Work and Fluid PressureWorkHooke's LawSlide 4Slide 5Winding CablePumping LiquidsPumping Liquids – GuidelinesPumping LiquidsFluid PressureSlide 11Assignment ACentroidSlide 14Slide 15Slide 16Slide 17Slide 18Centroid of Area Under a CurveCentroid of Region Between CurvesTry It Out!Assignment BWork and Fluid PressureLesson 6.5Work•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 done5012 ftHooke'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) xabxHooke'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( )baW F x dx=�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?4.5 0.1530( ) 30F k xkkF x x= �= �==0.5030W x dx=�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)( )2002 50W y dy= -�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)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 topabr20( )aW r b y dyr p= �� -�Pumping Liquids •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?8454.8 16Weight yp= �� �D(8 – y)8054.8 16 (8 )Work y yp= �� � - D�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( ) ( )dcF h y L y dyr= ��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 (-2,0)(2,0)(-4,2.5)(4,2.5)2.5 - yy = 1.25x – 2.5x = 0.8y + 22 (0.8y + 2)2.5 - y2.5062.4 (2.5 )(1.6 4)F y y dy= - +�Assignment A•Lesson 6.5A•Page 405•Exercises 5 – 21 oddCentroid•Center of mass for a systemThe point where all the mass seems to be concentratedIf the mass is of constant density this point is called the centroid 4kg6kg10kg•Centroid•Each mass in the system has a "moment"The product of the mass and the distance from the origin"First moment" is the sum of all the moments•The centroid is 4kg6kg10kg1 1 2 21 2m x m xxm m+=+Centroid•Centroid for multiple points•Centroid about x-axis11ni iiniim xxm===��First moment of the systemAlso notated My, moment about y-axisFirst moment of the systemAlso notated My, moment about y-axisTotal mass of the systemTotal mass of the system11ni iiniiyymm===��Also notated Mx,moment about x-axisAlso notated Mx,moment about x-axisAlso notated m, the total massCentroid•The location of the centroid is the ordered pair•Consider a system with 10g at (2,-1), 7g at (4, 3), and 12g at (-5,2)What is the center of mass? ( , )x yyxMMx ym m= =Centroid•Given 10g at (2,-1), 7g at (4, 3), and 12g at (-5,2)10g7g12g10 (2) 7 4 12 ( 5)10 ( 1) 7 3 12 210 7 12yxMMm= � + �+ �-= �- + �+ �= + +? ?x y= =Centroid•Consider a region under a curve of a material of uniform densityWe divide the region into rectanglesMass of each considered to be centered at geometric centerMass of each is the product of the density, ρ and the areaWe sum the products of distance and massabxD•Centroid of Area Under a Curve•First moment with respectto the y-axis•First moment with respectto the x-axis•Mass of the region( )byaM x f x dxr= ���[ ]21( )2bxaM f x dxr= ��( )bam f x dxr= ��Centroid of Region Between Curves•Moments•Massf(x)g(x)( )( ) ( )byaM x f x g x dxr= �� -�[ ] [ ]( )2 21( ) ( )2bxaM f x g x dxr= � -�[ ]( ) ( )bam f x g x dxr= � -�yxMMx ym m= =CentroidTry It Out!•Find the centroid of the plane region bounded by y = x2 + 16 and the x-axis over the interval 0 < x < 4Mx = ?My = ?m = ?Assignment B•Lesson 6.5B•Page 405•Exercises 23 – 41