Note that you can arrange some confusing arrangements that appear to
alter the arrangement but don't.
So:
- Tension in rope is supported_load / ropes_supporting load when ropes
are perpendicular to gravitational field
- Tension in rope is constant everywhere for a static load.
- if you build systems with ropes at angles to the gravitational
field then you need simple force- vector additions BUT the principle
is the same. (Sum of vertical forces applied to the load is zero. Some
of all horizontal forces applied to the load is zero).
> To make this very simple terms. Take the number of ropes attached to to
> the MOVEABLE pulleys divide that into the weight being lifted and that
> is the force required to lift the object.
>> To make this very simple terms. Take the number of ropes attached to to
>> the MOVEABLE pulleys divide that into the weight being lifted and that
>> is the force required to lift the object.
>
> As long as all ropes are vertical :-)
Well, only if it is lifting a free hanging mass.
But the principle is exactly the same for a block and tackle pulling
something up an incline, all the rope lengths going to the moving mass are
under equal tension.
>> To make this very simple terms. Take the number of ropes attached to to
>> the MOVEABLE pulleys divide that into the weight being lifted and that
>> is the force required to lift the object.
>
> As long as all ropes are vertical :-)
Well, only if it is lifting a free hanging mass.
But the principle is exactly the same for a block and tackle pulling
something up an incline, all the rope lengths going to the moving mass are
under equal tension.
> >> To make this very simple terms. Take the number of ropes attached to to
> >> the MOVEABLE pulleys divide that into the weight being lifted and that
> >> is the force required to lift the object.
> >
> > As long as all ropes are vertical :-)
>
> Well, only if it is lifting a free hanging mass.
>
> But the principle is exactly the same for a block and tackle pulling
> something up an incline, all the rope lengths going to the moving mass are
> under equal tension.
E&OE :-) ...
My point (not being quoted above) about ripes being vertical is that
ropes can be at an angle to the resisting force field so that only a
component of the rope tension supports or pulls the load. Such cases
complicate the simple examples for beginners.
eg - 2 horse pull a cart up a ramp by a rope fastened to a pulling
collar on each horse and passing through a single pulley in the front
middle of the cart. The horses are 6 feet apart and the pulling collar
attachment s are 10 feet in front of the cart.
If the drawbar pull on the cart is say 200 lbf the rope tension will
NOT be 200/2 = 100 lbf.
The up slope component will be 100 lbf but rope tension will be
sqrt(100+9)/10 x 100 lbf = 104 lbf.
Hardly any difference.
But if the horses were 10 feet apart it becomes sqrt(125)/10*100 = 112 lbf.
When/if the ropes get out to 45 degrees either side rope tension is
141 (100 x sqrt(2) lbf.
All the increase comes from pulling at an angle so that only a
component applies.
This increase has nothing to do with pulleys per se but can be
introduced into examples to complicate them without people being very
aware that it has happened.
As long as the ropes ascend vertically from YesNopeGus's head all is well.
> eg - 2 horse pull a cart up a ramp by a rope fastened to a pulling
> collar on each horse and passing through a single pulley in the front
> middle of the cart. The horses are 6 feet apart and the pulling collar
> attachment s are 10 feet in front of the cart.
Can't we just put the cart on the train leaving Boston at 2pm? :-)
> -----Original Message-----
> From: spam_OUTpiclist-bouncesTakeThisOuTmit.edu [.....piclist-bouncesKILLspam@spam@mit.edu] On Behalf
> Of Russell McMahon
> Sent: 18 May 2010 14:35
> To: Microcontroller discussion list - Public.
> Subject: Re: [TECH] The theory of pulleys
>
> > To make this very simple terms. Take the number of ropes attached to to
> > the MOVEABLE pulleys divide that into the weight being lifted and that
> > is the force required to lift the object.
>
> As long as all ropes are vertical :-)
'parallel' might cover more situations!
Mike
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> I want to pull the container down an incline that drops about 1 rod
> per 100 feet.
Rods do not compute.
Presumably you mean 40 millifurlong per 100 feet.
That's a ramp angle of about 160 milliradian, which is usefully steep.
Rollers and several helpers and it would probably be "on a roll" all
by itself.
> I am going to assume the incline won't make much difference..... it
> may assist me a bit.
> I will put some logs under it to reduce the drag friction.
Is your life and "accident" insurance up to date?
> I have winch that claims it can lift 2000 lbs
> My plan is to use a system of pulleys to move the container.
>
> The Truck is green. =A0The container is blue.
System is built wrongly.
You need to be able to pull on the end of the actual supporting rope
with the truck.
Look at the many 'block and tackle" references using Gargoyle images
eg
> Will the winch have a mechanical advantage of 4 due to the pulleys ?
No. System has ma of 1 as pull on main rope does not affect rope
attached to container.
> Will the winch be powerful enough ?
With a 4:1 system, once achieved, coefficient of sliding friction will
almost certainly be < 1 so force on winch will be < 2000 lbf.
