Question

in hydraulic pressure the area of the larger piston is half of the smaller piston.if the force of 4000N is applied on the smaller piston.what is the distance covered by the train ?​

Answer 1

The distance moved by the train is half the distance moved by the force

Step-by-step explanation:

Given that the force applied to the smaller piston, F₂ = 4,000 N

The pressure in the area of the larger piston, A₁ = 2 × The pressure in the smaller piston, A₂

∴ A₁ = 2 × A₂

We have;

`(A_1)/(A_2) = (2\cdot A_2 )/(A_2) =2 = (F_1)/(F_2)`

F₁ = 2·F₂ = 2 × 4,000 N = 8,000 N

F₁·d₁ = F₂·d₂

∴ 8,000 N ×d₁ = 4,000 N × d₂

d₁ = 1/2·d₂

∴ The distance moved by the train is half the distance moved by the force