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Chapter 13: Problem 20

(II) A hydraulic press for compacting powdered samples has a large cylinder which is \(10.0 \mathrm{~cm}\) in diameter, and a small cylinder with a diameter of \(2.0 \mathrm{~cm}\) (Fig. \(13-51) .\) A lever is attached to the small cylinder as shown. The sample, which is placed on the large cylinder, has an area of \(4.0 \mathrm{~cm}^{2}\). What is the pressure on the sample if \(350 \mathrm{~N}\) is applied to the lever?

Short Answer

Expert verified
The pressure on the sample is approximately 696.25 N/cm虏.

Step by step solution

01

Calculate Area of Cylinders

First, determine the area of the large and small cylinders using the formula for the area of a circle, \( A = \pi r^2 \), where \( r \) is the radius. - For the large cylinder: - Diameter = 10.0 cm, thus Radius = 5.0 cm. - Area \( A_L = \pi (5.0 \, \text{cm})^2 = 25 \pi \, \text{cm}^2 \). - For the small cylinder: - Diameter = 2.0 cm, thus Radius = 1.0 cm. - Area \( A_S = \pi (1.0 \, \text{cm})^2 = \pi \, \text{cm}^2 \).
02

Understand Pascal's Principle

According to Pascal's Principle, the pressure applied to the small cylinder is transferred to the large cylinder. The force exerted on the sample is the pressure times the area of the large cylinder.
03

Calculate Pressure on Small Cylinder

Using the given force on the lever of 350 N and the area of the small cylinder:\[ \text{Pressure on Small Cylinder} = \frac{350 \, \text{N}}{\pi \, \text{cm}^2} \approx 111.4 \, \text{N/cm}^2 \].
04

Calculate Force Exerted on Large Cylinder

Now, multiply the pressure on the small cylinder by the area of the large cylinder to find the force applied to the large cylinder:\[ \text{Force on Large Cylinder} = 111.4 \, \text{N/cm}^2 \times 25 \pi \, \text{cm}^2 \approx 2785 \text{ N} \].
05

Calculate Pressure on the Sample

The pressure on the sample is the force on the large cylinder divided by the area of the sample: \[ \text{Pressure on Sample} = \frac{2785 \, \text{N}}{4.0 \, \text{cm}^2} \approx 696.25 \, \text{N/cm}^2 \].

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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 is a fundamental concept in fluid mechanics. It states that pressure applied to a contained fluid is transmitted undiminished throughout the fluid. This principle has practical applications, such as in hydraulic systems like our hydraulic press example.
In this press, pressure exerted on the small cylinder is evenly distributed throughout the liquid connecting the cylinders. Because the press uses a fluid, the pressure from the small side is communicated to the larger side.
This allows a small force applied on the smaller piston to be translated into a much larger force on the bigger piston, enabling tasks like compacting samples with less effort.
Pressure Calculation
Calculating pressure involves understanding the relationship between force and area. Pressure is defined as the force exerted per unit area, given by the formula:

\[ \text{Pressure} = \frac{\text{Force}}{\text{Area}} \]

In the hydraulic press scenario, we start with the force applied to the small cylinder. Here, 350 N is applied on an area of the small cylinder, which is calculated using the radius derived from the cylinder's diameter. By dividing the force by this area, the pressure exerted on this cylinder is found. This pressure is then transferred to the large cylinder, showcasing the power of Pascal's Principle in action.
Area of Cylinders
Knowing the area of cylinders is crucial to solving the hydraulic press problem. The area of a circle, which is the shape of the cylinder's cross-section, is calculated using the formula:

\[ A = \pi r^2 \]

Here, \( r \) is the radius of the circle. The exercise involves two cylinders, each with different diameters. For the large cylinder with a 10.0 cm diameter, the radius is half of the diameter, 5.0 cm, resulting in an area of \( 25\pi \text{ cm}^2 \).
The small cylinder has a 2.0 cm diameter, leading to a radius of 1.0 cm and an area of \( \pi \text{ cm}^2 \). These area calculations are essential for determining pressure distribution and force transmission in hydraulic systems.

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Most popular questions from this chapter

A 2.8 -N force is applied to the plunger of a hypodermic needle. If the diameter of the plunger is 1.3 \(\mathrm{cm}\) and that of the needle 0.20 \(\mathrm{mm}\) , (a) with what force does the fluid leave the needle? (b) What force on the plunger would be needed to push fluid into a vein where the gauge pressure is 75 \(\mathrm{mm}\) -Hg? Answer for the instant just before the fluid starts to move.

A bucket of water is accelerated upward at 1.8 \(\mathrm{g} .\) What is the buoyant force on a 3.0 -kg granite rock \((\mathrm{SG}=2.7)\) submerged in the water? Will the rock float? Why or why not?

A hydraulic lift is used to jack a \(920-\mathrm{kg}\) car \(42 \mathrm{~cm}\) off the floor. The diameter of the output piston is \(18 \mathrm{~cm},\) and the input force is \(350 \mathrm{~N}\). \((a)\) What is the area of the input piston? \((b)\) What is the work done in lifting the car \(42 \mathrm{~cm} ?(c)\) If the input piston moves \(13 \mathrm{~cm}\) in each stroke, how high does the car move up for each stroke? \((d)\) How many strokes are required to jack the car up \(42 \mathrm{~cm} ?\) ( \(e\) ) Show that energy is conserved.

A raft is made of 12 logs lashed together. Each is \(45 \mathrm{~cm}\) in diameter and has a length of \(6.1 \mathrm{~m}\). How many people can the raft hold before they start getting their feet wet, assuming the average person has a mass of \(68 \mathrm{~kg}\) ? Do not neglect the weight of the logs. Assume the specific gravity of wood is \(0.60 .\)

(II) Because gasoline is less dense than water, drums containing gasoline will float in water. Suppose a 230 -L steel drum is completely full of gasoline. What total volume of steel can be used in making the drum if the gasoline-filled drum is to float in fresh water?
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