Solution 4.9a - Revision history

Ian at 10:20, 1 April 2010

2010-04-01T10:20:35Z

←Older revision		Revision as of 10:20, 1 April 2010
Line 50:		Line 50:

	<math>\begin{align}		<math>\begin{align}
-	& \mathbf{F}3=8\cos \left(-25{}^\circ \right)	+	& \mathbf{F}4=8\cos \left(-25{}^\circ \right)
	\mathbf{i}+8\sin \left(-25{}^\circ \right)		\mathbf{i}+8\sin \left(-25{}^\circ \right)
	\mathbf{j}=8\times 0\textrm{.}906\mathbf{i}+8\times \left(-0\textrm{.}423\right)\mathbf{j} \\		\mathbf{j}=8\times 0\textrm{.}906\mathbf{i}+8\times \left(-0\textrm{.}423\right)\mathbf{j} \\
	& =7\textrm{.}25\mathbf{i}-3\textrm{.}38\mathbf{j} \ \text{N}\\		& =7\textrm{.}25\mathbf{i}-3\textrm{.}38\mathbf{j} \ \text{N}\\
		+	\end{align}</math>
		+
		+	The resultant <math>\mathbf{R}</math> is the sum of these four forces.
		+
		+	<math>\begin{align}
		+	& \mathbf{R} =11\textrm{.}1\mathbf{i}+3\textrm{.}93\mathbf{j} \ \text{N}\\
		+	\end{align}</math>
		+
		+	Then if <math>R</math> is the magnitude of this resultant
		+
		+	<math>\begin{align}
		+	& {{R\ }^{2}}={{11\textrm{.}1}^{2}}+{{3\textrm{.}93}^{2}} \\
		+	& R=\sqrt{{{11\textrm{.}1}^{2}}+{{3\textrm{.}93}^{2}}}=11\textrm{.}8\ \text{N} \\
	\end{align}</math>		\end{align}</math>

Ian at 09:54, 1 April 2010

2010-04-01T09:54:28Z

←Older revision		Revision as of 09:54, 1 April 2010
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	& \mathbf{F}3=8\cos \left(-25{}^\circ \right)		& \mathbf{F}3=8\cos \left(-25{}^\circ \right)
	\mathbf{i}+8\sin \left(-25{}^\circ \right)		\mathbf{i}+8\sin \left(-25{}^\circ \right)
-	\mathbf{j}=8\times 0\mathbf{i}+8\times \left(-1\right)\mathbf{j} \\	+	\mathbf{j}=8\times 0\textrm{.}906\mathbf{i}+8\times \left(-0\textrm{.}423\right)\mathbf{j} \\
-	& =-1\textrm{.}85\mathbf{i}-5\textrm{.}71\mathbf{j} \ \text{N}\\	+	& =7\textrm{.}25\mathbf{i}-3\textrm{.}38\mathbf{j} \ \text{N}\\
	\end{align}</math>		\end{align}</math>

Ian at 09:42, 1 April 2010

2010-04-01T09:42:00Z

←Older revision		Revision as of 09:42, 1 April 2010
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	& \mathbf{F}3=6\cos \left(-108{}^\circ \right)		& \mathbf{F}3=6\cos \left(-108{}^\circ \right)
	\mathbf{i}+6\sin \left(-108{}^\circ \right)		\mathbf{i}+6\sin \left(-108{}^\circ \right)
-	\mathbf{j}=6\times 0\mathbf{i}+6\times \left(-1\right)\mathbf{j} \\	+	\mathbf{j}=6\times \left(-0\textrm{.}309\right)\mathbf{i}+6\times \left(-0\textrm{.}951\right)\mathbf{j} \\
-	& =-80\mathbf{j}\ \text{N}\\	+	& =-1\textrm{.}85\mathbf{i}-5\textrm{.}71\mathbf{j} \ \text{N}\\
	\end{align}</math>		\end{align}</math>

Line 53:		Line 53:
	\mathbf{i}+8\sin \left(-25{}^\circ \right)		\mathbf{i}+8\sin \left(-25{}^\circ \right)
	\mathbf{j}=8\times 0\mathbf{i}+8\times \left(-1\right)\mathbf{j} \\		\mathbf{j}=8\times 0\mathbf{i}+8\times \left(-1\right)\mathbf{j} \\
-	& =-30\textrm{.}8\mathbf{i}~~+84~~\textrm{.}6\mathbf{j} \ \text{N}\\	+	& =-1\textrm{.}85\mathbf{i}-5\textrm{.}71\mathbf{j} \ \text{N}\\
	\end{align}</math>		\end{align}</math>

