In the previous section, we saw the basics about surface tension. In this section we will see contact angle and capillarity. Some basic details about contact angle can be written in 9 steps:
1. Fig.10.53 below shows a drop of liquid on the surface of a solid
Fig.10.53
• In the fig., three different substances are interacting with each other. They are:
♦ The liquid substance of the drop
♦ The solid substance on which the liquid rests
♦ The air which surrounds both the drop and the solid
2. Consider a chain of liquid molecules in the periphery of the drop
• Let the length of the chain be k meters
♦ If k is very small, the chain can be considered to be straight
♦ We can then indicate the chain by a small straight line
♦ This is the yellow line shown in the fig.10.53
3. The molecules in the yellow line are liquid molecules
• They are acted upon by a 'force of attraction'
♦ This force of attraction is from the solid molecules
• This force will be tangential to the solid surface
♦ In other words, it will be parallel to the solid surface
• This force of attraction is indicated by the green arrow
♦ The green arrow is parallel to the solid surface
♦ This will be clear if we look from the side
♦ The side view is shown in fig.10.54 below:
Fig.10.54
• Let Ssl be the magnitude of this force per unit length
♦ Then the total force experienced by the line will be equal to (Ssl × k)
♦ The subscript ‘sl’ indicates that:
✰ It is the force due to the interaction between solid and liquid
4. The liquid molecules in the yellow line are acted upon by another force also
• It is the force of attraction from the air molecules
• This force will be tangential to the liquid surface
♦ The tangent should be drawn at:
✰ The point of contact between the yellow line and the liquid
• This force is denoted by the magenta arrow
• Let Sla be the magnitude of this force per unit length
♦ Then the total force experienced by the line will be equal to (Sla × k)
♦ The subscript ‘la’ indicates that:
✰ it is the force due to the interaction between liquid and air
■ It may be noted that:
♦ Sla is the ordinary surface tension which we saw earlier based on fig.10.48
♦ If the liquid is water, and the surrounding gas is atmospheric air, Sla will be equal to 0.0727 N m-1
5. There is one more force that we have to consider
• It is the force acting on the air molecules
♦ The air molecules close to the yellow line need to be considered
♦ Such a line of air molecules close to the yellow line is denoted by the grey line in figs.10.53 and 10.54
♦ It's length is also k
• The air molecules in the grey line are acted upon by the force of attraction
♦ It is the force of attraction by the solid molecules
• This force will be tangential to the solid surface
♦ The tangent should be drawn at:
✰ The point of contact between the grey line and the solid
✰ So it will be parallel to the solid surface
• This force is denoted by the red arrow
• Let Ssa be the magnitude of this force per unit length
♦ Then the total force experienced by the grey line will be equal to (Ssa × l)
♦ The subscript ‘sa’ indicates that
✰ It is the force due to the interaction between solid and air
6. So now we know the three forces. Let us see how they interact with each other
• Two different cases are possible:
Case 1. This can be explained in 3 steps:
(i) If the adhesion between air and solid is high, the air will try to spread over more and more solid surface
• The grey line will then push the yellow line more and more towards the right
(ii) In such a situation, if the liquid has greater surface tension, the magenta arrow will have a high magnitude
So the drop will roll upwards into a spherical shape
(iii) This will reduce the contact surface between the drop and the solid
• The area vacated by the liquid will be occupied by air
♦ This is shown in fig.10.55(a) below:
 |
| Fig.10.55 |
• In such a situation, we say that: the liquid is not able to wet the solid properly
♦ Action of water on wax is an example
Case 2. This can be explained in 2 steps:
(i) If the liquid has high adhesion to solid, the drop will spread out
• The yellow line will push the grey line more and more towards the left
• The liquid will wet more and more solid surface
(ii) The direction of Sla will change as shown in fig.10.55(b)
♦ The action of water on plastic is an example
7. So we see that, there is a competition between the three forces
• We can form a mathematical model to represent this competition
• For that, we introduce a new quantity: the angle of contact
♦ It is denoted by đ. This is shown in fig.10.55
8. The significance of đ can be written in 4 steps:
(i) In fig.10.55(a), the drop is in equilibrium
• So forces towards the left must be equal to the forces towards the right
• Thus we get: $\mathbf\small{\rm{S_{sa}k+S_{la}k \cos \theta=S_{sl}k}}$
⇒ $\mathbf\small{\rm{S_{sa}+S_{la} \cos \theta=S_{sl}}}$
• But cosine of any obtuse angle is negative
♦ So the Sla term will be added to the right side in the final result
(ii) In fig.10.53(b), the drop is in equilibrium
• So forces towards the left must be equal to the forces towards the right
• Thus we get: $\mathbf\small{\rm{S_{sa}k=S_{sl}k+S_{la}k \cos \theta}}$
⇒ $\mathbf\small{\rm{S_{sa}=S_{sl}+S_{la} \cos \theta}}$
