The area of the small piston of the hydraulic lift is S1 = 0.8 cm2, and the area of the large one is S2 = 40 cm2.
July 28, 2021 | education|
The area of the small piston of the hydraulic lift is S1 = 0.8 cm2, and the area of the large one is S2 = 40 cm2. What force must be applied to the small piston to lift a 7kN load?
The action of the force F1 on the piston with area S1 creates a pressure p under it equal to
p = F1 / S1.
This pressure acts on the piston of area S2, creating a force F2:
p = F2 / S2.
Let’s equate the two expressions and find the required value of the force F1:
F1 / S1 = F2 / S2;
F1 = F2 * S1 / S2;
F1 = 7 * 0.8 / 40 = 0.14 kN.
Thus, to lift a 7 kN load, a force of 0.14 kN must be applied to the small piston.
One of the components of a person's success in our time is receiving modern high-quality education, mastering the knowledge, skills and abilities necessary for life in society. A person today needs to study almost all his life, mastering everything new and new, acquiring the necessary professional qualities.