Simple Question regarding Levers
Joseph
I was wondering if one would dispose several levers in parallel
(having the same length), having them communicate the force between
the successive levers by means of wheels with linkages (teeth) – or a
teethed gear meshing with the end of each successive lever – in order
to lift a load located on the last lever, (noting that the fulcrum of
each successive lever would be positioned at the opposite end relative
to the previous lever, and that each levers would lift towards the
opposite end relative to its contiguous levers), does this
configuration use the leverage effect only once (based on the position
of the fulcrum of the last lever), or can one accumulate the leverage
effect of all the successive levers in order to lift the load, located
on the final lever.
Thank you for your time.
Regards,
Joseph
Raphanus
I'm not sure I picture your machine correctly so my statements will be
general (but applicable).
Levers function- assuming a "perfect" machine (no friction, etc.) -
because "work-in" = "work-out." Work is defined by force * distance.
Thus if you push on a lever with your hand and your hand moves 6 cm
and the "load" moves 1 cm - you have a 6-to-1 advantage. If the load
"weighs" 600 lb - you only have to use 100 lb. Look at your machine
with this idea in mind. "How much does your hand move versus how much
does the load move?"
Look at
http://www.math.nyu.edu/~crorres/Archimedes/Lever/LeverIntro.html
Joseph
>
> - Show quoted text -
Thank you Raphanus,
Meanwhile, this does not answer my original question (I understand the
principal of levers…). Basically, my question was to know if I
organize a series of levers in a chain, in such a manner that the work
done by one lifts the next lever in the chain, until the last one
which applies the work on the load, if the leverage effect would
correspond to the summation of all the levers involved in the chain.
After making further reading and thinking more about it, I believe
that the lever effect could be only considered for the last lever, and
therefore, there wouldn’t be any “multiplication” of the leverage
effect.
Thank you again for your time.
Regards,
Joseph
Joseph
quoted text -
>
> - Show quoted text -
Thank you Raphanus,
Meanwhile, this does not answer my original question (I understand the
principal of levers). Basically, my question was to know if I
organize a series of levers in a chain, in such a manner that the work
done by one lifts the next lever in the chain, until the last one
which applies the work on the load, if the leverage effect would be
the summation of all the levers involved in the chain. After making
further reading and thinking about it, I believe that the lever effect
could be only considered for the last lever, and therefore, there
wouldn't be any multiplication of the leverage effect.
Thank you again for your time.
Regards,
Joseph
Raphanus
The principle of F*x (in) = F*x(out) still applies regardless of how
many levers are in between or how they are connected.
Uncle Al
>
> On Sep 16, 12:01 pm, Raphanus <lester.we...@gmail.com> wrote:
> > On Sep 15, 1:01 am, Joseph <joseph.rl.fourn...@gmail.com> wrote:
> Meanwhile, this does not answer my original question (I understand the
> principal of levers…). Basically, my question was to know if I
> organize a series of levers in a chain, in such a manner that the work
> done by one lifts the next lever in the chain, until the last one
> which applies the work on the load, if the leverage effect would
> correspond to the summation of all the levers involved in the chain.
> After making further reading and thinking more about it, I believe
> that the lever effect could be only considered for the last lever, and
> therefore, there wouldn’t be any “multiplication” of the leverage
> effect.
Multiple block and tackle.
<http://en.wikipedia.org/wiki/Block_and_tackle>
<http://media-2.web.britannica.com/eb-media/09/6709-004-AD9F2874.gif>
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2
a student
>
> > - Show quoted text -
>
> Thank you Raphanus,
>
> Meanwhile, this does not answer my original question (I understand the
> principal of levers…). Basically, my question was to know if I
> organize a series of levers in a chain, in such a manner that the work
> done by one lifts the next lever in the chain, until the last one
> which applies the work on the load, if the leverage effect would
> correspond to the summation of all the levers involved in the chain.
> After making further reading and thinking more about it, I believe
> that the lever effect could be only considered for the last lever, and
> therefore, there wouldn’t be any “multiplication” of the leverage
> effect.
>
> Thank you again for your time.
>
> Regards,
>
>
> - Show quoted text -
Raphanus gave you all the info you need - which can be used to show
that, basically, the effect multiplies for levers used in series.
