Buoyancy

Understanding Buoyancy Using Archimedes’s Principle

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Understanding Buoyancy Using Archimedes’s Principle

Archimedes’ principle states that for a body wholly or partially immersed in a fluid, the upward buoyant force acting on the body is equal to the weight of the fluid it displaces.
Figure shows an object wholly immersed in a liquid. According to Archimedes’ principle:

Archimedes's Principle 1
Archimedes's Principle 2

Buoyancy of Objects

Archimedes's Principle 4

Figure shows four situations of objects in a liquid and will be used in the following discussion.

  • Figure (a) shows an object whose weight is smaller than the buoyant force. There is a net upward force acting on the object. Thus the object rises up.
    Archimedes's Principle 5
  • Figure (b) shows an object whose weight is bigger than the buoyant force. There is a net downward force acting on the object. Thus the object sinks.
    Archimedes's Principle 6
  • Figure (c) shows an object whose weight is equal to the buoyant force. The net force acting on the object is zero. Thus the object floats.
    Archimedes's Principle 7
  • Figure (d) shows an object whose weight is equal to the buoyant force. The net force acting on the object is zero. Thus the object floats.
    Archimedes's Principle 8
    Archimedes's Principle 9

Applications of Archimedes’ Principle

Submarine
Archimedes's Principle 16Archimedes's Principle 17

  • Figure (b) shows how a submarine dives and rises, which can be explained by applying Archimedes’ principle.
  • When the ballast tanks of a submarine are filled with water, the weight of the submarine becomes greater than the buoyant force. So, the submarine dives into the sea.
  • When compressed air is forced into the ballast tanks purging the water from them, the weight of the submarine becomes less than the buoyant force. So, the submarine rises.

Hydrometer
Archimedes's Principle 18

  • Figure shows how a hydrometer works based on Archimedes’ principle.
  • The higher the density of a liquid measured, the more the hydrometer floats (or the less the hydrometer submerges).
  • The hydrometer can be calibrated to measure the density of a liquid directly.

Hot air balloon
Archimedes's Principle 19

  • A hot air balloon displaces a large volume of air.
  • Hot air has a lower density than the surrounding air.
  • When the buoyant force is greater than the total weight of the balloon, the balloon will rise.

Airship

Archimedes's Principle 20

  • An airship is filled with helium, a gas with lower density than the surrounding air.
  • Because a large volume of air is displaced by the airship, a big buoyant force is created.
  • When the buoyant force is greater than the total weight of the airship, the airship will rise.

Archimedes Principle Example Problems with Solutions

Example 1. A concrete slab weighs 150 N. When it is fully submerged under the sea, its apparent weight is 102 N. Calculate the density of the sea water if the volume of the sea water displaced by the concrete slab is 4800 cm3, [g = 9.8 N kg-1]
Solution:
Archimedes's Principle 3

Example 2. A fishing boat sails from sea into a river. Figures (a) and (b) show when it is on the sea and on the river respectively.
Archimedes's Principle 10
From Figure, compare
(a) the buoyant forces acting on the boat,
(b) the weights of the sea water and the river water displaced by the boat,
(c) the densities of the sea water and the river water.
Solution:
(a) For a floating object, the weight of the object is equal to the buoyant force acting on it. Since it is the same boat for both cases, therefore the buoyant forces acting on the boat are the same.
(b) Since the buoyant force is equal to the weight of the water displaced, therefore the weight of sea water displaced is equal to the weight of river water displaced.
(c) As the weight of sea water displaced is equal to the weight of river water displaced,
Archimedes's Principle 11

Example 3. Figure shows a boat loaded with some goods floating on a sea. The density of the sea is 1020 kg-3.

Archimedes's Principle 12
(a) Calculate the weight of the boat.
(b) Figure shows the situation of maximum loading of the boat.
Archimedes's Principle 13
Calculate the additional weight of goods that has to be added to the boat to reach this situation.
Solution:
Archimedes's Principle 14
Example 4. A few balloons, which are filled with helium gas, are tied to a brick by a string.
Archimedes's Principle 15
Explain why when the string is cut, the balloons rise.
Solution:
Helium has a lower density than the surrounding air which the balloons displace. When the string is cut, the buoyant force acting on the balloons is now more than the total weight of the balloons. As a result, a net upward force pushes the balloons upward.

How Do You Find Buoyant Force

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How Do You Find Buoyant Force

Buoyancy
When a body is immersed in a liquid, the liquid exerts an upward force on the body called as the ‘upthrust‘ or ‘buoyant force.’

