/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 In the process of changing a fla... [FREE SOLUTION] | 91Ó°ÊÓ

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In the process of changing a flat tire, a motorist uses a hydraulic jack. She begins by applying a force of \(45 \mathrm{~N}\) to the input piston, which has a radius \(r_{1}\). As a result, the output plunger, which has a radius \(r_{2}\), applies a force to the car. The ratio \(r_{2} / r_{1}\) has a value of 8.3. Ignore the height difference between the input piston and output plunger and determine the force that the output plunger applies to the car.

Short Answer

Expert verified
The force applied to the car by the output plunger is approximately 3102.85 N.

Step by step solution

01

Understand the Problem

We need to determine the force applied to the car by the output plunger of a hydraulic jack. We know the input force is 45 N and the ratio of the radii of the output plunger to the input piston is 8.3.
02

Apply Pascal's Principle

According to Pascal's Principle, the pressure applied to a confined fluid is transmitted undiminished throughout the fluid. Therefore, the pressure on the input piston is equal to the pressure on the output plunger.
03

Establish the Relationship Between Force and Area

The pressure applied by the input force is given by \( P = \frac{F_1}{A_1} \) where \( A_1 = \pi r_1^2 \). Similarly, the pressure on the output plunger is \( P = \frac{F_2}{A_2} \) where \( A_2 = \pi r_2^2 \). Given \( P_1 = P_2 \), \( \frac{F_1}{\pi r_1^2} = \frac{F_2}{\pi r_2^2} \).
04

Use the Ratio of Radii

Since the ratio \( \frac{r_2}{r_1} = 8.3 \), substitute into the area relationship: \( \frac{F_1}{r_1^2} = \frac{F_2}{(8.3r_1)^2} \). Simplifying gives \( F_2 = F_1 \times 8.3^2 \).
05

Calculate the Output Force

Substitute the known value to find \( F_2 \) : \[ F_2 = 45 \times 8.3^2 \]. Calculating this gives \( F_2 \approx 3102.85 \text{ N} \).

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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 when pressure is applied to a confined fluid, the pressure change is transmitted equally throughout the entire fluid. This principle is vital for devices like hydraulic jacks, which are used to lift heavy loads with minimum effort. The pressure that we're mentioning is the same in all directions in a closed system, which is why a small force can exert a much larger force through the device allowing a motorist to lift a car easily. Understanding this principle helps us see how force is multiplied in hydraulic systems.
This principle can be applied in the context of the hydraulic jack, where the force exerted on the input piston is transmitted equally to the output plunger. Thus, the same pressure that the motorist applies on a small piston results in a significantly higher force output through the larger plunger, effectively lifting heavy objects.
Force Calculation
In hydraulic systems, calculating the force applied at different points involves understanding the relationship between force, pressure, and area. The formula used here is derived from Pascal's Principle and is given by:
  • Pressure, \( P = \frac{F}{A} \)
Where \( F \) is the force applied and \( A \) is the cross-sectional area.
  • Force on the input piston: \( F_1 \)
  • Area of the input piston: \( A_1 = \pi r_1^2 \)
  • Force on the output plunger: \( F_2 \)
  • Area of the output plunger: \( A_2 = \pi r_2^2 \)
By using the relation that pressure is transmitted equally, \( \frac{F_1}{A_1} = \frac{F_2}{A_2} \), you can find \( F_2 \) if \( F_1 \) is known. This relationship allows calculations to be simplified to solve for the unknown force. By rearranging and substituting known values, you can find that the output force \( F_2 \) is the input force times the square of the ratio of the radii of the plunger and piston. This effective control of force is what allows the jack to lift a car effortlessly.
Pressure and Area
The concept of pressure and its relation to area is pivotal in understanding how hydraulic jacks work. Pressure is defined as a force distributed over an area, mathematically expressed as \( P = \frac{F}{A} \). This relationship means that for a constant pressure, the force is directly proportional to the area.

In the hydraulic jack scenario, two pistons with different cross-sectional areas are involved. The input piston (or handle) applies a small force over a smaller area, creating a pressure in the fluid. The output piston, having a much larger area, converts this pressure into a larger force, due to the area multiplier.
  • The smaller area of the input piston allows for ease of applying a force.
  • The larger area of the output plunger allows it to exert significantly more force.
This directly illustrates how the multiplication effect in hydraulic systems allows for the lifting of heavy loads like a car, leveraging the principle of pressure distribution over larger areas.

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Most popular questions from this chapter

(a) The mass and radius of the sun are \(1.99 \times 10^{30} \mathrm{~kg}\) and \(6.96 \times 10^{8} \mathrm{~m}\). What is its density? (b) If a solid object is made from a material that has the same density as the sun, would it sink or float in water? Why? (c) Would a solid object sink or float in water if it were made from a material whose density was the same as that of the planet Saturn (mass \(=5.7 \times 10^{26} \mathrm{~kg},\) radius \(\left.=6.0 \times 10^{7} \mathrm{~m}\right)\) ? Provide a reason for your answer.

At a given instant, the blood pressure in the heart is \(1.60 \times 10^{4} \mathrm{~Pa}\). If an artery in the brain is \(0.45 \mathrm{~m}\) above the heart, what is the pressure in the artery? Ignore any pressure changes due to blood flow.

A uniform rectangular plate is hanging vertically downward from a hinge that passes along its left edge. By blowing air at \(11.0 \mathrm{~m} / \mathrm{s}\) over the top of the plate only, it is possible to keep the plate in a horizontal position, as illustrated in part \(a\) of the drawing. To what value should the air speed be reduced so that the plate is kept at a \(30.0^{\circ}\) angle with respect to the vertical, as in part \(b\) of the drawing? (Hint: Apply Bernoulli's equation in the form of Equation \(11.12 .)\)

A paperweight, when weighed in air, has a weight of \(\mathrm{W}=6.9 \mathrm{~N}\). When completely immersed in water, however, it has a weight of \(W_{\text {in water }}=4.3 \mathrm{~N}\). Find the volume of the paperweight.

A fountain sends a stream of water straight up into the air to a maximum height of 5.00 \(\mathrm{m} .\) The effective cross-sectional area of the pipe feeding the fountain is \(5.00 \times 10^{-4} \mathrm{~m}^{2}\) Neglecting air resistance and any viscous effects, determine how many gallons per minute are being used by the fountain. (Note: \(1 \mathrm{gal}=3.79 \times 10^{-3} \mathrm{~m}^{3}\).)

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