/*! 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 86 You are to use a long, thin wire... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 86

You are to use a long, thin wire to build a pendulum in a science mu- seum. The wire has an unstretched length of \(22.0 \mathrm{~m}\) and a cir- cular cross section of \(\begin{array}{lll}\text { diameter } & 0.860 & \mathrm{~mm}\end{array}\) it is made of an alloy that has a large breaking stress. One end of the wire will be attached to the ceiling, and a \(9.50 \mathrm{~kg}\) metal sphere will be attached to the other end. As the pendulum swings back and forth, the wire's maximum angular displacement from the vertical will be \(36.0^{\circ}\). You must determine the maximum amount the wire will stretch during this motion. So, before you attach the metal sphere, you suspend a test mass (mass \(m\) ) from the wire's lower end. You then measure the increase in length \(\Delta l\) of the wire for several different test masses. Figure \(\mathbf{P} 1 \mathbf{1} .86,\) a graph of \(\Delta l\) versus \(m\) shows the results and the straight line that gives the best fit to the data. The equation for this line is \(\Delta l=(0.422 \mathrm{~mm} / \mathrm{kg}) m\) (a) Assume that \(g=9.80 \mathrm{~m} / \mathrm{s}^{2},\) and use Fig. \(\mathrm{P} 11.86\) to calculate Young's modulus \(Y\) for this wire. (b) You remove the test masses, attach the \(9.50 \mathrm{~kg}\) sphere, and release the sphere from rest, with the wire displaced by \(36.0^{\circ} .\) Calculate the amount the wire will stretch as it swings through the vertical. Ignore air resistance.

Short Answer

Expert verified
The Young鈥檚 modulus for the wire is approximately \(1.19 \times 10^{9} \, \mathrm{Pa}\). The wire will stretch approximately \(0.0866 \, \mathrm{m}\) or \(86.6 \, \mathrm{cm}\) when the 9.50 kg sphere swings through the vertical.

Step by step solution

01

Calculate Young's modulus

Given that \(\Delta l = 0.422 \, \mathrm{mm/kg} \times m\) is the increase in length of the wire, let's take a unit mass, \(m = 1 \, \mathrm{kg}\), to simplify the calculation. Then, \(\Delta l = 0.422 \, \mathrm{mm} = 0.422 \times 10^{-3} \, \mathrm{m}\). We also know that the original length of the wire, \(L\) is 22.0 \, \mathrm{m}, the stress in the wire, \(\sigma\) equals \(F/A\), where \(F\) is the force applied to the wire and \(A\) is the cross-sectional area of the wire. Let's also take into account the force due to gravity, \(F = 1\, \mathrm{kg} \times 9.8 \, \mathrm{m/s}^2 = 9.8 \, \mathrm{N}\). The diameter of the wire is 0.860 \, \mathrm{mm} = 0.860 \times 10^{-3} \, \mathrm{m}, so the radius \(r = 0.430 \times 10^{-3} \, \mathrm{m}\) and \(A = \pi r^2\). Then, we apply the definition of Young's modulus, \(Y = \sigma / \epsilon\), to calculate \(Y\).
02

Calculate the force due to the sphere

First, calculate the force, \(F_s\) of the 9.50 kg metal sphere at its lowest point when it swings through the vertical: \(F_s = 9.50 \, \mathrm{kg} \times 9.80 \, \mathrm{m/s^2} = 93.1 \, \mathrm{N}\). This force will apply a stress to the wire, \(\sigma = F_s / A \). Then, using the Young's modulus obtained from the previous step, calculate the strain \(\epsilon = \sigma / Y\).
03

Calculate the stretch of the wire

The strain \(\epsilon\) is the fractional increase in length, defined as the ratio of the stretch \(\Delta l'\) to the original length \(L\): \(\Delta l' = \epsilon L\). By substituting the previously calculated values, find the stretch of the wire.

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Key Concepts

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

Pendulum Physics
Pendulum physics involves understanding the motion of a pendulum, which is a mass (referred to as the 'bob') suspended from a pivot so that it can swing freely back and forth. A pendulum exhibits periodic motion, with the time for one complete cycle, the period, being constant for small angular displacements.

When setting up the pendulum in the science museum, the maximum angular displacement and the mass of the metal sphere are important parameters. They influence the tension in the wire and consequently its stretch. The swinging motion is typically described by the laws of conservation of mechanical energy and can be approximated by simple harmonic motion for small angles.

The period of a pendulum is given by the formula \( T = 2\pi\sqrt{\frac{L}{g}} \) for small angles, where \(L\) is the length of the pendulum and \(g\) is the acceleration due to gravity. In this exercise, you will need to understand how the physics of a pendulum relates to the elastic properties of the wire from which it is suspended.
Material Stress and Strain
Material stress and strain examine how materials deform under various forces. Stress is defined as the force applied over an area \(\sigma = \frac{F}{A}\), and strain is the deformation or displacement it causes, relative to the material's original length \(\epsilon = \frac{\Delta l}{L}\).

For the pendulum wire, calculating the increase in length for a given test mass helps us understand these concepts. In your textbook exercise, you use test measurements to find the relationship between mass and the resulting stretch of the wire. Knowing the wire's dimensions and the amount of force exerted on it by different masses allows you to calculate the stress on the material. The ratio of stress to strain gives you Young's modulus, a measure of the material's stiffness or elastic property. It quantifies the material's ability to resist deformation and is crucial in ensuring the pendulum functions safely and effectively.
Elastic Properties of Materials
The elastic properties of materials are fundamental characteristics that determine how materials respond when subjected to forces. These properties are important in applications that involve bending, stretching, compressing, or twisting.

