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PHY 126 Chapter 13 Lecture Notes

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PHY 126 Chapter 13 Lecture Notes PHY 126 Chapter 13 Lecture Notes Chapter 13: Fluid Force & Pressure OBJECTIVES Define and determine buoyant force Define and determine the force of fluid pressure BUOYANCY What we weigh in air and what we weigh in water is actually different. This is due to buoyancy of the fluid that the object is being immersed in. Archimedes Principle: Any object placed in a fluid apparently loses weight equal to the weight of the displaced fluid. Objects that weigh more displace more water. For instance, a cargo ship that is empty will displace less water than a full cargo ship. (Draw a sketch). This principle is taken into consideration when constructing very heavy cargo carriers that use water as their routes of transportation. For example, a heavy barge will displace a great amount of water. To compensate for this, barges are built very long and wide so the water is displaced over a large area, thus keeping the carrier from being completely submerged and eventually sinking. Buoyant force is the upward force exerted on a submerged or partially submerged object in water. The buoyant force is equal to the weight of the water that is displaced by the object. Note that the buoyant force opposes the force of gravity acting on an object. What can found as a result of Archimedes Principle is that an object essentially weighs less in water than in air. The object’s weight in water is known as its “apparent weight”. WEIGHT IN WATER (known as “apparent weight) = WEIGHT IN AIR – BUOYANT FORCE (weight of displaced water) BUOYANT FORCE = WEIGHT IN AIR – WEIGHT IN WATER EXAMPLE 1: A piece of metal weighs 67 N in the air and 62 N in the water. Find the buoyant force of the water. BUOYANT FORCE = WEIGHT IN AIR – WEIGHT IN WATER = 67 N – 62 N BUOYANT FORCE = 5 N 1 | Susan S. Gorelick Lecture Notes P H Y 1 2 6 C H 1 3 PHY 126 Chapter 13 Lecture Notes To better understand the concept of buoyancy, density must be considered. The density of an object can be found by the following equation: 풎 푫 = 푽 The density of an object is the amount of mass it has per unit volume 푫 = the density of the fluid that the object is submerged in m = the mass of the object V = volume of the object The density of water is 1,000 kg/m3 Objects that are less dense than water are able to float, and objects that are denser than water will sink when immersed. This is due to the buoyant force of the water. EXAMPLE 2: A block of brass with a measured volume of 0.5m3 has a measured mass of approximately 4,350 kg. Find the density brass based on these findings. GIVEN FIND FORMULA m = 4,350 kg D = ? m D = V = 0.5m3 V 4,350 kg D = 0.5m3 풌품 푫 = ퟖ, ퟕퟎퟎ 풎ퟑ The density of brass is approximately 8,700 kg/m3. (Confirm w/ appendix D on pg. 682) EXAMPLE 3: (Page 682) Based on its density, should you expect the brass block to float or sink? How about a concrete block? What about a block made out of cork? Objects that are less dense than water are able to float, and objects that are denser than water will sink when immersed. Find the densities of these objects from appendix D. The brass and concrete block will sink, while the block made out of cork will float. 