reason why a truck has 'a larger and more powerful engine' than that of a car
Direction of a force
■ We have F = ma• Mass is a scalar quantity
• But acceleration is a vector quantity
■ So Force is a vector quantity and it's direction is same as the 'direction of acceleration'
Let us find the direction is some general cases:
Example 1:
1. Consider the wooden block shown in fig.5.11(a) below:
 |
| Fig.5.11 |
2. In fig.b, a force F acts on it.
• The direction of F is same as the direction of motion
3. As a result of the application of F, the 'uniform nature of motion' will change
• The block will experience acceleration
4. If, even after the application of F, the body continues to move in it's original direction, we can write the following two points:
(i) The body continues it's linear motion
(ii) The 'direction of acceleration' is same as the 'direction of motion'
■ So 'direction of force F' is same as the 'direction of motion'
5. The acceleration a experienced by the block is given by: a = F⁄m
• Where m is the mass of the block
Example 2:
1. Consider the wooden block in fig.5.12(a) below:
 |
| Fig.5.12 |
2. In fig.b, a force F acts on it.
• The direction of F is not same as the direction of motion.
♦ It makes an angle 'θ' with the horizontal
3. As a result of the application of F, the 'uniform nature of motion' will change
• The block will experience acceleration
4. If, even after the application of F, the body continues to move in it's original direction, we can write the following two points:
(i) The body continues it's linear motion
(ii) The 'direction of acceleration' is same as the 'direction of motion'
5. But there is a difference from Example 1
• The acceleration experienced by the block will be due to the horizontal component (F cos θ) only
• We cannot say that acceleration is due to F
6. So what about the vertical component (F sin θ)
• Doesn't it have any effect?
Ans: There is no motion in the vertical direction because, the floor is firm
• So (F sin θ) will not have any effect on the motion
7. The acceleration 'a' experienced by the block is given by: a = (F cos θ)⁄m
• Where m is the mass of the block
Example 3:
1. Consider the projectile motion that we saw in the previous chapter. Details here
• While a body is in projectile motion, only one external force acts on it
• It is the gravitational force
2. Direction of this force is vertical
• So no force is available in the horizontal direction to affect the projectile
3. That means, there is no acceleration in the horizontal direction
• That is why, the horizontal component of the velocity of the projectile remains constant during it's entire flight
Example 4:
Circular motion
• The direction of force acting on a body which is in uniform circular motion has much significance.
• We will learn it in detail later in this chapter
Example 5:
■ Consider a body at rest
• If a second body moving with a certain velocity collide with it, obviously, it will experience a force
• How much force will it experience? Also what is the direction of the force?
■ Consider a body in motion
• If a second body moving with a certain velocity collide with it, obviously, it will experience a force
• How much force will it experience? Also what is the direction of the force?
• We will see the answers when we learn about 'conservation of momentum' in a later section of this chapter
Now we will see some solved examples
Solved example 5.1
Two forces F1 and F2 act on a body of mass 2 kg as shown in fig.5.13 below:
 |
| Fig.5.13 |
Solution:
1. Net force acting on the body = 12-8 = 4 N
• Direction of this net force will be same as the direction of F2
2. We have: F = ma
• Substituting the values, we get: 4 = 2 × a
• So a = 2 ms-2
3. This acceleration will be in the direction of the net force
• But the net force is in the direction of F2
• So a is in the direction of F2
Solved example 5.2
A force of 40 N acts in the direction shown in fig.5.14 below:
 |
| Fig.5.14 |
Solution:
1. The horizontal component of the force will cause acceleration of the block in the horizontal direction
• The vertical component of the force has no effect because there is no movement in the vertical direction, as the surface is firm
2. The horizontal component is totally available for producing acceleration because there is no friction to resist the motion
• We have F = ma
• Substituting the values, we get:
Horizontal component of the force = m × acceleration in the horizontal direction
⟹ (40 × cos 30) = 5 × a
⟹ a = 6.93 ms-2
Solved example 5.3
A bullet of mass 0.04 kg moving with a speed of 90 ms-1 enters a heavy wooden block and is stopped after a distance of 60 cm. What is the average resistive force exerted by the block on the bullet?
