Calculate the tensile stress in a nylon fishing line of diameter 0.36 mm which a fish is pulling with a force of 20 N?

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Calculate the tensile stress in a nylon fishing line of diameter 0.36 mm which a fish is pulling with a force of 20 N?

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Answer 1 · 2018-05-14T12:43:57

$5 G P a$ $5 GPa$

Explanation:

$S t r e s s = \frac{F}{A}$ $Stress = F/A$

Find Area of the nylon fishing line using $A = π r^{2}$ $A=pir^2$ and half the diameter to get the radius and make sure you convert mm into m ( $\cdot 10^{- 3}$ $*10^-3$ ):

$A = π \cdot {(\frac{0.36 \cdot 10^{- 3}}{2})}^{2}$ $A=pi*((0.36*10^-3)/2)^2$
$A = 1.01787602 \cdot 10^{- 7} m^{2}$ $A= 1.01787602*10^-7 m^2$

Substitute this value into stress equation:

$S t r e s s = \frac{1.01787602 \cdot 10^{- 7}}{20}$ $Stress = (1.01787602*10^-7)/20$
$S t r e s s = 5 \cdot 10^{- 9} P a$ $Stress = 5*10^-9 Pa$
$S t r e s s = 5 G P a$ $Stress = 5 GPa$

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Calculate the tensile stress in a nylon fishing line of diameter 0.36 mm which a fish is pulling with a force of 20 N?

1 Answer

Explanation:

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