Factors affecting upthrust:
1. Larger the volume of the body submerged in the liquid, greater is the upthrust.
2. Larger the density of the liquid, greater is the upthrust.

Applying Archimedes’ Principle

Figure shows a boy transferring a boulder from the seabed to a beach.

(a) He finds that the boulder becomes heavier as it emerges from the surface of the sea.
(b) The boulder is lighter when it is immersed in the sea because of the existence of the buoyant force.
(c) The buoyant force is an upward force resulting from an object being wholly or partially immersed in a fluid.
(d) The bigger the volume of the boulder immersed in the sea, the bigger the buoyant force.
Applying Archimedes' Principle 1Figure shows the relationship between real weight and apparent weight of an object and the upward buoyant force acting on it.
Buoyant force = Actual weight – Apparent weight
Therefore,
Apparent weight = Actual weight – Buoyant force

Experiment:

Aim: To investigate the relationship between the weight of water displaced and the buoyant force.
Situation: A hawker immersed a watermelon into a tub filled with water. Water was displaced from the tub when the watermelon was submerged in it. He noticed that the submerged watermelon was lighter.
Applying Archimedes' Principle 5Problem: What is the relationship between the weight of water displaced and the buoyant force?
Materials: Plasticine, thread
Apparatus: Spring balance, electronic balance, eureka can, beaker
Method:
Applying Archimedes' Principle 6

  1. The mass of an empty beaker, m, is measured with an electronic balance and recorded.
  2. A plasticine is attached to a spring balance with a string as shown in Figure (a).
  3. The weight of the plasticine, W1 is measured and recorded.
  4. Water is poured into a eureka can until it flows out of the spout of the can.
  5. When the water has stopped dripping from the spout, the empty beaker is placed under it.
  6. The plasticine is slowly lowered into the eureka can as shown in Figure (b) until it is completely immersed in the water.
  7. The readings of the spring balance, W2 and the mass of the beaker that is filled with displaced water, m2 are taken and recorded.

Results:
Applying Archimedes' Principle 7
Discussion:
From the experiment, it is found that (m2 – m1) g = (W1 – W2).
Hence, the weight of water displaced is equal to the buoyant force.

Buoyant Force Example Problems with Solutions

Example 1. A body weighs 300 gmf in air and 260 gmf when completely immersed in water. Calculate the following
(i) loss in weight of the body
(ii) upthrust on the body.
Solution:    Given: Weight of body in air = 300 gmf
Weight of the body in water = 260 gmf
∴ Loss in weight of the body = 300 – 260 = 40 gmf
∴ Upthrust of the body = Loss in weight
= 40 gmf

Example 2. A solid block of volume 2 litres has a weight of 80 N. What will be its weight when immersed completely in water ?
Solution:     In order to calculate the weight of the block in water, first calculate the upthrust, i.e. the loss in weight of the body in water, then
Volume of the block = 2 litres = 2000 cc
∴ Volume of water displaced = 2000 cc
Weight of water displaced = 2000 gm
= 2.0 kgf
( Density of water = 1 gm/cc)
= 2.0 × 9.8 N = 19.6 N
∴ Upthrust of water = 19.6 N
Hence, weight of the body fully immersed in water = 80 N – 19.6 N = 60.4 N

Example 3. A solid block of density D has a weight W in air is fully immersed in a liquid of density d. Calculate its apparent weight when fully immersed in liquid.
Solution:    Weight of the block = W
Density of block = D
∴ Volume of the block = \(\frac { W }{ D }\) . d
∴ Upthrust on the block = \(\frac { W }{ D }\) .d
∴ Loss in weight of the block inside liquid = \(\frac { W }{ D }\) .d
Hence, apparent weight of the block when fully immersed in water
\(=\text{W}-\frac{\text{W}}{\text{D}}\text{d}=\text{W}\left( 1-\frac{\text{d}}{\text{D}} \right)\)

Example 4. The weight of a stone in air is 0.65 N. When it is completely submerged in water, its weight is 0.50 N.
Applying Archimedes' Principle 2What is the buoyant force acting on the stone when it is completely submerged in water?
Solution:
Buoyant force
= Actual weight – Apparent weight = 0.65 – 0.50 = 0.15 N

Example 5. Figure shows an empty oil drum floating on the surface of a pond.
Applying Archimedes' Principle 3
Draw and label the two forces acting on the oil drum.
Solution:
Applying Archimedes' Principle 4