In the previous section, we saw the details about Pascal’s law. In this section, we will see buoyancy and flotation
• A liquid of density ρl is taken in a trough
• Three cylinders of equal volume (same area of cross section) and same height) but different masses are placed in the liquid as shown in the fig.10.26 below
♦ The yellow cylinder sinks to the bottom of the trough
♦ The red cylinder remains stationary at a depth z below the liquid surface
♦ The blue cylinder floats with a height (h-x) above the liquid surface
• We want to know why the three cylinders behave differently
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| Fig.10.26 |
• First we consider the red cylinder. The explanation can be written in 9 steps
1. Let us put the red cylinder in a separate trough containing the same liquid
It is shown in fig.10.27 below
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| Fig.10.27 |
2. Let us write the three forces acting on the cylinder
(i) Pressure at the top of the cylinder = Pa + ρlgz
♦ Where Pa is the atmospheric pressure
• So force at the top of the cylinder = [Pa + ρlgz]A
♦ Where A is the area of cross section of the cylinder
• This force acts downwards and thus tends to push the cylinder downwards
♦ This is indicated by the upper green arrow
(ii) The pressure at the bottom of the cylinder = Pa + ρlg(z+h)
♦ So force at the bottom of the cylinder = [Pa + ρlg(z+h)]A
• This force acts upwards and thus tends to push the cylinder upwards
♦ This is indicated by the lower green arrow
(iii) The force mrg due to gravity
♦ Where mr is the mass of the red cylinder
• This force acts downwards and thus tends to pull the cylinder downwards
3. The cylinder is stationary. So net force acting on the cylinder is zero
• Let us assume that the upward forces are positive and downward forces are negative
• Then we can write:
-[Pa + ρlgz]A + [Pa + ρlg(z+h)]A - mrg = 0
⇒ -Pa A - ρlgzA + Pa A + ρlgzA + ρlghA - mrg = 0
⇒ ρlghA = mrg
4. based on this result, we can write:
• The red cylinder in fig.10.27 is stationary because:
♦ The weight mrg
♦ is balanced by
♦ The force ρlgAh
5. We know that mrg always acts in the downward direction
♦ So the force ρlgAh must be acting in the upward direction
♦ That means, the liquid exerts an upward force of ρlgAh on the red cylinder
• Let us analyze this force ρlgAh. It can be written in 4 steps:
(i) Ah is the volume of the cylinder
(ii) Since the cylinder is completely submerged in the liquid, we can write:
Volume of the liquid displaced is equal to the volume of the cylinder
• So Ah is the volume of the liquid displaced
(iii) Consequently, ρlAh is the mass of the liquid displaced
(iv) Consequently, ρlgAh is the weight of the liquid displaced
6. So based on step (4), we can write:
• The red cylinder in fig.10.27 is stationary because:
♦ The weight mrg
♦ is equal to
♦ The weight of the liquid displaced
■ Conversely, we can write:
If the weight of the body is equal to the weight of the liquid displaced, the body will not sink
7. Now we get an interesting relation between densities. It can be derived in 3 steps:
(i) If the red cylinder remains stationary as in fig.10.27, we have:
ρlgAh = weight of the cylinder = mrg
(ii) But mr = ρrAh
♦ Where ρr is the density of the red cylinder
(iii) Thus we get: ρlgAh = ρrgAh
■ That means:
If the cylinder remains stationary as in fig.10.27, the density of the cylinder will be equal to the density of the liquid
8. We can generalize the situation that we see in fig.10.27. We can write it in 3 steps:
(i) Take a body whose density is same as the density of the liquid
(ii) Place the body gently at an interior point of the liquid
(iii) The body will remain stationary at that point
9. This is just like a book resting on the table
• But it is very difficult to obtain such a perfect equilibrium
• There are three main reasons:
(i) If the surrounding temperature around the liquid changes, the density of the liquid will change
♦ In such a situation, the two densities will become different. The equilibrium will be lost
(ii) If any salts or minerals reach the liquid, they will get dissolved, resulting in a change in density of the liquid
♦ In such a situation, the two densities will become different. The equilibrium will be lost
(iii) If there is even the slightest movement in the liquid particles, the forces will not remain the same
♦ In such a situation also, the equilibrium will be lost
• We have seen a satisfactory explanation for the red cylinder in fig.10.26
• Now we will consider the yellow cylinder in that same fig.
