In the previous section we saw Newton's Third law of motion. In this section we will see an application of that law.
1. We have seen weighing scales used to find the weight of bodies.
• Let a person stand on it to find his own weight. The scale is placed on a level ground.
This is shown in fig.5.20(a) below:
2. Let his mass be m.
• The gravitational force pulls him down with a force of W
• We know that W = mg
3. While the weight is being measured, we observe that the person is stationary.
• So there is no net force acting on him
4. That means another force must be acting which is canceling W
• This ‘another force’ is none other than the reaction ‘R’ from the scale
• We can write:
■ The scale is compelled to apply R in order to support the person
• This is shown in fig.5.20(b) above
5. Since the person is observed to be at rest, R must be equal and opposite to W
• That is., R = -W
■ The reaction R which the scale is compelled to apply, appears as the reading on that scale.
• We will denote this reading as Reading0
• This reading indicates the normal weight of that person
• Now, let the person stand on the same scale kept on the floor of a lift.
• The reading on the scale will depend upon the type of motion of the lift. We have seven cases possible:
Case 1: The lift is stationary
• The scale is placed on the floor of the lift and the person stands on it
• Let us denote the reading as Reading1
■ If the lift is stationary, Reading1 will be obviously equal to Reading0
• So Reading1 is equal to Reading0, and it indicates the normal weight of that person
Case 2: The lift moves upwards (or downwards) with a uniform velocity
• The scale is placed on the floor of the lift and the person stands on it
• If we look at the reading while the lift is in uniform motion (upwards or downwards), we will find that it is the same Reading1
Let us see the reason:
1. When the lift is in uniform motion, the person will also be in uniform motion
• Uniform motion means no acceleration
• No acceleration means no net force
• ‘No net force’ means that, all the forces acting are canceling each other out.
2. The force acting on the person is the Weight 'W' which is downwards
• This W being canceled out
3. That means another force must be acting which is canceling W
• This ‘another force’ is none other than the reaction ‘R’ from the scale
4. We can write:
• The scale is compelled to apply R in order to support the person
• Since the person is observed to be in uniform motion, R must be equal and opposite to W
• That is., R = -W
5. The reaction R which the scale is compelled to apply, appears as the reading on that scale.
• Let us denote the reading as Reading2
6. But from (4), we see that: R is numerically equal to W
• In case 1 also, R is numerically equal to W
• So Reading2 will be the same Reading1
Case 3: The lift moves upwards with an acceleration ‘a’
• The lift starts from rest.
• After some time, it attains a velocity 'v'.
• Once it acquires 'v', it will continue to travel with that velocity
• So after acquiring the required velocity, the lift will be in uniform motion
♦ This is represented by the blue horizontal line in the velocity time graph in fig.5.21 below:
• We have seen that if we take the reading at any instant when the lift is in uniform motion, it will be Reading1. This we saw in case 2
• Now consider the time interval between 0 and t1
• During this time, the lift has an acceleration of ‘a’.
♦ This is shown by the magenta line which slope upwards
• The lift can attain the required velocity 'v' only if it is given an acceleration from rest
■ So acceleration is inevitable during the initial stage of upward motion.
• What about the reading if it is taken between 0 and t1?
Let us analyse:
• We will take upward direction as positive and downward direction as negative
1. The person is experiencing an acceleration ‘a’
• So a net force is acting on him
2. We know that W acts downwards. So we have -W
• We know that R from the scale acts upwards. So we have +R
• So the net force F must be equal to (-W+R)
3. Now we apply Newton’s second law: F = ma
• That is., if we see a body moving with acceleration ‘a’, the product of mass of that body and ‘a’ will be equal to the net force acting on that body
4. Since the ‘a’ is in upward direction, we have +a
• So the net force F = (-W+R) = ma
5. We want R. So we rearrange the above equation:
• R = (W + ma) = (mg + ma) = m(g+a)
6. The scale is compelled to provide this R.
• The reaction R which the scale is compelled to apply, appears as the reading on that scale.
• Let us denote the reading as Reading3
7. Now we compare Reading3 and Reading1
We have:
Reading3 = mg + ma
Reading1 = mg
• That means, the scale will show a greater reading than normal.
