The diameters of a small piston and a large piston of a hydraulic jack are 3 cm and 10 cm respectively. A force of 80 N is applied on the small piston. Find the load lifted by the large piston when

Hydraulic Jack Force Calculation

🔧 Problem Statement

The diameters of a small piston and a large piston of a hydraulic jack are 3 cm and 10 cm respectively. A force of 80 N is applied on the small piston. Find the load lifted by the large piston when:
(a) the pistons are at the same level.
(b) the small piston is 40 cm above the large piston.
The density of the liquid in the jack is given as 1000 kg/m³.

📊 Given Data

Diameter of small piston, $d = 3 \, \text{cm} = 0.03 \, \text{m}$
Diameter of large piston, $D = 10 \, \text{cm} = 0.1 \, \text{m}$
Force on small piston, $F = 80 \, \text{N}$
Height difference (for part b), $h = 40 \, \text{cm} = 0.4 \, \text{m}$
Density of liquid, $\rho = 1000 \, \text{kg/m}^3$
Acceleration due to gravity, $g = 9.81 \, \text{m/s}^2$

⚙️ Solution

1. Calculate Piston Areas

First, we calculate the areas of the small piston ($a$) and the large piston ($A$) in SI units (m²).

$$ a = \frac{\pi}{4} d^2 $$ $$ a = \frac{\pi}{4} (0.03 \, \text{m})^2 $$ $$ a \approx 0.0007068 \, \text{m}^2 $$

$$ A = \frac{\pi}{4} D^2 $$ $$ A = \frac{\pi}{4} (0.1 \, \text{m})^2 $$ $$ A \approx 0.007854 \, \text{m}^2 $$

(a) Pistons at the Same Level

When the pistons are at the same level, the pressure is transmitted equally according to Pascal's Law.

$$ P_{\text{small}} = P_{\text{large}} $$ $$ \frac{F}{a} = \frac{W}{A} $$

Rearrange to solve for the load lifted, $W$.

$$ W = F \times \frac{A}{a} $$ $$ W = 80 \, \text{N} \times \frac{0.007854 \, \text{m}^2}{0.0007068 \, \text{m}^2} $$ $$ W = 80 \times 11.11 $$ $$ W \approx 888.96 \, \text{N} $$

(b) Small Piston 40 cm Above Large Piston

When the small piston is higher, the pressure at the large piston's level is the sum of the pressure from the applied force AND the pressure from the 0.4 m column of liquid.

The pressure equation is:

$$ P_{\text{large piston level}} = P_{\text{from force}} + P_{\text{from height}} $$ $$ \frac{W}{A} = \frac{F}{a} + \rho g h $$

Calculate the pressure due to the height ($P_{\text{head}}$).

$$ P_{\text{head}} = \rho g h $$ $$ P_{\text{head}} = 1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2 \times 0.4 \, \text{m} $$ $$ P_{\text{head}} = 3924 \, \text{N/m}^2 $$

Now, solve for $W$.

$$ W = \left( \frac{F}{a} + P_{\text{head}} \right) \times A $$ $$ W = \left( \frac{80 \, \text{N}}{0.0007068 \, \text{m}^2} + 3924 \, \text{N/m}^2 \right) \times 0.007854 \, \text{m}^2 $$ $$ W = (113186 + 3924) \times 0.007854 $$ $$ W = 117110 \times 0.007854 $$ $$ W \approx 919.7 \, \text{N} $$

🎯 Final Results:

(a) When pistons are level, the load lifted is $ \approx 889.0 \, \text{N} $.

(b) When the small piston is 40 cm higher, the load lifted is $ \approx 919.7 \, \text{N} $.

💡 Explanation of Principles

This problem demonstrates two key principles of fluid mechanics:

1. Pascal's Law: This 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 part (a), the pressure $F/a$ is transmitted directly to the large piston, resulting in a multiplied force $W$.

2. Hydrostatic Pressure: This is the pressure exerted by a fluid at rest due to gravity ($p = \rho g h$). In part (b), the level of the large piston is 40 cm below the small piston. Therefore, it experiences additional pressure from the weight of the 40 cm column of liquid. This extra pressure assists in lifting the load.

🔬 Physical Meaning

The results clearly show the effect of both force multiplication and hydrostatic pressure. The base lifting force due to the area difference is approximately 889 N.

In the second scenario, by raising the small piston by 40 cm, an additional 30.7 N of lifting force is generated ($919.7 - 889.0$). This extra force comes from the weight of the fluid column itself, which adds to the pressure exerted on the large piston. This shows that the configuration of a hydraulic system can be adjusted to take advantage of the fluid's own weight to improve performance.