• It can be written in 5 steps:
1. In the case of the red cylinder we saw this:
♦ The liquid exerts a force ρlgAh (in the upward direction) on the red cylinder
♦ The weight of the red cylinder mrg is equal to this force. So it does not sink
2. The liquid will exert the same force on the yellow cylinder also
♦ Because ρl, A and h are the same
• Even then, the yellow cylinder sinks
• The reason is obvious:
The upward force is less than the weight of the yellow cylinder
3. So we can write:
The yellow cylinder sinks because: ρlgAh < myg
♦ Where my is the mass of the yellow cylinder
4. But my = ρyAh
♦ Where ρy is the density of the yellow cylinder
• Thus we get: ρlgAh < ρygAh
⇒ ρl < ρy
■ That means:
If the yellow cylinder sinks as in fig.10.26, the density of the cylinder will be greater than the density of the liquid
5. We can generalize this situation. We can write it in 3 steps:
(i) Take a body whose density is greater than the density of the liquid
(ii) Place the body gently inat an interior point of the liquid
(iii) The body will sink to the bottom
Next we will consider the blue cylinder in fig.10.26 . It can be written in 11 steps:
1. In the case of the red cylinder, we saw this:
♦ The liquid exerts a force ρlgAh (in the upward direction) on the red cylinder
♦ The weight of the red cylinder is equal to this force. So it does not sink
2. The liquid will exert the same force ρlgAh on the blue cylinder also
♦ Because ρl, A and h are the same
• Even then, the blue cylinder rises to the surface
• The reason is obvious:
The upward force is greater than the weight of the blue cylinder
3. So we can write:
The blue cylinder rises because: ρlgAh > mbg
♦ Where mb is the mass of the blue cylinder
4. But mb = ρbAh
♦ Where ρb is the density of the blue cylinder
• Thus we get: ρlgAh > ρbgAh
⇒ ρl > ρb
■ That means:
If the blue cylinder rises as in fig.10.26, the density of the cylinder will be less than the density of the liquid
5. We can generalize this situation. We can write it in 3 steps:
(i) Take a body whose density is less than the density of the liquid
(ii) Place the body gently at an interior point of the liquid
(iii) The body will rise to the surface of the liquid
6. Now another question arises.The question can be elaborated in 3 steps:
(i) When the blue cylinder is completely immersed in the liquid
♦ the upward force exerted by the liquid
♦ is greater than
♦ the weight of the cylinder
(ii) So there will be a net upward force
• As a result, the blue cylinder will move upwards
(iii) But it cannot move up indefinitely. So how far up will it move?
• The following steps from (7) to (11) will give the answer
7. In fig.10.26, a distance x is marked for the blue cylinder
♦ It indicates the ‘portion which is submerged’
• The 'portion above x' is above the surface of the liquid
• We want to find this x
8. The blue cylinder is initially, completely submerged
• As it moves up, it begins to emerge out of the liquid surface
♦ More and more portion emerges
♦ Less and less portion is submerged
9. 'Less submerged portion' means: 'less water is displaced'
• We have seen that, the upward force exerted by the liquid is equal to the weight of the liquid displaced
♦ Now, the displaced water is becoming lesser and lesser
♦ So the upward force will also become lesser and lesser
10. When the upward force becomes equal to the weight of the blue cylinder, the motion will stop
• The cylinder will no longer rise up
11. Based on (10), we can find x. It can be written in steps:
(i) When the blue cylinder is at equilibrium, we have:
Volume of liquid displaced = Ax
(ii) So upward force at equilibrium = weight of liquid displaced = ρlgAx
• This must be equal to the weight of the cylinder
• So we can write: ρlgAx = ρbgAh
⇒ ρlx = ρbh
• Thus we get:
Eq.10.3: $\mathbf\small{\rm{x=\left(\frac{\rho}{\rho_l}\right )h}}$
♦ Where ρ is the density of the floating body and ρl is the density of the liquid
Solved example 10.15
An ice cube of side 4 cm floats in water. What height of the cube will be visible above the water surface? Given that: Density of water and ice are 1000 kg m-3 and 916.7 kg m-3 respectively
Solution:
1. Let x be the depth below water
• We have Eq.10.3: $\mathbf\small{\rm{x=\left(\frac{\rho}{\rho_l}\right )h}}$
2. Substituting the known values, we get:
x = (916.7⁄1000)×4 = 3.67 cm
• So the height above water = (4-3.67) = 0.33 cm
3. Consider the fraction (916.7⁄1000) that we see in step (2)
• This fraction can be written as: (91.67⁄100)
• That means, when ice floats in water, 91.67% of the total height will be hidden below the water surface
• Consider the red cylinder that we saw in fig.10.27 at the beginning of this section
• After completing the analysis of that cylinder, we noted that:
It is very difficult to obtain such a perfect equilibrium
• We saw the reasons also
■ But we see submarines and other mechanical devices which remain at a constant level under water
♦ Some images can be seen here
• This can be explained in 4 steps:
1. Consider a submarine at an interior point below water surface
♦ Let the density of the submarine be: 𝛒s
♦ Let the density of the water be: 𝛒w
2. We know that:
If 𝛒s = 𝛒w, the submarine will remain at the same level
3. If the submarine begins to experience a downward force, it means that, 𝛒w is lesser than 𝛒s
• The solution for this problem can be written in 4 steps:
(i) The submarine will let out some water using special valves
(ii) When water is let out, the mass of the submarine decreases
(iii) But there is no change in the volume. So the density decreases
(iv) When the density becomes equal to the density of the surrounding water, the submarine will no longer experience the downward force. Then the valves are closed.
4. If the submarine begins to experience an upward force, it means that, 𝛒w is larger than 𝛒s
• The solution for this problem can be written in 4 steps:
(i) The submarine will take in some water using special valves
(ii) When water is taken in, the mass of the submarine increases
(iii) But there is no change in the volume. So the density increases
(iv) When the density becomes equal to the density of the surrounding water, the submarine will no longer experience the upward force. Then the valves are closed
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