Pembahasan
Diketahui
F = 810π N
T𝑔 = 40N/Nm²
Ditanya diameter paku keling (d)
Dijawab \begin{aligned}d&=&\sqrt{\frac{4\bullet F}{\pi\bullet T_g}}\\ d&=&\sqrt{\frac{4\bullet810\bullet\pi}{\pi\bullet40}}\\ d&=&\sqrt{81}\\ d&=&9\ mm \end{aligned}
Pembahasan
Diketahui
F = 810π N
T𝑔 = 40N/Nm²
Ditanya diameter paku keling (d)
Dijawab \begin{aligned}d&=&\sqrt{\frac{4\bullet F}{\pi\bullet T_g}}\\ d&=&\sqrt{\frac{4\bullet810\bullet\pi}{\pi\bullet40}}\\ d&=&\sqrt{81}\\ d&=&9\ mm \end{aligned}
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