• If we know the value of m and a, we can easily predict the reading
8. This completes our analysis of case 3
• But before taking up case 4, let us write some simple logical steps also:
(i) The lift and the person are initially at rest
• From that state, the lift gradually accelerates upwards
♦ This is shown in fig.5.22(a) below:
(ii) But due to inertia, the body of the person does not want to move up
• So the floor of the lift (and the scale) will have to apply some extra force on the person
• This extra force increases the reading.
(iii) It may be note that, the person himself will feel more weight during this upward acceleration
Case 4: The lift moves upwards with an acceleration ‘-a’
• Consider the time interval between t2 and t3
• During this time, the lift has an acceleration of ‘-a’
♦ In fig.5.21, this is shown by the white line which slope downwards
• The lift has to reduce speed gradually and come to rest
• It can attain the zero velocity only if it is given a negative acceleration
• So negative acceleration is inevitable even while moving upwards.
• What about the reading if it is taken between t2 and t3?
Let us analyse:
• As usual, we will take upward direction as positive and downward direction as negative
1. The person is experiencing an acceleration ‘a’ in the downward direction
• In other words, the person is experiencing a retardation while moving upwards
• That is., the person is experiencing an acceleration of ‘-a’
• Since there is acceleration, we can say: A net force is acting on him
2. We know that W acts downwards. So we have -W
• We know that R from the scale acts upwards. So we have +R
• So the net force F must be equal to (-W+R)
3. Now we apply Newton’s second law: F = ma
• That is., if we see a body moving with acceleration ‘a’, the product of mass of that body and ‘a’ will be equal to the net force acting on that body
4. Since the ‘a’ is retardation in the upward direction, we have -a
• So The net force F = (-W+R) = -ma
5. We want R. So we rearrange the above equation:
• R = (W - ma) = (mg - ma) = m(g-a)
6. The scale is compelled to provide this R.
• The reaction R which the scale is compelled to apply, appears as the reading on that scale.
• Let us denote the reading as Reading4
7. Now we compare Reading4 and Reading1
We have:
Reading3 = mg - ma
Reading1 = mg
• That means, the scale will show a lesser reading than normal.
• If we know the value of m and a, we can easily predict the reading
8. This completes our analysis of case 4
• But before taking up case 5, let us write some simple logical steps also:
(i) The lift and the person are initially moving with uniform velocity
• From that state, the lift gradually decelerates upwards
This is shown in fig.5.22 (b) above
[Note that, the following two items are equivalent:
♦ Applying a negative acceleration (that is., deceleration) in the upward direction
♦ Applying a positive acceleration in the downward direction]
(ii) But due to inertia, the body of the person does not want to stop. It tends to continue the upward motion.
• So there is a slight loss of contact between the feet and the scale
• Thus, the floor of the lift (and the scale) will need to apply a lesser force only
• This reduction in the force decreases the reading.
(iii) It may be note that, the person himself will feel less weight during this upward deceleration
Case 5: The lift moves downwards with an acceleration ‘-a’
• Here we consider the return journey of the lift from a top floor to a bottom floor
• The lift starts from rest. After some time, it attains a velocity. Once it acquires the required velocity, it will continue to travel with that velocity
• So after acquiring the required velocity, the lift will be in uniform motion
♦ This is represented by the blue horizontal line in the velocity time graph in fig.5.23 below
• We have seen that if we take the reading at any instant between t1 and t2, it will be Reading1. This we saw in case 2
• Now consider the time interval between 0 and t1
• During this time, the lift has an acceleration of ‘a’
• But this acceleration is in the downward direction
• So we can write: During the time interval 0 to t1 in the downward motion of the lift, it travels with an acceleration of ‘-a’
• It can attain the required velocity only if it is given an acceleration from rest
■ So acceleration is inevitable during the downward motion also
• What about the reading if it is taken while the lift is accelerating between 0 and t1?
Let us analyse:
• As usual, we will take upward direction as positive and downward direction as negative
1. The person is experiencing an acceleration ‘-a’
• So a net force is acting on him
2. We know that W acts downwards. So we have -W
• We know that R from the scale acts upwards. So we have +R
• So the net force F must be equal to (-W+R)
3. Now we apply Newton’s second law: F = ma
• That is., if we see a body moving with acceleration ‘a’, the product of mass of that body and ‘a’ will be equal to the net force acting on that body
4. Since the ‘a’ is in downward direction, we have -a
• So The net force F = (-W+R) = -ma
5. We want R. So we rearrange the above equation:
• R = (W - ma) = (mg - ma) = m(g-a)
6. The scale is compelled to provide this R
• The reaction R which the scale is compelled to apply, appears as the reading on that scale.
