Answer:
(a) the input force is 36.56 N
(b) the input force is 37.49 N
Explanation:
Given;
density of hydraulic oil, ρ = 8.53 x 10² kg/m³
radius of plunger, r₁ = 0.135 m
radius of piston, r₂ = 5.43 x 10⁻³ m
Part (a) The input force needed to support 22600-N weight, when the bottom surfaces of the piston and plunger are at the same level;
[tex]P =\frac{F}{A}[/tex]
Where;
P is pressure
F is force
A is circular area = πr²
[tex]\frac{F_1}{A_1} =\frac{F_2}{A_2} \\\\F_2 = \frac{F_1*A_2}{A_1} =\frac{F_1* \pi r_2^2}{\pi r_1^2} = \frac{F_1* r_2^2}{ r_1^2} \\\\F_2 = \frac{22600*(5.43*10^{-3})^2 }{(0.135)^2}\\\\F_2 = 36.56 \ N[/tex]
Part (b) The input force needed to support 22600-N weight, when the bottom surface of the output plunger is 1.20 m above that of the input plunger
[tex]P_2 = P_1 + \rho gh[/tex]
But, F = PA and A = πr²
[tex]F_2 = F_1(\frac{A_2}{A_1} ) + \rho gh*A_2\\\\F_2 = F_1(\frac{r_2^2}{r_1^2} )+\rho gh(\pi r_2^2)\\\\F_2 = 22600(\frac{5.43*10^{-3}}{0.135})^2 \ + 853*9.8*1.2*\pi (5.43*10^{-3})^2\\\\F_2=36.56 + 0.93\\\\F_2 = 37.49 \ N[/tex]
