B How hydraulic pressure can change the magnitude of a force?

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Hydraulic systems utilize fluid pressure to amplify force, allowing a smaller piston to exert significantly more force on a larger piston, exemplified by a 1:10 ratio delivering ten times the force. This principle mirrors the mechanics of levers, converting a small force over a large distance into a larger force over a shorter distance while conserving energy. However, the larger piston moves slower than the smaller one, demonstrating a trade-off between force and displacement; for instance, a small piston moving 10mm results in a larger piston moving only 1mm. Power, defined as the rate of doing work, differs from force, as it accounts for both force and velocity. Ultimately, while hydraulic systems provide a force advantage, they require more input motion to achieve the desired output motion.
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how liquid based systems e.g. hydraulic systems, can change the force which applied on it is smaller pistons, and transfer much force to a larger piston in area, my textbooks says 1:10 ratio hydraulic system can deliver ten times the force applied on the smaller piston to the large piston. it would be a machine that deliver more output power than the power it takes in.
 
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It's a similar principle to using a lever. You can change a small force moving through a large distance into a large force moving through a small distance. Since the work done is ##Fd##, this means that energy is conserved.
 
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inuka00123 said:
my textbooks says 1:10 ratio hydraulic system can deliver ten times the force applied on the smaller piston to the large piston. it would be a machine that deliver more output power than the power it takes in.
Power and force are different things:

Power = Force * Velocity

So if output velocity is less than input velocity, then you can have more output force than input force, despite output power being less than input power due to losses.

In your case the bigger piston will move slower than the small piston, because the liquid volume is roughly constant.
 
The volume of fluid that flows in metres cubed, multiplied by the linear distance moved by the piston, is the energy in joules, transferred during the operation.

The pressure difference of the liquid in pascals, multiplied by the volume of hydraulic fluid that flows in cubic metres per second, is the power in joules per second, called watts.
 
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Baluncore said:
The volume of fluid that flows in metres cubed, multiplied by the linear distance moved by the piston, is the energy in joules, transferred during the operation.
Units don't look right here - did you miss out the pressure?
 
Ibix said:
Units don't look right here - did you miss out the pressure?
The volume of fluid that flows in metres cubed, multiplied by the pressure in pascals, is the energy in joules, transferred during the operation.
 
inuka00123 said:
how liquid based systems e.g. hydraulic systems, can change the force which applied on it is smaller pistons, and transfer much force to a larger piston in area, my textbooks says 1:10 ratio hydraulic system can deliver ten times the force applied on the smaller piston to the large piston. it would be a machine that deliver more output power than the power it takes in.
 
No, hydraulic systems work on fluid pressure. You push on the small piston, which pressurizes the hydraulic fluid. That same pressure is transferred through the fluid to the larger piston. Both pistons have the same hydraulic pressure on different areas In this case, 10 times the piston area yield 10 times the force that was applied to the small piston. Power is something else again see below.

But the large piston only moves 1/10 of the displacement of the smaller piston.
10mm on the small piston moves the larger piston only 1mm.
You get a force increase at the expense of a displacement decrease. This means that while a hydrolic jack can lift a heavy object like a car, you have to pump the small piston a lot of times in the same 10:1 ratio. ie if 1 stroke of the small piston is 10mm then you need to pump it 10 times to move the large piston 10 mm.

In short: A force advantage = a displacement disadvantage.

This has nothing to do with comparative power, which is the time rate of doing work.
power = work/time
P = W/t

Work = Force x Distance
W = F x S

Therefore: P = (F x S)/t
 
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