Re: Hydraulic Presses and Force
- From: "Dennis B" <Utopian@xxxxxxxx>
- Date: 30 Jun 2006 04:36:58 -0700
PD wrote:
Dennis B wrote:
A hydraulic press is supposedly capable of amplifying force. For
example, 1 dyne of input force can supposedly be converted into 10
dynes or even 100 dynes of output force. In a hydraulic press the
supposed increase of force is offset by a decrease in the distance the
output piston travels. The output is dependent upon the (exposed)
surface area of the piston (and input force). As the volume or capacity
of the output piston chamber increases (and the exposed surface area of
the piston as well), the distance which the output piston travels will
decrease in response to the input force (presuming the input remains
constant). Force is determined by measuring the displacement of a mass
(as well as the time involved for the displacement to occur). Since the
distance a piston travels is dependent upon the volume or capacity of
the piston chamber (as well as input force), the distance travelled by
two different (output) pistons as a result of a given input force (1
dyne for example) will not be the same if the volume or capacity of the
output piston chambers are different (for example if the exposed piston
surface area of one is 10 cm^2 and 100 cm^2 in the other). Yet, the
total volume of fluid displaced (1 cm^3) is the same for both pistons.
In other words, the amount of mass (i.e., the water) displaced in
response to a given input force (1 dyne in this example) is the same.
Thus, wouldn't the output force of both pistons be the same,
considering that force is measured by the displacement (or
acceleration) of a mass (such as the water inside the piston in this
case)? Undeniably, the total energy remains the same (ignoring losses
due to friction, etc.), although the energy density must decrease at
the output for the same reason that the total energy density of a fluid
accelerating through a tube of varying diameter changes, while the
total energy remains the same (ignoring losses). The definition of
force is that which causes acceleration of a body. Force would
therefore appear to be synonymous with the definition of "energy". So,
how is it that the input force increases while the input energy density
decreases in the case of a hydraulic press? One may be inclined to
argue that the energy density does not decrease as the volume/capacity
of the output piston chamber (or circumference of the tubing)
increases. Yet, the energy density does in fact decrease as the volume
(capacity) of the piston chamber or circumference of the tubing
decreases, for the energy must expand to occupy a larger volume of
space. Thus, I would expect that the force would decrease as well, just
as the ram pressure (and velocity) of an accelerating fluid decreases
as the circumference of the tube in which it flows increases. To use an
example, I would expect that a force of 1 dyne applied to a piston with
an exposed surface area of 1 cm^2 would be converted into a force of .1
dyne cm^2 in an output piston with an exposed surface area of 10 cm^2.
-Dennis B
You're a little confused. The *work* is the same, but the force is
amplified.
This is much the same way that a pulley system amplifies force, though
conserving work.
In a pulley system, like the gears on your bike or a block & tackle or
the chain linkage that helps you raise a heavy door, lower force
applied comes at the expense of requiring larger displacement. For
example, you can raise a piano with a block and tackle by exerting only
1/5 the weight of the piano, but you have to pull 5 ft of rope through
the pulley for every foot you want to raise the piano. You can lower
the force by a factor of 3 by changing gears on your bike, but you have
to crank three times as fast in the lower gear. In both cases, the work
(force x displacement) is the same at the engine and at the load
(ignoring friction), though the force may be changed quite a bit.
The same thing is true in a hydraulic system. At the engine, the force
applied might be low (F1 = P * A1), but you'll have to push a long
column of fluid (d1), compared to the large force applied at the load
(F2 = P * A2), where a much shorter column of fluid comes out (d2).
Note that the work done is identical at both engine and at load (P * A1
* d1 = P * A2 * d2). In this case, the volume of the fluid pushed is
the same at the engine and at the load (A1 * d1 = A2 * d2).
PD
Force is defined as mass times acceleration. In other words, force is
essentially a measure of the total energy it takes to move a certain
mass a certain distance within a certain amount of time. Work is
defined as force times distance and is essentially the same as force
minus the measure of time. In other words, it is a measure of the total
energy required to displace the mass a certain distance without regard
to the time involved.
If the input piston (having an exposed surface area of 1 cm^2) travels
1 cm as a result of 1 dyne of force, and the output piston (having an
exposed surface area of 10 cm^2) travels only .1 cm, then the force
involved to displace 1 cm^3 of fluid (or water) is the same for both
the input and the output because the same volume of fluid is displaced
at either piston. If one takes into consideration the mass of the
pistons, this also partially explains why the output piston traverses
less distance than the input piston. It would take 10 times the input
energy (or 10 dynes) to displace the output piston by 1 cm, using the
body of water in between as a mediator of the applied force. An input
force of 1 dyne can only displace the more massive output piston a
distance of .1 cm. In other words, to reverse the situation, if 1 dyne
of force were applied to the larger "output" piston, it would move it
the same distance of .1 cm, and also displace the smaller "input"
piston by 1 cm. Thus, the input and output force must be the same.
Because work is a measure of force or energy, determined by the
displacement of a mass without regard for the time involved, the work
applied to the input piston must be the same as the work of the output
piston.
A weight scale measures weight by the distance the scale plate
traverses as a result of the gravitational force (or acceleration)
between the earth and the mass being weighed (presuming the measurement
is performed on Earth) according to the following equation:
F = ma = mg = w
Key:
F = force
m = mass
a = acceleration
g = gravitional acceleration
w = weight
....The greater the mass being measured, the greater the gravitational
force between the mass and the Earth and therefore the greater the
distance the scale's plate will traverse. Although weight is a measure
of force, and force is defined as essentially an accelerating mass (or
energy since energy and mass are equivalent, mass being made of
energy)...the fact is that a scale measures the force of weight without
measuring the acceleration or time involved to displace the mass.
Rather, a weight scale measures only the WORK involved in displacing
the scale plate. Yet, it does indeed measure the (total) force (or
energy) which displaces the scale plate. Thus, implying the following
equation:
F = ma = mg = w = W
Key:
W = work
Yet, if one uses a scale to measure the force at the output of a
hydraulic press, the scale will actually measure less force or work as
the volume of the output piston chamber increases, because the piston
will traverse increasingly less distance as the volume of the output
piston chamber expands.
Dennis B
.
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