Solution 4.7a - Revision history

Ian at 15:04, 24 March 2010

Ian — Wed, 24 Mar 2010 15:04:17 GMT

←Older revision		Revision as of 15:04, 24 March 2010
Line 2:		Line 2:
	& \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>		\end{align}</math>
-

	<math>F1=100\ \text{N}</math>		<math>F1=100\ \text{N}</math>
Line 9:		Line 8:
	</math>		</math>
	this gives		this gives
-

	<math>\begin{align}		<math>\begin{align}
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	& =86\textrm{.}6\mathbf{i}+50\mathbf{j}\ \text{N}\\		& =86\textrm{.}6\mathbf{i}+50\mathbf{j}\ \text{N}\\
	\end{align}</math>		\end{align}</math>
-
-

	<math>F2=90\ \text{N}</math>		<math>F2=90\ \text{N}</math>
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	</math>		</math>
	this gives		this gives
-

	<math>\begin{align}		<math>\begin{align}
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	\mathbf{j}=90\times \left(-0\textrm{.}342 \right)\mathbf{i}+90\times 0\textrm{.}94\mathbf{j} \\		\mathbf{j}=90\times \left(-0\textrm{.}342 \right)\mathbf{i}+90\times 0\textrm{.}94\mathbf{j} \\
	& =-30\textrm{.}8\mathbf{i}+84\textrm{.}6\mathbf{j} \ \text{N}\\		& =-30\textrm{.}8\mathbf{i}+84\textrm{.}6\mathbf{j} \ \text{N}\\
		+	\end{align}</math>
		+
		+	<math>F3=80\ \text{N}</math>
		+	and
		+	<math>\alpha 3=-90{}^\circ
		+	</math>
		+	this gives
		+
		+	<math>\begin{align}
		+	& \mathbf{F}3=80\cos \left(-90{}^\circ \right)
		+	\mathbf{i}+80\sin \left(-90{}^\circ \right)
		+	\mathbf{j}=80\times 0\mathbf{i}+80\times \left(-1\right)\mathbf{j} \\
		+	& =-80\mathbf{j}\ \text{N}\\
	\end{align}</math>		\end{align}</math>

Ian at 14:15, 24 March 2010

Ian — Wed, 24 Mar 2010 14:15:59 GMT

←Older revision		Revision as of 14:15, 24 March 2010
Line 20:		Line 20:


-	<math>F2=50\ \text{N}</math>	+	<math>F2=90\ \text{N}</math>
	and		and
-	<math>\alpha 2=-30{}^\circ	+	<math>\alpha 2=180{}^\circ-70{}^\circ=110{}^\circ
	</math>		</math>
	this gives		this gives
Line 28:		Line 28:

	<math>\begin{align}		<math>\begin{align}
-	& \mathbf{F}2=50\cos ~~\left( -30~~	+	& \mathbf{F}2=90\cos 110 {}^\circ
-	~~\right)~~ {}^\circ	+	\mathbf{i}+90\sin 110 {}^\circ
-	\mathbf{i}+50\sin ~~\left( -30 \right)~~ {}^\circ	+	\mathbf{j}=90\times \left(-0\textrm{.}342 \right)\mathbf{i}+90\times 0\textrm{.}94\mathbf{j} \\
-	\mathbf{j}=50\times 0\textrm{.}~~866~~\mathbf{i}~~-50~~\times 0\textrm{.}50\mathbf{j} \\	+	& =-30\textrm{.}8\mathbf{i}+84\textrm{.}6\mathbf{j} \ \text{N}\\
-	& =43\textrm{.}3\mathbf{i}+25\mathbf{j} \ \text{N}\\	+
	\end{align}</math>		\end{align}</math>

Ian: New page: $\begin{align} & \mathbf{F}=F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j} \end{align}$ $F1=100\ \text{N}$ and $\alpha 1=30{}^\circ$ this gives <...

Ian — Wed, 24 Mar 2010 13:37:48 GMT

New page: <math>\begin{align} & \mathbf{F}=F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j} \end{align}</math> <math>F1=100\ \text{N}</math> and <math>\alpha 1=30{}^\circ </math> this gives <...

New page

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

<math>F1=100\ \text{N}</math>
and
<math>\alpha 1=30{}^\circ
</math>
this gives

<math>\begin{align}
& \mathbf{F}1=100\cos 30{}^\circ
\mathbf{i}+100\sin 30{}^\circ
\mathbf{j}=100\times 0\textrm{.}866\mathbf{i}+100\times 0\textrm{.}5\mathbf{j} \\
& =86\textrm{.}6\mathbf{i}+50\mathbf{j}\ \text{N}\\
\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\textrm{.}866\mathbf{i}-50\times 0\textrm{.}50\mathbf{j} \\
& =43\textrm{.}3\mathbf{i}+25\mathbf{j} \ \text{N}\\
\end{align}</math>

Solution 4.7a - Revision history

Ian at 15:04, 24 March 2010

Ian at 14:15, 24 March 2010

Ian: New page: \begin{align} & \mathbf{F}=F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j} \end{align} F1=100\ \text{N} and \alpha 1=30{}^\circ this gives <...

Ian: New page: $\begin{align} & \mathbf{F}=F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j} \end{align}$ $F1=100\ \text{N}$ and $\alpha 1=30{}^\circ$ this gives <...