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Chapter 14: Problem 28

A piston of cross-sectional area \(a\) is used in a hydraulic press to exert a small force of magnitude \(f\) on the enclosed liquid. A connecting pipe leads to a larger piston of cross-sectional area \(A\) (Fig. 14-36). (a) What force magnitude \(F\) will the larger piston sustain without moving? (b) If the piston diameters are \(3.80 \mathrm{~cm}\) and \(53.0 \mathrm{~cm}\), what force magnitude on the small piston will balance a \(20.0 \mathrm{kN}\) force on the large piston?

Short Answer

Expert verified
(a) Force on larger piston: \( F = \frac{fA}{a} \). (b) Force on small piston: approximately 102 N.

Step by step solution

01

Understand the Problem

We have a hydraulic press system with a small piston and a larger piston. We need to find the amount of force that can be exerted by the larger piston given an initial force applied on the smaller one. This involves Pascal's principle which states that pressure applied to an enclosed fluid is transmitted undiminished to every part of the fluid and to the walls of its container.
02

Apply Pascal鈥檚 Principle

The pressure exerted by the small piston can be calculated as the force divided by the area: \( P = \frac{f}{a} \). In accordance with Pascal鈥檚 principle, the pressure is the same across the fluid, so \( P = \frac{F}{A} \) for the larger piston, where \( F \) is the force on the larger piston.
03

Equate the Pressures

Set the pressure expressions equal to each other from Step 2: \( \frac{f}{a} = \frac{F}{A} \). From this, the relationship between the forces is \( F = \frac{f}{a} \times A \). This equation allows us to solve for the force \( F \) exerted by the larger piston.
04

Solve for Force F (Part a)

To find the force \( F \) in terms of \( f \), \( a \), and \( A \), rearrange the equation \( F = \frac{f}{a} \times A \). Here, \( F \) will be the force the larger piston sustains without moving.
05

Calculate Force Magnitude on Small Piston (Part b)

Given the diameters of the pistons, calculate their cross-sectional areas: \( a = \pi \left(\frac{3.80}{2}\right)^2 \) cm虏 and \( A = \pi \left(\frac{53.0}{2}\right)^2 \) cm虏. The force \( F = 20.0 \) kN on the large piston can be balanced by solving \( \frac{f}{a} = \frac{20000}{A} \), giving the necessary force on the small piston \( f = \frac{20000 \times a}{A} \).
06

Calculate the Areas

Convert the diameters to radii in meters and find the areas: \( a = \pi \left(\frac{3.80 \times 10^{-2}}{2}\right)^2 \) and \( A = \pi \left(\frac{53.0 \times 10^{-2}}{2}\right)^2 \). Calculate \( a \) and \( A \) using these formulas.
07

