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A horizontal rectangular platform is suspended by four identical wires, one at each of its corners. The wires are \(3.0 \mathrm{~m}\) long and have a diameter of \(2.0 \mathrm{~mm}\). Young's modulus for the material of the wires is \(180 \mathrm{GPa}\). How far will the platform drop (due to elongation of the wires) if a \(50-\mathrm{kg}\) load is placed at the center of the platform?

Short Answer

Expert verified
The platform will drop approximately 0.261 mm.

Step by step solution

01

Understand the Problem

We need to find out how much the platform will drop when a weight is added. This dropping is due to the elongation of the wires. Elongation depends on the applied force, the original length of the wires, their cross-sectional area, and the material's Young's modulus.
02

Identify Relevant Formula

The elongation of a wire can be calculated using the formula: \[ \Delta L = \frac{F \cdot L_0}{A \cdot Y} \]where \( F \) is the force applied, \( L_0 \) is the original length of the wire, \( A \) is the cross-sectional area, and \( Y \) is Young's modulus.
03

Calculate Force Applied

The force applied by the 50-kg mass is due to gravity, thus:\[ F = m \cdot g = 50 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 490.5 \, \text{N} \]
04

Calculate Cross-sectional Area

The cross-sectional area of a wire with a circular cross-section is:\[ A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{2.0 \, \text{mm}}{2}\right)^2 = \pi \left(1.0 \, \text{mm}\right)^2 \ \text{Convert mm to m:} A = \pi \times (1.0 \times 10^{-3} \, \text{m})^2 = \pi \times 10^{-6} \, \text{m}^2 \]
05

Calculate Elongation for One Wire

Using the formula for elongation:\[ \Delta L = \frac{490.5 \, \text{N} \cdot 3.0 \, \text{m}}{\pi \times 10^{-6} \, \text{m}^2 \times 180 \times 10^9 \, \text{Pa}} = \frac{1471.5}{0.56\pi} \approx 2.61 \times 10^{-4} \, \text{m} \]
06

Account for Four Wires

The load is equally shared by the four wires, so the elongation for the whole system is same as that calculated for one wire, i.e.,\[ \Delta L_{total} = \Delta L = 2.61 \times 10^{-4} \text{ m} \]

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Key Concepts

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

Elongation Formula
The elongation of a material refers to how much it stretches under an applied force. When you add weight to a wire, it pulls and extends slightly. The amount it stretches can be found using an important formula. This formula helps us calculate elongation (\(\Delta L\)) by considering several factors:
  • \( F \): the force applied to the wire
  • \( L_0 \): the wire's original length
  • \( A \): the wire's cross-sectional area
  • \( Y \): Young's modulus of the material
By plugging these variables into the formula \(\Delta L = \frac{F \cdot L_0}{A \cdot Y}\), you can calculate just how much a wire will elongate when a specific force is applied. Each term in the equation plays a crucial role in determining the final result.
Cross-Sectional Area
The cross-sectional area of a wire is an essential factor in calculating elongation. It describes the size of the wire's surface when it is cut perpendicular to its length. For circular wires, like the ones in the exercise, you calculate it using the radius of the wire:The formula used is:
  • \( A = \pi \left(\frac{d}{2}\right)^2 \)
where \( d \) is the diameter of the wire. In metric units, it's common to convert dimensions into meters for any calculation. For example, if the diameter is given in millimeters, you first divide by two to get the radius and then convert to meters by multiplying by \(10^{-3}\).Understanding the cross-section's size helps account for how the force is distributed throughout the material, influencing how much it will stretch.
Applied Force
To find out how much an object stretches a wire, you need to determine the applied force on the wire. This force is typically due to the weight of the object being suspended, calculated using its mass. Gravity pulls it downwards. The force due to gravity is calculated as:
  • \( F = m \cdot g \)
where:
  • \( m \) is the mass of the object in kg
  • \( g \) is the acceleration due to gravity (approximately 9.81 \( \text{m/s}^2 \) on Earth)
For example, a 50 kg object applies a force of 490.5 N on the platform. This force is shared by the wires supporting it. Calculating applied force correctly is vital to assessing how much the material will extend under load.
Material Properties
Material properties significantly impact how much a wire will elongate when force is applied. One key property is the Young's modulus, a measure of the material's stiffness or elasticity. It tells us how much it will stretch for a given applied force.Young's modulus (\( Y \)) is defined as:
  • Stress divided by strain within the material's elastic limit
In most practical problems, such as the exercise here, this value is provided since it's a characteristic of the material used.A higher Young's modulus indicates a stiffer material, meaning it will stretch less under the same force compared to a material with a lower Young's modulus. For example, in the exercise, the wires have a Young's modulus of 180 GPa. This indicates a relatively rigid material which is critical for maintaining structure under loads.

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Most popular questions from this chapter

A solid cube of aluminum is \(2.00 \mathrm{~cm}\) on each edge. The density of aluminum is \(2700 \mathrm{~kg} / \mathrm{m}^{3}\). Find the mass of the cube. $$ \text { Mass of cube }=\rho V=\left(2700 \mathrm{~kg} / \mathrm{m}^{3}\right)(0.0200 \mathrm{~m})^{3}=0.0216 \mathrm{~kg}=21.6 \mathrm{~g} $$

The compressibility of water is \(5.0 \times 10^{-10} \mathrm{~m}^{2} / \mathrm{N}\). Find the decrease in volume of \(100 \mathrm{~mL}\) of water when subjected to a pressure of \(15 \mathrm{MPa}\).

A thin sheet of gold foil has an area of \(3.12 \mathrm{~cm}^{2}\) and a mass of \(6.50 \mathrm{mg}\). How thick is the sheet? The density of gold is \(19300 \mathrm{~kg} / \mathrm{m}^{3}\). One milligram is \(10^{-6} \mathrm{~kg}\), so the mass of the sheet is \(6.50 \times 10^{-6} \mathrm{~kg}\). Its volume is $$ V=(\text { area }) \times(\text { thickness })=\left(3.12 \times 10^{-4} \mathrm{~m}^{2}(d)\right. $$ where \(d\) is the thickness of the sheet. We equate this expression for the volume to \(m / \rho\) to get $$ \left(3.12 \times 10^{-4} \mathrm{~m}^{2}\right)(d)=\frac{6.50 \times 10^{-6} \mathrm{~kg}}{19300 \mathrm{~kg} / \mathrm{m}^{3}} $$ from which \(d=1.08 \times 10^{-6} \mathrm{~m}=1.08 \mu \mathrm{m}\).

A \(60-\mathrm{kg}\) motor sits on four cylindrical rubber blocks. Each cylinder has a height of \(3.0 \mathrm{~cm}\) and a crosssectional area of \(15 \mathrm{~cm}^{2}\). The shear modulus for this rubber is \(2.0 \mathrm{MPa} .(a)\) If a sideways force of \(300 \mathrm{~N}\) is applied to the motor, how far will it move sideways? (b) With what frequency will the motor vibrate back and forth sideways if disturbed?

Determine the fractional change in volume as the pressure of the atmosphere \(\left(1 \times 10^{5} \mathrm{~Pa}\right)\) around a metal block is reduced to zero by placing the block in vacuum. The bulk modulus for the metal is \(125 \mathrm{GPa}\).

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