Question

A hydraulic lift consists of two pistons that connect to each other by an incompressible fluid. If one piston has an area of 0.15 m^2 and the other an area of 6.0 m^2, how large a mass can be raised by a force of 130 N exerted on the smaller piston?

Answer 1

Using the hydraulic principle, a 130 N force applied on a small piston with an area of 0.15 m^2 can raise a mass of 530.61 kg when the system's larger piston has an area of 6.0 m^2.

Step-by-step explanation:

The question relates to the principle of a hydraulic lift and involves using the concept of hydraulic pressure to determine the maximum mass that can be lifted by applying a certain force on one of the pistons. Given that a force of 130 N is exerted on the smaller piston with an area of 0.15 m2, we can calculate the pressure applied to the system since pressure is equal to force divided by area (P = F/A).

The pressure exerted on the smaller piston is:

P = F / A1

P = 130 N / 0.15 m2

P = 866.67 Pa (Pascal)

Since the fluid is incompressible and the system is assumed to be frictionless, the pressure throughout the system is constant. Therefore, the same pressure is exerted on the larger piston with an area of 6.0 m2. The force on the larger piston (F2) is the pressure multiplied by its area (F2 = P x A2).

F2 = 866.67 Pa x 6.0 m2

F2 = 5200 N

Finally, to find the mass that this force can lift, we use the relationship between force and mass (F = m x g, where g is the acceleration due to gravity, 9.8 m/s2).

m = F2 / g

m = 5200 N / 9.8 m/s2

m = 530.61 kg

Thus, a mass of 530.61 kg can be raised by a force of 130 N exerted on the smaller piston of the hydraulic lift.

Answer 2

You have:

A1= 0.15 m²
A2=6.0 m²
F1=130.0 N

To solve this problem you must use the pressure formula, as below:

P1=P2
F1/A1=F2/A2

Then, you have:

m2=A2xF1/A1xg

g=9.81 m/s² (This is the acceleration of gravity)

When you substitute the values in m2=A2xF1/A1xg, you obtain:

m2=(0.60 m² x 130.0 N)/(0.15 m² x 9.81 m/s²)

The answer is:

m2=530 Kg