• Let us denote the reading as Reading5
7. Now we compare Reading5 and Reading1
We have:
Reading5 = mg - ma
Reading1 = mg
• That means, the scale will show a lesser reading than normal.
• If we know the value of m and a, we can easily predict the reading
8. This completes our analysis of case 5
• But before taking up case 6, let us write some simple logical steps also:
(i) The lift and the person are initially at rest
• From that state, the lift gradually accelerates downwards
♦ This is shown in fig.5.22(c) above
(ii) But due to inertia, the body of the person does not want to move down
• The floor and the scale starts downward motion, whether or not, the person wants to come along
• So there is a slight loss of contact between the scale and the feet
• This reduces the reading.
(iii) It may be noted that, the person himself will feel less weight during this downward acceleration
Case 6: The lift moves downwards with an acceleration ‘a’
• Consider the time interval between t2 and t3
• During this time, the lift has an acceleration of ‘-a’
♦ In fig.5.23, this is shown by the white line which slopes downwards
• The lift has to reduce speed gradually and come to rest
• It can attain the zero velocity only if it is given a negative acceleration
• This negative acceleration is applied in the downward direction. So it becomes positive
• So positive acceleration is inevitable even while moving downwards.
• What about the reading if it is taken while the lift is experiencing this ‘a’ between t2 and t3?
Let us analyse:
• As usual, we will take upward direction as positive and downward direction as negative
1. The person is experiencing an acceleration ‘-a’ in the downward direction
• That is., the person is experiencing an acceleration of ‘+a’
• Since there is acceleration, we can say: A net force is acting on him
2. We know that W acts downwards. So we have -W
• We know that R from the scale acts upwards. So we have +R
• So the net force F must be equal to (-W+R)
3. Now we apply Newton’s second law: F = ma
• That is., if we see a body moving with acceleration ‘a’, the product of mass of that body and ‘a’ will be equal to the net force acting on that body
• Since the ‘-a’ is in the downward direction, we have +a
• So The net force F = (-W+R) = +ma
5. We want R. So we rearrange the above equation:
• R = (W + ma) = (mg + ma) = m(g+a)
6. The scale is compelled to provide this R.
• The reaction R which the scale is compelled to apply, appears as the reading on the scale
• Let us denote this reading as Reading6
7. Now we compare Reading6 and Reading1
We have: Reading6 = mg + ma
Reading1 = mg
• That means, the scale will show a greater reading than normal.
• If we know the value of m and a, we can easily predict the reading
8. This completes our analysis of case 6
• Before taking up case 7, let us write some simple logical steps also:
(i) The lift and the person are initially moving with uniform velocity
• From that state, the lift gradually decelerates downwards
This is shown in fig.5.22(d) above
(ii) But due to inertia, the body of the person does not want to stop. It tends to continue the downward motion.
• The floor of the lift (and the scale) is slowing down
• This increases the contact between the feet of the person and the scale. The feet presses down on the scale. So an extra force comes into play
(iii) It may be noted that, the person himself will feel greater weight during this downward deceleration
Case 7: The cable which pulls the lift breaks
• This is similar to Case 5: The lift moves downwards with an acceleration ‘-a’
• But instead of 'a', we use 'g'
♦ That is., the lift moves down wards with acceleration '-g'
• Both the lift and the person will be in 'free fall'
• In free fall, the velocity does not depend on the mass
• So both the person and the lift will be falling with the same velocity
• So we can say that, the person's feet and the scale will always be in contact
• We want the reaction R in such a situation
2. We know that W acts downwards. So we have -W
• We know that R from the scale acts upwards. So we have +R
• So the net force F must be equal to (-W+R)
3. Now we apply Newton’s second law: F = ma
• That is., if we see a body moving with acceleration ‘a’, the product of mass of that body and ‘a’ will be equal to the net force acting on that body
• In this case, instead of 'a', we have 'g'
• Since the ‘g’ is in the downward direction, we have -g
• So The net force F = (-W+R) = -mg
4. We want R. So we rearrange the above equation:
• R = (W - mg) = (mg - mg) = 0
5. So the scale need not provide any reaction. The reading will be zero
7. Here also we can write a logical explanation:
• The feet may be in contact with the scale during free fall
• But since, the body and the scale are moving with the same velocity, there will not be any contact force. So the reading will be zero
■ Upward motion
• Accelerating: Reading = m(g+a)
• Decelerating: Reading = m(g-a)
■ Downward motion
• Accelerating: Reading = m(g-a)
• Decelerating: Reading = m(g+a)
■ Free fall
• Reading = 0
[While using the above equations, we must input the numerical values only. The sign (direction) is already taken into account while deriving the equations]
Apparent weight in a lift
We will write the steps:1. We have seen weighing scales used to find the weight of bodies.
