rvanveshana.com
t = 0.00s
LOADF₁ = 50NF₂ = 500N
Input Force (F_1) Output Force (F_2)

Live Telemetry & Summary

Observe the live values changing in real-time. Notice that mechanical work (F * d) remains conserved on both pistons.

Pressure (P):50.0 kPa
Mechanical Advantage:10.0x
Piston 1 Displacement:0.0 cm
Piston 2 Displacement:0.00 cm
Work Done (W_1 / W_2):0.00 J

Variable Adjuster

Input Force (F₁)50N
10200
Piston 1 Area (A₁)10cm²
525
Piston 2 Area (A₂)100cm²
50250

Pascal's Hydraulic Lift Lab

LIFT

Pascal's Principle states that pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the walls of the containing vessel. In a hydraulic press, applying a force F₁ on a small piston A₁ generates a pressure P which transmits to a larger piston A₂, resulting in an amplified output force F₂.

P=F1A1=F2A2⇒F2=F1(A2A1),W=F1d1=F2d2P = \frac{F_1}{A_1} = \frac{F_2}{A_2} \Rightarrow F_2 = F_1 \left(\frac{A_2}{A_1}\right), \quad W = F_1 d_1 = F_2 d_2

Whiteboard Solver Steps

Step 1

Pascal's Uniform Pressure Law

Pressure applied to a confined fluid is transmitted undiminished throughout the fluid, acting perpendicular to all surfaces.

P=F1A1=50 N0.0010 m2=50000 PaP = \frac{F_1}{A_1} = \frac{50 \text{ N}}{0.0010 \text{ m}^2} = 50000 \text{ Pa}
Step 2

Force Amplification on Output Piston

Since Piston 2 has a larger surface area, the uniform fluid pressure generates a larger upward force. The mechanical advantage is A_2 / A_1 = 10.0.

F2=P⋅A2=50000⋅0.0100=500.0 NF_2 = P \cdot A_2 = 50000 \cdot 0.0100 = 500.0 \text{ N}
Step 3

Work & Distance Tradeoff

Due to conservation of energy, the work input equals the work output. The force is amplified, but the output piston moves a much smaller distance.

d2=d1⋅A1A2=0⋅10100=0.00 cmd_2 = d_1 \cdot \frac{A_1}{A_2} = 0 \cdot \frac{10}{100} = 0.00 \text{ cm}