/*! 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 50 Figure 12-61 represents an insec... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Figure 12-61 represents an insect caught at the midpoint of a spider-web thread. The thread breaks under a stress of \(8.20 \times 10^{8} \mathrm{~N} / \mathrm{m}^{2}\) and a strain of 2.00. Initially, it was horizontal and had a length of \(2.00 \mathrm{~cm}\) and a cross-sectional area of \(8.00 \times\) \(10^{-12} \mathrm{~m}^{2}\). As the thread was stretched under the weight of the insect, its volume remained constant. If the weight of the insect puts the thread on the verge of breaking, what is the insect's mass? (A spider's web is built to break if a potentially harmful insect, such as a bumble bee, becomes snared in the web.)

Short Answer

Expert verified
The insect's mass is approximately \(6.69 \times 10^{-4} \, \text{kg}\).

Step by step solution

01

Understand the Problem

We need to find the mass of the insect that causes the thread to be on the verge of breaking. The thread's properties given include its stress capacity, strain, original length, and cross-sectional area.
02

Calculate Stress

Stress is defined as the force per unit area, so we can say: \( \text{Stress} = \frac{F}{A} = 8.20 \times 10^8 \, \text{N/m}^2 \). Here, \(F\) is the force due to the insect's weight, and \(A\) is the cross-sectional area.
03

Calculate Force

Since stress is \( \frac{F}{A}\), we rearrange to find \(F\): \( F = \text{Stress} \times A = 8.20 \times 10^8 \, \text{N/m}^2 \times 8.00 \times 10^{-12} \, \text{m}^2 = 6.56 \times 10^{-3} \, \text{N} \).
04

Calculate Insect's Mass

The force \(F\) exerted by the insect is equal to its weight, which is \(mg\), where \(m\) is the mass and \(g\) is the acceleration due to gravity (\(9.8 \, \text{m/s}^2\)). Thus, \( m = \frac{F}{g} = \frac{6.56 \times 10^{-3} \, \text{N}}{9.8 \, \text{m/s}^2} \approx 6.69 \times 10^{-4} \, \text{kg} \).
05

Finalize the Answer

The calculations show that the mass of the insect is approximately \(6.69 \times 10^{-4} \, \text{kg}\). Ensure that all assumptions like constant volume and properties given (stress, strain) align with translating stress and force relationships.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Stress
Stress is a fundamental concept in physics and engineering that describes how an internal force is distributed within a material. It is essentially the measure of force exerted per unit area within materials. In the context of the spider-web exercise, the stress is given as the critical amount before the web breaks. This is represented mathematically as: - Stress (\( ext{σ}\)) = Force (\(F\)) divided by Cross-sectional Area (\(A\)): \[\sigma = \frac{F}{A}\],This equation implies that for a material to maintain its structural integrity under force, the internal stress must not exceed its maximum threshold, which in this case is \(8.20 \times 10^8 \text{ N/m}^2\). Stress can be:
  • Tensile, when it pulls apart.
  • Compressive, when it pushes together.
  • Shear, when forces are parallel but opposite.
Balancing these forces in engineering ensures safety and efficiency in design.
Strain
Strain details the deformation or elongation of a material when stress is applied. It is calculated as the ratio of change in length to the original length. Unlike stress, which focuses on force per area, strain relates to the material's change in shape or size. In this case, even when the thread is about to break, it has a design limit strain of 2.00. This reveals how much the spider-web can stretch before failure. Mathematically, strain (\(\epsilon\)) is given by:\[\epsilon = \frac{\Delta L}{L_0}\]Where \(\Delta L\) is the change in length, and \(L_0\) is the original length.The spider-web remains one piece (without breaking) until the strain reaches 2.00, a helpful safety feature nature has designed in webs to prevent disaster when larger prey is caught.
Spider-Web Mechanics
The mechanics of spider-webs are fascinating. Spider silk is a marvel of natural material engineering, exhibiting remarkable properties considering its lightweight and high tensile strength. Here’s how it relates to stress and strain:
  • **Tensile Strength:** Spider silk can withstand extensive force without breaking.
  • **Elasticity:** It can stretch significantly which helps in warning (through strain) when close to breaking point.
  • **Volume Conservation:** In the exercise, despite the stretching, the volume stays constant.
This implies that as the web gets longer, due to catching an insect, it becomes narrower (because volume = area × length = constant). These properties make spider-webs incredibly resilient and perfect for trapping insects.
Force Calculations
Understanding the principle of force calculations is key to solving physics problems like the spider-web scenario. Here, force is directly related to the insect's weight acting on the web. Given the stress, we first need to restate it in terms of force.Using the formula:\[F = \sigma \times A\]Where \(\sigma\) is stress and \(A\) is area, we calculate the force that the spider web can hold before breaking. Substituting the values, \(F = 8.20 \times 10^8 \, \text{N/m}^2 \times 8.00 \times 10^{-12} \, \text{m}^2 \approx 6.56 \times 10^{-3} \text{ N}\)Finally, the insect's mass (\(m\)) is deduced by dividing the force by the gravitational acceleration \(g\), as \(F = mg\), or:\[\m = \frac{F}{g} = \frac{6.56 \times 10^{-3} \text{ N}}{9.8 \, \text{m/s}^2} \approx 6.69 \times 10^{-4} \, \text{kg}\]This step-by-step method showcases how to connect the physics of a scenario to its mathematical calculations efficiently.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A uniform cube of side length \(8.0 \mathrm{~cm}\) rests on a horizontal floor. The coefficient of static friction between cube and floor is \(\mu .\) A horizontal pull \(\vec{P}\) is applied perpendicular to one of the vertical faces of the cube, at a distance \(7.0 \mathrm{~cm}\) above the floor on the vertical midline of the cube face. The magnitude of \(\vec{P}\) is gradually increased. During that increase, for what values of \(\mu\) will the cube eventually (a) begin to slide and (b) begin to tip? (Hint: At the onset of tipping, where is the normal force located?)

A mine elevator is supported by a single steel cable \(2.5 \mathrm{~cm}\) in diameter. The total mass of the elevator cage and occupants is \(670 \mathrm{~kg}\). By how much does the cable stretch when the elevator hangs by (a) \(12 \mathrm{~m}\) of cable and (b) \(362 \mathrm{~m}\) of cable? (Neglect the mass of the cable.)

A \(73 \mathrm{~kg}\) man stands on a level bridge of length \(L .\) He is at distance \(L / 4\) from one end. The bridge is uniform and weighs \(2.7 \mathrm{kN}\). What are the magnitudes of the vertical forces on the bridge from its supports at (a) the end farther from him and (b) the nearer end?

A \(75 \mathrm{~kg}\) window cleaner uses a \(10 \mathrm{~kg}\) ladder that is \(5.0 \mathrm{~m}\) long. He places one end on the ground \(2.5 \mathrm{~m}\) from a wall, rests the upper end against a cracked window, and climbs the ladder. He is \(3.0 \mathrm{~m}\) up along the ladder when the window breaks. Neglect friction between the ladder and window and assume that the base of the ladder does not slip. When the window is on the verge of breaking, what are (a) the magnitude of the force on the window from the ladder, (b) the magnitude of the force on the ladder from the ground, and (c) the angle (relative to the horizontal) of that force on the ladder?

A uniform cubical crate is \(0.750 \mathrm{~m}\) on each side and weighs \(500 \mathrm{~N}\). It rests on a floor with one edge against a very small, fixed obstruction. At what least height above the floor must a horizontal force of magnitude \(350 \mathrm{~N}\) be applied to the crate to tip it?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.