♦ Here the Sla term is already on the right side
(iii) So the results in (i) and (ii) are basically, the same. We need not worry about the directions. The cosine of đ will take care of the sign
■ We can write the general equation:
Eq.10.19: $\mathbf\small{\rm{S_{sa}=S_{sl}+S_{la} \cos \theta}}$
(iv) The angle will depend on the properties of the liquid and the solid
• Comparison has to be made between two items:
(i) The cohesion between liquid molecules
(ii) The adhesion between liquid molecules and solid molecules
• If the first item is larger, the liquid molecules will group together to form a spherical shape. It will not wet the solid
♦ In this case, đ will be obtuse
• If the first item is smaller, the liquid molecules will spread out on the solid surface. It will wet the solid
♦ In this case, đ will be acute
• So a manufacturer of water proof materials will choose the solid material in such a way that, đ is obtuse
9. Now we can write the definition for đ. It can be written in 5 steps:
(i) Mark the boundary between the solid and the liquid
♦ In fig.10.55(a), the boundary is just a point P
♦ In fig.10.55(b), the boundary is the line PQ
(ii) What ever be the shape of the boundary, within that boundary, there will be no contact between liquid and air
• So we take the point just outside the boundary
♦ This is the first point at which the liquid makes contact with the air
♦ Recall that, the grey line in fig. corresponds to this point
(iv) Draw a tangent at this point
• It must be tangential to the liquid surface at that point
♦ In fig.10.55, this tangent is the force Sla k
(v) Mark the angle between this tangent and the solid surface
■ This angle is the angle of contact
■ It is important to always mark the angle inside the liquid
• Next we will see some practical applications of contact angle
• First we will see capillary rise
• In fig.10.56(a) below, a capillary tube is inserted vertically into an open vessel of water
 |
| Fig.10.56 |
• Water rises up to a height h inside the tube. This is known as capillary rise
• The word 'capilla' in Latin means hair
• If the inner diameter of the capillary tube is very small as in the case of a hair strand, the height of water h will be large
• The explanation for capillary rise can be written in 10 steps:
1. The angle of contact đ between glass and water is acute
• So the surface of water will be curved downwards
♦ When looked from above, the surface will have a concave shape
■ The curved surface of a liquid is called meniscus
• If đ is 90o, the surface will be flat
• If đ is obtuse, the surface will be convex
2. Fig.10.56(b) shows an enlarged view of the point of contact
• The surface tension of water (Sla) makes an angle đ with the glass surface
♦ For convenience, we will denote 'Sla' as simply 'S'
♦ This S is indicated in magenta color
♦ Recall that, we used the same magenta color to indicate 'Sla' in the earlier figs.10.53, 10.54 and 10.55
• The S indicates the tendency of water to spread over the glass surface
♦ This tendency is due to the adhesion between water and glass molecules
3. We see two đ angles in the fig.10.56(b)
(i) The lower đ is the one drawn according to the definition that we wrote in step (9) where we discussed contact angle
(ii) The upper đ will be the same because, they are opposite angles
4. Since S is inclined, it can be resolved into horizontal and vertical components
♦ The horizontal component is S sin đ
♦ The vertical component is S cos đ
♦ These are shown in fig.10.56(c)
5. The vertical component will lift the water upwards
• We see that, the vertical component acts at the tip of the meniscus
• But what appear as a tip is actually a ring. It is the periphery of the meniscus
♦ This is clear from the 3D view shown below:
 |
| Fig.10.57 |
• The horizontal components will all cancel each other
6. S is the force per unit length
• So S cos đ is also the force per unit length
• So the total upward force = (S cos đ × Peripheral length of the meniscus) = (S cos đ × 2đšr)
♦ Where r is the radius of the capillary tube
7. This total upward force will support the weight of the water column
• We have: weight of the water column = (Ī × đšr2hg)
♦ Where Ī is the density of water
8. Equating the results in (6) and (7), we get:
(S cos đ × 2đšr) = (Ī × đšr2hg)
■ Thus we get Eq.10.20: $\mathbf\small{\rm{h=\frac{2S\cos \theta}{\rho gr}}}$
9. When water is in contact with glass, the angle đ is very small. It is nearly equal to zero
• When đ is zero, cos đ is 1
■ So, when the liquid inside the capillary tube is water, we have:
Eq.10.21: $\mathbf\small{\rm{h=\frac{2S}{\rho gr}}}$
10. From Eqs.10.20 and 10.21, it is clear that, more water will rise when radius of the tube decreases
• Let us calculate the rise of water in a glass tube of radius 0.05 cm:
♦ Using Eq.10.21, we get: $\mathbf\small{\rm{h=\frac{2S}{\rho gr}=\frac{2(0.073)}{(10^3)(9.8) (5\times 10^{-4})}}}$ = 2.98 cm
✰ Where S of water is 0.073 N m-1, which is obtained from the data book
• In fig.10.58(a) below, a capillary tube is inserted vertically into an open vessel of mercury
 |
| Fig.10.58 |
• Mercury falls by a height h inside the tube. This is known as capillary depression
