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1 Physics 131: Lecture 3 Today s Agenda Description of Fluids at Rest Pressure vs Depth Pascal s Principle: hydraulic forces Archimedes Principle: objects in a fluid Bernoulli s equation Physics 01: Lecture 1, Pg 1 Announcements Today: Last new material Wednesday in class (Dr. Harlow and I will review the semester) Final Exam: Thursday 1/13 at :00pm Review Session: Thursday 1/6 10:00am-1pm 1pm in SS118 (Dr. Harlow and I) Exam Jam My extra office hours: 1/11 9am 1 and 5pm 1/1 9am 1 and 5pm Drop in center: Check link on portal Conceptual Questions (mastering physics) Physics 01: Lecture 1, Pg States of Matter Fluids Solid Hold Volume Hold Shape Liquid Hold Volume Adapt Shape Gas Adapt Volume Adapt Shape Physics 01: Lecture 1, Pg 3 Page 1

2 Fluids What parameters do we use to describe fluids? Density Pressure m V P F A These letters look similar! Be careful. Physics 01: Lecture 1, Pg 4 You are my density The density of a material is its mass per unit volume: Physics 01: Lecture 1, Pg 5 Pressure Pressure = Force/Area Units SI Pascal = Newtons/m If I apply 30 Pa of pressure over an area of 0 m, how much force do I apply? F = P A = 30 N/m 0 m = 600 Newtons Physics 01: Lecture 1, Pg 6 Page

3 Pressure The same force applied over a smaller area results in greater pressure think of poking a balloon with your finger and then with a needle. Physics 01: Lecture 1, Pg 7 Pressure Physics 01: Lecture 1, Pg 8 Atmospheric Pressure Basically weight of atmosphere! Air molecules are colliding with you right now! Pressure = 1x10 5 N/m = 14.7 lbs/in! Example: Magdeburg sphere w/ r = 0.1 m A = 4 p r = 0.15 m F = 1,000 Newtons (over,500 lbs)! Physics 01: Lecture 1, Pg 9 Page 3

4 Pressure vs. Depth Physics 01: Lecture 1, Pg 10 Clicker Question 1: What can you say about the pressures at points 1,, and 3? A. p 1 = p = p 3. B. p 1 = p > p 3. C. p 3 > p 1 = p. D. p 3 > p 1 > p. E. p 1 = p 3 > p. Physics 01: Lecture 1, Pg 11 Gauge Pressure Many pressure gauges, such as tire gauges and the gauges on air tanks, measure not the actual or absolute pressure p but what is called gauge pressure p g. where 1 atm = kpa. Physics 01: Lecture 1, Pg 1 Page 4

5 Pascal s Principle Pascal s principle: An external pressure applied to an enclosed fluid is transmitted unchanged to every point within the fluid. Physics 01: Lecture 1, Pg 13 Pascal s Principle Consider the system shown: A downward force F 1 is applied to the piston of area A 1. This force is transmitted through the liquid to create an upward force F. Pascal s Principle says that increased pressure from F 1 (F 1 /A 1 ) is transmitted throughout the liquid. d 1 F1 F A 1 A d F1 F A F F1 A1 A A1 Check that F d is the same on both sides. Energy is conserved! Physics 01: Lecture 1, Pg 14 Clicker Question : The small piston of a hydraulic lift has a cross sectional area of 3 cm. You need to lift a 15,000 N weight by applying a force of only 00 N. What is the area of the large piston, assuming both pistons are at the same height, as shown in the drawing? (a) 75 cm (b) 15 cm (c) 5 cm Physics 01: Lecture 1, Pg 15 Page 5

6 Clicker Question 3: If you push the small piston down a distance of 5 cm. How far up does the weight move? (a) (b) (c) 0.3 cm 8. cm 5 cm Physics 01: Lecture 1, Pg 16 Clicker Question 4: When the cylinder is lowered into the water, how will the scale readings for the cylinder and the water change? a) cylinder s scale will decrease, but water s scale will increase b) cylinder s scale will increase, but water s scale will decrease c) cylinder s scale will decrease, but water s scale will not change d) both scales will decrease W W e) both scales will increase 1? Physics 01: Lecture 1, Pg 17 Archimedes Principle Suppose we weigh an object in air and in water. Since the pressure at the bottom of the object is greater than that at the top of the object, the water exerts a net upward force, the buoyant force, on the object. The buoyant force is equal to the difference in the pressures times the area. p F 1 B F F 1 F ( p ) B p1 A gha F B liquid gvliquid Mliquid g Wliquid W 1 F y 1 1 y A Therefore, the buoyant force is equal to the weight of the fluid displaced. W? Physics 01: Lecture 1, Pg 18 F p Page 6

