Solution 4.6a - Revision history

Ian at 11:39, 27 March 2011

Ian — Sun, 27 Mar 2011 11:39:20 GMT

←Older revision		Revision as of 11:39, 27 March 2011
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>

Ian — Tue, 23 Mar 2010 15:22:32 GMT

←Older revision		Revision as of 15:22, 23 March 2010
Line 14:		Line 14:
	& \mathbf{F}1=40\cos 20{}^\circ		& \mathbf{F}1=40\cos 20{}^\circ
	\mathbf{i}+40\sin 20{}^\circ		\mathbf{i}+40\sin 20{}^\circ
-	\mathbf{j}=40\times 0.94\mathbf{i}+40\times 0.342\mathbf{j} \\	+	\mathbf{j}=40\times 0\textrm{.}94\mathbf{i}+40\times 0\textrm{.}342\mathbf{j} \\
-	& =37.6\mathbf{i}+13.7\mathbf{j} \\	+	& =37\textrm{.}6\mathbf{i}+13\textrm{.}7\mathbf{j}\ \text{N}\\
	\end{align}</math>		\end{align}</math>

Line 31:		Line 31:
	\right) {}^\circ		\right) {}^\circ
	\mathbf{i}+50\sin \left( -30 \right) {}^\circ		\mathbf{i}+50\sin \left( -30 \right) {}^\circ
-	\mathbf{j}=50\times 0.866\mathbf{i}-50\times 0.50\mathbf{j} \\	+	\mathbf{j}=50\times 0\textrm{.}866\mathbf{i}-50\times 0\textrm{.}50\mathbf{j} \\
-	& =43.3\mathbf{i}+25\mathbf{j} \\	+	& =43\textrm{.}3\mathbf{i}+25\mathbf{j} \ \text{N}\\
	\end{align}</math>		\end{align}</math>

Ian — Tue, 23 Mar 2010 15:21:02 GMT

←Older revision		Revision as of 15:21, 23 March 2010
Line 1:		Line 1:
	<math>\begin{align}		<math>\begin{align}
	& \mathbf{F}=F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j}		& \mathbf{F}=F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j}
		+	\end{align}</math>
		+
		+
		+	<math>F1=40\ \text{N}</math>
		+	and
		+	<math>\alpha 1=20{}^\circ
		+	</math>
		+	this gives
		+
		+
		+	<math>\begin{align}
		+	& \mathbf{F}1=40\cos 20{}^\circ
		+	\mathbf{i}+40\sin 20{}^\circ
		+	\mathbf{j}=40\times 0.94\mathbf{i}+40\times 0.342\mathbf{j} \\
		+	& =37.6\mathbf{i}+13.7\mathbf{j} \\
		+	\end{align}</math>
		+
		+
		+
		+	<math>F2=50\ \text{N}</math>
		+	and
		+	<math>\alpha 2=-30{}^\circ
		+	</math>
		+	this gives
		+
		+
		+	<math>\begin{align}
		+	& \mathbf{F}2=50\cos \left( -30
		+	\right) {}^\circ
		+	\mathbf{i}+50\sin \left( -30 \right) {}^\circ
		+	\mathbf{j}=50\times 0.866\mathbf{i}-50\times 0.50\mathbf{j} \\
		+	& =43.3\mathbf{i}+25\mathbf{j} \\
	\end{align}</math>		\end{align}</math>

Ian — Tue, 23 Mar 2010 15:16:33 GMT

New page: <math>\begin{align} & \mathbf{F}=F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j} \end{align}</math>

New page

<math>\begin{align}
& \mathbf{F}=F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j}
\end{align}</math>