PHY130 – FUNDAMENTAL OF PHYSICS
CHAPTER 8 MATTER
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8.1 Elasticity • The shape and volume of a solid can be changed by applying external forces. • When the external forces are removed, the object tends to return to its original shape and size.
• This is called elastic behavior. Elasticity The property by which a body returns to its original size and shape when the forces that deformed it are removed.
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• The elastic properties of solid are discussed in terms of stress and strain. Stress The force per unit area causing a deformation.
đ??š đ?‘†đ?‘Ąđ?‘&#x;đ?‘’đ?‘ đ?‘ = đ??´
(1)
Unit : Pascal (Pa) & Nm-2 Strain
A measure of the amount of the deformation. 3
• Stress is proportional to strain. Stress = Elastic modulus x Strain
(2)
• Elastic modulus is analogous to spring constant.
• Can be referred as the stiffness of material. • A material having a large elastic modulus is very stiff and hard to
deform.
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• There are three relationships having the form of Equation (2),
corresponding to the tensile, shear and bulk deformation and all of them satisfy an equation similar to Hooke’s Law for springs: đ??š = đ?‘˜đ?‘Ľ
(3)
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Figure 1 Hooke’s Law : L applied force.
Figure 2 Applied force vs. elongation for a typical metal under tension. 6
8.1.1
Stress – strain graph.
Figure 3 Stress vs. strain curve for an elastic solid. 7
8.1.1
Stress – strain graph.
• It is possible to exceed the elastic limit by applying a sufficiently great stress. • A material when given a stress beyond it’s elastic limit ordinarily doesn’t return to it’s original length when the external force is removed.
• When the stress is increased further, it surpasses the ultimate strength.
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Ultimate Strength The greatest stress the substance can withstand without breaking.
• The breaking point for a brittle material is just beyond the ultimate strength.
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8.1.2 Stress and strain. 8.1.2.1 Tensile deformation. The Young’s Modulus (Tensile Modulus, Y) • Describe the length elasticity of a material. • Suppose a wire or rod of original length đ??żđ?‘œ and cross sectional area, đ??´ elongates an amount ď „đ??ż under a stretching force, đ??š applied to it’s end. Figure 4 10
∆đ??ż đ?‘‡đ?‘’đ?‘›đ?‘ đ?‘–đ?‘™đ?‘’ đ?‘†đ?‘Ąđ?‘&#x;đ?‘Žđ?‘–đ?‘› = đ??żđ?‘œ
đ??š đ?‘‡đ?‘’đ?‘›đ?‘ đ?‘–đ?‘™đ?‘’ đ?‘†đ?‘Ąđ?‘&#x;đ?‘’đ?‘ đ?‘ = đ??´ đ??š ∆đ??ż =YĂ— đ??´ đ??żđ?‘œ đ??šđ??żđ?‘œ đ?‘Œ= đ??´âˆ†đ??ż
Unit : Pascal (Pa) & Nm-2
(4)
Value of Y depends only on the material of the wire or rod and not on its dimension.
https://youtu.be/yenIZ2ssHBA 11
8.1.2.2 Shear deformation. The Shear Modulus, S • Describe the shape elasticity of a material. • As shown in Figure 5, equal and opposite tangential forces, đ??š act on a rectangular block. • These shearing forces distort the block but the volume remains unchanged. Figure 5
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𝑇𝑎𝑛𝑔𝑒𝑛𝑡𝑖𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝑎𝑐𝑡𝑖𝑛𝑔 𝐹 𝑆ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑆𝑡𝑟𝑒𝑠𝑠 = = 𝐴𝑟𝑒𝑎 𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑏𝑒𝑖𝑛𝑔 𝑠ℎ𝑒𝑎𝑟𝑒𝑑 𝐴 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑠ℎ𝑒𝑎𝑟𝑑 ∆𝐿 𝑆ℎ𝑒𝑎𝑟𝑖𝑛𝑔 𝑆𝑡𝑟𝑎𝑖𝑛 = = 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑠𝑢𝑟𝑓𝑎𝑐𝑒𝑠 𝐿𝑜 𝐹𝐿𝑜 𝑆= 𝐴∆𝐿
(5) https://youtu.be/RP8IyG8QI-Y
Unit : Pascal (Pa) & Nm-2 13
8.1.2.3
Volume deformation.
The Bulk Modulus, B • Describe the volume elasticity of a material. • In Figure 6, a uniformly distributed compressive force acts on the surface of an object and is directed perpendicular to the surface at all points. Figure 6
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đ??š đ?‘ƒđ?‘&#x;đ?‘’đ?‘ đ?‘ đ?‘˘đ?‘&#x;đ?‘’ đ?‘œđ?‘› đ?‘ đ?‘˘đ?‘&#x;đ?‘“đ?‘Žđ?‘?đ?‘’ đ?‘Žđ?‘&#x;đ?‘’đ?‘Ž, đ??´ = đ?‘? = đ??´
Unit : Pascal (Pa) & Nm-2
• Suppose that the pressure on an object of original volume, đ?‘‰ is increased by an amount of ď „đ?‘?.
