In the previous section we saw the acceleration due to gravity 'g'. We obtained it's value as 9.8 m s-2. In this section we will see some of it's applications. Also later in this section we will see thrust and pressure.
• The mass of a particular object is a constant. Consider a wooden block of weight 3 kg. If we put this wooden block on one side of a weighing balance, we will have to put an exact 3 kg standard weighing block on the other side of the balance.
• The mass of that wooden block will be the same 3 kg, where ever we take it in the universe.
• Mass is used in another situation also. We have seen that it is a measure of inertia. That is., greater the mass, greater is the inertia of that object. So we can say this: Mass is the resistance shown by the body to change it's state of rest or state of uniform motion.
• If we want to lift it from the floor, we will have to apply some force. This force that we need to apply is equal and opposite to the force applied on the block by the earth.
• We know that this force is mg. Where m is the mass of the block and g is the acceleration due to gravity.
• If there are two blocks, one of 3 kg and the other of 5 kg, we would prefer to lift the 3 kg block. This is because lesser force is sufficient to lift it when compared to the 5 kg block.
• So we need a method to specify the force exerted by the earth on an object.
• To describe this force, we use the term ‘weight’. So we can write: Weight of an object is the force exerted by the earth on that object.
• For a 3 kg wooden block, the weight will be 3 × 9.8 = 29.4. Now we need a unit to specify weight.
• Since weight is the force, the unit newton that we use for force is used for weight also.
• So we can write: The weight of a 3 kg wooden block is 29.4 N.
• The symbol for weight is W. So we can write in this way also:
For a 3 kg wooden block, W = 29.4 N
Now we will see an interesting problem:
Solved example 3.6
Weight of an object on earth is 4 kg. How much would it weigh on the moon?
Given: Mass of the moon = 7.36 × 1022 kg, Radius of the moon = 1.74 × 106 m
Solution:
• On the earth, the object is placed on one side of the balance. Then on the other side, we will have to place a 4 kg standard weighing block. Then only it will balance.
• Now the same object is taken to the moon. On landing on the moon, we will notice that it is easier to lift the same object. That is., we do not have to use the same amount of force, that we needed to lift it on earth.
What could be the reason? Also, we want to know it’s exact weight on the moon. We can use the following steps:
1. Mass of 4 kg has not changed when the object was taken to the moon
2. But the force with which the object is attracted to the centre of the moon has changed.
3. Let us calculate the force Fm with which any object on the surface of the moon would be attracted to it's centre. For this calculation we will use the universal law of gravitation (Details here):
(i) Let m1 be the mass of the moon (given as 7.36 × 1022 kg) and m2 the mass of any object placed on the surface of the moon.
(ii) Let R be the radius of the moon (given as 1.74 × 106 m).
(iii) Then we have: Fm = G(m1m2⁄R2). Substituting the given values we get:
So we have: Force of attraction on an object of mass m2 on the surface of the moon = Fm = 1.63 m2.
4. The mass of our object = 4 kg. So Fm = Weight of the object on the moon = 1.63 × 4 = 6.52 N
Note: The final equation in step 3(iii) is: Fm = 1.63 m2.
• Compare this with the equation: Force = mass × acceleration
• Comparing the right sides, it is obvious that 1.63 is the acceleration. That is., acceleration due to gravity on the moon. So we can write:
■ Acceleration due to gravity on the moon = 1.63 m s-2.
• Now, acceleration due to gravity on the earth = 9.8 m s-2.
• Let us take the ratio. We get: 9.8⁄1.63 = 6.01 = 6 (approximately)
So we can write:
■ Acceleration due to gravity on the earth is 6 times that on the moon
■ Consequently, weight of any object on the moon will be one sixth of it's weight on the earth
Solved example 3.7
Mass of an object is 10 kg. What is it's weight on the earth?
Solution:
1. Weight of an object on the earth = The force exerted by earth on that object = m × g
Where m is the mass of the object (given as 10 kg) and g is the acceleration due to gravity (which is equal to 9.8 m s-2)
2. So weight = 10 × 9.8 = 98 N
Solved example 3.8
An object weighs 10 N when measured on the surface of the earth. What would be its weight when measured on the surface of the moon?
Solution:
1. We know that, weight of any object on the moon will be one sixth of it's weight on the earth.
2. So we get: Weight of the given object on the moon = 1⁄6 × 10 N = 1.67 N
• So now we know what weight is. It is a force. Even an object which is simply placed on the floor is experiencing a force. This force is it’s weight W.
• Now, according to Newton’s third law, every force has an equal and opposite reaction. So which is the reaction here?
• The object placed on the floor is trying to move downwards and reach the centre of the earth. It is exerting a force on the floor. This force is the action and is equal to it's weight W
• The floor obstructs the downward movement of the object, by applying an equal and opposite force. This is the reaction.
