/*! 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 30 A spring of spring constant \(3.... [FREE SOLUTION] | 91Ó°ÊÓ

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A spring of spring constant \(3.75 \times 10^{4} \mathrm{~N} / \mathrm{m}\) is between a rigid beam and the output piston of a hydraulic lever. An empty container with negligible mass sits on the input piston. The input piston has area \(A_{i}\), and the output piston has area \(18.0 A_{2}\). Initially the spring is at its rest length. How many kilograms of sand must be (slowly) poured into the container to compress the spring by \(5.00 \mathrm{~cm}\) ?

Short Answer

Expert verified
About 10.62 kg of sand is needed.

Step by step solution

01

Determine Spring Force

The first step is to determine the force exerted by the spring when compressed. We use Hooke's Law for this, which states that the force of the spring \( F_s \) is given by \( F_s = k x \), where \( k \) is the spring constant and \( x \) is the compression length. For this problem, \( k = 3.75 \times 10^4 \mathrm{~N/m} \) and \( x = 5.00 \mathrm{~cm} = 0.050 \mathrm{~m} \). Thus, \( F_s = 3.75 \times 10^4 \times 0.050 = 1875 \mathrm{~N} \).
02

Apply Pascal's Principle

Pascal's Principle relates the force applied on the input piston to the force on the output piston through the ratio of the areas: \( \frac{F_i}{A_i} = \frac{F_o}{A_o} \). In this case, \( A_o = 18 A_i \). Therefore, \( F_o = F_s = 1875 \mathrm{~N} \) and \( F_i = \frac{F_s}{18} = \frac{1875}{18} \mathrm{~N} = 104.17 \mathrm{~N} \).
03

Find Mass of Sand Required

The gravitational force due to the sand, \( F_g \), is equal to \( F_i \) and is given by the formula \( F_g = m g \), where \( m \) is the mass of the sand and \( g \) is the acceleration due to gravity \( 9.81 \mathrm{~m/s^2} \). Solving for \( m \), \( m = \frac{F_i}{g} = \frac{104.17}{9.81} \mathrm{~kg} \approx 10.62 \mathrm{~kg} \).

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Key Concepts

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

Hooke's Law
Hooke's Law is fundamental when dealing with springs and elastic materials. This law tells us how much force is needed to compress or stretch a spring by a certain amount. The formula for Hooke's Law is:
  • \( F_s = kx \)
Here, \( F_s \) is the force exerted by the spring, \( k \) is the spring constant, and \( x \) is the distance that the spring is compressed or stretched from its rest position.
For the exercise in question, the spring constant is given as \( 3.75 \times 10^4 \, \text{N/m} \). When you compress this spring by \( 5.00 \, \text{cm} \) or \( 0.050 \, \text{m} \), the force needed is \( 1875 \, \text{N} \). This is calculated by multiplying the spring constant by the compression distance. Understanding this helps us predict how springs will behave in different situations.
Hydraulic Lever
A hydraulic lever is a fantastic application of fluid mechanics principles. It allows us to amplify force using fluid pressure in a confined space. In this exercise, the setup involves input and output pistons connected by a fluid-filled chamber.
  • The input piston is where force is applied.
  • The output piston experiences a transmitted force.
The input and output pistons have different areas, which is a key feature of how a hydraulic lever works. In this scenario, the area of the output piston is \(18\) times that of the input piston. This means that any force applied on the input piston will be distributed over a much larger area on the output piston, effectively increasing the force by a factor of \(18\). This multiplication of force is what makes hydraulic systems so powerful and widely used in machines, like car brakes and excavators.
Pascal's Principle
Pascal's Principle is a crucial concept for understanding the behavior of fluids and is the cornerstone of hydraulic systems. This principle states that any change in pressure applied to an enclosed fluid is transmitted equally throughout the fluid. In simpler terms:
  • If you apply pressure to one part of a fluid, it gets transferred in all directions.
In a hydraulic lever, this principle allows force applied on the input piston to be transferred efficiently to the output piston. In our exercise, the pressure applied by the force \( F_i \) on the smaller input area \( A_i \) is transferred to the larger output area \( A_o = 18 A_i \).
Hence, when the input piston experiences a certain force, the output piston experiences an amplified force due to its larger area, allowing the spring to be compressed by the desired amount efficiently.
Gravitational Force
Gravitational force is the attractive force between any two masses. On Earth, this manifests as the weight of objects. The gravitational force acting on an object is calculated using the formula:
  • \( F_g = mg \)
Where \( F_g \) is the gravitational force (or weight), \( m \) is the mass of the object, and \( g \) is the acceleration due to gravity, approximately \( 9.81 \, \text{m/s}^2 \) on Earth. In the exercise, the gravitational force of sand poured into the container is needed to balance with the force exerted by the spring.
The required mass of sand was found to be about \( 10.62 \, \text{kg} \), which corresponds to the gravitational force balancing the hydraulic system's force requirements. Understanding gravitational force helps illuminate why a particular mass is necessary to create balance in a system based on weight.

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Most popular questions from this chapter

Water flows through a horizontal pipe and then out into the atmosphere at a speed \(v_{1}=23.0 \mathrm{~m} / \mathrm{s}\). The diameters of the left and right sections of the pipe are \(5.00 \mathrm{~cm}\) and \(3.00 \mathrm{~cm}\). (a) What volume of water flows into the atmosphere during a \(20.0 \mathrm{~min}\) period? In the left section of the pipe, what are (b) the speed \(v_{2}\) and (c) the gauge pressure?

Suppose that two tanks, 1 and 2, each with a large opening at the top, contain different liquids. A small hole is made in the side of each tank at the same depth \(h\) below the liquid surface, but the hole in tank 1 has half the cross- sectional area of the hole in tank 2. (a) What is the ratio \(\rho_{1} / \rho_{2}\) of the densities of the liquids if the mass flow rate is the same for the two holes? (b) What is the ratio \(R_{V 1} / R_{V 2}\) of the volume flow rates from the two tanks? (c) At one instant, the liquid in tank 1 is \(16.0 \mathrm{~cm}\) above the hole. If the tanks are to have equal volume flow rates, what height above the hole must the liquid in tank 2 be just then?

A water pipe having a \(2.5 \mathrm{~cm}\) inside diameter carries water into the basement of a house at a speed of \(0.90 \mathrm{~m} / \mathrm{s}\) and a pressure of \(190 \mathrm{kPa}\). If the pipe tapers to \(1.2 \mathrm{~cm}\) and rises to the second floor \(7.6 \mathrm{~m}\) above the input point, what are the (a) speed and (b) water pressure at the second floor?

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 \(6.00 \mathrm{~cm}\) diameter and is to be tested with water flowing past it at \(2.00 \mathrm{~m} / \mathrm{s}\). (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?

Two streams merge to form a river. One stream has a width of \(8.2 \mathrm{~m}\), depth of \(3.4 \mathrm{~m}\), and current speed of \(2.3 \mathrm{~m} / \mathrm{s}\). The other stream is \(6.8 \mathrm{~m}\) wide and \(3.2 \mathrm{~m}\) deep, and flows at \(2.6 \mathrm{~m} / \mathrm{s}\). If the river has width \(10.5 \mathrm{~m}\) and depth \(4.5 \mathrm{~m}\), what is its speed?

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