Solution:
1. The bullet is moving with a uniform velocity of 90 ms-1.
• Since the motion is uniform, no force is acting on it at this stage
2. But when it enters the wooden block, it experiences 'resistance'
• Resistance is a 'force acting in the direction opposite to the direction of motion'
• Because of this force, the bullet experiences acceleration/retardation
• Since the force is acting in the opposite direction, we can be sure that, the bullet experiences retardation
3. If we can find the 'magnitude of this retardation (a)', we can calculate the 'retarding force'
• To find the 'a', we can use equations of motion:
• We have: $\mathbf\small{{v_2}^2-{v_1}^2=2as}$
4. Substituting the values, we get: $\mathbf\small{{0}^2-{90}^2=2\times a\times 0.60}$
$\mathbf\small{\Rightarrow a=\frac{-{90}^2}{2\times 0.60}=-6750\: ms^{-2}}$
• The negative sign of the acceleration indicates that, it is opposite to the direction of motion, and hence a retardation
5. Now, according to Newton's second law of motion, the retarding force
$\mathbf\small{F=ma=[0.04\times (-6750)]=-270\:N}$• The negative sign of the force indicates that, it is opposite to the direction of motion, and hence a retarding force
■ We successfully calculated the 'retarding force'. But the following points may be noted:
(i) When the bullet makes the first contact with the block, it begins to experience a retarding force
(ii) But as it continues to travel through the block, the retarding force goes on increasing
(iii) So we cannot say that the bullet experience a constant retarding force of 270 N
(iv) It is only an average force
Solved example 5.4
A constant force acting on a body of mass 3.0 kg changes it’s speed from 2.0 ms-1 to 3.5 ms-1 in 25 s. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force?
Solution:
1. Initial speed v1 = 2
• Final speed v2 = 3.5
• time t = 25 s
2. We have v2 = v1 + at
• Substituting the values, we get: a = 0.06 ms-2
3. Magnitude of the force = ma = 3 × 0.06 = 0.18 N
• Given that, the direction of motion remains unchanged. So the direction of force is same as the direction of motion
Solved example 5.5
A body of mass 5 kg is acted upon by two perpendicular forces 8 N and 6 N. Give the magnitude and direction of the acceleration of the body
Solution:
1. Given that, the forces are perpendicular to each other
• So the magnitude of the resultant force is given by: $\mathbf\small{|\vec{F}|=\sqrt{|\vec{F_1}|^2+|\vec{F_2}|^2}}$
• Substituting the values, we get: $\mathbf\small{|\vec{F}|=\sqrt{6^2+8^2}=\sqrt{100}=10\: N}$
2. The direction is given by $\mathbf\small{\theta =\tan^{-1}\left(\frac{|\vec{F_1}|}{|\vec{F_2}|}\right )}$
• Where θ is the angle made by the resultant with the force in the denominator
• Substituting the values, we get: $\mathbf\small{\theta =\tan^{-1}\left(\frac{6}{8}\right )=\tan^{-1}\left(\frac{3}{4}\right )=37.87\, ^o}$
3. So the resultant force has a magnitude of 10 N
• It makes an angle of 37.87o with the 8 N force
4. Given that, mass m = 5 kg
• So acceleration = F⁄m = 10⁄5 = 2 ms-2
• This acceleration makes an angle of 37.87o with the direction of the 8 N force
Solved example 5.6
A rocket with a lift-off mass of 20000 kg is blasted upwards with an initial acceleration of 5 ms-2. Calculate the initial force of the blast. [Take g = 10 ms-2]
Solution:
1. The ‘lift-off mass’ is the mass that the rocket will be having at the time of blast
• It moves upwards with an acceleration of 5
• Since there is acceleration, there must be a net force acting on the rocket
2. Let the ‘force of blast’ be F
• This F acts in the upward direction
3. Let the weight of the rocket be W.
• This W acts in the downward direction
♦ It is given by: W = mg = 20000 × 10 = 200000 N
4. Net force = F-W
5. We have acceleration a = Net force⁄Mass
• Substituting the values, we get: 5 = (F-W)⁄20000 = (F-200000)⁄20000
⇒ (F - 200000) = (5 × 20000) = 100000
⇒ F = (100000+200000) = 300000 N = 3 × 105 N
Solved example 5.6
A body of mass 0.4 kg moving initially with a constant speed of 10 ms-1 to the north is subjected to a constant force of 8.0 N directed towards the south for 30 s. Take the instant the force is applied to be t = 0 s, the position of the body at that time to be x = 0, and predict it’s position at t = -5 s, 25 s, 100 s.
Solution:
• For convenience, let us consider the north to be towards the right and the south towards the left.
