Answer:
The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.
Explanation:
By means of the Pascal's Principle, the hydraulic lift can be modelled by the following two equations:
Hydraulic Lift - Before change
[tex]P = \frac{F}{A}[/tex]
Hydraulic Lift - After change
[tex]P + \Delta P = \frac{F + \Delta F}{A}[/tex]
Where:
[tex]P[/tex] - Hydrostatic pressure, measured in pascals.
[tex]\Delta P[/tex] - Change in hydrostatic pressure, measured in pascals.
[tex]A[/tex] - Cross sectional area of the hydraulic lift, measured in square meters.
[tex]F[/tex] - Hydrostatic force, measured in newtons.
[tex]\Delta F[/tex] - Change in hydrostatic force, measured in newtons.
The additional weight is obtained after some algebraic handling and the replacing of all inputs:
[tex]\frac{F}{A} + \Delta P = \frac{F}{A} + \frac{\Delta F}{A}[/tex]
[tex]\Delta P = \frac{\Delta F}{A}[/tex]
[tex]\Delta F = A\cdot \Delta P[/tex]
Given that [tex]\Delta P = 100\,Pa[/tex] and [tex]A = 25\,m^{2}[/tex], the additional weight is:
[tex]\Delta F = (25\,m^{2})\cdot (100\,Pa)[/tex]
[tex]\Delta F = 2500\,N[/tex]
The additional mass needed for the additional weight is:
[tex]\Delta m = \frac{\Delta F}{g}[/tex]
Where:
[tex]\Delta F[/tex] - Additional weight, measured in newtons.
[tex]\Delta m[/tex] - Additional mass, measured in kilograms.
[tex]g[/tex] - Gravitational constant, measured in meters per square second.
If [tex]\Delta F = 2500\,N[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then:
[tex]\Delta m = \frac{2500\,N}{9.807\,\frac{m}{s^{2}} }[/tex]
[tex]\Delta m = 254.92\,kg[/tex]
The additional weight and mass needed for lifting the other piston slowly is 2500 N and 254.92 kg, respectively.
