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Since Work = Force x Distance, you can substitute Force x Distance on each side of the work equation. Thus:
in which FI = effort applied, in pounds S1 = distance through which effort moves, in feet F2 = resistance overcome, in pounds S2 = distance resistance is moved, in feet Now substitute the known values, and you get: 1OO x S1 = 300 x I S1 = 3 feet The advantage of using the lever is not that it makes any less work for you, but it allows you to do the job with the force at your command. Youd probably have some difficulty lifting 300 pounds directly upward without a machine to help you!
Figure 7-5 - A block and tackle makes work easier A block and tackle also makes work easier. Like any other machine, it can t decrease the total amount of work to be done. With a rig like the one shown in figure 7-5, the sailor has a mechanical advantage of 5, neglecting friction. Notice that five parts of the rope go to and from the movable block. To raise the 600-pound load 20 feet, he needs to exert a pull of only one-fifth of 600or 120 pounds. He is going to have to pull more than 20 feet of rope through his hands to do this. Use the formula again to figure why this is so: Work input = work output
And by substituting the known values: 120 x S1 = 600 x 20 S1 = 100 feet.
This means that he has to pull 100 feet of rope through his hands to raise the load 20 feet. Again, the advantage lies in the fact that a small force operating through a large distance can move a big load through a small distance. The sailor busy with the big piece of machinery in figure 7-6 has his work cut out for him. He is trying to seat the machine squarely on its foundations. He must shove the rear end over one-half foot against a frictional
Figure 7-6.A big push. resistance of 1,500 pounds. The amount of work to be done is 1,500 x 1/2, or 750 foot-pounds. He will have to apply at least this much force on the jack he is using. If the jack has a 2 1/2-foot handle R = 2 1/2 feetand the pitch of the jack screw is one-fourth inch, he can do the job with little effort. Neglecting friction, you can figure it out this way:Work input = work output In which F1 = force in pounds applied on the handle; S1 = distance in feet that the end of the handle travels in one revolution; F2 = resistance to overcome; S2 = distance in feet that the head of the jack advanced by one revolution of the screw, or, the pitch of the screw. And, by substitution, Fl x 2 x 3.14 x 21/2 = 1,500 x 1/48 since 1/4 inch = 1/48 of a foot F1 x 2 x 2 1/2 = 1,5000 x 1/48 F1 = 2 poundsThe jack makes it theoretically possible for the sailor to exert a 1,500-pound push with a 2-pound effort. Look at the distance through which he must apply that effort. One complete turn of the handle represents a distance of 15.7 feet. That 15.7-foot rotation advances the piece of machinery only one-fourth of an inch, or one-forty-eighth of a foot. You gain force at the expense of distance. |
 
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