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A piston of cross-scctional 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 \(A\) (lig. 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 magnitude \( F = f \frac{A}{a} \). (b) Requires \( 101.79 \, N \) on the small piston to balance a 20 kN force on the large piston.

Step by step solution

01

Understanding the Relationship Between Force and Area in a Hydraulic System

In a hydraulic press, the pressure exerted by the small piston is equal to the pressure exerted by the larger piston due to Pascal's principle. The pressure \( P \) can be defined as \( P = \frac{F}{A} \), where \( F \) is the force and \( A \) is the area. For the system to be in balance, the pressure applied on the small piston should equal the pressure on the large piston: \[\frac{f}{a} = \frac{F}{A}\] This equation will allow us to solve for the force magnitude \( F \) on the larger piston in terms of the force \( f \) on the smaller piston.
02

Calculating Force on the Larger Piston

From the equation \( \frac{f}{a} = \frac{F}{A} \), rearranging gives \( F = f \frac{A}{a} \). Here, the force \( f \) is applied by the small piston with area \( a \), and \( A \) is the area of the larger piston. This equation shows that a small force on the small piston can balance a much larger force on the large piston, due to the difference in cross-sectional areas.
03

Compute Cross-sectional Areas from Diameters

The cross-sectional area \( a \) of the small piston is \( \pi \left( \frac{d_1}{2} \right)^2 \) and the area \( A \) of the larger piston is \( \pi \left( \frac{d_2}{2} \right)^2 \). Given the diameters of the pistons, \( d_1 = 3.80 \) cm and \( d_2 = 53.0 \) cm, convert these to meters: \( d_1 = 0.038 \) m and \( d_2 = 0.53 \) m. Using these, compute \( a \) and \( A \).
04

Calculate the Force on the Small Piston

From the balance condition \( \frac{f}{a} = \frac{F}{A} \) or \( f = F \frac{a}{A} \), substitute the known \( F = 20,000 \, \text{N} \), and the calculated \( a \) and \( A \) to find \( f \). Calculate:\[ a = \pi \left( \frac{0.038}{2} \right)^2 = \pi \times 0.000361 = 0.00113 \, \text{m}^2\]\[ A = \pi \left( \frac{0.53}{2} \right)^2 = \pi \times 0.0706 = 0.2218 \, \text{m}^2\]Plug into:\[ f = 20,000 \times \frac{0.00113}{0.2218} \approx 101.79 \, N\] Thus, a force of approximately \( 101.79 \, \text{N} \) on the small piston balances the \( 20,000 \, \text{N} \) force on the large piston.

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Key Concepts

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

Pascal's Principle
Pascal's principle plays a crucial role in hydraulic systems, including the hydraulic press. This principle, formulated by Blaise Pascal, states that a change in pressure applied to an enclosed fluid is transmitted undiminished throughout the fluid. This means that pressure changes at one point in a closed system are felt evenly at all other points in the system. In the case of a hydraulic press, this principle allows a small force applied to a small piston to be transmitted into a larger force on a larger piston.

In practical terms, when you exert a force on the smaller piston, it increases the pressure in the fluid beneath it. According to Pascal's principle, this increase in pressure is experienced throughout the entire fluid, ultimately pushing up against the larger piston. This is why hydraulic systems enable us to lift heavy objects with relatively small forces. It's fascinating how this principle allows force magnification through hydraulic machines.
Pressure
Pressure is at the heart of understanding how a hydraulic press operates. In physics, pressure is defined as the force applied per unit area. Mathematically, it is expressed as \( P = \frac{F}{A} \), where \( P \) is the pressure, \( F \) is the force, and \( A \) is the area over which the force is distributed.

In a hydraulic press, the pressure generated by the small piston equates to the pressure experienced by the large piston, due to the properties of the fluid and Pascal's principle. This equal pressure ensures that the balance of forces between the two pistons is maintained. By considering the ratios of force and area, a small piston can exert a significant influence on a larger piston, resulting in increased force output on the larger piston.

Think of it as spreading the effort over the entire area of the larger piston by utilizing the incompressible nature of the fluid to hold that increased force effectively.
Force Calculation
Force calculation in a hydraulic press involves interpreting the relationship between force, pressure, and area. The basic equation that guides us is \( F = P \cdot A \), and for two pistons in a hydraulic system, the balance condition \( \frac{f}{a} = \frac{F}{A} \) is utilized. The equation can be rearranged to find the unknown variables depending on the data given.

Let's consider a practical scenario: if you know the force exerted on the large piston (\( F \)) and the areas of the two pistons, you can calculate the necessary force on the small piston (\( f \)) to balance \( F \). With values from the exercise:
  • The large piston force is given as 20,000 N.
  • Calculations of areas \( a \) and \( A \) provide us with the needed numerical values.
Substitute these into the formula \( f = F \frac{a}{A} \), plug in the known values, and solve for the unknown force, resulting in quite a small force on the small piston being sufficient to balance a much larger force on the large piston.
Cross-sectional Area
Cross-sectional area is a critical factor when analyzing the function of a hydraulic press. The area of a piston determines how the force is distributed across it when pressure is applied.

To calculate the cross-sectional area of a piston, you use the formula for the area of a circle: \( A = \pi \left(\frac{d}{2}\right)^2 \), where \( d \) is the diameter of the piston. By plugging in the diameter values of the pistons, you convert them from centimeters to meters to maintain standard units and compute their areas accurately.

The role of cross-sectional area becomes prominent when it comes to force distribution. A larger piston area means that a relatively small pressure can support a larger force. Hence, the hydraulic press can amplify force based on the difference between the cross-sectional areas of the two pistons. This simple yet elegant solution is what makes hydraulic presses so powerful and versatile in practical applications.

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