Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 10, Problem 23P

(a)

To determine

The diameter and tensile stress in the wire.

(a)

Expert Solution
Check Mark

Answer to Problem 23P

The diameter and tensile stress in the wire is respectively 1.3mm and 8.4×107N/m2.

Explanation of Solution

The length of the wire is 3.0m, the change in length (elongation) of the wire is 2.1mm, the young’s modulus of the copper wire is 120×109Pa and the force or weight at the end is 120N.

Write the ex

Write the expression to calculate the power output of the pump.

P=mgyt

Here, P is the power, m is the mass of the water, g is the acceleration due to gravity, y is the depth of the well.

Write the expression for the force and young’s modulus of the wire.

FA=YΔLL (I)

Here, F is the force at the end, A is the cross-sectional area of the wire, Y is the young’s modulus, L is the length of the wire and ΔL is the change in length.

Write the expression to calculate the cross-sectional area of the wire.

A=π(d2)2

Here, d is the diameter of the wire.

Use the above expression in (I) to rewrite.

Fπ(d2)2=YΔLL

Rewrite the above equation in terms of d.

d=2FYπ(LΔL)

Substitute 3.0m for L, 2.1mm for ΔL, 120×109Pa for Y and 120N for F in the above equation to calculate d.

d=2(120N120×109Pa(π))(3.0m2.1mm(103m1mm))d=0.0013m(103mm1m)=1.3mm

Write the expression for the tensile stress.

FA=YΔLL

Here, FA is the tensile stress.

Substitute 3.0m for L, 2.1mm for ΔL, 120×109Pa for Y in the above equation to calculate FA.

FA=(120×109Pa)2.1mm(103m1mm)3.0m=8.4×107N/m2

Conclusion:

Therefore, the diameter and tensile stress in the wire is respectively 1.3mm and 8.4×107N/m2.

(b)

To determine

The maximum weight in order to hang the wire.

(b)

Expert Solution
Check Mark

Answer to Problem 23P

The maximum weight in order to hang the wire is 531N.

Explanation of Solution

The tensile strength is 400MPa.

Write the expression for the tensile stress.

WA=F

Here, F is the tensile strength and W is the maximum weight.

Rewrite the above equation using A=π(d2)2 in terms of W.

Wπ(d2)2=FW=πFd24

Substitute 400MPa for F and 1.3mm for d in the above equation to calculate W.

W=π(400MPa)(106Pa1MPa)(1.3mm(103m1mm))24=531N

Conclusion:

Therefore, the maximum weight in order to hang the wire is 531N.

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Chapter 10 Solutions

Physics

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