The small piston of a hydraulic press with an area of 2 cm2 under the action of force lowered by 16 cm

| May 26, 2021 | education

The small piston of a hydraulic press with an area of 2 cm2 under the action of force lowered by 16 cm; the area of a large piston is 8 cm2. Determine to what height the load is lifted?

Sm = 2 cm2 = 0.0002 m2.

Lm = 16 cm = 0.16 m.

Sb = 8 cm2 = 0.0008 m2.

Lb -?

Let us express the working volume Vm of the small piston during movement: Vm = Sm * Lm, where Sm is the area of the small piston, Lm is the working stroke of the small piston.

The same working volume will have a large piston Vm = Vb.

Let us express the working volume of the large piston: Vb = Sb * Lb, where Sb is the area of the large piston, Lb is the working stroke of the large piston.

Lb = Vm / Sb = Sm * Lm / Sb.

Lb = 0.0002 m2 * 0.16 m / 0.0008 m2 = 0.04 m.

Answer: the load was lifted by a large piston to a height of Lb = 0.04 m.



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