Solids and Fluids Chapter 13 PHYS-110 PDF

Solids and Fluids Source: livescience.com 1  Outline 13.2 States of Matter Chapter (13) 13.4 Pressure Pressure-Depth Relation 2  Le...

Solids and Fluids Source: livescience.com 1  Outline 13.2 States of Matter Chapter (13) 13.4 Pressure Pressure-Depth Relation 2  Learning Outcomes After studying this chapter, you will be able to: Differentiate between different states of matter: gas, liquid, and solid. Identify that pressure is a force per unit area. Identify density. Know the pressure- depth relation and apply it to solve problems. 3  States of Matter Gas Liquid Solid  Is a system in which When a liquid is placed in Does not require a each atom or molecule a container, it fills only container but defines moves through space as the volume corresponding its own shape. a free particle. to its initial volume.  Is compressible and fills Is nearly incompressible. Is incompressible. its container. Fluids 4  States of Matter Other States of Matter There are other states of matter that do not fit into the solid/liquid/gas classification. For example, foams, gels and matter in stars are in the form of plasma. The state of a certain matter depends on its temperature. Ice - solid Water - liquid Steam - gas 5  Pressure We will consider the pressure of the liquids and gases (fluids) at rest. Pressure is a scalar quantity and is defined as the force per unit area: The SI unit of pressure is N/m2, which has been given the name pascal, abbreviated Pa: The average pressure of the Earth’s atmosphere, 1 atm, is: 1 atm = 1.01 × 105 𝑃𝑎 Atmospheric pressure is often quoted in non-SI units: 6  Density Every material has a certain density associated with it. Density is a measure of mass per unit of volume: 𝒎 𝝆= 𝑽 Density is a scalar quantity which has magnitude only. The SI unit of density is: kg/m3. For example the density of water is 997 kg/m3 7  External example A stiletto heel has a surface area of approximately 2.5 cm × 2.0 cm. If a lady puts all of her 60 kg 'weight' on it, what pressure is created on the floor surface. SOLUTION: Stiletto area = (2.5/100)×(2.0/100) = 0.0005 m2 Weight = magnitude of gravitational force = mg = (60 kg)×(9.8 m/s2) = 588 N 𝐹 588 𝑁 Stiletto pressure: P = = = 1176000 N/m2 = 1.2 × 106 Pa 𝐴 0.0005 m2 8  Pressure-depth relationship In an open container, the pressure at any given height h consists of: 1) the atmospheric pressure po which is the pressure at y = 0. 2) the pressure due to the fluid which is equal to 𝛒 g h. Therefore, the pressure p at depth h where 𝛒 is the density of the fluid g is the acceleration of gravity= 9.8 m/𝐬𝟐 h is the depth. 9 Concept Check Three containers of water have different shapes. The distance between the top surface of the water and the bottom of each container is the same. Which of the following statements correctly characterizes the pressure at the bottom of the containers? A. The pressure at the bottom of container 1 is the highest. B. The pressure at the bottom of container 2 is the highest. C. The pressure at the bottom of container 3 is the highest. D. The pressure at the bottom of all three containers is the same. 10  External example A swimming pool of width 9.0 m and length 24.0 m is filled with water to a depth of 3.0 m. a) Calculate pressure on the bottom of the pool? b) Calculate the force on the bottom of the pool ? c) Calculate pressure on the bottom of the pool due to the water? Solution: 𝑝𝑜 = 1.01325 × 105 𝑃𝑎 𝜌 = 1000 kg/m3 𝑔 = 9.8m/s 2 ℎ = 3m a) The pressure at the bottom of the pool → 𝑝 = 𝑝𝑜 + 𝜌𝑔ℎ 𝑝 = 𝑝𝑜 + 𝜌𝑔ℎ = 1.01325 × 105 + [1000 × 9.8 × 3] =1.31× 105 Pa 𝐹 b) The force on the bottom of the pool 𝑃 = 𝐴 → 𝐹 = 𝑃 𝐴 = 1.31 × 105 × 216 = 2.82 × 107 𝑁 𝒘=𝟗𝐦 𝒍 = 𝟐𝟒 𝐦 c) The pressure at the bottom of the pool due to water → 𝑃 = 𝜌𝑔ℎ 𝜌 = 1000 kg/m3 , 𝑔 = 9.8 m/s2 ℎ = 3m 𝒉=3m P = (1000)×(9.8)×(3) = 29400 Pa 11  Equation summary Pressure (1) (2) Pressure at a given depth of a liquid: Density (3) 𝒎 𝝆= 𝑽 12 The END OF CHAPTER (13 ) 13