♦ If the floor is well built, with materials like stone or concrete, it can resist the object easily.
♦ But if the floor is of materials like loose sand, the object will penetrate some distance into the floor.
• Another example: A book rests on a table. The book tries to move downwards and reach the centre of the earth. It’s weight W is the action. The table provides an equal and opposite reaction. So the downward movement of the book is prevented.
1. A person stands on loose sand, his feet would go deep into the sand.
2. But if he lay down on the sand, his body will not go that deep.
3. Note that the same downward force is acting in both cases. Because, in both cases, it is the weight W of the person. There is no other force. How is the sand able to provide better resistance when the person lay down?
• The answer is that, the resisting capacity of the sand did not increase or decrease. It remains the same. In the latter case, a larger area of sand could take part in providing the resistance. It is like dividing a task among more number of workers. This can be explained as follows:
♦ Two workers are given the task of lifting a timber log
♦ The same log can be lifted more easily by 7 workers
• In the same way, more sand was able to take part in the task of 'resisting the weight of the body'.
• Now, how did ‘more sand’ come into the picture?
Ans: The contact area between the body and sand increased when the person lay down.
• So ‘contact area’ has an important role to play here.
• Let the same force F act on two areas A1 and A2. Let A2 be greater than A1. Then the effect of the force F will be greater on the area A1.
■ So we need a new quantity to measure the effect of a force. This new quantity is called pressure.
• Pressure is the ratio of force to the area on which that force acts.
• So to obtain pressure, we will be dividing force by the area. When we do that division, what we get is the 'force per unit area'.
■But we have to note one point: The force under consideration must be acting in an exact perpendicular direction to the area. This is shown in the fig.3.5 below:
• In the fig.3.5 above,
♦ The green surface is perfectly horizontal
♦ The red surface is a sloping surface
♦ The blue arrow shows a perfectly vertical force
♦ The yellow arrow is a sloping force, but is perpendicular to the red surface
• Consider the green surface and the blue force.
♦ We have a vertical force on a horizontal surface. That means the force is acting perpendicular to the area. So the blue force can be considered for calculating the pressure on the green surface
♦ Similarly, the yellow force can be considered for calculating the pressure on the red surface
• By the same argument:
♦ The yellow force cannot be considered for calculating the pressure on the green surface. Because yellow force is not perpendicular to the green surface
♦ Similarly, the blue force cannot be considered for calculating the pressure on the red surface. Because blue force is not perpendicular to the red surface
• So we cannot use ‘any force’ to calculate pressure. We must use only ‘thrust’.
• Thus the definition of pressure becomes:
■ Pressure is the thrust per unit area.
Mathematically we write:
Eq.3.4:
• Pressure = Thrust⁄Area
• Substituting the units of thrust and area, we will get the unit of pressure. Thus:
• Unit of pressure = N⁄m2 = N m-2
• This N m-2 is given a special name ‘pascal’ in honour of the scientist Blaise Pascal. The short form is pa
• So we can use either N m-2 or pa. Both are same.
A block of wood is kept on a tabletop. The mass of wooden block is 5 kg and its dimensions are 40 cm × 20 cm × 10 cm. Find the pressure exerted by the wooden block on the table top if it is made to lie on the table top with its sides of dimensions (a) 20 cm × 10 cm and (b) 40 cm × 20 cm.
Solution:
• Mass of the wooden block = 5 kg
• Force exerted on the table top = Weight of the block = mg = 5 × 9.8 = 49 N
• The block is kept on the table top in two different ways as shown in the fig.3.6 below:
Case (i):
1. In fig.3.6(a), the block is resting on a 20 × 10 face
• So area = 20 × 10 = 200 cm2 = 200⁄10000 m2 = 0.02 m2.
2. The force, which is the weight, is acting in a vertical direction. So it is perpendicular to the top surface of the table.
• So the force can be considered as the thrust
3. Thus we get: pressure = Thrust⁄Area = 49⁄0.02 = 2450 Nm-2.
Case (ii):
1. In fig.3.6(a), the block is resting on a 40 × 20 face
• So area = 40 × 20 = 800 cm2 = 800⁄10000 m2 = 0.08 m2.
2. In this case also, the force, which is the weight, is acting in a vertical direction. So it is perpendicular to the top surface of the table.
• So the force can be considered as the thrust
3. Thus we get: pressure = Thrust⁄Area = 49⁄0.08 = 612.5 Nm-2.
■ Comparing the two cases, we can see that, when area increases pressure decreases
In the next section, we will see Work, Power and Energy.
