Physics Form 4 Notes : CHAPTER THREE – FLOATING AND SINKING

CHAPTER THREE

FLOATING AND SINKING

Any object in a liquid, whether floating or submerged, experiences an upward force from the liquid; this force is known as upthrust force. Upthrust force is also called buoyant force and is denoted by the letter ‘u’.

Archimedes’ principle

Archimedes, a Greek scientist, carried out the first experiments to measure upthrust on an object in liquid in the third century. Archimedes’ principle states that “When a body is wholly or partially immersed in a fluid (liquid or gas), it experiences an upthrust equal to the weight of the displaced fluid.”

Experiment: To demonstrate Archimedes’ principle

Procedure

1. Pour water into an overflow can (eureka can) until it starts to flow out, then wait until it stops dripping.
2. Tie a suitable solid body securely and suspend it on a spring balance. Determine its weight in air.
3. Lower the body slowly into the overflow can while still attached to the spring balance, then read off its weight when fully submerged.
4. Weigh the displaced water collected in a beaker. Record your readings as follows:
   • Weight of body in air = W₁
   • Weight of body in water = W₂
   • Weight of empty beaker = W₃
   • Weight of beaker and displaced liquid = W₄
   • Upthrust of the body = W₁ – W₂
   • Weight of displaced water = W₄ – W₃

   

Discussion

The upthrust on the solid body will be found to be equal to the weight of displaced water, thereby demonstrating Archimedes’ principle.

Example

A block of metal of volume 60 cm³ weighs 4.80 N in air. Determine its weight when fully submerged in a liquid of density 1,200 kg/m³.

Solution

Volume of liquid displaced = 60 cm³ = 6.0 × 10^-5 m³.

Weight of the displaced liquid = volume × density × gravity = v × ρ × g

= 6.0 × 10^-5 × 1200 × 10

= 0.72 N

Upthrust = weight of the liquid displaced.

Weight of the block in the liquid = (4.80 – 0.72) = 4.08 N.

Floating objects

Objects that float in a liquid are less dense than the liquid in which they float. We need to determine the relationship between the weight of the displaced liquid and the weight of the body.

Experiment: To demonstrate the law of floatation

Procedure

1. Weigh the block in air and record its weight as W₁.
2. Put water into the overflow can (eureka can) up to the level of the spout.
3. Collect displaced water in a beaker. Record the weight of the beaker first in air and record as W₂. Weigh both the beaker and the displaced water and record as W₃.
4. Record the same procedure with kerosene and record your results as shown below.

   	W1	W2	W3	W3 – W2
   Water
   Kerosene

5. What do you notice between W₁ and W₃ – W₂?

Discussion

The weight of the displaced liquid is equal to the weight of the block in air. This is consistent with the law of floatation which states “A body displaces its own weight of the liquid in which it floats.” Mathematically, the following relation can be deduced:

Weight = volume × density × gravity = v × ρ × g, therefore

W = v_d × ρ × g where v_d is the volume of displaced liquid.

NOTE – Floatation is a special case of Archimedes’ principle because a floating body sinks until the upthrust equals the weight of the body.

Example

A wooden block of dimensions 3 cm × 3 cm × 4 cm floats vertically in methylated spirit with 4 cm of its length in the spirit. Calculate the weight of the block. (Density of methylated spirit = 8.0 × 10² kg/m³).

Solution

Volume of the spirit displaced = (3 × 3 × 4) = 36 cm³ = 3.6 × 10^-5 m³.

Weight of the block = v_d × ρ × g = (3.6 × 10^-5) × 8.0 × 10² × 10 = 2.88 × 10^-1 N.

Relative density

We have established relative density as the ratio of the density of a substance to the density of water. Since by the law of floatation an object displaces a fluid equal to its own weight, the following mathematical expressions can be established:

Relative density = density of substance / density of water.

= weight of substance / weight of equal volume of water.

= mass of substance / mass of equal volume of water.

Applying Archimedes’ principle, the relative density ‘d’ is:

d = weight of substance in air / upthrust in water or d = W / u

Since upthrust is given by (W₂ – W₁) where W₂ is weight in air, and W₁ is weight when submerged.

Hence d = W / u = W / (W₂ – W₁). The actual density, ρ, of an object can be obtained as follows:

ρ of an object = d × 1,000 kg/m³.

Relative density of a floating body

Experiment: To determine the relative density of a cork

Procedure

1. Select a sinker which is heavy enough to make the cork sink.
2. Attach the cork and the sinker as shown:

3. Record the results obtained as follows:
   • Weight of the sinker in water = W₁
   • Weight of the sinker in water and cork in air = W₂
   • Weight of the sinker and cork in water = W₃
   • Weight of the cork in air = W₂ – W₁
   • Upthrust on the cork = W₂ – W₃

The relative density of the cork in air is determined as follows:

d = weight of the cork in air / upthrust on the cork.

Applications of Archimedes’ principle and relative density

1. Ships – Steel, which is used to make ships, is 6-7 times denser than water, but a ship is able to float on water because it is designed to displace more water than its volume. Load lines called plimsoll marks are marked on the side to indicate the maximum load at different seasons to avoid overloading.
2. Submarines – They are made of steel and consist of ballast tanks which contain water when they have to sink and are filled with air when they have to float. This allows submarines to balance their weight and be able to rise upwards.
3. Balloons – When filled with helium gas, balloons become lighter and the upthrust on the balloon becomes greater than their weight, allowing them to rise upwards.
4. Hydrometers – They are used to measure the relative densities of liquids quickly and conveniently. Various types of hydrometers are made to measure different ranges of densities, e.g., lactometer – for measuring milk density (range 1.015 – 1.045), battery acid tester – used to test the charge in a lead-acid battery.

Examples

1. A solid of mass 1.0 kg is suspended using a thread and then submerged in water. If the tension on the thread is 5.0 N, determine the relative density of the solid.

   Solution

   Mass of solid = 1.0 kg

   Weight of solid W = mg = 10 N

   Tension on the string (T) = 5 N

   Upthrust on solid (u) = W – T = 10 – 5 = 5 N

   Relative density (d) = W / u = 10 / 5 = 2.

2. A balloon made of fabric weighing 80 N has a volume of 1.0 × 10⁷ cm³. The balloon is filled with hydrogen of density 0.9 kg/m³. Calculate the greatest weight, in addition to that of the hydrogen and the fabric, which the balloon can carry in air of average density 1.25 kg/m³.

   Solution

   Upthrust = weight of the air displaced

   = volume of air × density × gravity

   = (1.0 × 10⁷ × 10^-6) × (1.25 × 10)

   = 10 × 1.25 × 10 = 125 N

   Weight of hydrogen = 10 × 0.09 × 10 = 9 N

   Total weight of hydrogen and fabric = 80 + 9 = 89 N

   Total additional weight to be lifted = 125 – 89 = 36 N.

3. A material of density 8.5 g/cm³ is attached to a piece of wood of mass 100 g and density 0.2 g/cm³. Calculate the volume of material X which must be attached to the piece of wood so that the two just submerge beneath a liquid of density 1.2 g/cm³.

   Solution

   Let the volume of the material be V cm³.

   The mass of the material is 8.5 V grams.

   Volume of wood = 100 g / 0.2 g/cm³ = 500 cm³.

   To have an average density of 1.2 g/cm³ = total mass / total volume,

   Therefore, (100 + 8.5V) / (500 + V) = 1.2 g/cm³.

   Hence, V = 68.5 cm³.