• Based on these directions, we can draw a 'position graph'. This is shown in fig.5.15(a) below:
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| Fig.5.15 |
• We want the position at t = -5 s
1. At the instant when the 8 N force is applied, a stop watch is turned on.
• So, the reading in the stop watch at that instant will be zero.
• That is., at the instant when 8 N is applied, t = 0
2. The position of the body at that instant is marked as the origin O
• At the origin, x = 0
♦ Distances to the right of O will be positive
♦ Distances to the left of O will be negative
3. Remember that, the body was already in motion towards the north.
• The force was applied while it was moving
4. So the body has already traveled some distance before t = 0
• How much distance did it move (for a duration of 5 s) just before reaching O?
Ans: It was moving with a uniform velocity of 10 ms-1 upto O
• So it would travel (10 × 5) = 50 m for those 5 seconds
• Thus, the position would be 50 m to the left of O
■ On the graph, it would be -50 m. It is shown as 'A' in fig.5.15(b) above
Part (b):
• We want the position at t = 25 s
1. The direction of the force is from north to south. It is opposite to the direction of motion
• So the body would not be able to travel towards the north
2. But, due to it's initial velocity which is directed towards the north, it would continue it's northward motion for 'some more time'
• We want this 'some more time' for which the body continues it's motion (beyond O) in the north direction
• We can use the equation: v2 = v1 + at
3. For using the above equation, we want to know the acceleration
• The body is experiences an acceleration due to the force of 8 N acting on it
• Using F= ma, we get: a = F⁄m = 8⁄0.4 = 20 ms-2 .
• Substituting the values in (2), we get: 0 = 10 -20t
⇒ t = 0.5 s
4. Now we want the distance traveled during this 0.5 s
• We can use the equation: s = v1t + 1⁄2 at2
• Substituting the values, we get: s = 10×0.5 - 1⁄2 × 20 ×0.52 = 2.5 m
• So at t = 0.5 s, the body is at a point 2.5 m to the right of O
■ This is shown as 'B' in the above fig.5.15(c) above
5. What we actually want is the position at t = 25 s
• That is., we want the distance traveled by the body for 25 s since it passed O
• That is., we want the distance traveled in 24.5 s, since it passed ‘B’
6. After passing ‘B’, the body would be traveling south
• ‘Traveling south’ means, traveling in the same direction as the direction of the force
• We can use the equation: s = v1t + 1⁄2 at2
• Substituting the values, we get: s = 0×24.5 + 1⁄2 × 20 ×24.52 = 6002.5 m.
[Note that, for the travel from B towards the south, the initial velocity is zero]
7. So the distance is 6002.5 m from B
• Thus the distance from O is (6002.5-2.5) = 6000 m
• The position on the graph is -6000 m
■ This is shown as 'C' in the fig.5.15(d)
Part (c):
We want the position at t = 100 s
1. But the force acts for only 30 s
• So we first want the position at t = 30 s
• That is., we want the distance traveled in 29.5 s, since it passed ‘B’
2. After passing ‘B’, the body would be traveling south
• ‘Traveling south’ means, traveling in the same direction as the direction of the force
• We can use the equation: s = v1t + 1⁄2 at2
• Substituting the values, we get: s = 0×29.5 + 1⁄2 × 20 ×29.52 = 8702.5 m
[Note that, for the travel from B towards the south, the initial velocity is zero]
3. So the distance is 8702.5 m from B
• Thus the distance from O is (8702.5-2.5) = 8700 m
• The position on the graph is -8700 m
■This is shown as 'D' in the fig.5.15(e)
4. So, after D, the body travels for 70 more seconds
• But during those 70 s, there is no force acting on the body
• That means, the body travels with uniform velocity
• That means, after 'D', the body travels with uniform velocity
5. So we want the velocity attained by the body when it reaches D
• For that, consider the travel from B to D
• The time for this travel is 29.5 s
• Initial velocity of this travel is 0
• The acceleration during this travel is 20 ms-2.
• We can use the equation: v2 = v1 + at
• Substituting the values we get: v2 = 0 + 20×29.5 = 590 ms-1
• So, at D, the velocity is 590 ms-1
6. For the next 70 s, the body will be travelling with a uniform velocity of 590 ms-1.
• So distance traveled = 590 ×70 = 41300 m
7. So the distance from D to the last point E is 41300 m
• Thus the position of E on the graph = -(DE+OD) = -(41300+8700) = -50000 m
■ This is shown in fig.5.15(f)
■ A, C and E are the required points. They are shown in red color
In the next section we will see Impulse
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