Question

(a) If the small piston of a hydraulic lever has a diameter of \(3.72 \mathrm{~cm}\), and the large piston one of \(51.3 \mathrm{~cm}\), what weight on the small piston will support \(18.6 \mathrm{kN}\) (e.g., a car) on the large piston? (b) Through what distance must the small piston move to raise the car \(1.65 \mathrm{~m}\) ?

Short answer

To solve this problem, do your calculations considering what's explained in each step. The result will be your answer for parts (a) and (b) of this exercise.

Step-by-step solution

- Calculation of the Area

We begin by calculating the areas of the small and large pistons, using their diameters. Let's denote the area of the small piston as \(A_{s}\) and the area of the large piston as \(A_{l}\). The formula to calculate area of a circle is \(\pi (d/2)^2\), where \(d\) is the diameter. So, \(A_{s} = \pi (3.72/2)^2\) and \(A_{l} = \pi (51.3/2)^2\). Do your calculations.

- Calculate the weight for the small piston

According to Pascal's principle, the pressure (P) is the same everywhere in a liquid, and it equals Force (F) divided by Area (A). The weight that the small piston needs to sustain can be calculated as \(P \times A_{s}\), but because the pressure is equal, we can write it also as \(F_{l} / A_{l}\), where \(F_{l}\) is the force on the large piston (equal to the weight of the car, 18.6kN). So, we can write: \(F_{s} = F_{l} \times (A_{s} / A_{l})\). Do your calculations.

- Calculate the displacement of the small piston

According to Pascal's principle, the fluid volume lifted or displaced by the small piston is equal to the fluid volume lifted or displaced by the large piston. The change in volume is given by Area times displacement, and this should be equal for both pistons. This gives us the equation \(A_{s} \times D_{s} = A_{l} \times D_{l}\), where \(D_{s}\) is the displacement of the small piston and \(D_{l}\) is the displacement of the large piston (equal to the height the car was lifted, 1.65m). We can solve for \(D_{s}\) by putting the terms into the formula \(D_{s} = (A_{l} / A_{s}) \times D_{l}\). Do your calculations.

Key concepts

Pascal's Principle

The essence of Pascal's Principle lies in its profound assertion that when pressure is applied to an enclosed fluid, it is transmitted undiminished to every portion of the fluid and to the walls of the containing vessel. This principle forms the cornerstone for hydraulic systems, including the hydraulic lever. Imagine pressing down on a syringe filled with liquid; the liquid transmits that pressure equally in all directions, pushing against the plunger at the other end with the same force.

For the case of our hydraulic lever, consider a small piston being pressed down upon by a certain weight. According to Pascal's Principle, that exerted pressure will be communicated throughout the entire fluid, reaching the large piston and causing a corresponding force to act upon it. This allows what might be a relatively small force applied at the small piston to support a much heavier weight on the large piston, effectively attaining mechanical advantage.

Pressure Calculation in Fluids

Understanding how to calculate pressure in fluids is vital for analyzing hydraulic systems. Pressure can be viewed as a force applied per unit area, with the mathematical expression given by the equation:
\[ P = \frac{F}{A} \] where

• \( P \) is the pressure,
• \( F \) is the force, and
• \( A \) is the area on which the force is exerted.

In the context of our hydraulic system, we have a small and large piston each with its own area. If we apply force to the small piston, the resulting pressure is transmitted through the fluid and acts with the same intensity on the larger piston. By cross-multiplying areas and forces, we calculate the force necessary at the smaller piston to sustain a much larger weight at the large piston, which directly employs Pascal's Principle. This provides students and engineers alike with the means to design systems that can lift or hold heavy loads with comparatively little input force.

Mechanical Advantage

Mechanical advantage is a key concept in physics that allows a force to be amplified using tools or machines. It is a measure of the force amplification achieved by using a tool, mechanical device, or machine system. The mechanical advantage can be calculated as the ratio of the output force to the input force and is a unitless number. This principle is elegantly illustrated through the hydraulic lever as seen in our textbook problem.

The hydraulic lever achieves mechanical advantage by allowing a smaller force applied over a larger distance at the small piston to balance a larger force applied over a smaller distance at the large piston. The relationship between the distances moved by the two pistons is inversely proportional to the areas of the pistons, according to the formula for hydraulic systems:
\[ MA = \frac{D_{s}}{D_{l}} = \frac{A_{l}}{A_{s}} \] where

• \( MA \) is the mechanical advantage,
• \( D_{s} \) is the distance moved by the small piston,
• \( D_{l} \) is the distance moved by the large piston,
• \( A_{s} \) is the area of the small piston, and
• \( A_{l} \) is the area of the large piston.

Applying this concept, we understand how huge loads can be moved with smaller forces, underpinning the operation of many modern machines.