• Let a person stand on it to find his own weight. The scale is placed on a level ground.
This is shown in fig.5.20(a) below:
 |
| Fig.5.20 |
• The gravitational force pulls him down with a force of W
• We know that W = mg
3. While the weight is being measured, we observe that the person is stationary.
• So there is no net force acting on him
4. That means another force must be acting which is canceling W
• This ‘another force’ is none other than the reaction ‘R’ from the scale
• We can write:
■ The scale is compelled to apply R in order to support the person
• This is shown in fig.5.20(b) above
5. Since the person is observed to be at rest, R must be equal and opposite to W
• That is., R = -W
■ The reaction R which the scale is compelled to apply, appears as the reading on that scale.
• We will denote this reading as Reading0
• This reading indicates the normal weight of that person
• Now, let the person stand on the same scale kept on the floor of a lift.
• The reading on the scale will depend upon the type of motion of the lift. We have seven cases possible:
Case 1: The lift is stationary
• The scale is placed on the floor of the lift and the person stands on it
• Let us denote the reading as Reading1
■ If the lift is stationary, Reading1 will be obviously equal to Reading0
• So Reading1 is equal to Reading0, and it indicates the normal weight of that person
Case 2: The lift moves upwards (or downwards) with a uniform velocity
• The scale is placed on the floor of the lift and the person stands on it
• If we look at the reading while the lift is in uniform motion (upwards or downwards), we will find that it is the same Reading1
Let us see the reason:
1. When the lift is in uniform motion, the person will also be in uniform motion
• Uniform motion means no acceleration
• No acceleration means no net force
• ‘No net force’ means that, all the forces acting are canceling each other out.
2. The force acting on the person is the Weight 'W' which is downwards
• This W being canceled out
3. That means another force must be acting which is canceling W
• This ‘another force’ is none other than the reaction ‘R’ from the scale
4. We can write:
• The scale is compelled to apply R in order to support the person
• Since the person is observed to be in uniform motion, R must be equal and opposite to W
• That is., R = -W
5. The reaction R which the scale is compelled to apply, appears as the reading on that scale.
• Let us denote the reading as Reading2
6. But from (4), we see that: R is numerically equal to W
• In case 1 also, R is numerically equal to W
• So Reading2 will be the same Reading1
Case 3: The lift moves upwards with an acceleration ‘a’
• The lift starts from rest.
• After some time, it attains a velocity 'v'.
• Once it acquires 'v', it will continue to travel with that velocity
• So after acquiring the required velocity, the lift will be in uniform motion
♦ This is represented by the blue horizontal line in the velocity time graph in fig.5.21 below:
 |
| Fig.5.21 |
• Now consider the time interval between 0 and t1
• During this time, the lift has an acceleration of ‘a’.
♦ This is shown by the magenta line which slope upwards
• The lift can attain the required velocity 'v' only if it is given an acceleration from rest
■ So acceleration is inevitable during the initial stage of upward motion.
• What about the reading if it is taken between 0 and t1?
Let us analyse:
• We will take upward direction as positive and downward direction as negative
1. The person is experiencing an acceleration ‘a’
• So a net force is acting on him
2. We know that W acts downwards. So we have -W
• We know that R from the scale acts upwards. So we have +R
• So the net force F must be equal to (-W+R)
3. Now we apply Newton’s second law: F = ma
• That is., if we see a body moving with acceleration ‘a’, the product of mass of that body and ‘a’ will be equal to the net force acting on that body
4. Since the ‘a’ is in upward direction, we have +a
• So the net force F = (-W+R) = ma
5. We want R. So we rearrange the above equation:
• R = (W + ma) = (mg + ma) = m(g+a)
6. The scale is compelled to provide this R.
• The reaction R which the scale is compelled to apply, appears as the reading on that scale.
