The components and their dimensions of a hydraulic jack are shown in the figure. The specific gravity of the fluid in the jack is SG=0.8. A force F=400 N is applied at the left end of the handle, which is hinged at the right end. Determine:
(a) The fluid pressure p1 on the lower surface of the small piston.
(b) The fluid pressure p2 on the lower surface of the large piston.
(c) The weight W of the load that can be supported.
Pressure at A=50,000N/m^2
Pressure at B=65696N/m^2
Pressure at C=7883.52N/m^2
To determine the fluid pressures and weight supported by the hydraulic jack, we need to understand the principles of Pascal's law and the equilibrium conditions for the system.
Pascal's law states that the pressure in an enclosed fluid is transmitted equally in all directions. In the case of a hydraulic jack, this means that the pressure exerted on the small piston is the same throughout the fluid and will be transmitted to the larger piston.
Let's start by calculating the fluid pressure on the lower surface of the small piston, p1.
(a) Fluid pressure on the lower surface of the small piston (p1):
We can use the formula for fluid pressure:
p = ρgh
Where:
p1 = fluid pressure on the lower surface of the small piston
ρ = density of the fluid
g = acceleration due to gravity
h = height difference between the small piston and the fluid surface
Given:
ρ = specific gravity × ρwater
= 0.8 × 1000 kg/m³ (density of water)
= 800 kg/m³
h = height difference between small piston and fluid surface = 40 mm = 0.04 m
Using the formula for fluid pressure:
p1 = ρgh
Substituting the given values:
p1 = 800 kg/m³ × 9.8 m/s² × 0.04 m
p1 = 313.6 Pa
Therefore, the fluid pressure on the lower surface of the small piston (p1) is 313.6 Pascal (Pa).
Next, let's determine the fluid pressure on the lower surface of the large piston, p2.
(b) Fluid pressure on the lower surface of the large piston (p2):
Since the pressure is transmitted equally in all directions according to Pascal's law, the pressure p2 on the lower surface of the large piston is equal to p1.
Therefore, p2 = p1 = 313.6 Pa.
Finally, let's find the weight of the load that can be supported by the hydraulic jack.
(c) Weight of the load supported (W):
The hydraulic jack works based on the principle of force amplification. According to Pascal's law, the force applied on the small piston is amplified on the larger piston, allowing us to support a larger weight.
Using the equation for force:
F1/A1 = F2/A2
Where:
F1 = Force exerted on the small piston
A1 = Area of the small piston
F2 = Force exerted on the large piston
A2 = Area of the large piston
Given:
F1 = 400 N
A1 = πr1² (Area of small piston)
A2 = πr2² (Area of large piston)
r1 = radius of the small piston = 24 mm = 0.024 m
r2 = radius of the large piston = 60 mm = 0.06 m
Substituting the given values and solving for F2:
F1/A1 = F2/A2
400 N / (π × (0.024 m)²) = F2 / (π × (0.06 m)²)
F2 = (400 N × (π × (0.06 m)²)) / (π × (0.024 m)²)
F2 = 400 N × (0.06 m / 0.024 m)²
F2 = 400 N × (2.5)²
F2 = 400 N × 6.25
F2 = 2500 N
Therefore, the weight of the load that can be supported (W) is 2500 Newtons (N).