Final Calculation for Part b

Using the calculated areas, find the force \( f \) by substituting back into the equation: \( f = \frac{20000 \times a}{A} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydraulic Systems
Hydraulic systems are fascinating engineering marvels that utilize liquids to transmit force. Thanks to Pascal's Principle, these systems can amplify force and perform tasks that require significant power. A hydraulic system typically includes two pistons connected by a pipe filled with a liquid, commonly oil.
  • The small piston (input) receives a force and applies pressure on the fluid.
  • The larger piston (output) experiences this pressure and produces a larger force.
The system allows for an effortless transfer of mechanical energy via fluid pressure, enabling operations like lifting cars in a garage with minimal initial force.
Pressure Calculation
Pressure plays a central role in understanding hydraulic systems. In such setups, pressure is defined as the force applied per unit area. For any given piston, the pressure is calculated as:\[ P = \frac{F}{A} \]where \( P \) is the pressure, \( F \) is the force, and \( A \) is the cross-sectional area.
In hydraulic systems, the pressure exerted by a smaller piston is equal to the pressure experienced by the larger piston due to Pascal's Principle:
  • The pressure remains constant throughout the fluid.
  • This allows the calculation of forces between interconnected pistons using their respective cross-sectional areas.
Thus, hydraulic systems exemplify the practical smoothing of energy through consistent pressure application.
Force Balance
Force balance in hydraulic systems is crucial for understanding how the system maintains equilibrium. According to Pascal鈥檚 Principle, the force exerted on a fluid in a closed container transmits uniformly throughout the fluid. This means the pressures at different points in the fluid are equal.
To maintain balance, the forces on the pistons must equate via pressure:\( \frac{f}{a} = \frac{F}{A} \)
  • \( f \) is the force on the smaller piston, \( a \) is its area.
  • \( F \) is the force on the larger piston, \( A \) is its area.
This relation allows engineers to determine the force needed on the small piston to balance a desired force on the larger piston, making hydraulic systems both flexible and powerful.
Cross-Sectional Area
The cross-sectional area is a vital parameter in the functionality of hydraulic systems. It determines how pressure translates into force. Calculating the area of a piston involves understanding its diameter. For a circular piston, the formula is:\[ A = \pi \left(\frac{d}{2}\right)^2 \]where \( d \) is the diameter of the piston.
In the exercise, pistons with diameters of 3.80 cm and 53.0 cm dictate their respective cross-sectional areas.
  • Smaller piston area influences the initial pressure calculated from initial force.
  • Larger piston area converts this pressure back into a usable force.
Having these calculated areas allows precise force manipulation within hydraulic machinery, optimizing it for various tasks.

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Most popular questions from this chapter

Models of torpedoes are sometimes tested in a horizontal pipe of flowing water, much as a wind tunnel is used to test model airplanes. Consider a circular pipe of internal diameter \(25.0 \mathrm{~cm}\) and a torpedo model aligned along the long axis of the pipe. The model has a \(5.00 \mathrm{~cm}\) diameter and is to be tested with water flowing past it at \(2.50 \mathrm{~m} / \mathrm{s}\). (a) With what speed must the water flow in the part of the pipe that is unconstricted by the model? (b) What will the pressure difference be between the constricted and unconstricted parts of the pipe?

In an experiment, a rectangular block with height \(h\) is allowed to float in four separate liquids. In the first liquid, which is water, it floats fully submerged. In liquids \(A, B\), and \(C\), it floats with heights \(h / 2,2 h / 3\), and \(h / 4\) above the liquid surface, respectively. What are the relative densities (the densities relative to that of water) of (a) \(A,(\mathrm{~b}) B\), and \((\mathrm{c}) C?\)

Three liquids that will not mix are poured into a cylindrical container. The volumes and densities of the liquids are \(0.50 \mathrm{~L}, 2.6 \mathrm{~g} / \mathrm{cm}^{3}\); \(0.25 \mathrm{~L}, 1.0 \mathrm{~g} / \mathrm{cm}^{3} ;\) and \(0.40 \mathrm{~L}, 0.80 \mathrm{~g} / \mathrm{cm}^{3} .\) What is the force on the bot- tom of the container due to these liquids? One liter \(=1 \mathrm{~L}=1000\) \(\mathrm{cm}^{3}\). (Ignore the contribution due to the atmosphere.)

The maximum depth \(d_{\max }\) that a diver can snorkel is set by the density of the water and the fact that human lungs can function against a maximum pressure difference (between inside and outside the chest cavity) of \(0.050 \mathrm{~atm}\). What is the difference in \(d_{\text {max }}\) for fresh water and the water of the Dead Sea (the saltiest natural water in the world, with a density of \(1.5 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) )?

A block of wood has a mass of \(3.67 \mathrm{~kg}\) and a density of \(600 \mathrm{~kg} / \mathrm{m}^{3}\). It is to be loaded with lead \(\left(1.14 \times 10^{4} \mathrm{~kg} / \mathrm{m}^{3}\right)\) so that it will float in water with \(0.900\) of its volume submerged. What mass of lead is needed if the lead is attached to (a) the top of the wood and (b) the bottom of the wood?
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