From Mechanics

Revision as of 15:22, 23 March 2010 (edit) Ian (Talk | contribs) ← Previous diff		Current revision (11:39, 27 March 2011) (edit) (undo) Ian (Talk | contribs)
Line 32:		Line 32:
	\mathbf{i}+50\sin \left( -30 \right) {}^\circ		\mathbf{i}+50\sin \left( -30 \right) {}^\circ
	\mathbf{j}=50\times 0\textrm{.}866\mathbf{i}-50\times 0\textrm{.}50\mathbf{j} \\		\mathbf{j}=50\times 0\textrm{.}866\mathbf{i}-50\times 0\textrm{.}50\mathbf{j} \\
-	& =43\textrm{.}3\mathbf{i}+25\mathbf{j} \ \text{N}\\	+	& =43\textrm{.}3\mathbf{i}-25\mathbf{j} \ \text{N}\\
	\end{align}</math>		\end{align}</math>

---

Current revision

\displaystyle \begin{align} & \mathbf{F}=F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j} \end{align}

\displaystyle F1=40\ \text{N} and \displaystyle \alpha 1=20{}^\circ this gives

\displaystyle \begin{align} & \mathbf{F}1=40\cos 20{}^\circ \mathbf{i}+40\sin 20{}^\circ \mathbf{j}=40\times 0\textrm{.}94\mathbf{i}+40\times 0\textrm{.}342\mathbf{j} \\ & =37\textrm{.}6\mathbf{i}+13\textrm{.}7\mathbf{j}\ \text{N}\\ \end{align}

\displaystyle F2=50\ \text{N} and \displaystyle \alpha 2=-30{}^\circ this gives

\displaystyle \begin{align} & \mathbf{F}2=50\cos \left( -30 \right) {}^\circ \mathbf{i}+50\sin \left( -30 \right) {}^\circ \mathbf{j}=50\times 0\textrm{.}866\mathbf{i}-50\times 0\textrm{.}50\mathbf{j} \\ & =43\textrm{.}3\mathbf{i}-25\mathbf{j} \ \text{N}\\ \end{align}