Mass of an object
• Mass of an object is the amount of matter in that object. For example, if we consider a wooden block, it’s mass is the amount of wood in that block.• The mass of a particular object is a constant. Consider a wooden block of weight 3 kg. If we put this wooden block on one side of a weighing balance, we will have to put an exact 3 kg standard weighing block on the other side of the balance.
• The mass of that wooden block will be the same 3 kg, where ever we take it in the universe.
• Mass is used in another situation also. We have seen that it is a measure of inertia. That is., greater the mass, greater is the inertia of that object. So we can say this: Mass is the resistance shown by the body to change it's state of rest or state of uniform motion.
Weight of an object
• Consider a wooden block kept on the floor. The earth is attracting the block towards it’s centre.• If we want to lift it from the floor, we will have to apply some force. This force that we need to apply is equal and opposite to the force applied on the block by the earth.
• We know that this force is mg. Where m is the mass of the block and g is the acceleration due to gravity.
• If there are two blocks, one of 3 kg and the other of 5 kg, we would prefer to lift the 3 kg block. This is because lesser force is sufficient to lift it when compared to the 5 kg block.
• So we need a method to specify the force exerted by the earth on an object.
• To describe this force, we use the term ‘weight’. So we can write: Weight of an object is the force exerted by the earth on that object.
• For a 3 kg wooden block, the weight will be 3 × 9.8 = 29.4. Now we need a unit to specify weight.
• Since weight is the force, the unit newton that we use for force is used for weight also.
• So we can write: The weight of a 3 kg wooden block is 29.4 N.
• The symbol for weight is W. So we can write in this way also:
For a 3 kg wooden block, W = 29.4 N
Now we will see an interesting problem:
Solved example 3.6
Weight of an object on earth is 4 kg. How much would it weigh on the moon?
Given: Mass of the moon = 7.36 × 1022 kg, Radius of the moon = 1.74 × 106 m
Solution:
• On the earth, the object is placed on one side of the balance. Then on the other side, we will have to place a 4 kg standard weighing block. Then only it will balance.
• Now the same object is taken to the moon. On landing on the moon, we will notice that it is easier to lift the same object. That is., we do not have to use the same amount of force, that we needed to lift it on earth.
What could be the reason? Also, we want to know it’s exact weight on the moon. We can use the following steps:
1. Mass of 4 kg has not changed when the object was taken to the moon
2. But the force with which the object is attracted to the centre of the moon has changed.
3. Let us calculate the force Fm with which any object on the surface of the moon would be attracted to it's centre. For this calculation we will use the universal law of gravitation (Details here):
(i) Let m1 be the mass of the moon (given as 7.36 × 1022 kg) and m2 the mass of any object placed on the surface of the moon.
(ii) Let R be the radius of the moon (given as 1.74 × 106 m).
(iii) Then we have: Fm = G(m1m2⁄R2). Substituting the given values we get:
4. The mass of our object = 4 kg. So Fm = Weight of the object on the moon = 1.63 × 4 = 6.52 N
Note: The final equation in step 3(iii) is: Fm = 1.63 m2.
• Compare this with the equation: Force = mass × acceleration
• Comparing the right sides, it is obvious that 1.63 is the acceleration. That is., acceleration due to gravity on the moon. So we can write:
■ Acceleration due to gravity on the moon = 1.63 m s-2.
• Now, acceleration due to gravity on the earth = 9.8 m s-2.
• Let us take the ratio. We get: 9.8⁄1.63 = 6.01 = 6 (approximately)
So we can write:
■ Acceleration due to gravity on the earth is 6 times that on the moon
■ Consequently, weight of any object on the moon will be one sixth of it's weight on the earth
We will now see some solved examples
Solved example 3.7
Mass of an object is 10 kg. What is it's weight on the earth?
Solution:
1. Weight of an object on the earth = The force exerted by earth on that object = m × g
Where m is the mass of the object (given as 10 kg) and g is the acceleration due to gravity (which is equal to 9.8 m s-2)
2. So weight = 10 × 9.8 = 98 N
Solved example 3.8
An object weighs 10 N when measured on the surface of the earth. What would be its weight when measured on the surface of the moon?
Solution:
1. We know that, weight of any object on the moon will be one sixth of it's weight on the earth.
2. So we get: Weight of the given object on the moon = 1⁄6 × 10 N = 1.67 N
• So now we know what weight is. It is a force. Even an object which is simply placed on the floor is experiencing a force. This force is it’s weight W.
• Now, according to Newton’s third law, every force has an equal and opposite reaction. So which is the reaction here?
• The object placed on the floor is trying to move downwards and reach the centre of the earth. It is exerting a force on the floor. This force is the action and is equal to it's weight W
• The floor obstructs the downward movement of the object, by applying an equal and opposite force. This is the reaction.
♦ If the floor is well built, with materials like stone or concrete, it can resist the object easily.
