Find the net force per unit length on z.

Z e r o

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40 \times 10^{- 5}

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20 \times 10^{- 5}

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10 \times 10^{- 5}

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Solution

The correct option is B $40 \times 10^{- 5}$

$F_{2}^{l} = F_{2 y}^{l} - F_{2 x}^{l}$
$= \frac{M_{0}}{2 π} [\frac{600}{20 \times 10^{- 2}} - \frac{300}{30 \times 15^{2}}]$
$= \frac{2 \times 10^{- 7}}{10^{- 2}} [30 - 10]$
$= 40 \times 10^{- 5} N m^{- 1}$

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