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[TowerTalk] Ginpole for 20 foot sections

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Subject: [TowerTalk] Ginpole for 20 foot sections
From: (David Robbins)
Date: Wed, 13 Jun 2001 16:27:41 +0100 wrote:
> The pulley at the top of a gin pole merely changes
> the direction of the rope.  Please explain how a
> block and tackle "Magically" reduces the force on
> the rope to the top pulley and how to install the
> block and tackle to achieve this magical reduction.

start with this, a single pulley at B with a 300lb load at C and a 300lb person
at A holding it off the ground.  there is 300lb of tension in the rope to hold
the load up, so at B there is 300lb pulling down going to C and 300lb more
pulling down going to A so the total on the support at B is 600lb... no magic
here, the pulley at B only changes the direction of the force.

       /   \    
      /     \   
     /       \  
    /         \ 
   /           \
   A           C|

> One more point.  One pulley can give a 2:1 mechanical
> advantage if one end of the rope is attached to a FIXED
> support.  If both ends of the rope are moving, there is
> NO mechanical advantage, only a direction change
> plus loss due to friction.

ok, so you understand the 2:1 advantage of the single moving pulley.. Fix the
rope to a support at D, run through the pulley at C that is attached to the
load, and pull up at B.

          B           /
           \         /
            \       /
             \     /
              \   /

so now to lift the load at C you are pulling up at B.  Since there are two
segments of the rope supporting the weight at C each of them must carry half
this load.  so for a 300lb load the force on each rope C-B and C-D, so each of
them has to support 1/2 the load weight or 150lb.  so as you expect the lifter
at B only exerts 150lb of lift for the 2:1 advantage.  the other 150lb is
pulling down at D, so the load on the support is only 1/2 the weight... this is
the magic!  you have divided the load to 2 different supports thereby reducing
your effort by 1/2, though you do have to pull twice as much rope to move the
load the same distance as before.

now take the line you are pulling up on to lift something with that single
pulley and run it through another pulley so you can pull down on it while
lifting the load attached to the moving pulley.  note, the new pulley does only
change the direction of the pull as you would expect for a gin pole pulley.

so now you have an arrangement like this with a pulley 'C' holding the load, and
a pulley 'B' attached to the ceiling(or gin pole), where the end 'D' is also

        B|             /D
        /O\           /
       /   \         /
      /     \       /
     /       \     /
    /         \   /
   /           \O/
   A           C|

now, lets put that 300lb load at C.  you will agree that the person pulling on
the rope at A only needs to exert 150lbs on the rope.  so looking at point B
there is 150lb on the rope from A and 150lb on the rope going to C for a total
of 300lb on the support at B.  the rope going up to the fixed end at D also has
150lb of tension so the force on D is 150lb.  so the total downward force on the
top support is now 450lb.  at C there is 150lb upward on the rope going to B and
150lb upward on the rope going to D so you have the 300lb total needed to keep
the 300lbs at C in the air.

of course if B is the top pulley on the gin pole you would attach the dead end
shown above at D to the top of the pole, and then have the moving pulley C below
it... it gets messy to draw in text but would look something like this:

       | | |
       | | |
       | | |
       | | |
       | | |
       | \O/C
       |  |

getting a 3:1 advantage would require another fixed pulley at the top of the
pole to reverse the direction again and then attach the dead end to the lifting
pulley.  this results in 3 segments of rope supporting the load so each one only
supports 1/3 of the weight of the load and the lifting force needed at A would
be 1/3 of the weight.  so you still have the full load weight being supported by
the ropes between B and C.  then to find the total load that B must support you
add the load weight to the rope tension exerted at A.

David Robbins K1TTT
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