Young's modulus is one of these properties and represents the ability of a material to withstand changes in length when under lengthwise tension or compression. It is defined as the ratio of stress \(\sigma\) to strain \(\epsilon\): \(Y = \frac{\sigma}{\epsilon}\). It allows us to predict how much a material will stretch or compress under a given force, and is a measure of the wire's stiffness in your exercise.

Calculating Young's modulus requires understanding both the stress applied to a material and how much it deforms in response. In the pendulum example, by measuring the change in length of the wire for different test masses, you create a basis from which to calculate the modulus. This calculation is critical when selecting materials for construction to ensure they can support the loads without excessive deformation.

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Most popular questions from this chapter

A loaded cement mixer drives onto an old drawbridge, where it stalls with its center of gravity threequarters of the way across the span. The truck driver radios for help, sets the handbrake, and waits. Meanwhile, a boat approaches, so the drawbridge is raised by means of a cable attached to the end opposite the hinge (Fig. \(\mathbf{P} 1 \mathbf{1 . 5 6}\) ). The drawbridge is \(40.0 \mathrm{~m}\) long and has a mass of \(18,000 \mathrm{~kg} ;\) its center of gravity is at its midpoint. The cement mixer, with driver, has mass \(30,000 \mathrm{~kg}\). When the drawbridge has been raised to an angle of \(30^{\circ}\) above the horizontal, the cable makes an angle of \(70^{\circ}\) with the surface of the bridge. (a) What is the tension \(T\) in the cable when the drawbridge is held in this position? (b) What are the horizontal and vertical components of the force the hinge exerts on the span?

A 3.00-m-long, \(190 \mathrm{~N}\), uniform rod at the \(\mathrm{zoo}\) is held in a horizontal position by two ropes at its ends (Fig. E11.21). The left rope makes an angle of \(150^{\circ}\) with the rod, and the right rope makes an angle \(\theta\) with the horizontal. A \(90 \mathrm{~N}\) howler monkey (Alouatta seniculus) hangs motionless \(0.50 \mathrm{~m}\) from the right end of the rod as he carefully studies you. Calculate the tensions in the two ropes and the angle \(\theta\). First make a free-body diagram of the rod.

(a) In Fig. P11.64 a \(6.00-\mathrm{m}\) -long, uniform beam is hanging from a point \(1.00 \mathrm{~m}\) to the right of its center. The beam weighs \(140 \mathrm{~N}\) and makes an angle of \(30.0^{\circ}\) with the vertical. At the right-hand end of the beam a \(100.0 \mathrm{~N}\) weight is hung; an unknown weight \(w\) hangs at the left end. If the system is in equilibrium, what is \(w ?\) You can ignore the thickness of the beam. (b) If the beam makes, instead, an angle of \(45.0^{\circ}\) with the vertical, what is \(w ?\)

A door \(1.00 \mathrm{~m}\) wide and \(2.00 \mathrm{~m}\) high weighs \(330 \mathrm{~N}\) and is supported by two hinges, one \(0.50 \mathrm{~m}\) from the top and the other \(0.50 \mathrm{~m}\) from the bottom. Each hinge supports half the total weight of the door. Assuming that the door's center of gravity is at its center, find the horizontal components of force exerted on the door by each hinge.

You need to measure the mass \(M\) of a \(4.00-\mathrm{m}\) -long bar. The bar has a square cross section but has some holes drilled along its length, so you suspect that its center of gravity isn't in the middle of the bar. The bar is too long for you to weigh on your scale. So, first you balance the bar on a knife-edge pivot and determine that the bar's center of gravity is \(1.88 \mathrm{~m}\) from its left-hand end. You then place the bar on the pivot so that the point of support is \(1.50 \mathrm{~m}\) from the left-hand end of the bar. Next you suspend a \(2.00 \mathrm{~kg}\) mass \(\left(m_{1}\right)\) from the bar at a point \(0.200 \mathrm{~m}\) from the left-hand end. Finally, you suspend a mass \(m_{2}=1.00 \mathrm{~kg}\) from the bar at a distance \(x\) from the left-hand end and adjust \(x\) so that the bar is balanced. You repeat this step for other values of \(m_{2}\) and record each corresponding value of \(x\). The table gives your results. $$ \begin{array}{l|llllll} m_{2}(\mathrm{~kg}) & 1.00 & 1.50 & 2.00 & 2.50 & 3.00 & 4.00 \\ \hline x(\mathrm{~m}) & 3.50 & 2.83 & 2.50 & 2.32 & 2.16 & 2.00 \end{array} $$ (a) Draw a free-body diagram for the bar when \(m_{1}\) and \(m_{2}\) are suspended from it. (b) Apply the static equilibrium equation \(\Sigma \tau_{z}=0\) with the axis at the location of the knife-edge pivot. Solve the equation for \(x\) as a function of \(m_{2}\). (c) Plot \(x\) versus \(1 / m_{2}\). Use the slope of the best-fit straight line and the equation you derived in part (b) to calculate that bar's mass \(M .\) Use \(g=9.80 \mathrm{~m} / \mathrm{s}^{2} .\) (d) What is the \(y\) -intercept of the straight line that fits the data? Explain why it has this value.
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