Mecanica Geral

kopkiook

Para encontrarmos o momento, realizaremos os cálculos abaixo:

\(\begin{align} & {{M}_{A}}={{r}_{C/A}}\times F \\ & \\ & {{r}_{C/A}}={{r}_{C}}-{{r}_{A}} \\ & {{r}_{C/A}}=0,3i+0,08k \\ & \\ & F=F\lambda \\ & F=200\left( \frac{{{r}_{D/C}}}{r} \right) \\ & F=200\left( \frac{-0,3i+0,24j-0,32k}{0,5} \right) \\ & F=-120i+96j-128k \\ & \\ & M=\left( \begin{matrix} i & j & k \\ 0,3 & 0 & 0,08 \\ -120 & 96 & -128 \\ \end{matrix} \right) \\ & M=\det \left( \begin{matrix} i & j & k \\ 0,3 & 0 & 0,08 \\ -120 & 96 & -128 \\ \end{matrix} \right)\begin{matrix} i & j \\ 0,3 & 0 \\ -120 & 96 \\ \end{matrix} \\ & M=\left( -7,68i+28,8j+28,8k \right)Nm \\ \end{align}\ \)

Portanto, o momento será: \(\begin{align} & M=\left( -7,68i+28,8j+28,8k \right)Nm \\ \end{align}\ \).