Chapter 3 Force and Pressure Solutions

Column 1

Column 2

Column 3

1. Pressure

Force/area

Decreases with increase in area

2. Density

Mass/volume

Useful to determine the purity of a substance

3. Atmos­pheric pressure

The pascal

Decreases with increase in height above the sea level

4. Relative density

No unit

Specific gravity

Answer the following questions in one sentence:

Answer the following questions:

Use your brainpower!
Make a list of some more examples in which contact and non contact forces are applied. Write the types of force.
Answer:
1. Some examples in which contact forces are applied:

2. Some examples in which non contact forces are applied:

Try this:
Take two plastic bottles with rectangular shape. Close their openings by fitting the lids tightly. Keep two small bar magnets on them and fix them neatly using a sticking tape.

Fill a big plastic tray with water and leave the two bottles floating with magnets at the top. Take one bottle near the other. If the north pole of the magnet is near the south pole of the other magnet, the bottles will head towards each other, because unlike poles attract each other.

Observe what will happen when the directions of the bottles are changed. We can observe change in the motion of the bottles without any direct contact. This means that there exists a non contact force between the two magnets.

Observation: When the directions of the bottles are changed, if the north pole of one magnet is near the north pole of the other magnet (or the south pole of one magnet is near the south pole of the other magnet), the bottles will move away from each other because like poles repel each other.

Use your brain power!

Question.
You have learnt about static electricity in the previous standard. Electrostatic force is a non contact force. To verify this, which experiment will you perform?
Answer:

Do not switch on the fan in the room. Keep very small pieces of paper on the table. Rub a plastic comb against hair and bring it near the bits of paper. You will find that the bits of paper are attracted by the comb. The comb, on rubbing acquires electrostatic charge. It induces opposite charges on the bits of paper. Hence, the bits of paper are attracted by the comb.

Try this:
Balanced forces and unbalanced force:

Take a cardboard box, tie thick string to its two sides and keep it on a smooth table as shown in Fig. Take the strings on both sides of the table. Tie weighing pans to the two ends. Keep equal masses in both the pans. The box does not move on the table.

If more mass is kept in one of the pans than in the other, the box starts moving in the direction of that pan. Equal gravitational force acts on both the pans when equal masses are kept in them. This means balanced forces act on the box, with effective force equal to zero as these are acting in opposite directions.

On the contrary, if more mass is kept in one pan than in the other, the box starts moving in the direction of the pan with more mass. When unequal forces are applied to the box on the two sides, an unbalanced force acts on the box resulting in imparting motion to the box.

Children playing tug of war pull the rope in their respective directions. If the pull of the force is equal on the two sides, the rope does not move. If the force is more on one side, the rope moves in that direction. This means that initially, the two forces are balanced; the rope moves in the direction of higher force when the forces become unbalanced.

Let us see one more example. When a big grain storage container is required to slide on the ground, it becomes easier if two persons push it rather than one person. When the force is applied by both in the same direction, the movement is easy. You may have experienced this. What do we understand from this example?

Force is a vector quantity. If more than one force are acting on a body, then the effect on the body is due to the net force. When a force is applied on a stationary object it moves, its speed and direction change. Similarly, a force is required to stop an object in motion.

An object can change its shape due to force. While kneading a dough made from flour, the dough changes its shape when a force is applied. A potter applies a force in a specific direction while shaping the pot. Rubber, when stretched, expands. There are many such examples.,

Consider a body moving in a straight line, If we apply a force in the direction of motion of the body, the speed of the body increases. On the contrary, if we apply a force in the direction opposite to that of motion of the body, the speed of the body decreases. These ( examples show that force has magnitude as well as direction, i.e., force is a vector quantity.

For example, a body moving with constant speed along a circular path in a horizontal plane will fly tangentially in the sense of motion if the centripetal force (the force directed towards the centre of the circle) is removed completely.

Always remember:

Types of inertia:

1. Inertia of the state of rest: An object in the state of rest cannot change its state of rest due to its inherent property. This property is called the inertia of the state of rest.

2. Inertia of motion: The inherent property of an object due to which its state of motion cannot change, is called its inertia of motion. For example, a revolving’ electric fan continues to revolve even after it is switched off, passengers sitting in the running bus get aerk in the forward direction if the bus suddenly stops.

3. Directional inertia: The inherent property of an object due to which the object cannot change the direction of its motion, is called directional inertia. For example, if a vehicle in motion along a straight line suddenly turns, the passengers sitting in it are thrown opposite to the direction of turning.

