Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 10, Problem 22P

(a)

To determine

The stress in the column.

(a)

Expert Solution
Check Mark

Answer to Problem 22P

The stress is 2.8×107Pa.

Explanation of Solution

The cross sectional area is 25cm2, load is 7.0×104N, Young’s modulus is 60GPa, and the compressive strength is 200MPa.

Write the equation for stress.

S=FA

Here, S is the stress, F is the applied force, and A is the area of cross section.

Conclusion:

Substitute 7.0×104N for F and 25cm2 for A in the above equation to find S.

S=7.0×104N25cm2(104m21cm2)=2.8×107N

Therefore, the stress is 2.8×107Pa.

(b)

To determine

The strain in the column.

(b)

Expert Solution
Check Mark

Answer to Problem 22P

The strain is 4.7×104.

Explanation of Solution

The cross sectional area is 25cm2, load is 7.0×104N, Young’s modulus is 60GPa, and the compressive strength is 200MPa.

Write the equation for strain.

St=FYA

Here, St is the strain and Y is the young’s modulus.

Conclusion:

Substitute 7.0×104N for F, 60GPa for Y, and 25cm2 for A in the above equation to find S.

S=7.0×104N(60GPa(109Pa1GPa))(25cm2(104m21cm2))=4.7×104N

Therefore, the strain is 4.7×104.

(c)

To determine

The elongation due to the supporting load.

(c)

Expert Solution
Check Mark

Answer to Problem 22P

The elongation is 9.3×104m.

Explanation of Solution

The height of column is 2.0m, cross sectional area is 25cm2, load is 7.0×104N, Young’s modulus is 60GPa, and the compressive strength is 200MPa.

Write the equation to find the elongation due to the supporting load.

ΔL=FLYA

Conclusion:

Substitute 7.0×104N for F, 2.0m for L, 60GPa for Y, and 25cm2 for A in the above equation to find ΔL.

ΔL=(7.0×104N)(2.0m)(60GPa(109Pa1GPa))(25cm2(104m21cm2))=9.3×104m

Therefore, the strain is 9.37×104m.

(d)

To determine

The maximum weight that can be supported by the load.

(d)

Expert Solution
Check Mark

Answer to Problem 22P

The maximum load is 5.0×105N.

Explanation of Solution

The height of column is 2.0m, cross sectional area is 25cm2, load is 7.0×104N, Young’s modulus is 60GPa, and the compressive strength is 200MPa.

When the column supports the maximum possible weight, the compressive stress will be equal to the stress.

Write the equation for the maximum weight that can be supported by the load.

Wmax=ScomA

Here, Wmax is the maximum weight and Scom is the compressive strength.

Conclusion:

Substitute 200MPa for Scom and 25cm2 for A in the above equation to find Wmax.

Wmax=(200MPa(106Pa1MPa))(25cm2(104m21cm2))=5.0×105N

Therefore, the maximum load is 5.0×105N.

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Chapter 10 Solutions

Physics

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