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3 years ago

10

in hydraulic pressure the area of the larger piston is half of the smaller piston.if the force of 4000N is applied on the smalle

r piston.what is the distance covered by the train ?

1 answer:

irina1246 [14]3 years ago

7 0

Answer:

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

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;

$\dfrac{A_1}{A_2} = \dfrac{2\cdot A_2 }{A_2} =2 = \dfrac{F_1}{F_2}$

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

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