2 | Susan S. Gorelick Lecture Notes P H Y 1 2 6 C H 1 3 PHY 126 Chapter 13 Lecture Notes To find the buoyant force exerted on an object immersed in a fluid: 푭푩 = 푫 ∙ 푽 ∙ 품 The buoyant force of exerted on submerged object is equal to the product of the objects (mass) density times its volume times the acceleration due to gravity. FB = buoyant force 푫 = the density of the fluid that the object is submerged in V = the volume of the water displaced by the immersed object g = acceleration due to gravity The density of water is 1,000 kg/m3 EXAMPLE 4: A large metal block displaces 0.25 m3 of water. Find the buoyant force of the water exerted on the block. GIVEN FIND FORMULA 3 Dwater = 1,000 kg/m FB =? FB = D ∙ V ∙ g V = 16.8 m3 g = 9.80 m/s2 kg m F = (1,000 ) (0.25 m3)(9.80 ) B m3 s2 푭푩 = ퟐ, ퟒퟓퟎ 푵 A buoyant force of 2,450 N is exerted on the block. 3 | Susan S. Gorelick Lecture Notes P H Y 1 2 6 C H 1 3 PHY 126 Chapter 13 Lecture Notes EXAMPLE 5: A 75 kg rock lies at the bottom of a pond. Its volume is 0.01m3. A) What is the buoyant force exerted on the rock? GIVEN FIND FORMULA 3 Dwater = 1,000 kg/m FB =? FB = D ∙ V ∙ g V = 0.01 m3 g = 9.80 m/s2 kg m F = (1,000 ) (0.01 m3)(9.80 ) m = 9.80 m/s2 B m3 s2 푭푩 = ퟗퟖ 푵 A buoyant force of 98 N is exerted on the block. B) What is the rock’s apparent weight in the water? APPARENT WEIGHT IN WATER = WEIGHT IN AIR – BUOYANT FORCE APPARENT WEIGHT IN WATER =735 N – 98 N APPARENT WEIGHT IN WATER =637 N C) How much force is needed to lift it while it is in the water? APPARENT WEIGHT IN WATER =637 N, so it must take at least 637 N of force to lift the object while in water. HYDROSTATIC PRESSURE Pressure is defined as the amount of force applied per unit area. 푭 푷 = 푨 P = the pressure applied to an object F = the force applied to an object A = the area of the surface (cross-sectional area) where the force is applied Pressure is measured in N/m2 (Pa) or lb/in2 (psi) Hydrostatic pressure is the amount of pressure liquid exerts on a submerged object. 4 | Susan S. Gorelick Lecture Notes P H Y 1 2 6 C H 1 3 PHY 126 Chapter 13 Lecture Notes As the depth of the water increases, so does the pressure. This is similar to stacking weights on top of each other. The bottom weight feels much more pressure than the top weight due to more bricks stacked on top of it. This is the same for water. The deeper you go, the more water molecules are stacked on top of you, so the more pressure you feel. The hydraulic (Pascal’s) principle states that “The pressure applied to a confined liquid is transmitted without measurable loss throughout the entire liquid to all inner surfaces of that container. Liquid are pretty much non-compressible. This means that when a specific amount of pressure is applied to a liquid, the liquid will transmit that same pressure. Pressure in (P1) Pressure out (P2) 푷풓풆풔풔풖풓풆 풊풏 = 푷풓풆풔풔풖풓풆 풐풖풕 푷ퟏ = 푷ퟐ 푭 푭 ퟏ = ퟐ 푨ퟏ 푨ퟐ What this means is that we can “step up” or “step down” force by changing the area at which the force is applied. (Look the picture of a hydraulic lift in textbook Chapter 13.) EXAMPLE 1: A small force of 5 lb is applied to a small piston with a cross-sectional area of 2 in2. How much force can be exerted by a large piston with an area of 30 in2? GIVEN FIND FORMULA F1 = 5 lb F2 =? 푭ퟏ 푭ퟐ 2 = A1 = 2 in 2 푨ퟏ 푨ퟐ A2 = 30 in P = P 1 2 ퟓ 풍풃 푭ퟐ = ퟐ 풊풏ퟐ ퟑퟎ 풊풏ퟐ ퟓ 풍풃 ퟐ 푭ퟐ = (ퟑퟎ 풊풏 ) ퟐ 풊풏ퟐ 푭ퟐ = ퟕퟓ 풍풃 The large piston can exert a force of 75 lb. EXAMPLE 2: A hospital bed weighs 1600 N and is raised by a large piston with a cross-sectional area of 75 cm2. What force must be exerted on the small piston with a cross-sectional area of 3.75 cm2 to lift the bed? 5 | Susan S. Gorelick Lecture Notes P H Y 1 2 6 C H 1 3 PHY 126 Chapter 13 Lecture Notes GIVEN FIND FORMULA F1 = 1,600 N F2 =? 푭ퟏ 푭ퟐ 2 = A1 = 75 cm 2 푨ퟏ 푨ퟐ A2 = 3.75 cm P = P 1 2 ퟏퟔퟎퟎ 푵 푭ퟐ = ퟕퟓ 풄풎ퟐ ퟑ. ퟕퟓ 풄풎ퟐ ퟏퟔퟎퟎ 푵 푭 = (ퟑ. ퟕퟓ 풄풎ퟐ) ퟐ ퟕퟓ 풄풎ퟐ 푭ퟐ = ퟖퟎ 푵 The small piston must exert a force of 80 N to lift the bed. 6 | Susan S. Gorelick Lecture Notes P H Y 1 2 6 C H 1 3 .
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