25 lbs 12 5 lbs the work done in moving the bottom

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·25 lbs.=12.5 lbs. , the work done in moving the bottom half of the chain5 ft. upward isW1=(12.5 lbs. )·(5 ft. )=62.5 ft.-lbs.25
It remains to examine the second14of the chain. Here, each piece of the chain islifted a distance equal to twice its original distance from the pointPlocated14of theway down the chain (eg., the pointPdoes not move at all, but the midpoint of the chainstarts 2.5 ft. belowP, and ends up 2.5 ft. aboveP, which is 2·2.5 ft.=5 ft. aboveits original position). Therefore, if we setxto be the starting distance belowP, thenwe letxrange over 0x2.5 ft., and each piece of thicknessdxis moved upward adistance 2x. The weight density of the chain is given by25 lbs.10 ft.=2.5 lbs./ft.This means that a piece of chain of lengthdxwill have weight 2.5dxlbs., and since itis moved a distance 2x, the work done is given by the integralW2=integraldisplay2.50(2.5dx)(2x)=integraldisplay2.505xdx=2.5x2|x=2.5x=0=2.5·(2.5)20=15.625 ft.-lbs.Putting the two parts of the work together, the total work done isW1+W2=62.5 ft.-lbs.+15.625 ft.-lbs.=78.125 ft.-lbs.26
Problem 11A granary has the shape of a half cylinder lying on its rectangular side(the cut). The cylinder’s height is 10m, and the radius of the base is 2m. If the granaryis full of barley, with density 600kg/m3, how much work is done in removing all thegrain via an opening at the top of the granary?Solution:Step 1— DRAW A PICTURE!! (see accompanying diagrams).We divide the granary into horizontal slices, each at a heightzabove the base ofthe granary (so thatzruns from 0 to 2m). For eachz, the slice at heightzis a rectangleof length 10m (the “height” of the sideways cylinder), and thicknessdz, whose widthstretches from one side of the semi-circular cross-section to the opposite side. A simpleapplication of the Pythagorean Theorem shows that this width is 24z2. Thus thevolume of the slice at heightzis given bydV=(10 m)(2radicalbig4z2m)(dzm)=20radicalbig4z2dzm3Since the density of the barley is 600 kg/m3, the mass of barley in this slice isdm=(600 kg/m3)(dV)=600·20radicalbig4z2dzkg=12,000radicalbig4z2dzkgFrom the lawF=mg=m·9.8, we have a forcedF=dm·9.8=(9.8)(12,000radicalbig4z2dz)=117,600radicalbig4z2dzNacting on the slice at heightz.Next, observe that the slice at heightzis moved a distance 2zupward to get outthe top of the granary. Therefore,W=integraldisplay20(117,600radicalbig4z2dz)·(2z)=117,600integraldisplay20(2z)radicalbig4z2dzThis integral can be split into two parts. The first part isW1=117,600integraldisplay202radicalbig4z2dz=235,200integraldisplay20radicalbig22z2dz27
Observe that this integral is a candidate for the trigonometric substitutionz=2 sin(u),withdz=2 cos(u)du, so thatintegraldisplayz=2z=0radicalbig22z2dx=integraldisplayu=π/2u=0radicalBig2222sin2(u)(2 cos(u)du)=integraldisplayπ/204 cos2(u)dusinceradicalBig22(1sin2(u))=2radicalbigcos2(u)=cos(u)for 0uπ/

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