Question

The hydraulic oil in a car lift has a density of 8.53 x 102 kg/m3. The weight of the input piston is negligible. The radii of the input piston and output plunger are 5.43 x 10-3 m and 0.135 m, respectively. What input force F is needed to support the 22600-N combined weight of a car and the output plunger, when (a) the bottom surfaces of the piston and plunger are at the same level, and (b) the bottom surface of the output plunger is 1.20 m above that of the input plunger

Answer 1

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;

$P =\frac{F}{A}$

Where;

P is pressure

F is force

A is circular area = πr²

$\frac{F_1}{A_1} =\frac{F_2}{A_2} \\\\F_2 = \frac{F_1*A_2}{A_1} =\frac{F_1* \pi r_2^2}{\pi r_1^2} = \frac{F_1* r_2^2}{ r_1^2} \\\\F_2 = \frac{22600*(5.43*10^{-3})^2 }{(0.135)^2}\\\\F_2 = 36.56 \ N$

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

$P_2 = P_1 + \rho gh$

But, F = PA  and  A = πr²

$F_2 = F_1(\frac{A_2}{A_1} ) + \rho gh*A_2\\\\F_2 = F_1(\frac{r_2^2}{r_1^2} )+\rho gh(\pi r_2^2)\\\\F_2 = 22600(\frac{5.43*10^{-3}}{0.135})^2 \ + 853*9.8*1.2*\pi (5.43*10^{-3})^2\\\\F_2=36.56 + 0.93\\\\F_2 = 37.49 \ N$