Question

A hydraulic lift is used to lift a Car of weight = 16,000 N. What is the external input force F applied through small piston. Given area of small piston is 50 cm² and area of bigger piston is 4000 cm²​

Answer 1

Answer:

Do it by yourself this is your problem

Answer 2

        $\text{Force applied on the outpu}\text{t side}(F_1) = 16000\:N$

        $\text{Area of cross-section of the outpu}\text{t side}(A_2) = 4000 \: cm^2$

        $\text{Area of cross-section of the inpu}\text{t side}(A_2) = 50\:cm^2$

        $\text{Force applied on the inpu}\text{t side} = F_2$

        $\text{We know that in a hydraulic lift the pressure on both sides are equal}$

$\implies \: P_1 = P_2$

        $\text{We know that }P = \dfrac{F}{A}$

$\implies \: \dfrac{F_1}{A_1} = \dfrac{F_2}{A_2}$

$\implies \: \dfrac{16000}{4000} = \dfrac{F_2}{50}$

$\implies \: 40 = \dfrac{F_2}{50}$

$\implies \: \boxed{2000\:N = F_2}$