BUT static force to get going is unknown. Rollers will help muchly.
Any sliding surface that stops container front edge digging in will also he=
lp.
If required a system of small rollers (eg water pipe) and some timber
to make levers to lift container to inswrt rollers initially, would
allow say 2 or 3 people to "easily enough" do this without a truck.
Over any distance it gets annoying.
> I expect the container moves
> with less than an
> 8000 lb pull.
As above.
With rollers no b&t should be needed and will complicate things.
With an N:1 advantage the pulling vehicle etc moves N times as far as
the pulled object. SO here the truck moves 4 ramp lengths with a 4:1
B&T. May need several bites depending on available room.
> Okay what I am actually planning to is:
> I have a shipping container that weighs 8000 pounds.
> I want to pull the container down an incline that drops about
> 1 rod per 100 feet. I am going to assume the incline won't
> make much difference..... it may assist me a bit.
> I will put some logs under it to reduce the drag friction.
If the surface is hard and you have enough logs, then you may
have a very interesting time as the container goes careening
down the incline.
> see image at
> dent-the-universe.com/tikiwiki/tiki-list_file_gallery.php?galleryId=4
>
> I have winch that claims it can lift 2000 lbs
> My plan is to use a system of pulleys to move the container.
>
> The Truck is green. The container is blue.
> The pulleys are red circles. The winch is a violet square.
> The tree is a green line.
>
> Will the winch have a mechanical advantage of 4 due to the pulleys ?
Yes.
> Will the winch be powerful enough ?
Maybe. Ensure it is rated for a dead lift of 2000 lbs (which
may leave you with no margin for error (due to stiction) ).
> I expect the container moves with less than an 8000 lb pull.
Maybe, maybe not. Depends on the surface, if a corner digs in,
if it snags on a tree root, etc.
I'd be _VERY_ carefull about the ratings of the various
components. Where are you going to find a rope (i.e. blue
rope) with 8 tons breaking strengh? Steel wire rope? With
this much tension, if something breaks the resulting shrapnel
could maime or kill. Quick list:
- the winch mounting (and truck structure under the winch
mount on the truck).
- the 3 pulleys near the green truck, the red rope, and
the red rope's anchor on the truck. They will all have
roughly 2000 lbs of force applied.
- the yoke by truck (2 pulleys & blue rope attachment),
blue rope itself, the pulley by the tree, and whatever
mechanism you devise to anchor the pulley to the tree;
all will have roughly 8000 lbs force.
- if the ground is soft/soggy, the force may not break
the tree, but it could pull it over -- a falling tree's
top can reach out farther than you estimate
I did a similar project about 2 years ago but objects were
under 400 lbs each. It was hard to find sufficiently robust
pulleys and slings to handle those loads (with some safety
margin) [well, within my limited budget for a one-off]. I
would be real concerned about scaling it all up by a factor
of 10 to 20.
If I were doing it, I'd mount the yoke (2 pulleys by truck)
to the tree, loose the "blue rope", and move the "red rope"
anchor from truck to the container. Then the rope would
only need a breaking strengh of 2 tons (yes, factor of 2
safety margin). And there would be fewer points/parts
that take a full 8000 lbs stress.
Russell McMahon wrote:
> System is built wrongly.
> You need to be able to pull on the end of the actual supporting rope
> with the truck.
I took the diagram to mean that the truck was merely a anchor and that the
winch would do the work. If he expects the truck to do the work, then
you're right, the pulleys are in the wrong place to provide mechanical
advantage. However, what would be the point of having a winch if it's not
intended to do the actual pulling?
Lee Jones wrote:
> - the yoke by truck (2 pulleys & blue rope attachment),
> blue rope itself, the pulley by the tree, and whatever
> mechanism you devise to anchor the pulley to the tree;
> all will have roughly 8000 lbs force.
No. The tree and the pulley attached to it will see double that force. He
needs to check the tree's datasheet very carefully to make sure it is rated
for 16,000 pounds. That's a lot, even for a decent size tree. You
certainly wouldn't want to do that to a tree you liked.
> Russell McMahon wrote:
>
>> System is built wrongly.
>> You need to be able to pull on the end of the actual supporting rope
>> with the truck.
>>
>
> I took the diagram to mean that the truck was merely a anchor and that the
> winch would do the work. If he expects the truck to do the work, then
> you're right, the pulleys are in the wrong place to provide mechanical
> advantage. However, what would be the point of having a winch if it's not
> intended to do the actual pulling?
>
>
>
>> I want to pull the container down an incline that drops about 1 rod
>> per 100 feet.
> There is no need to deliberately use obscure units just because you can.
> No, I'm not going to look up how long a rod is so that I can compare it to
> 100 feet.
I said:
>> Presumably you mean 40 millifurlong per 100 feet.
but I transposed some brain cells.
In fact 1 rod* = 25 millifurlongs.
>> That's a ramp angle of about 160 milliradian, which is usefully steep.
I think I got that right.
