In a hydraulic system, the area of the large piston is hundred times of the small piston. If
15 N force is applied to the small piston, how much mass can the large piston carry?
In a hydraulic system, the pressure exerted on the fluid is the same throughout the system. Therefore, the force exerted on the large piston is equal to the force applied on the small piston.
Let's assume the area of the small piston is A and the area of the large piston is 100A.
Since pressure = force / area, the pressure exerted on the small piston is 15 N / A, and the pressure exerted on the large piston is 15 N / (100A).
Since the pressure is the same, we can equate the two expressions for pressure:
15 N / A = 15 N / (100A)
By cross-multiplying and solving for A, we find that A = 1/100, so the area of the small piston is 1/100th of the area of the large piston.
Now, we know that force = mass * acceleration, and since the acceleration due to gravity is constant, we can simplify this equation to force = mass * gravity, where gravity is approximately 9.8 m/s^2.
Therefore, the mass that the large piston can carry is:
force of large piston = mass * gravity
15 N = mass * 9.8 m/s^2
Solving for mass:
mass = 15 N / 9.8 m/s^2
mass ≈ 1.53 kg
Therefore, the large piston can carry approximately 1.53 kg of mass.