• The pressure increased cause a volume change ď „đ?‘‰, where ď „đ?‘‰ will be negative.
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𝑉𝑜𝑙𝑢𝑚𝑒 𝑆𝑡𝑟𝑒𝑠𝑠 = ∆𝑝 ∆𝑉 𝑉𝑜𝑙𝑢𝑚𝑒 𝑆𝑡𝑟𝑎𝑖𝑛 = − 𝑉 https://youtu.be/kr_Gipe158M
∆𝑉 ∆𝑝 = 𝐵 × − 𝑉 ∆𝑝𝑉 𝐵=− ∆𝑉
(6) 16
Example (1) A metal wire 75 cm long and 0.130 cm in diameter stretches 0.035 cm when a load of 8.0 kg is hung on its end. Find the stress, the strain and the Young’s modulus for the material of the wire. Answer : 1.27 x 1011 Pa Example (2) The bulk modulus of water is 2.1 GPa. Compute the volume contraction of 100 mL of water when subjected to a pressure of 1.5 Mpa.
Answer : - 0.07 mL 17
Example (3) A box-shaped piece of gelatin desert has a atop area 15 cm2 and a height of 3 cm. When a shearing force of 0.50 N is applied to the upper surface, the upper surface displaces 4 mm relative to the bottom surface. What are the shearing stress, the shearing strain and the shear modulus for the gelatin?
Answer : 333 Pa, 0.133, 2503.76 Pa
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8.2 Pressure in fluid Pressure
=
Force Surface area
Force
= Due to the weight of the column of liquid above it. F
= W
Figure 7 Calculating the pressure at a depth h in a liquid.
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P
=
W
=
A
mg A
=
( V) g A
=
Ahg A
𝑃 = 𝜌𝑔ℎ
(7) 20
8.3 Pascal Principle Pascal Principle When the pressure on any part of a confined fluid is changed, the pressure on every part of the fluid is also changed by the same amount.
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A. Hydraulic lift • A small diameter cylinder is connected to larger cylinder. • It shows that a small force is used to exert a large force by making the area of the output piston larger than the area of the input piston. đ??šđ?‘–đ?‘› đ??šđ?‘œđ?‘˘đ?‘Ą = đ??´đ?‘–đ?‘› đ??´đ?‘œđ?‘˘đ?‘Ą
(8)
Figure 8 Application of Pascal’s Principle : Hydraulic lift.
https://youtu.be/UVEkJ7h1_Oo 22
Example (4) A hydraulic jack with a small cylinder with cross sectional are of 100 mm2 and a big cylinder with cross sectional of 10 000 mm2. If a force 10 N is exerted to the small cylinder, what is the force produced on the big cylinder?
Answer : 1000 N
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B. Manometer • It is used to measure gas/air pressure. • Consist of U-shaped tube partially filled with a liquid (mercury or water). • When both hands are exposed to đ?‘ƒđ?‘Žđ?‘Ąđ?‘š, liquid level at A and B are equal. đ?‘ƒđ??´ = đ?‘ƒđ??ľ
(Same level)
C
B
A
Figure 9 Open tube manometer. 24
• To measure pressure:
ď‚Ž Right hand right-hand tube connected to gas supply ď‚Ž Gas exert pressure to liquid at point A ď‚Ž Liquid level at point B will increase until đ?‘ƒđ?‘? = đ?‘ƒđ??´ đ?‘ƒđ??ś = đ?‘ƒđ??´ đ?‘ƒđ?‘”đ?‘Žđ?‘ = đ?‘ƒđ?‘Žđ?‘Ąđ?‘š + â„Žđ?œŒđ?‘” https://youtu.be/2P5_J5JEHTQ
https://youtu.be/-P1EvVuuPoI
Atmospheric pressure = 1 atm
= 1.013 x 1015 N m-2 = 101.3 kPa = 76 cm Hg 25
Example (5) A gas is pumped into the left arm of a mercury manometer. The pressure in the left and right arm will be in equilibrium when the difference in mercury height is 30 cm. Calculate the gas pressure. Given the density of a mercury is 13600 kgm-3 and the atmospheric pressure is 76 cm Hg.
Answer : 61.3 kPa
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C. Barometer • It is used to measure atmospheric pressure, đ?‘ƒđ?‘Žđ?‘Ąđ?‘š. • Consist of a glass tube that is completely filled with mercury and it is inverted into a bowl of mercury.
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đ?‘ƒđ??´ = Atmospheric pressure đ?‘ƒđ??ľ = Vacuum pressure + ď ˛đ?‘”â„Ž ď œ đ?‘ƒđ??´ = đ?‘ƒđ??ľ = đ?‘ƒđ?‘Žđ?‘Ąđ?‘š = 0
+ ď ˛đ?‘”â„Ž
+ (13600)(9.8)
B A
Figure 10 Mercury Barometer.