To keep it simple, consider three seesaws (fulcrums somewhere between
force and load) in series:
\ / \
A downwards force F1 is applied to the left end of the leftmost
seesaw; this applies a force F2 upwards on the left end of the middle
seesaw; and in turn this applies a force F3 downwards on the rightmost
seesaw, which, finally applies an upwards force F4 on some load. You
want to know the relation between F1 and F4.
Suppose that the length from the left end of the first seesaw to its
fulcrum is L1, and the length from the fulcrum to the right end of the
first seesaw is R1. Similarly define L2, R2, L3 and R3. one then
has, applying geometry to Raphanus's hint (and noting action and
reaction are equal and opposite), that
F1 L1 = F2 R1
F2 L2 = F3 R2
F3 L3 = F4 R3,
and so one obtains the efficiency
F4/F1 = (L1/R1) (L2/R2) (L3/R3),
i.e., it is the product of the individual efficiencies.
Joseph
context to only the parts you are replying to. -ik ]
> > >http://www.math.nyu.edu/~crorres/Archimedes/Lever/LeverIntro.html-Hidequoted text -
> i.e., it is the product of the individual efficiencies.- Hide quoted text -
>
> - Show quoted text -
Hi,
Thank you both for your answers. Meanwhile, the description regarding
Gear Ratios found at the site hereafter appears to indicate that the
effect wouldn’t multiply, contrary to your explanation, if one setup a
series of levers in parallel. Although this text refers to gears and
not levers, shouldn’t the logic applicable to gears be the same as for
levers?
http://en.wikipedia.org/wiki/Gear_ratio
Raphanus
Although this text refers to gears and
> not levers, shouldn't the logic applicable to gears be the same as for
> levers?
>
> http://en.wikipedia.org/wiki/Gear_ratio
>
> Thank you again for your time.
Any simple machine composed of 1) levers, 2) gears, 3) block and
tackle (pulleys), and 4) inclines must obey "work-in"="work-out" where
work = force * distance. The machine could be hidden in a black box
so that all you see is the amount of work-in and the work-out and the
principle holds - the details of the machine doesn't matter. Of
course, you can't have electrical cords, hydraulic lines or steam
pipes going into the black box. No form of external energy. This
also assumes frictionless behavior - otherwise "work-out" < "work-
in" (2nd law of thermodynamics)
So yes, the logic applied to gears is the same as that applied to
levers. So....?
a student
> > Raphanus gave you all the info you need - which can be used to show
> > that, basically, the effect multiplies for levers used in series.
>
> > To keep it simple, consider three seesaws (fulcrums somewhere between
> > force and load) in series:
> > \ / \
> > A downwards force F1 is applied to the left end of the leftmost
> > seesaw; this applies a force F2 upwards on the left end of the middle
> > seesaw; and in turn this applies a force F3 downwards on the rightmost
> > seesaw, which, finally applies an upwards force F4 on some load. You
> > want to know the relation between F1 and F4.
>
> > Suppose that the length from the left end of the first seesaw to its
> > fulcrum is L1, and the length from the fulcrum to the right end of the
> > first seesaw is R1. Similarly define L2, R2, L3 and R3. one then
> > has, applying geometry to Raphanus's hint (and noting action and
> > reaction are equal and opposite), that
> > F1 L1 = F2 R1
> > F2 L2 = F3 R2
> > F3 L3 = F4 R3,
> > and so one obtains the efficiency
> > F4/F1 = (L1/R1) (L2/R2) (L3/R3),
> > i.e., it is the product of the individual efficiencies.
>
>
> Thank you both for your answers. Meanwhile, the description regarding
> Gear Ratios found at the site hereafter appears to indicate that the
> effect wouldn't multiply, contrary to your explanation, if one setup a
> series of levers in parallel. Although this text refers to gears and
> not levers, shouldn't the logic applicable to gears be the same as for
> levers?
Gears are circular. L1=R1, L2=R2, L3=R3. So F4=F1. No
contradiction.
Can talk about torque though - if the first gear has radius R1 and cog
number N1, and similarly the last gear has radius R3 and cog number
N2, then noting T=RF and F1=F4, the torque ratio is
T3/T1 = R3/R1 = N3/N1.
Joseph
Thank you all for your clarifications and explanations.
I wish you all a great day.
Regards,
Joseph Fournier