Try this:
Activity 1:
Take a postcard and keep it on a glass. Keep a 5 Rupee coin on it. Now skilfully push the card. The coin straight away falls in the glass. Have you ever done this?
Answer:
Yes. (Explanation: The postcard moves forward due to the applied force and then falls due to the earth’s gravitational force. In the absence of adequate frictional force between the coin and the postcard, the coin does not move forward with the postcard, but straightaway falls in the glass due to the earth’s gravitational force.)

Activity 2: Hang a half a kg mass from a stand, with a string 1. Tie another string 2 to the mass and keep it hanging. Now pull the string 2 with a jerk. The string 2 breaks but the mass does not fall. Heavy mass does not move. Now pull the string 2 slowly. The string 1 breaks and the mass fall down. This is because of the tension developed in the string 2 due to the mass.

(Explanation: (1) As no force acts on the mass, it remains at rest due to inertia. (2) The transmission of force results in the tension in the string 1. As the string 1 cannot withstand it, it breaks and the mass falls down.)

2. The upper parts of their bodies, however, continue to be in a state of motion due to inertia. Hence, the passengers are pushed in the forward direction.

Try this:

Activity 3: Take some sharp pointed nails and push them into a wooden plank by hammering on their heads. Now take? a nail and hold it with its head on the plank and hammer it down from the pointed end. When pressing the drawing pins into a drawing board, they get into the board easily. By applying a force using the thumb one can push the pins into the boards. On the contrary, while pressing ordinary pins into the board with a thumb, the thumb may get hurt.

What does this simple experiment tell?
The nail easily penetrates into wood from its pointed end. From this you will notice that when a force is applied on the head of the nail, it is easy to hammer it into the plank.
Explanation: The less the area of the surface on which the force is applied, the greater is the effect of the force.

Use your brain power!

This shows that pressure varies inversely as the area of the surface on which the force is applied, if the force remains constant. If the same force is applied to surfaces having different areas, the pressure is more on the surface having a smaller area.

[Note: The bottom surface of a camel’s feet is broad. Hence, the camel’s weight acts on a large surface area. This reduces the pressure on the sand. Hence, the camel’s feet do not penetrate deep into the sand. Therefore, it becomes easier for the camel to walk on sand.]

Clean the brick and place it on the dough with one of its faces measuring 10 cm x 5 cm in contact with the dough. Observe how deep the brick penetrates into the dough. You will find that the brick penetrates deeper in this case than that in the first case.

Try this:

Use your brainpower!

[Note: The pressure exerted by a liquid (or gas or fluid) at a depth h below the free surface of the liquid = hpg, where p is the density of the liquid (or gas or fluid) and g is the acceleration due to gravity.]

[Note: Liquids and gases together are called fluids. Gases have very low viscosity compared to liquids. A liquid with low viscosity flows easily. A liquid with high viscosity does not flow easily.]

[Note: The pressure exerted by a fluid is a scalar quantity.]

The pressure exerted by air or the atmosphere surrounding the earth is known as the atmospheric pressure. It is the ratio of the weight of the air to the area of the surface of the earth. It decreases with altitude as the density of air decreases with altitude and also the weight of the air column above a given place.
[Note: At sea level the atmospheric pressure is about 105 Pa. We do not feel it because the pressure of blood and other fluids in our body balances it.]

Use your brain power!

Question ii.
Some people feel breathless as they climb higher and higher on a mountain.
Answer:
Atmospheric pressure decreases with altitude. Hence, some people feel breathless as they climb higher and higher on a mountain.

Question iii.
We can enjoy a cold-drink or fruit juice with the help of a straw but can we imagine drinking a cold-drink or fruit juice on the moon using a straw?
Answer:
When we suck air in the straw, the pressure of the air in the straw becomes less than that of the outside air on the cold drink or fruit juice in the bottle (or the glass). Hence, the cold drink or fruit juice rises in the straw and enters our mouth. We can then drink it. There is no atmosphere on the moon. Hence, it is not possible to enjoy a cold drink or fruituice on the moon by using a straw.

Question iv.
People are often advised not to carry fountain pens while travelling by air.
Answer:
The ink in a fountain pen (filled at sea level at atmosphere pressure) may come out through its mouth while travelling by air if the outside pressure becomes less than the pressure in the ink holder of the pen. This can spoil the clothes/purse/bag. Hence, people are often advised not to carry fountain pens while travelling by air.

As a result, there arises a difference in the internal and external pressures on the walls of the cells and blood capillaries. If the difference is large, it may cause the cell wall and the blood capillaries to burst. Thus, the capillaries in the nose (and ear) may burst causing bleeding.