• If the inner diameter of the capillary tube is very small as in the case of a hair strand, the depression h will be large
• The explanation for capillary depression can be written in 10 steps:
1. The angle of contact đ between glass and mercury is obtuse
• So the surface of mercury will be curved downwards
♦ When looked from above, the surface will have a convex shape
■ The curved surface of a liquid is called meniscus
• If đ is 90o, the surface will be flat
• If đ is acute the surface will be concave
2. Fig.10.58(b) shows an enlarged view of the point of contact
• The surface tension of mercury (Sla) makes an angle đ with the glass surface
♦ For convenience, we will denote 'Sla' as simply 'S'
♦ This S is indicated in magenta color
♦ Recall that, we used the same magenta color to indicate 'Sla' in the earlier figs.10.53, 10.54 and 10.55
• The S indicates the tendency of mercury to repel away from the glass surface
♦ This tendency arises because:
✰ The cohesive force between mercury molecules
✰ is greater than
✰ the adhesion between mercury and glass molecules
✰ So the cohesive forces will try to pull mercury molecules into a sphere
3. We see two đ angles in the fig.10.58(b)
(i) The lower đ is the one drawn according to the definition that we wrote in step (9) where we discussed contact angle
(ii) The upper đ will be the same because, they are opposite angles
4. Since S is inclined, it can be resolved into horizontal and vertical components
Note that, the angle with the vertical is (180-đ)
♦ The horizontal component = S sin (180-đ) = S sin đ
♦ The vertical component is S cos (180-đ) = -S cos đ
♦ These are shown in fig.10.58(c)
✰ The -ve sign indicates that the vertical component acts downwards
5. The vertical component will pull the mercury downwards
• We see that, the vertical component acts at the tip of the meniscus
• But what appear as a tip is actually a ring. It is the periphery of the meniscus
♦ This is clear from the 3D view shown in fig.10.57(b) above
• The horizontal components will all cancel each other
6. S is the force per unit length
• So S cos đ is also the force per unit length
• So the total downward force = (S cos đ × Peripheral length of the meniscus) = (S cos đ × 2đšr)
♦ Where r is the radius of the capillary tube
7. This total downward force will keep the weight of the mercury column suppressed
• We have: weight of the mercury column = (Ī × đšr2hg)
♦ Where Ī is the density of mercury
8. Equating the results in (6) and (7), we get:
(S cos đ × 2đšr) = (Ī × đšr2hg)
■ Thus we get the same Eq.10.20: $\mathbf\small{\rm{h=\frac{2S\cos \theta}{\rho gr}}}$
9. When mercury is in contact with glass, the angle đ is 130o
10. From Eqs.10.20, it is clear that, more mercury will fall when radius of the tube decreases
• Let us calculate the fall of mercury in a tube of radius 1.25 mm:
♦ Using Eq.10.20, we get: $\mathbf\small{\rm{h=\frac{2S \cos \theta}{\rho gr}=\frac{2(0.4355)}{(13.6 \times 10^3)(9.8) (1.25\times 10^{-3})}}}$ = -0.00336 m = -3.36 mm
✰ Where S of mercury is 0.4355 N m-1, which is obtained from the data book
Solved example 10.34
What is the maximum mass of a needle which can rest on the surface of water. The length of the needle is 4cm. Surface tension of water = 0.073 N m-1
Solution:
1. Fig.10.59(a) shows the 3D view of a needle resting on the surface of water
 |
| Fig.10.59 |
2. The weight of the needle acts on the surface of water
♦ Due to the weight of the needle, there will be a small depression
♦ Due to the depression, there will be a contact angle
♦ This is shown in fig.10.60(a) below:
 |
| Fig.10.60 |
3. If there is no depression, there will not be a contact angle
♦ This is clear from fig.10.60(b)
4. We see two đ angles in the fig.10.60(a)
(i) The lower đ is the one drawn according to the definition that we wrote in step (9) where we discussed contact angle
(ii) The upper đ will be the same because, they are opposite angles
5. Since S is inclined, it can be resolved into horizontal and vertical components
♦ The horizontal component is S sin đ
♦ The vertical component is S cos đ
6. The vertical component will support the weight of the needle
♦ We see that, the vertical component acts at the tip of the meniscus
♦ But what appear as a tip is actually a line. It is the line of the needle
♦ In fact, there are two lines. One on either side of the needle
♦ This is shown in the earlier fig.10.59(b)
• So the total upward force = 2Sl cos đ
♦ Where l is the length of the needle
7. When the mass of the needle increases, the water surface become more and more depressed as shown in fig.10.60(c)
• As a result, the angle đ becomes smaller and smaller
• The angle đ can decrease to such a level that, it is very close to zero
• But đ cannot become equal to zero because, it would then mean that:
♦ S is vertical and
♦ there is no horizontal component
• Surface tension cannot become vertical
8. However, for the extreme case (maximum possible mass of the needle), we can take đ to be very close to zero and hence cos đ as 1
• So result in (6) becomes:
♦ Total upward force = 2Sl
9. Equating this to the maximum possible weight, we get:
2Sl = m(max)g
⇒ $\mathbf\small{\rm{m_{max}=\frac{2Sl}{g}=\frac{(2)(0.073)(0.04)}{(9.8)}}}$ = 0.0006 kg = 0.6 grams
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