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1 year ago

15

Consider a steel cable with a diameter of 8.0 mm. Calculatethe stress in the cable when it holds a person weighing 850 N.Report

your answer in MPa (106 Pa = 1 MPa).

1 answer:

pishuonlain [190]1 year ago

7 0

Answer:

stress = 16.9 MPa

Explanation:

The stress in the cable can be calculated as:

$\text{Stress = }\frac{F}{A}$

Where F is the force and A is the area. So, the area can be calculated as:

$A=\pi^{}\cdot r^2$

Where r is the radius. Since the radius is half the diameter, the radius is 4.0 mm and the area will be equal to:

$\begin{gathered} A=3.14(4\operatorname{mm})^2 \\ A=50.24\operatorname{mm}^2 \end{gathered}$

Then, replacing the force F by 850 N, and A by 50.24 mm², we get that the stress is equal to:

$\text{stress = }\frac{850N}{50.24mm^2}=16.9\text{MPa}$

Therefore, the answer is 16.9 MPa

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Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

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$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

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In other words, the force between the two point charges would be maximized when the charge is evenly split:

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