When the piston is moved up, the coloured water rises in the part of the cylinder below the piston due to the atmospheric pressure. Finally, the inner pressure equals the atmospheric pressure and no more coloured water enters in or comes out.

To spray the coloured water, the tube is taken out and the piston is moved towards the opening of the tube. As the inner pressure is now greater than the atmospheric pressure, the coloured water gushes out of the narrow opening of the tube.

When the tip of the narrow tube is dipped in coloured water (or any other liquid) and the piston is pushed towards the tip, up to the bottom, most of the air in the cylinder escapes through the tube, reducing the pressure. When the piston is moved up, the coloured water rises in the part of the cylinder below the piston due to the atmospheric pressure.

Finally, the inner pressure equals the atmospheric pressure and no more coloured water enters in or comes out. To spray the coloured water, the tube is taken out and the piston is moved towards the opening of the tube. As the inner pressure is now greater than the atmospheric pressure, the coloured water gushes out of the narrow opening of the tube.

The tip of a syringe is fitted with a very fine and hollow needle. The required quantity of medicine can be taken in the syringe with the help of the piston. The medicine can then be injected into the body of a patient using the needle and the piston.

Try this:

Buoyant force:
Take a plastic bottle and fix the lid tightly. Now place it in wafer and see. It will float on water. Try and push it into the water. Even when pushed, it continues to float. This experiment can also be done with a plastic hollow ball. Now fill the bottle with water to the fullest capacity and close the lid, and release in water. The bottle will float inside the water. Why does this happen?

The empty plastic bottle floats on the surface of water. On the contrary, the bottle full of water floats inside water but does not go to the bottom. The weight of the empty bottle is negligible as compared with the weight of the water inside.

Such a bottle with water neither floats on the surface, nor does it go to the bottom. This means the force due to gravity acting downwards (fg), must have been balanced by an opposing force in the upward direction (fb) on the bottle filled with water. This force must have originated from the water surrounding the bottle. The upward force acting on the object in water or other fluid or gas is called buoyant force (fg).

Use your brain power!

Question.
While pulling a bucket from a well, the bucket full of water immersed fully in water appears to weigh less than when it has been pulled out of water. Why?
Answer:
1. When a bucket full of water is immersed in water, the net force acting on the bucket = weight of the bucket full of water-the buoyant force exerted by the water on the bucket. The buoyant force due to a fluid is proportional to the density of the fluid.

2. The density of water is much greater than that of air. Therefore, the buoyant force acting on a bucket full of water while it is in water is much greater than that when it is out of water, i.e., in air. Hence, it appears to weigh less, while it is in water than when it has been pulled out of water.

Try this:

Do you know?

How is it decided that an object left in a liquid will get sink in the liquid, will float on the surface, or will float inside the liquid?
1. The object floats if the buoyant force is larger than its weight.
2. The object sinks if the buoyant force is smaller than its weight.
3. The object floats inside the liquid if the buoyant force is equal to its weight. Which forces are unbalanced in the above cases?
Answer:
Unbalanced force:
1. Magnitude of the unbalanced force acting on the object = magnitude of the buoyant force on the object-magnitude of the weight of the object. The direction of the unbalanced force is the direction of the buoyant force.

2. Magnitude of the unbalanced force on the object = magnitude of the weight of the object – magnitude of the buoyant force on the object. The direction of the unbalanced force is the direction of the weight of the object.

3. Here, the unbalanced force is zero.

Try this:

Take a long rubberband and cut it at one point. At one of its ends tie a clean washed stone or a 50 g weight as shown in Figure
Now hold the other end of the rubberband and make a mark there. Keep the stone hanging in air and measure the length of the rubberband from the stone to the mark made earlier.

Now take water in a pot and hold the rubberband at such a height that the stone sinks in it. Again measure the length of the rubberband up to the mark. What is observed? This length is shorter than the earlier length. While dipping the stone in water, length of the stretched rubber gets slowly reduced and is minimum when it sinks completely. What could be the reason for a shorter length of the rubberband in water?

When the stone is sunk in water, a buoyant force acts on it in the upward direction. The weight of the stone acts downwards. Therefore, the force which acts on it in the downward direction is effectively reduced.
(This decreases the downard force on the rubberband. Hence, its length decreases.)

2. If the density of the fluid is greater than that of the body, the magnitude of the buoyant force on the body will be greater than that of the weight of the body. Therefore, the body will float in the fluid.

3. If the density of the fluid is equal to that of the body, the magnitude of the buoyant force on the body will be equal to that of the weight of the body. Therefore, the body will remain in equilibrium anywhere within the fluid.

