Question

Two pistons of hydraulic press have diameter of 30.0 cm and 2.5 cm. Find the force exerted by on the longer piston when 50.0 kg wt. is placed on smaller piston. If the stroke of smaller piston is 4.0 cm, find the distance through which the longer piston would move after 10 strokes.

Answer 1

Here,

F₁ = ?

d₁ = 30 cm = 0.3 m

Area of the first piston A₁ = π (d₁/2)² = 3.142×(0.30/2)² = 0.07 sq.m

F₂ = 50 kg.wt

d₂ = 2.5 cm = 0.025 m

Displacement in first stroke = 0.04 m

Area of the first piston A₂ = π (d₂/2)² = 3.142×(0.025/2)² = 5×10^-4 sq.m

By pascal’s law

F₂/A₂ = F₁/A₁

=> F₁/A₁ = 50/5×10^-4

=> F₁ = 50×0.07 /5×10^-4

=> F₁ = 7000 kg.wt

Thus, required force on the larger piston = 7000 kg.wt

Again,

Total number of strokes = 10

Total displacement S₂ = 10×0.04 = 0.4 m

Force on smaller piston F₂ = 50 kg.wt

Total displacement of larger piston be = S₁

Total work done = F₂S₂ = 50×0.4= 20 J

Work done by the 2^nd piston will be F₁S₁

Assuming no loss in work in friction etc,

F₂S₂ = F₁S₁

=> S₁ = F₂S₂/F₁

=> S₁ = 20/7000

=> S₁ = 20/7000 ≈ 0.003 m = 3 mm