(I) A nylon string on a tennis racket is under a tension of 275 N. If its diameter is 1.00 mm, by how much is it lengthened from its untensioned length of 30.0 cm?

Short Answer

Expert verified

The nylon string is lengthened by \(3.50 \times {10^{ - 2}}\;{\rm{m}}\)from its untensioned length of 30.0 cm.

Step by step solution

01

Young’s modulus

When a force (F) is applied to stretch a uniform wire made of a certain material, the length of the wire gets changed. The change in length of the wire is proportional to the original length (\({l_0}\)), and inversely proportional to the cross-sectional area (A), i.e.,

\(\Delta l = \frac{1}{E}\frac{F}{A}{l_0}\).

Here, E is the constant of proportionality and is termed as the elastic modulus or Young’s modulus. The value of Young’s modulus depends on the material of the wire.

In this problem, the value of Young’s modulus of nylon string is \(E{\bf{ = 3 \times 1}}{{\bf{0}}^{\bf{9}}}\;{\bf{N/}}{{\bf{m}}^{\bf{2}}}\).

02

Given information

Tension acting on a nylon string is T = 275 N.

The diameter of nylon string is,\(d = 1.00\;{\rm{mm}} = 1.00 \times 1{{\rm{0}}^{ - 3}}\;{\rm{m}}\).

The radius of nylon string is calculated as follows:

\(\begin{array}{c}r = \frac{d}{2}\\ = \frac{{1.00 \times 1{{\rm{0}}^{ - 3}}\;{\rm{m}}}}{2}\\ = 0.50 \times 1{{\rm{0}}^{ - 3}}\;{\rm{m}}\end{array}\)

The original length of the nylon string is\({l_{\rm{o}}} = 30.0\;{\rm{cm}} = 30.{\rm{0}} \times {\rm{1}}{{\rm{0}}^{ - 2}}\;{\rm{m}}\).

03

Determination of change in length of the string

The area of the cross-section of the nylon string is calculated as follows:

\(\begin{array}{c}A = \pi {r^2}\\ = 3.14 \times {\left( {0.50 \times 1{{\rm{0}}^{ - 3}}\;{\rm{m}}} \right)^2}\\ = 0.785 \times 1{{\rm{0}}^{ - 6}}\;{{\rm{m}}^2}\end{array}\)

The change in length of the nylon string due to tension applied on it is as follows:

\(\begin{array}{c}\Delta l = \frac{1}{E}\frac{F}{A}{l_0}\\ = \frac{1}{{\left( {3 \times {{10}^9}\;{\rm{N/}}{{\rm{m}}^{\rm{2}}}} \right)}}\frac{{\left( {275\;{\rm{N}}} \right)}}{{\left( {0.785\;{{\rm{m}}^2}} \right)}}\left( {30.{\rm{0}} \times {\rm{1}}{{\rm{0}}^{ - 2}}\;{\rm{m}}} \right)\\ = 3.50 \times {10^{ - 2}}\;{\rm{m}}\end{array}\)

Thus, the length of the nylon string is increased by \(3.50 \times {10^{ - 2}}\;{\rm{m}}\) from its untensioned length of 30.0 cm.

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Most popular questions from this chapter

Calculate \({F_{\rm{A}}}\) and \({F_{\rm{B}}}\) for the beam shown in Fig. 9–56. The downward forces represent the weights of machinery on the beam. Assume that the beam is uniform and has a mass of 280 kg.


A uniform meter stick with a mass of 180 g is supported horizontally by two vertical strings, one at the 0-cm mark and the other at the 90-cm mark (Fig. 9–82). What is the tension in the string (a) at 0 cm? (b) at 90 cm?


A parking garage is designed for two levels of cars. To make more money, the owner decides to double the size of the garage in each dimension (length, width, and the number of levels). For the support columns to hold up four floors instead of two, how should he change the columns' diameter?

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(e) He doesn't need to increase the diameter of the columns

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