Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 10, Problem 101P

(a)

To determine

The strain in the cable.

(a)

Expert Solution
Check Mark

Answer to Problem 101P

The strain in the cable is 8×104_.

Explanation of Solution

Consider the segment of the rope between the fixed end and the walker’s position. A right triangle can be constructed with the non-stretched segment and stretched segment of the cable as shown in Figure 1.

Physics, Chapter 10, Problem 101P , additional homework tip  1

Write the expression for the strain in the cable.

strain=ΔLL=LLL (I)

Here, ΔL is the change in length, L is the length of non-stretched segment of the cable, and is the stretched length of the segment of the cable.

Write the expression for L from the right triangle in Figure 1.

L=Lcosθ (II)

Use equation (II) in (I).

strain=LLcosθLcosθ=1cosθ1 (III)

Conclusion:

Substitute 0.040rad for θ in equation (III) to find the strain.

strain=1cos(0.040rad)1=8×104

Therefore, the strain in the cable is 8×104_.

(b)

To determine

The tension in the cable when the walker is at the middle of the rope.

(b)

Expert Solution
Check Mark

Answer to Problem 101P

The tension in the cable when the walker is at the middle of the rope is 8.0kN_.

Explanation of Solution

The free-body diagram of the walker is given in Figure 2.

Physics, Chapter 10, Problem 101P , additional homework tip  2

Write the equilibrium conditions on the forces in y direction.

Fy=0 (IV)

Apply the condition in equation (I) to the free-body diagram.

2TsinθW=0 (V)

Here, T is the tension in the cable, and W is the weight of the walker.

Solve equation (V) for T.

T=W2sinθ (VI)

Conclusion:

Substitute 640N for W, and 0.040rad for θ in equation (VI) to find T.

T=640N2sin(0.040rad)=8002N=8002N×1kN1000N=8.0kN

Therefore, the tension in the cable when the walker is at the middle of the rope is 8.0kN_.

(c)

To determine

The cross sectional area of the cable.

(c)

Expert Solution
Check Mark

Answer to Problem 101P

The cross sectional area of the cable is 5×105m2_.

Explanation of Solution

Equation (VI) gives the tension in the cable.

T=W2sinθ

Divide equation (VI) by the cross sectional area of the cable A.

TA=W2Asinθ (VII)

Write the expression of Hooke’s law when force equal to the tension T applied to the cable.

TA=YΔLL (VIII)

Here, Y is the Young’s modulus (for steel Y=200×10Pa)

Use equation (I) in (VIII).

TA=Y(strain) (IX)

Equate the right hand sides of the equations (VII) and (IX) and solve for A.

W2Asinθ=Y(strain)A=W2Y(strain)sinθ (X)

Conclusion:

Substitute 640N for W, 200×109Pa for Y, 8×104 for strain, and 0.040rad for θ in equation (X) to find A.

A=640N2(200×109Pa)(8×104)sin(0.040rad)=640N2(200×109Pa)(8×104)sin(0.040rad)=5×105m2

Therefore, the cross sectional area of the cable is 5×105m2_.

(d)

To determine

Whether the cable stretched beyond its elastic limit.

(d)

Expert Solution
Check Mark

Answer to Problem 101P

The cable did not stretched beyond its elastic limit.

Explanation of Solution

Write the expression for the stress in terms of the tension in the cable.

stress=TA (XI)

Conclusion:

Substitute 8.0kN for T, and 5×105m2 for A in equation (XI) to find the stress in the cable.

stress=8.0kN5×105m2=8.0kN×1000N1kN5×105m2=1.6×108Pa<2.5×108Pa

Therefore, the cable did not stretched beyond its elastic limit.

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Chapter 10 Solutions

Physics

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