Ch 11: Equilibrium & Elasticity

Young & Freedman Calc - University Physics 14th Edition

Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook

Young & Freedman Calc 14th Edition

Ch 11: Equilibrium & Elasticity

Problem 25

Chapter 11, Problem 25

A circular steel wire 2.00 m long must stretch no more than 0.25 cm when a tensile force of 700 N is applied to each end of the wire. What minimum diameter is required for the wire?

Verified step by step guidance

First, identify the relevant formula for the problem. The formula for the elongation of a wire under tensile stress is given by:

\frac{ΔL}{L} = \frac{F}{A E}

, where

ΔL

is the change in length,

L

is the original length,

F

is the force applied,

A

is the cross-sectional area, and

E

is the Young's modulus of the material.

Next, rearrange the formula to solve for the cross-sectional area

A

A = \frac{F L}{E ΔL}

. This will allow you to find the area needed to ensure the wire stretches no more than the specified amount.

Substitute the known values into the equation. The force

F

is 700 N, the original length

L

is 2.00 m, and the change in length

ΔL

is 0.25 cm (which should be converted to meters). The Young's modulus

E

for steel is approximately 2.1 x 10¹¹ N/m².

Calculate the cross-sectional area

A

using the substituted values. This will give you the minimum area required to ensure the wire does not stretch beyond the specified limit.

Finally, use the formula for the area of a circle

A = \frac{π d^2}{4}

to solve for the diameter

d

. Rearrange the formula to find

d

and substitute the calculated area to find the minimum diameter required.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Young's Modulus

Young's Modulus is a measure of the stiffness of a material, defined as the ratio of tensile stress to tensile strain. It is a constant for a given material and is used to predict how much a material will deform under a given load. In this problem, it helps determine how much the steel wire will stretch when a force is applied.

Tensile Stress and Strain

Tensile stress is the force applied per unit area of a material, while tensile strain is the deformation or elongation per unit length. These concepts are crucial for understanding how materials respond to forces, and they are used to calculate the change in length of the wire when a specific force is applied.

Cross-sectional Area of a Wire

The cross-sectional area of a wire is important in determining how it will respond to tensile forces. It is calculated using the diameter of the wire, and a larger area means the wire can withstand more force without stretching. In this problem, finding the minimum diameter involves ensuring the wire's cross-sectional area is sufficient to limit its elongation under the given force.

Recommended video:

Guided course

08:10

Calculating the Vector (Cross) Product Using Components

Related Practice

Textbook Question

A solid gold bar is pulled up from the hold of the sunken RMS Titanic. The bulk modulus of lead is one-fourth that of gold. Find the ratio of the volume change of a solid lead bar to that of a gold bar of equal volume for the same pressure change.

views

Textbook Question

Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50 cm in diameter. The combination is subjected to a tensile force with magnitude 4000 N. For each rod, what are (a) the strain and (b) the elongation?

views

Textbook Question

A 15,000-N crane pivots around a friction-free axle at its base and is supported by a cable making a 25° angle with the crane (Fig. E11.18). The crane is 16 m long and is not uniform, its center of gravity being 7.0 m from the axle as measured along the crane. The cable is attached 3.0 m from the upper end of the crane. When the crane is raised to 55° above the horizontal holding an 11,000-N pallet of bricks by a 2.2-m, very light cord, find the tension in the cable. Start with a free-body diagram of the crane.

views

Textbook Question

A nonuniform beam 4.50 m long and weighing 1.40 kN makes an angle of 25.0° below the horizontal. It is held in position by a frictionless pivot at its upper right end and by a cable 3.00 m farther down the beam and perpendicular to it (Fig. E11.20). The center of gravity of the beam is 2.00 m down the beam from the pivot. Lighting equipment exerts a 5.00-kN downward force on the lower left end of the beam. Find the tension T in the cable and the horizontal and vertical components of the force exerted on the beam by the pivot. Start by sketching a free-body diagram of the beam.

views

Textbook Question

A 9.00 m-long uniform beam is hinged to a vertical wall and held horizontally by a 5.00 m-long cable attached to the wall 4.00 m above the hinge (Fig. E11.17). The metal of this cable has a test strength of 1.00 kN, which means that it will break if the tension in it exceeds that amount. Find the horizontal and vertical components of the force the hinge exerts on the beam. Is the vertical component upward or downward?

views

Textbook Question

A nylon rope used by mountaineers elongates 1.10 m under the weight of a 65.0 kg climber. If the rope is 45.0 m in length and 7.0 mm in diameter, what is Young's modulus for nylon?

views