Use your brain power!

Question.
Explain the observations in the earlier experiments according to the Archimedes’ principle.
Answer:
The increase in the length of the rubberband (y) is proportional to the downward force (f) acting on it.
1. When the stone tied to the rubberband is hanging in air, the magnitude of the buoyant force exerted by air on the stone is negligible compared to that of the weight of the stone. Hence, ignoring it, we have f = fg = mg, where m is the mass of the stone and g is the acceleration due to gravity.

2. When the stone is immersed partially in water, f = fg – fb, where fb is the magnitude of the buoyant force exerted by water on the stone.
Now, fg = Vρg and fb = xVρwg, where V = volume of the stone, ρ = density of the stone, xV= volume of the water displaced by the part of the stone immersed in water = volume of the part of the stone immersed in water (x < 1) and ρ = density of water.

This shows that as the stone is gradually lowered in water, x goes on increasing and hence/goes on decreasing. Therefore, elongation (y) of the rubberband goes on decreasing, i.e., the length of the rubberband goes on decreasing.

3. When the stone is completely immersed in water, x becomes maximum, equal to 1. Here, f = fg . This is the minimum value of. Here, the elongation of the rubberband is minimum, i.e., the length of the rubberband is minimum.

Hence, the volume of the liquid displaced by a floating body is inversely proportional to the density of the liquid. As a result, when a piece of wood floats in kerosine, it displaces greater volume of kerosine compared to the volume of water displaced when the piece of wood floats in water. Hence, it sinks more in kerosine than in water.
[Note: When a body floats in a liquid fb = fg
∴ Vρ1g = Vbρbg
∴ The volume of the liquid displaced by the body is V = Vb , where Vb is the volume of the body, ρb is the density of the material of the body and ρ1 is the density of the liquid. Thus, V ∝ 1 /ρ1]

When ice in water melts, this difference equals the volume of the water formed when the part of ice above the surface of water melts. Therefore, the water level remains the same. Hence, there is no overflow of water when the ice melts.

Therefore, the ball will start moving upward. As it comes up with part of the ball above the water surface, the volume of the water displaced by the ball becomes less and hence at a certain stage, the buoyant force and the weight balance each other. Then the ball continues to remain in that state, as the net force on the ball becomes zero.
[Note: Initially, the ball moves slightly up and down near the water surface. The force due to friction with water, opposing the motion of the ball, finally makes the ball steady.]

Write short notes:

2. The resultant of all these forces acts upward. It is called the upthrust or buoyant force.

3. The buoyant force is proportional to (i) the volume of the liquid displaced by the body (ii) the density of the liquid (iii) the acceleration due to gravity. Its magnitude equals the magnitude of the weight of the liquid displaced by the body.

2. Archimedes’ principle is used in design of ships and submarines. A submarine is provided with large tanks at the front and the back. Its weight can be increased by filling the tanks with sea water or air from compressed air reservoirs. The weight can be decreased by pumping out water from the tanks by forcing compressed air in them. By controlling the weight, it can be made to sink or rise to the surface as desired.

3. The density of a body that floats or sinks in water or kerosine can be determined by. Archimedes’ principle.
4. The density of kerosine can be determined by Archimedes’ principle, using a body of material that is not affected by water and kerosine.

Give scientific reasons:

[Note: If there were no tiles, the feet will come in direct contact with the slushy ground. The area of the feet being less, the weight of the person will act over a smaller area. Therefore. there will be more pressure and hence the feet will sink into the slushy ground.]

Solve the following examples:

Problem 1.
(i) Calculate the pressure exerted by the wooden block when it is kept in the vertical position.
Given: The length of the wooden block is 80 cm, the breadth is 50 cm, the thickness is 20 cm and the weight is 500 N
(ii) Also calculate the pressure when the wooden block is kept in the horizontal position with its surface 80 cm × 50 cm touching the floor.

Solution:
Data: F = W = 500 N, l = 80 cm = 0.8 m, b = 50 cm = 0.5 m. h = 20 cm = 0.2 m
(i) A = bh = 0.5 m × 0.2 m = 0.1 m²

The pressure exerted in the vertical position of the block = 5000 N/m²or 5000 Pa.

(ii) A = lb = 0.8 m × 0.5 m = 0.4 m²

The pressure exerted in the horizontal position of the block = 1250 N/m²or 1250 Pa.