76 100
= 1 Ă— 105 đ?‘ƒđ?‘Ž https://youtu.be/3rIWdTYSfDE 28
8.4 Buoyancy and Archimedes’ Principle • Figure 10 below showing a cylinder of height h whose top and bottom ends have an area A and it is completely submerged in a fluid of density . Buoyant force occurs because: • pressure in fluid increases with depth
So, • the upward pressure (bottom surface) is greater than the downward pressure (top surface). Figure 11 Determination of the buoyant force
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Buoyant force
= đ??š2 − đ??š1 = (đ?‘ƒ2 Ă— đ??´2) − (đ?‘ƒ1 Ă— đ??´1) = (â„Ž2 đ?œŒđ?‘”đ??´) − (â„Ž1 đ?œŒđ?‘”đ??´) = (â„Ž2 − â„Ž1) đ?œŒđ?‘”đ??´ = đ?‘‰đ?œŒđ?‘”
V = volume of displaced fluid
= (đ?œŒđ?‘‰)đ?‘”
= đ?‘šđ?‘”
m = mass of displaced fluid
ď œ Buoyant force = weight of fluid displaced by the cylinder 30
Archimedes’ Principle It states that the buoyant force on a body immersed (fully or partly immersed) in a fluid is equal to the weight of the fluid displaced by that object.
đ??šđ??ľ = đ?‘Šđ?‘‘đ?‘–đ?‘ đ?‘?đ?‘™đ?‘Žđ?‘?đ?‘’đ?‘‘
Where:
(9)
đ?‘đ?‘Š
= Buoyant force
đ?‘žđ?’…đ?’Šđ?’”đ?’‘đ?’? = Weight of displaced fluid
đ?‘Š − đ?‘Š ′ = đ?‘Šđ?‘‘đ?‘–đ?‘ đ?‘?đ?‘™đ?‘Žđ?‘?đ?‘’đ?‘‘
(10)
đ?‘ž
= True weight (Weight in air)
�’
= Apparent weight (Weight in fluid)
https://youtu.be/YxYXSoj2wpU 31
Principle of Flotation It states that an object that floats on a surface of fluid will displaced fluid with weight that is equal to the weight of object that is immersed in the fluid.
Figure 12 An object floating in equilibrium : đ??šđ??ľ = đ?‘šđ?‘”
https://youtu.be/pjJIubb8nwM 32
Example (6) Given an object with mass 200 g and volume 25 cm3. When it is fully submerged in the liquid, the weight of the object becomes 0.17 gN. Calculate the density for the liquid. Answer : 1200 kgm-3
Example (7) An ice cube with boundaries of 2.0 cm is floating in a cup of tea. It is obtain that one of the surface; 0.20 cm is on the surface of the tea. Calculate the density for the tea. Given ice density = 920 kgm-3 Answer : 10222.2 kgm-3 33
8.5 Fluid in Motion Flow Rate and the Equation of Continuity • Hydrodynamics or Fluid Dynamics is the study of fluids in motion. • There are two types of flow : Laminar flow & Turbulent flow.
Figure 13 (a) Streamline or laminar flow; (b) turbulent flow
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• The flow is said to be streamline or laminar flow when the neighboring layers of fluid slide by each other smoothly (Figure 13a).
• Each particle of the fluids follow a smooth path called a streamline and they do not cross over one another. • Above certain speed, the flow becomes turbulent. • Turbulent flow is characterized by erratic, small, whirlpool-like circles called eddy currents or eddies (Figure 13b).
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• Lets consider the steady laminar flow of a fluid through an enclosed tube or pipe as shown in Figure 14.
Figure 14 Fluids flow through a pipe of varying diameter.
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• The mass flow rate is the mass that passes a given point per unit time. • The flow rates at any two points must be equal, as long as no fluid is being added or taken away. • This gives us the equation of continuity:
đ?œŒ1 đ??´1 đ?‘Ł1 = đ?œŒ2 đ??´2 đ?‘Ł2
(11)
• If the density doesn’t change – typical for liquids – this simplifies to:
Where the pipe is wider, the flow is slower. 37
Bernoulli’s Principle
It states where the velocity of a fluid is high, the pressure is low, and when the velocity is low, the pressure is high
• Bernoulli’s equation for the steady flow of a continuous fluid: 1 2 1 2 đ?‘ƒ1 + đ?œŒđ?‘Ł1 + â„Ž1 đ?œŒđ?‘” = đ?‘ƒ2 + đ?œŒđ?‘Ł2 + â„Ž2 đ?œŒđ?‘” 2 2 Where đ?œŒ1 = đ?œŒ2 = đ?œŒ and đ?‘” is due to gravity. https://youtu.be/UJ3-Zm1wbIQ Figure 15 Fluid flow: for derivation of Bernoulli’s equation
https://youtu.be/brN9citH0RA 38
Example (8) A horizontal pipe has a constriction in it, as shown in figure below. At point 1 the diameter is 6 cm, while at point 2 it is only 2 cm. At point 1, đ?‘Ł1 = 2 đ?‘šđ?‘ −1 and đ?‘ƒ1 = 180 đ?‘˜đ?‘ƒđ?‘Ž. Calculate đ?‘Ł2 and đ?‘ƒ2. Answer : 18 ms-1, 20 kPa
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