Question

The hydraulic oil in a car lift has a density of 8.30 102 kg/m³. The weight of the input piston is negligible. The radii of the input piston and output plunger are 6.30 10⁻³ m and 0.125 m, respectively. What input force F is needed to support the 27,000-N combined weight of a car and the output plunger under the following conditions?
(a) the bottom surfaces of the piston and plunger are at the same level, and
(b) the bottom surface of the output plunger is 1.30 m above that of the input piston?

Answer 1

Answer:

(a) $F_i=68.58\ N$

(b) $F_i=69.903\ N$

Explanation:

Given:

• density of hydraulic oil, $\rho=830\ kg.m^{-3}$

• radius of input piston, $r_i=6.3\times 10^{-3}\ m$

• radius of output plunger, $r_o=0.125\ m$

• force to be supported, $F_o=27000\ N$

(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:

$\frac{F_i}{A_i} =\frac{F_o}{A_o}$

putting respective values

$\frac{F_i}{\pi\times r_i^2} =\frac{27000}{\pi\times r_o^2}$

$\frac{F_i}{\pi\times (6.3\times 10^{-3})^2} =\frac{27000}{\pi\times 0.125^2}$

$F_i=68.58\ N$

(b)

Condition:The bottom surface of the output plunger is 1.30 m above that of the input piston.

Given:

$h=1.3\ m$

Now,

$P_i=P_o+\rho.g.h$

$\frac{F_i}{\pi\times (6.3\times 10^{-3})^2} =\frac{27000}{\pi\times 0.125^2} +830\times 9.8\times 1.3$

$F_i=69.903\ N$