The small piston of a hydraulic lift has a
cross-sectional area of 3.6 cm2
and the large
piston has an area of 86 cm2.
What force F must be applied to the small
piston to maintain the load of 52 kN at a
constant elevation?
To determine the force F required to maintain the load at a constant elevation, we can use Pascal's Law, which states that pressure applied to a fluid is transmitted equally in all directions.
First, we need to calculate the pressure on the small piston. Pressure (P) is defined as the force per unit area. The force on the small piston (F1) can be calculated by multiplying the pressure (P1) by the cross-sectional area (A1) of the small piston.
F1 = P1 * A1
Similarly, we can calculate the pressure on the large piston and the force on the large piston.
F2 = P2 * A2
According to Pascal's Law, the pressure in a fluid is the same throughout. Therefore, P1 = P2.
Since the pistons are connected by a hydraulic system, the force on the small piston (F1) and the force on the large piston (F2) are equal. So we can write:
F1 = F2
Now, we can rearrange the equation to solve for F1:
F1 = F2 = P2 * A2 = P1 * A1
We are given the load of 52 kN, which is the force acting on the large piston (F2). We can convert this to Newtons by multiplying it by 1000 (1 kN = 1000 N).
F2 = 52 kN = 52,000 N
We are also given the cross-sectional areas of both pistons:
A1 = 3.6 cm^2
A2 = 86 cm^2
To maintain equilibrium, the force on the small piston (F1) must balance the load on the large piston (F2). Therefore,
F1 = F2 = P2 * A2 = P1 * A1
Rearranging the equation to solve for F1:
F1 = F2 * (A1 / A2)
Substituting the known values:
F1 = 52,000 N * (3.6 cm^2 / 86 cm^2)
Now, we can calculate F1:
F1 = 52,000 N * (0.04186)
F1 ≈ 2,173.12 N
Therefore, the force F that must be applied to the small piston to maintain the load of 52 kN at a constant elevation is approximately 2,173.12 Newtons.