|
| |
Back
Figure 7-3.—No motion, no work. | Up
Basic Machines - Intro to machines and motion theories | Next
Friction |
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 600—or 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
F1 x S1 =FZXSZ
And by substituting the known
SI = 100 feet.
values:
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 7-5.—A block and tackle makes work easier.
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 feet—and 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
FI =
S1 =
Fz =
S2 =
force in pounds applied on the handle;
distance in feet that the end of the handle
travels in one revolution;
resistance to overcome;
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
FI x 2 x 2 1/2 = 1,5000 x 1/48
F1 = 2 pounds
The 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
7-3
|