• Let us denote the reading as Reading3
7. Now we compare Reading3 and Reading1
We have:
Reading3 = mg + ma
Reading1 = mg
• That means, the scale will show a greater reading than normal.
• If we know the value of m and a, we can easily predict the reading
8. This completes our analysis of case 3
• But before taking up case 4, let us write some simple logical steps also:
(i) The lift and the person are initially at rest
• From that state, the lift gradually accelerates upwards
♦ This is shown in fig.5.22(a) below:
 |
| Fig.5.22 |
• So the floor of the lift (and the scale) will have to apply some extra force on the person
• This extra force increases the reading.
(iii) It may be note that, the person himself will feel more weight during this upward acceleration
Case 4: The lift moves upwards with an acceleration ‘-a’
• Consider the time interval between t2 and t3
• During this time, the lift has an acceleration of ‘-a’
♦ In fig.5.21, this is shown by the white line which slope downwards
• The lift has to reduce speed gradually and come to rest
• It can attain the zero velocity only if it is given a negative acceleration
• So negative acceleration is inevitable even while moving upwards.
• What about the reading if it is taken between t2 and t3?
Let us analyse:
• As usual, we will take upward direction as positive and downward direction as negative
1. The person is experiencing an acceleration ‘a’ in the downward direction
• In other words, the person is experiencing a retardation while moving upwards
• That is., the person is experiencing an acceleration of ‘-a’
• Since there is acceleration, we can say: A net force is acting on him
2. We know that W acts downwards. So we have -W
• We know that R from the scale acts upwards. So we have +R
• So the net force F must be equal to (-W+R)
3. Now we apply Newton’s second law: F = ma
• That is., if we see a body moving with acceleration ‘a’, the product of mass of that body and ‘a’ will be equal to the net force acting on that body
4. Since the ‘a’ is retardation in the upward direction, we have -a
• So The net force F = (-W+R) = -ma
5. We want R. So we rearrange the above equation:
• R = (W - ma) = (mg - ma) = m(g-a)
6. The scale is compelled to provide this R.
• The reaction R which the scale is compelled to apply, appears as the reading on that scale.
• Let us denote the reading as Reading4
7. Now we compare Reading4 and Reading1
We have:
Reading3 = mg - ma
Reading1 = mg
• That means, the scale will show a lesser reading than normal.
• If we know the value of m and a, we can easily predict the reading
8. This completes our analysis of case 4
• But before taking up case 5, let us write some simple logical steps also:
(i) The lift and the person are initially moving with uniform velocity
• From that state, the lift gradually decelerates upwards
This is shown in fig.5.22 (b) above
[Note that, the following two items are equivalent:
♦ Applying a negative acceleration (that is., deceleration) in the upward direction
♦ Applying a positive acceleration in the downward direction]
(ii) But due to inertia, the body of the person does not want to stop. It tends to continue the upward motion.
• So there is a slight loss of contact between the feet and the scale
• Thus, the floor of the lift (and the scale) will need to apply a lesser force only
• This reduction in the force decreases the reading.
(iii) It may be note that, the person himself will feel less weight during this upward deceleration
Case 5: The lift moves downwards with an acceleration ‘-a’
• Here we consider the return journey of the lift from a top floor to a bottom floor
• The lift starts from rest. After some time, it attains a velocity. Once it acquires the required velocity, it will continue to travel with that velocity
• So after acquiring the required velocity, the lift will be in uniform motion
♦ This is represented by the blue horizontal line in the velocity time graph in fig.5.23 below
 |
| Fig.5.23 |
• Now consider the time interval between 0 and t1
• During this time, the lift has an acceleration of ‘a’
• But this acceleration is in the downward direction
• So we can write: During the time interval 0 to t1 in the downward motion of the lift, it travels with an acceleration of ‘-a’
• It can attain the required velocity only if it is given an acceleration from rest
■ So acceleration is inevitable during the downward motion also
• What about the reading if it is taken while the lift is accelerating between 0 and t1?