7 Clicker Question 5: Imagine holding two identical bricks in place underwater. Brick 1 is just beneath the surface of the water, and brick is held about feet down. The force needed to hold brick in place is: a) greater b) the same c) smaller 1 Physics 01: Lecture 1, Pg 19 Archimedes Principle So the force the water pushes up with is equal to weight of the water displaced by the object. F B = W fluid F B = m fluid g= ( Vg) fluid Physics 01: Lecture 1, Pg 0 Clicker Question 5.5: Two blocks are of identical size. One is made of lead and sits on the bottom of a pond; the other is of wood and floats on top. Upon which is the buoyant force greater? A. On the lead block. B. On the wood block. C. They both experience the same buoyant force. Physics 01: Lecture 1, Pg 1 Page 7

8 Will it Float? Object is in equilibrium B liquid g Vdispl. F ( mg) object object g Vobject F mg B y V displ. Vobject object liquid The Tip of The Iceberg: What fraction of an iceberg is submerged? V 3 water displ. ice 917 kg/m 90% V 3 ice water 104 kg/m Physics 01: Lecture 1, Pg Archimedes Principle An object made of material that is denser than water can float only if it has indentations or pockets of air that make its average density less than that of water. Physics 01: Lecture 1, Pg 3 Clicker Question 6: A plastic cube (ρ = 780 kg/m 3, V = 0.71 m 3 ) is suspended by a string such that half the volume is submersed in water (ρ = 1000 kg/m 3 ) as shown in the figure. What is the tension T in the string? (a) (b) (c) (d) (e) 37 N 48 N 931 N 1950 N 380 N Physics 01: Lecture 1, Pg 4 Page 8

9 Clicker Question 7: Suppose you float a large ice-cube in a glass of water, and that after you place the ice in the glass the level of the water is at the very brim. When the ice melts, the level of the water in the glass will: A. Go up, causing the water to spill out of the glass. B. Go down. C. Stay the same. Physics 01: Lecture 1, Pg 5 Clicker Question 8: Which weighs more: A. A large bathtub filled to the brim with water. B. A large bathtub filled to the brim with water with a battle-ship floating in it. C. They will weigh the same. Tub of water Overflowed water Physics 01: Lecture 1, Pg 6 Fluids in motion Physics 01: Lecture 1, Pg 7 Page 9

10 Example Water comes out of a hose which has a radius of 1 cm. You are holding your finger over the hose and the area the hose can flow through is m. If water in hose has speed 1m/s, what is speed when water passes you finger? Assume the water is incompressible. 1 A 1 v 1 = A v 1 = A 1 v 1 = A v (.01 m) (1 m/s) = ( m ) v v = 1.56 m/s Physics 01: Lecture 1, Pg 8 Bernoulli s equation When a fluid moves from a wider area of a pipe to a narrower one, its speed increases; therefore, work has been done on it. W = F d = P(Ad) = ( P)V Physics 01: Lecture 1, Pg 9 Bernoulli s equation W =½mv f -½mv i W = F d = P(Ad) = ( P)V So: (P P 1 )V = ½mv -½mv 1) f i m= V (P P 1 )V = ½ Vv f -½ Vv i (P P 1 ) = ½ v 1 -½ v P + ½ v = P 1 + ½ v 1 P + gy + ½ v = P 1 + gy 1 + ½ v 1 Physics 01: Lecture 1, Pg 30 Page 10

11 Bernoulli s equation The kinetic energy of a fluid element is: Equating the work done to the increase in kinetic energy gives: Physics 01: Lecture 1, Pg 31 Bernoulli s equation The general case, where both height and speed may change, is described by Bernoulli s equation: This equation is essentially a statement of conservation of energy in a fluid. Physics 01: Lecture 1, Pg 3 Bernoulli s equation (at constant height) P + ½ v = P 1 + ½ v 1 P + ½ v This quantity must always remain the same So fluids with faster velocity will exert less pressure! This has many interesting applications! Physics 01: Lecture 1, Pg 33 Page 11

12 Applications of Bernoulli s equation The Bernoulli effect is simple to demonstrate all you need is a sheet of paper. Hold it by its end, so that it would be horizontal if it were stiff, and blow across the top. The paper will rise, due to the higher speed, and therefore lower pressure, above the sheet. Demos! Physics 01: Lecture 1, Pg 34 Applications of Bernoulli s equation: Airplane wing Physics 01: Lecture 1, Pg 35 Applications of Bernoulli s equation This lower pressure at high speeds is what rips roofs off houses in hurricanes and tornadoes, and causes the roof of a convertible to expand upward. It even helps prairie dogs with air circulation! Physics 01: Lecture 1, Pg 36 Page 1

13 Clicker Question 9: Oil (ρ o = 90 kg/m 3 ) flows through a horizontal cylindrical pipe of radius R 1 = 1 cm with a speed of v 1 = m/s. Along the flow direction the pipe widens to a radius R = 4 cm (as shown in the figure), the speed of the oil is then equal to v. What is the pressure difference P = P P 1 between the wide and narrow parts of the pipe? (a) P = 660 Pa (b) P = 1185 Pa (c) P = 150 Pa (d) P = 175 Pa (e) P = 1950 Pa Physics 01: Lecture 1, Pg 37 Page 13

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