A man can never have enough milliradians - especially when rolling
heavy containers downhill.
R
* Rods, I'm pleased to say, have never made it out here to the
beginning of every day and the edge of the Empire that the sun never
sets on. Except, perhaps, for cleaning drains. Arguably some people
have hot ones, but we are blessed to not have acquired the 25
millifurlong variety.
> * Rods, I'm pleased to say, have never made it out here to the
> beginning of every day and the edge of the Empire that the sun never
> sets on. Except, perhaps, for cleaning drains. Arguably some people
> have hot ones, but we are blessed to not have acquired the 25
> millifurlong variety.
And here was me believing that surveyors used to use them in that part of
the world, as in 'rod, pole or perch' all being the same unit of different
names, when measuring out the dimensions of real estate.
But even so does it need to be connected to the cables or ropes.
Also where I grew up in Southern Minnesota all farm boys knew that a ROD
was 16.5 feet long.
the survey of my property also has rod measurements. Many lots are 66
feet (4 rods) wide. and that is in the city.
>> On May 19, 2010, at 7:26 AM, BOB wrote:
>>
>> Why is the truck even needed? Just secured the blue line to the base
>> of the tree.
>>
>> OLD BOB
>>
>
> The truck provides power for the winch
>
>
>> Why is the truck even needed? Just secured the blue line
>> to the base of the tree.
>>
>> OLD BOB
> The truck provides power for the winch
>From your drawing, it appears that the winch & 4-fold force
multiplier pulley arrangement is anchored to the truck.
Recall -- For each action, there is an opposite reaction.
How heavy is the truck? A light pickup truck could easily
weigh less than an 8000 lbs container. And the truck could
have much less friction sitting on 4 smallish round tire
contact patches than the break-away friction of the flat
surface of the bottom of a large container.
Lee Jones wrote:
> And the truck could have much less friction sitting on 4 smallish round
> tire contact patches than the break-away friction of the flat surface of
> the bottom of a large container.
Non-issue. If you work it out, you'll see that contact area cancels out.
The available lateral force is simply the weight of the truck multiplied
by the coefficient of friction between the rubber and the earth. Same
thing for the container.
Yes, there are generally two different numbers (static vs. sliding
coefficients of friction) for each pair of materials, so you need to
verify all of the combinations.
Lee Jones wrote:
>> And the truck could have much less friction sitting on 4 smallish
>> round tire contact patches than the break-away friction of the
>> flat surface of the bottom of a large container.
> Non-issue. If you work it out, you'll see that contact area
> cancels out. The available lateral force is simply the weight
> of the truck multiplied by the coefficient of friction between
> the rubber and the earth. Same thing for the container.
I agree with your mathematics & physics. And we may be talking
about a truck that weighs 1/3 of the container's dead weight.
Now let me posit a possible scenario based on items that have
been mentioned in previous posts.
one end of rope -- 8000 lb square cornered container caught on
a tree root in soft earth road.
other end of rope -- 2500 lb smallish pickup truck sitting on
4 round wheels (i.e. no corners to dig in) of which only the
rear two are locked into a non-rotational state (i.e. parking
brake on); front wheels are free to rotate.
Maybe you think this won't happen. And maybe you're right.
Personally, I would want to be a safe distance away from such
a light truck when the winch is turned on.
[ Oops, I apologize, I forgot to trim headers before sending. ]
Lee Jones wrote:
>> And the truck could have much less friction sitting on 4 smallish
>> round tire contact patches than the break-away friction of the
>> flat surface of the bottom of a large container.
> Non-issue. If you work it out, you'll see that contact area
> cancels out. The available lateral force is simply the weight
> of the truck multiplied by the coefficient of friction between
> the rubber and the earth. Same thing for the container.
I agree with your mathematics & physics. And we may be talking
about a truck that weighs 1/3 of the container's dead weight.
Now let me posit a possible scenario based on items that have
been mentioned in previous posts.
one end of rope -- 8000 lb square cornered container caught on
a tree root in soft earth road.
other end of rope -- 2500 lb smallish pickup truck sitting on
4 round wheels (i.e. no corners to dig in) of which only the
rear two are locked into a non-rotational state (i.e. parking
brake on); front wheels are free to rotate.
Maybe you think this won't happen. And maybe you're right.
Personally, I would want to be a safe distance away from such
a light truck when the winch is turned on.
> I want to pull the container down an incline that drops
> about 1 rod per 100 feet. I am going to assume the
> incline won't make much difference..... it may assist
> me a bit.
Wrong assumption, 1 to 7 incline would assist you greatly. If you cut
the container loose, it would do the job on its own.
> I will put some logs under it to reduce the drag friction.
Some grease instead would be better and don't forget about placing a
lot of sand at the end of the path.
WARNING: USE CERTIFIED PROFESSIONALS TO DO THE ABOVE. DOING IT ON YOUR
OWN YOU ARE UNDER THE RISK OF YOUR AND OTHERS HEALTH/PROPERTY DAMAGE.
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