Let us analyse:
• As usual, we will take upward direction as positive and downward direction as negative
1. The person is experiencing an acceleration ‘-a’
• So a net force is acting on him
2. We know that W acts downwards. So we have -W
• We know that R from the scale acts upwards. So we have +R
• So the net force F must be equal to (-W+R)
3. Now we apply Newton’s second law: F = ma
• That is., if we see a body moving with acceleration ‘a’, the product of mass of that body and ‘a’ will be equal to the net force acting on that body
4. Since the ‘a’ is in downward direction, we have -a
• So The net force F = (-W+R) = -ma
5. We want R. So we rearrange the above equation:
• R = (W - ma) = (mg - ma) = m(g-a)
6. The scale is compelled to provide this R
• The reaction R which the scale is compelled to apply, appears as the reading on that scale.
• Let us denote the reading as Reading5
7. Now we compare Reading5 and Reading1
We have:
Reading5 = mg - ma
Reading1 = mg
• That means, the scale will show a lesser reading than normal.
• If we know the value of m and a, we can easily predict the reading
8. This completes our analysis of case 5
• But before taking up case 6, let us write some simple logical steps also:
(i) The lift and the person are initially at rest
• From that state, the lift gradually accelerates downwards
♦ This is shown in fig.5.22(c) above
(ii) But due to inertia, the body of the person does not want to move down
• The floor and the scale starts downward motion, whether or not, the person wants to come along
• So there is a slight loss of contact between the scale and the feet
• This reduces the reading.
(iii) It may be noted that, the person himself will feel less weight during this downward acceleration
Case 6: The lift moves downwards with an acceleration ‘a’
• Consider the time interval between t2 and t3
• During this time, the lift has an acceleration of ‘-a’
♦ In fig.5.23, this is shown by the white line which slopes downwards
• The lift has to reduce speed gradually and come to rest
• It can attain the zero velocity only if it is given a negative acceleration
• This negative acceleration is applied in the downward direction. So it becomes positive
• So positive acceleration is inevitable even while moving downwards.
• What about the reading if it is taken while the lift is experiencing this ‘a’ between t2 and t3?
Let us analyse:
• As usual, we will take upward direction as positive and downward direction as negative
1. The person is experiencing an acceleration ‘-a’ in the downward direction
• That is., the person is experiencing an acceleration of ‘+a’
• Since there is acceleration, we can say: A net force is acting on him
2. We know that W acts downwards. So we have -W
• We know that R from the scale acts upwards. So we have +R
• So the net force F must be equal to (-W+R)
3. Now we apply Newton’s second law: F = ma
• That is., if we see a body moving with acceleration ‘a’, the product of mass of that body and ‘a’ will be equal to the net force acting on that body
• Since the ‘-a’ is in the downward direction, we have +a
• So The net force F = (-W+R) = +ma
5. We want R. So we rearrange the above equation:
• R = (W + ma) = (mg + ma) = m(g+a)
6. The scale is compelled to provide this R.
• The reaction R which the scale is compelled to apply, appears as the reading on the scale
• Let us denote this reading as Reading6
7. Now we compare Reading6 and Reading1
We have: Reading6 = mg + ma
Reading1 = mg
• That means, the scale will show a greater reading than normal.
• If we know the value of m and a, we can easily predict the reading
8. This completes our analysis of case 6
• Before taking up case 7, let us write some simple logical steps also:
(i) The lift and the person are initially moving with uniform velocity
• From that state, the lift gradually decelerates downwards
This is shown in fig.5.22(d) above
(ii) But due to inertia, the body of the person does not want to stop. It tends to continue the downward motion.
