In the previous section, we saw the basics about buoyancy. In this section, we will continue that discussion. Later in this section, we will see Archimedes principle
Solved example 10.16
A wooden block has a length 6 m, width 2.5 m and height 1.5 m. What height will be exposed above the surface when it floats in water? What is the upward force that it will experience? Given: density of wood = 650 kg m-3
Solution:
Part (a):
1. The arrangement is shown in fig.10.28 below:
2. We have Eq.10.3: $\mathbf\small{\rm{x=\left(\frac{\rho}{\rho_l}\right )h}}$
Substituting the values, we get:
x = (650⁄1000)×1.5 = 0.975 m
• So the height above water = (1.5-0.975) = 0.525 m
Part (b):
1. Volume submerged = (x × 2.5 × 6) = (0.975 × 2.5 × 6) = 14.625 m3
2. So volume of water displaced = 14.625 m3
3. So weight of water displaced = 14.625 × 1000 × 9.8 = 143325 N
4. Thus we get:
Upward force = Weight of water displaced = 143325 N
• Eq.10.3 gives us the 'submerged depth' x
• Just as we derived Eq.10.3, we can derive another equation for the 'submerged volume'
• It can be written in 3 steps:
1. We have:
• Weight of the liquid displaced = Weight of the floating body
• Using this, we wrote: ρlgAx = ρgAh
2. But 'Ax' is the submerged volume
• Also 'Ah' is the total volume
• So we get: ρlVsub = ρVtot
♦ Where Vsub is the submerged volume and Vtot is the total volume
3. So we get:
Eq.10.4: $\mathbf\small{\rm{V_{sub}=\left(\frac{\rho}{\rho_l}\right )V_{tot}}}$
♦ Where ρ is the density of the floating body
• This equation is applicable to irregular shapes also
• If we apply this equation in the earlier solved example 10.15, we will get:
♦ 91.67% of the total volume of ice will remain hidden below the water surface
Now we will see Archimedes principle. It can be explained in 4 steps:
1. We have seen that, when immersed in a liquid:
Any body will experience an upward force
♦ If this upward force is less than the weight of the body, the body will sink
♦ If this upward force is equal to the weight of the body, the body will remain stationary
♦ If this upward force is greater than the weight of the body, the body will float
■ In all three cases, the object is certain to feel an upward force
• The upward force is certain to act because:
♦ The pressure at top
♦ is less than
♦ The pressure at bottom
(See steps (2) and (3) just below fig.10.27 of the previous section)
2. Due to this upward force, the weight of the body will appear to have become less
• This can be explained in 5 steps:
(i) In fig.10.29(a) below, the weight of the red block is taken in air
(ii) In fig.10.29(b), the weight of the red block is taken when it is completely immersed in a liquid
(iii) We know that, the reading in fig.b will be lesser than that in fig.a
(iv) This is because, the liquid exerts an upward force on the red block
(v) The 'net downward force' will be seen as the reading in fig.b
• So we can write:
Actual weight – Upward force = Observed weight
⇒ Actual weight – Observed weight = Upward force
3. But (Actual weight – Observed weight) is the ‘loss of weight’
• Also, upward force is the weight of the liquid displaced
• So we can write:
Loss of weight = Weight of the liquid displaced
■ This is known as Archimedes principle
The principle can be summarized as:
The loss of weight of a body submerged (partially or fully) in a fluid is equal to the weight of the fluid displaced
4. The upward force exerted by the fluid is called buoyant force or buoyancy
Solved example 10.17
A stone weighs 40 kg in air and 20 kg in water. Calculate the volume of the stone and it’s density
Solution:
♦ Where 'g' is the acceleration due to gravity
2. So upward force experienced by the stone (buoyancy) = 20×g
4. But weight of the water displaced = V × 𝛒water × g
♦ Where:
✰ V is the volume of water displaced which is equal to Volume of the stone
✰ 𝛒water is the density of water which is equal to 1000 kg m-3
5. Equating (3) and (4), we get:
20 × 9.8 = V × 1000 × 9.8
V = 0.02 m3
6. Weight in air = (mass × density × g) = (V × 𝛒stone × 9.8)
• Substituting the known values, we get:
40 × 9.8 = 0.02 × 𝛒stone × 9.8
⇒ 𝛒stone = 2000 kg m-3
A body of dimensions 1.5 m × 1.0 m × 2 m weighs 200 kg in water. Find it's weight in air. What is it's density?
Solution:
1. Volume of water displaced = 1.5×1×2 = 3 m3
2. Weight of water displaced = 3×1000×g = 3000×g N
♦ Where 'g' is the acceleration due to gravity
3. Buoyancy = Weight of water displaced = 3000×g N
4. Also buoyancy = (Weight in air – weight in water) = (Weight in air – 200×g)
5. So equating (3) and (4), we get: 3000×g = Weight in air – 200×g
⇒ Weight in air = (3000 + 200)×g = 3200×g N
• So mass = 3200 kg
Solved example 10.19
A gold ornament weighs 50 g in air. In water it weighs 46 g. How much mass of copper is present in that ornament? Given:
Density of gold = 20000 kg m-3
Density of copper = 10000 kg m-3
Solution:
1. Loss of weight = (50-46) = 4 g
2. So buoyant force = (4 × 10-3 × 9.8) N
4. But weight of water displaced
= Volume of water displaced × Density of water × 9.8
⇒ (4×10-3×9.8) N = Volume of water displaced × Density of water × 9.8
= volume of water displaced = 4 × 10-6 m3
♦ Let VG be the actual volume of gold in the ornament
♦ Actual mass of gold in the ornament
✰ = Volume of gold × Density of gold
✰ = VG × 20000 kg
✰ = Volume of copper × Density of copper
✰ = VC × 10000 kg
• So we get: (VG × 20000 + VC × 10000) = 50 × 10-3
VG = (4 × 10-6 - VC)
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