A block and tackle also makes work easier. Like anyother machine, it can’ t decrease the total amount of workto be done. With a rig like the one shown in figure 7-5,the sailor has a mechanical advantage of 5, neglectingfriction. Notice that five parts of the rope go to and fromthe movable block. To raise the 600-pound load 20 feet,he needs to exert a pull of only one-fifth of 600—or 120pounds. He is going to have to pull more than 20 feet ofrope through his hands to do this. Use the formulaagain to figure why this is so:Work input = work outputF1x S1 =FZXSZAnd by substituting the knownSI = 100 feet.values:This means that he has to pull 100 feet of ropethrough his hands to raise the load 20 feet. Again, theadvantage lies in the fact that a small force operatingthrough a large distance can move a big load through asmall distance.The sailor busy with the big piece of machinery infigure 7-6 has his work cut out for him. He is trying toseat the machine squarely on its foundations. He mustshove the rear end over one-half foot against a frictionalFigure 7-5.—A block and tackle makes work easier.Figure 7-6.—A big push.resistance of 1,500 pounds. The amount of work to bedone is 1,500 x 1/2, or 750 foot-pounds. He will haveto apply at least this much force on the jack he isusing. If the jack has a 2 1/2-foot handle— R =2 1/2 feet—and the pitch of the jack screw is one-fourthinch, he can do the job with little effort. Neglectingfriction, you can figure it out this way:Work input = work outputIn whichFI =S1 =Fz =S2 =force in pounds applied on the handle;distance in feet that the end of the handletravels in one revolution;resistance to overcome;distance in feet that the head of the jackadvanced by one revolution of the screw, or,the pitch of the screw.And, by substitution,Flx 2 x 3.14 x 21/2 = 1,500 x 1/48since1/4 inch = 1/48 of a footFIx 2 x 2 1/2 = 1,5000 x 1/48F1 = 2 poundsThe jack makes it theoretically possible for thesailor to exert a 1,500-pound push with a 2-pound effort.Look at the distance through which he must apply thateffort. One complete turn of the handle represents adistance of 15.7 feet. That 15.7-foot rotation advancesthe piece of machinery only one-fourth of an inch, or7-3

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