• The floor of the lift (and the scale) is slowing down
• This increases the contact between the feet of the person and the scale. The feet presses down on the scale. So an extra force comes into play
(iii) It may be noted that, the person himself will feel greater weight during this downward deceleration
Case 7: The cable which pulls the lift breaks
• This is similar to Case 5: The lift moves downwards with an acceleration ‘-a’
• But instead of 'a', we use 'g'
♦ That is., the lift moves down wards with acceleration '-g'
• Both the lift and the person will be in 'free fall'
• In free fall, the velocity does not depend on the mass
• So both the person and the lift will be falling with the same velocity
• So we can say that, the person's feet and the scale will always be in contact
• We want the reaction R in such a situation
2. We know that W acts downwards. So we have -W
• We know that R from the scale acts upwards. So we have +R
• So the net force F must be equal to (-W+R)
3. Now we apply Newton’s second law: F = ma
• That is., if we see a body moving with acceleration ‘a’, the product of mass of that body and ‘a’ will be equal to the net force acting on that body
• In this case, instead of 'a', we have 'g'
• Since the ‘g’ is in the downward direction, we have -g
• So The net force F = (-W+R) = -mg
4. We want R. So we rearrange the above equation:
• R = (W - mg) = (mg - mg) = 0
5. So the scale need not provide any reaction. The reading will be zero
7. Here also we can write a logical explanation:
• The feet may be in contact with the scale during free fall
• But since, the body and the scale are moving with the same velocity, there will not be any contact force. So the reading will be zero
Let us write a summary:
• Accelerating: Reading = m(g+a)
• Decelerating: Reading = m(g-a)
■ Downward motion
• Accelerating: Reading = m(g-a)
• Decelerating: Reading = m(g+a)
■ Free fall
• Reading = 0
[While using the above equations, we must input the numerical values only. The sign (direction) is already taken into account while deriving the equations]
Solved example 5.9
A man of mass 70 kg stands on a weighing scale in a lift which is moving
(a) upwards with a uniform speed of 10 ms-1,
(b) downwards with a uniform acceleration of 5 ms-2,
(c) upwards with a uniform acceleration of 5 ms-2.
What would be the readings on the scale in each case?
(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity ?
[Take g = 10 ms-2]
Solution:
Part (a):
• When in uniform motion, the reading would show the normal weight. That is., the weight corresponding to 70 kg
• The weight W corresponding to 70 kg = mg = 70 × 10 = 700 N
• If the scale shows the weight in newton itself, the reading will be 700 N
• On a regular scale available in the market, the unit 'kg' will be used.
• The reading in such a scale will be 700⁄10 = 70 kg
(∵ W = mg ⟹ m = W⁄g)
Part (b):
• The weight will be less
• Reading = m(g-a) = 70 × (10-5) = 70 × 5 = 350 N
• If the scale shows the weight in newton itself, the reading will be 350 N
• On a regular scale available in the market, the unit 'kg' will be used.
• The reading in such a scale will be 350⁄10 = 35 kg
Part (c):
• The weight will be more
• Reading = m(g+a) = 70 × (10+5) = 70 × 15 = 1050 N
• If the scale shows the weight in newton itself, the reading will be 1050 N
• On a regular scale available in the market, the unit 'kg' will be used.
• The reading in such a scale will be 1050⁄10 = 105 kg
Part (d):
When the lift falls under gravity, the reading will be zero in all types of scales
A man of mass 70 kg stands on a weighing scale in a lift which is moving
(a) upwards with a uniform speed of 10 ms-1,
(b) downwards with a uniform acceleration of 5 ms-2,
(c) upwards with a uniform acceleration of 5 ms-2.
What would be the readings on the scale in each case?
(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity ?
[Take g = 10 ms-2]
Solution:
Part (a):
• When in uniform motion, the reading would show the normal weight. That is., the weight corresponding to 70 kg
• The weight W corresponding to 70 kg = mg = 70 × 10 = 700 N
• If the scale shows the weight in newton itself, the reading will be 700 N
• On a regular scale available in the market, the unit 'kg' will be used.
• The reading in such a scale will be 700⁄10 = 70 kg
(∵ W = mg ⟹ m = W⁄g)
Part (b):
• The weight will be less
• Reading = m(g-a) = 70 × (10-5) = 70 × 5 = 350 N
• If the scale shows the weight in newton itself, the reading will be 350 N
• On a regular scale available in the market, the unit 'kg' will be used.
• The reading in such a scale will be 350⁄10 = 35 kg
Part (c):
• The weight will be more
• Reading = m(g+a) = 70 × (10+5) = 70 × 15 = 1050 N
• If the scale shows the weight in newton itself, the reading will be 1050 N
• On a regular scale available in the market, the unit 'kg' will be used.
• The reading in such a scale will be 1050⁄10 = 105 kg
Part (d):
When the lift falls under gravity, the reading will be zero in all types of scales
In the next section we will see Conservation of Momentum
No comments:
Post a Comment