1 / 12

Work and Fluid Pressure

Work and Fluid Pressure. Lesson 7.7. 50. Work. Definition The product of The force exerted on an object The distance the object is moved by the force When a force of 50 lbs is exerted to move an object 12 ft. 600 ft. lbs. of work is done. 12 ft. a. b. x. Hooke's Law.

zanta
Télécharger la présentation

Work and Fluid Pressure

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Work and Fluid Pressure Lesson 7.7

  2. 50 Work • DefinitionThe product of • The force exerted on an object • The distance the object is moved by the force • When a force of 50 lbs is exerted to move an object 12 ft. • 600 ft. lbs. of work is done 12 ft

  3. a b x Hooke's Law • Consider the work done to stretch a spring • Force required is proportional to distance • When k is constant of proportionality • Force to move dist x = k • x = F(x) • Force required to move through i thinterval, x • W = F(xi) x

  4. Hooke's Law • We sum those values using the definite integral • The work done by a continuous force F(x) • Directed along the x-axis • From x = a to x = b

  5. Hooke's Law • A spring is stretched 15 cm by a force of 4.5 N • How much work is needed to stretch the spring 50 cm? • What is F(x) the force function? • Work done?

  6. Winding Cable • Consider a cable being wound up by a winch • Cable is 50 ft long • 2 lb/ft • How much work to wind in 20 ft? • Think about winding in y amt • y units from the top  50 – y ft hanging • dist = y • force required (weight) =2(50 – y)

  7. Pumping Liquids • Consider the work needed to pump a liquid into or out of a tank • Basic concept: Work = weight x dist moved • For each V of liquid • Determine weight • Determine dist moved • Take summation (integral)

  8. r b a Pumping Liquids – Guidelines • Draw a picture with thecoordinate system • Determine mass of thinhorizontal slab of liquid • Find expression for work needed to lift this slab to its destination • Integrate expression from bottom of liquid to the top

  9. Pumping Liquids 4 • Suppose tank has • r = 4 • height = 8 • filled with petroleum (54.8 lb/ft3) • What is work done to pump oil over top • Disk weight? • Distance moved? • Integral? 8 (8 – y)

  10. Fluid Pressure • Consider the pressure of fluidagainst the side surface of the container • Pressure at a point • Density x g x depth • Pressure for a horizontal slice • Density x g x depth x Area • Total force

  11. 2.5 - y Fluid Pressure • The tank has cross sectionof a trapazoid • Filled to 2.5 ft with water • Water is 62.4 lbs/ft3 • Function of edge • Length of strip • Depth of strip • Integral (-4,2.5) (4,2.5) (2,0) (-2,0) y = 1.25x – 2.5x = 0.8y + 2 2 (0.8y + 2) 2.5 - y

  12. Assignment • Lesson 7.7a • Page 307 • Exercises 1 – 13 odd, 21 • Lesson 7.7b • Page 307 • Exercises 23 – 35 odd

More Related