Problem 2.
Measure the length, breadth, height and mass of a rectangular tiffin box. Find the weight of the box and calculate the pressure in two different positions as in Ex. (1) above.
Solution:
Let l = 0.25 m, 6 = 0.1 m, h = 0.05 m, F = W= 0.5 N
(i) A = bh = 0.1 m × 0.05 m = 0.005 m²

(ii) A = lb = 0.25 m × 0.1 m = 0.025 m²

Problem 3.
A force of 1000 N is applied over an area 50 cm × 20 cm. Find the corresponding pressure.
Solution:
Data: F = 1000 N,
A = 50 cm × 20 cm = 0.5 m × 0.2 m = 0.1 m², pressure = ?

The pressure = 10 N/m².

Problem 4.
A metal block has thmensions 10 cm × 5 cm × 2 cm and the density of the metal is 8 × 10³kg/m³. It is kept on a table with the face 10 cm × 5 cm in contact with the table. Find the force and pressure exerted by the block on the table. (g = 9.8 m/s²)
Solution :
Data : 1 = 10 cm, b = 5 cm,
h = 2 cm, p= 8 × 10 kg/m3, g = 9.8 m/s²,
A = lb = 10cm × 5cm = 50cm²= 50 × 104m²
= 5 × 10^-5m², force =?, pressure = ?
Volume of the block = lbh =
10 cm × 5 cm × 2 cm= 100 cm³
= 100 × 10^-6m³= 1 × 10^-4m³
Mass of the block = volume × density
(∵ density = mass/volume)
∴Mass of the block,
m = 1 × 10^-4m³× 8 × 10³kg/m³= 0.8 kg
Weight of the block = mg = 0.8 kg × 9.8 m/s²= 7.84 N
∴ The force exerted by the block on the table = 7.84 N.
force 7.84 N

= 1.568 × 10³N/m²or 1.568 × 10³Pa
The pressure exerted by the block on the table = 1.568 × 10 N/m2 or 1.568 × 10³Pa.

Problem 5.
A body of volume loo cm³is immersed completely in water. Find the weight of the water displaced by the body. 1g = 9.8 m/s². p (water) = kg/m³]
Solution :
Data: V = 100 cm³= 100 × 10^-6m³
= 1 × 10 m³, p(water) = 10 kg/m³
g = 9.8 m/s², weight of the displaced water ?
Density =
∴ Mass = volume × density
Volume of the water displaced by the body = l × 10^-4m³
∴Mass of the water displaced,
m = l × 10^-4m³× 10 kg/m³= 0.1 kg
∴Weight of the water displaced
= mg = 0.1 kg × 9.8 m/s²= 0.98 N.
The weight of the water displaced by the body = 0.98 N.

Problem 6.
A body of mass 200 g and volume 50 cm³is put in a bucket containing water. Will it float or sink ?
[ρ(water) = 1 g/cm³]
Solution:
Data: m = 200 g, V= 50 cm³,
ρ (water) = 1 g/cm³Density(ρ) =
∴ ρ (body) =
It is greater than the density of water.
Hence, the body will sink in water.

Problem 7.
A body of mass 200 g and volume 400 cm³is put in a bucket containing water. Will it float or sink?
[ρ (water)=1 g/cm³]
Solution:
Proceed as above.
ρ (body) = = 0.5 g/cm³
It is less than the density of water.
∴ The body will float in water.

Problem 8.
The mass of a tile is 500 g. If the density of the tile is 2.5 g/cm³, what will be the weight of the tile when it is completely immersed in water?
(g = 9.8 m/s², ρ(water) = 1000 kg/m³)
Solut10n:
Data: m = 500, g = 0.5 kg, ρ
(tile) = 2.5 g/cm³= 2500 kg/m³, g = 9.8 m/s²,
ρ (water) = 1000 kg/m³, weight of the tile when completely immersed in water (also called the apparent weight) = ?
ρ =

∴ Volume of water displaced by the tile
= 2 × 10^-4m³
∴ Mass of water displaced by the tile
m’ = ρ (water) V = 1000 kg/m³× 2 × 10^-4m³
= 0.2kg
∴ Magnitude of the weight of this water
= mg = 0.2 kg × 9.8 m/s²= 1.96 N
∴ Buoyant force exerted on the tile = 1.96 N
Magnitude of the weight of the tile =
mg = 0.5 kg × 9.8 m/s²= 4.9 N
∴ Weight of the tile when completely immersed in water (apparent weight) = weight of the tile in air-buoyant force on the tile
= 4.9 N – 1.96 N = 2.94 N(downward)

Examples For Practice:

[g = 9.8 m/s², ρ (water) = 10³kg/m³= 1 g/cm³]