♦ But if the floor is of materials like loose sand, the object will penetrate some distance into the floor.
• Another example: A book rests on a table. The book tries to move downwards and reach the centre of the earth. It’s weight W is the action. The table provides an equal and opposite reaction. So the downward movement of the book is prevented.
Thrust and Pressure
Consider the following situation:1. A person stands on loose sand, his feet would go deep into the sand.
2. But if he lay down on the sand, his body will not go that deep.
3. Note that the same downward force is acting in both cases. Because, in both cases, it is the weight W of the person. There is no other force. How is the sand able to provide better resistance when the person lay down?
• The answer is that, the resisting capacity of the sand did not increase or decrease. It remains the same. In the latter case, a larger area of sand could take part in providing the resistance. It is like dividing a task among more number of workers. This can be explained as follows:
♦ Two workers are given the task of lifting a timber log
♦ The same log can be lifted more easily by 7 workers
• In the same way, more sand was able to take part in the task of 'resisting the weight of the body'.
• Now, how did ‘more sand’ come into the picture?
Ans: The contact area between the body and sand increased when the person lay down.
• So ‘contact area’ has an important role to play here.
• Let the same force F act on two areas A1 and A2. Let A2 be greater than A1. Then the effect of the force F will be greater on the area A1.
■ So we need a new quantity to measure the effect of a force. This new quantity is called pressure.
• Pressure is the ratio of force to the area on which that force acts.
• So to obtain pressure, we will be dividing force by the area. When we do that division, what we get is the 'force per unit area'.
■But we have to note one point: The force under consideration must be acting in an exact perpendicular direction to the area. This is shown in the fig.3.5 below:
 |
| Fig.3.5 |
♦ The green surface is perfectly horizontal
♦ The red surface is a sloping surface
♦ The blue arrow shows a perfectly vertical force
♦ The yellow arrow is a sloping force, but is perpendicular to the red surface
• Consider the green surface and the blue force.
♦ We have a vertical force on a horizontal surface. That means the force is acting perpendicular to the area. So the blue force can be considered for calculating the pressure on the green surface
♦ Similarly, the yellow force can be considered for calculating the pressure on the red surface
• By the same argument:
♦ The yellow force cannot be considered for calculating the pressure on the green surface. Because yellow force is not perpendicular to the green surface
♦ Similarly, the blue force cannot be considered for calculating the pressure on the red surface. Because blue force is not perpendicular to the red surface
• We give a special name for the 'force in the perpendicular direction'. It is the thrust. So thrust can be defined as follows:
■ Force acting on a body in a direction perpendicular to it’s surface is called thrust.• So we cannot use ‘any force’ to calculate pressure. We must use only ‘thrust’.
• Thus the definition of pressure becomes:
■ Pressure is the thrust per unit area.
Mathematically we write:
Eq.3.4:
• Pressure = Thrust⁄Area
• Substituting the units of thrust and area, we will get the unit of pressure. Thus:
• Unit of pressure = N⁄m2 = N m-2
• This N m-2 is given a special name ‘pascal’ in honour of the scientist Blaise Pascal. The short form is pa
• So we can use either N m-2 or pa. Both are same.
Now we will see some solved examples
Solved example 3.6A block of wood is kept on a tabletop. The mass of wooden block is 5 kg and its dimensions are 40 cm × 20 cm × 10 cm. Find the pressure exerted by the wooden block on the table top if it is made to lie on the table top with its sides of dimensions (a) 20 cm × 10 cm and (b) 40 cm × 20 cm.
Solution:
• Mass of the wooden block = 5 kg
• Force exerted on the table top = Weight of the block = mg = 5 × 9.8 = 49 N
• The block is kept on the table top in two different ways as shown in the fig.3.6 below:
 |
| Fig.3.6 |
1. In fig.3.6(a), the block is resting on a 20 × 10 face
• So area = 20 × 10 = 200 cm2 = 200⁄10000 m2 = 0.02 m2.
2. The force, which is the weight, is acting in a vertical direction. So it is perpendicular to the top surface of the table.
• So the force can be considered as the thrust
3. Thus we get: pressure = Thrust⁄Area = 49⁄0.02 = 2450 Nm-2.
Case (ii):
1. In fig.3.6(a), the block is resting on a 40 × 20 face
• So area = 40 × 20 = 800 cm2 = 800⁄10000 m2 = 0.08 m2.
2. In this case also, the force, which is the weight, is acting in a vertical direction. So it is perpendicular to the top surface of the table.
• So the force can be considered as the thrust
3. Thus we get: pressure = Thrust⁄Area = 49⁄0.08 = 612.5 Nm-2.
■ Comparing the two cases, we can see that, when area increases pressure decreases
In the next section, we will see Work, Power and Energy.
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