Respuesta :
Answer:
(a) [tex]F_i=68.58\ N[/tex]
(b) [tex]F_i=69.903\ N[/tex]
Explanation:
Given:
- density of hydraulic oil, [tex]\rho=830\ kg.m^{-3}[/tex]
- radius of input piston, [tex]r_i=6.3\times 10^{-3}\ m[/tex]
- radius of output plunger, [tex]r_o=0.125\ m[/tex]
- force to be supported, [tex]F_o=27000\ N[/tex]
(a)
Condition:The bottom surfaces of piston and plunger at the same level.
According to Pascal's law the pressure of a fluid is exerted equally in all directions against the walls of its container.
Mathematically:
[tex]\frac{F_i}{A_i} =\frac{F_o}{A_o}[/tex]
putting respective values
[tex]\frac{F_i}{\pi\times r_i^2} =\frac{27000}{\pi\times r_o^2}[/tex]
[tex]\frac{F_i}{\pi\times (6.3\times 10^{-3})^2} =\frac{27000}{\pi\times 0.125^2}[/tex]
[tex]F_i=68.58\ N[/tex]
(b)
Condition:The bottom surface of the output plunger is 1.30 m above that of the input piston.
Given:
[tex]h=1.3\ m[/tex]
Now,
[tex]P_i=P_o+\rho.g.h[/tex]
[tex]\frac{F_i}{\pi\times (6.3\times 10^{-3})^2} =\frac{27000}{\pi\times 0.125^2} +830\times 9.8\times 1.3[/tex]
[tex]F_i=69.903\ N[/tex]