Ian at 09:26, 1 April 2010

2010-04-01T09:26:29Z

←Older revision		Revision as of 09:26, 1 April 2010
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	& \mathbf{F}2=9\cos 90 {}^\circ		& \mathbf{F}2=9\cos 90 {}^\circ
	\mathbf{i}+9\sin 90 {}^\circ		\mathbf{i}+9\sin 90 {}^\circ
-	\mathbf{j}=9\times ~~\left(-~~0~~\textrm{.}342 \right)~~\mathbf{i}+9\times 0\textrm{.}94\mathbf{j} \\	+	\mathbf{j}=9\times 0\mathbf{i}+9\times 1\textrm{.}0\mathbf{j} \\
-	& =~~-30~~\textrm{.}~~8\mathbf{i}+84\textrm{.}6~~\mathbf{j} \ \text{N}\\	+	& =9\textrm{.}0\mathbf{j} \ \text{N}\\
	\end{align}</math>		\end{align}</math>

Ian at 09:20, 1 April 2010

2010-04-01T09:20:43Z

←Older revision		Revision as of 09:20, 1 April 2010
Line 12:		Line 12:
	& \mathbf{F}1=7\cos 35{}^\circ		& \mathbf{F}1=7\cos 35{}^\circ
	\mathbf{i}+7\sin 35{}^\circ		\mathbf{i}+7\sin 35{}^\circ
-	\mathbf{j}=7\times 0\textrm{.}~~866~~\mathbf{i}+7\times 0\textrm{.}5\mathbf{j} \\	+	\mathbf{j}=7\times 0\textrm{.}819\mathbf{i}+7\times 0\textrm{.}574\mathbf{j} \\
-	& =86\textrm{.}6\mathbf{i}+50\mathbf{j}\ \text{N}\\	+	& =5\textrm{.}73\mathbf{i}+4\textrm{.}02\mathbf{j}\ \text{N}\\
	\end{align}</math>		\end{align}</math>

Ian at 11:51, 31 March 2010

2010-03-31T11:51:54Z

←Older revision		Revision as of 11:51, 31 March 2010
Line 3:		Line 3:
	\end{align}</math>		\end{align}</math>

-	<math>F1=~~100~~\ \text{N}</math>	+	<math>F1=7\ \text{N}</math>
	and		and
-	<math>\alpha 1=30{}^\circ	+	<math>\alpha 1=35{}^\circ
	</math>		</math>
	this gives		this gives

	<math>\begin{align}		<math>\begin{align}
-	& \mathbf{F}1=~~100~~\cos 30{}^\circ	+	& \mathbf{F}1=7\cos 35{}^\circ
-	\mathbf{i}+~~100~~\sin 30{}^\circ	+	\mathbf{i}+7\sin 35{}^\circ
-	\mathbf{j}=~~100~~\times 0\textrm{.}866\mathbf{i}+~~100~~\times 0\textrm{.}5\mathbf{j} \\	+	\mathbf{j}=7\times 0\textrm{.}866\mathbf{i}+7\times 0\textrm{.}5\mathbf{j} \\
	& =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>
	and		and
-	<math>\alpha 2=~~180{}^\circ-70{}^\circ=110~~{}^\circ	+	<math>\alpha 2=90{}^\circ
	</math>		</math>
	this gives		this gives

	<math>\begin{align}		<math>\begin{align}
-	& \mathbf{F}2=90\cos ~~110~~ {}^\circ	+	& \mathbf{F}2=9\cos 90 {}^\circ
-	\mathbf{i}+90\sin ~~110~~ {}^\circ	+	\mathbf{i}+9\sin 90 {}^\circ
-	\mathbf{j}=90\times \left(-0\textrm{.}342 \right)\mathbf{i}+90\times 0\textrm{.}94\mathbf{j} \\	+	\mathbf{j}=9\times \left(-0\textrm{.}342 \right)\mathbf{i}+9\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>		\end{align}</math>

-	<math>F3=80\ \text{N}</math>	+	<math>F3=6\ \text{N}</math>
	and		and
-	<math>\alpha 3=-90{}^\circ	+	<math>\alpha 3=-180{}^\circ+72{}^\circ=-108{}^\circ
	</math>		</math>
	this gives		this gives

	<math>\begin{align}		<math>\begin{align}
-	& \mathbf{F}3=80\cos \left(-90{}^\circ \right)	+	& \mathbf{F}3=6\cos \left(-108{}^\circ \right)
-	\mathbf{i}+80\sin \left(-90{}^\circ \right)	+	\mathbf{i}+6\sin \left(-108{}^\circ \right)
-	\mathbf{j}=80\times 0\mathbf{i}+80\times \left(-1\right)\mathbf{j} \\	+	\mathbf{j}=6\times 0\mathbf{i}+6\times \left(-1\right)\mathbf{j} \\
	& =-80\mathbf{j}\ \text{N}\\		& =-80\mathbf{j}\ \text{N}\\
		+	\end{align}</math>
		+
		+
		+	<math>F4=8\ \text{N}</math>
		+	and
		+	<math>\alpha 3=-25{}^\circ
		+	</math>
		+	this gives
		+
		+	<math>\begin{align}
		+	& \mathbf{F}3=8\cos \left(-25{}^\circ \right)
		+	\mathbf{i}+8\sin \left(-25{}^\circ \right)
		+	\mathbf{j}=8\times 0\mathbf{i}+8\times \left(-1\right)\mathbf{j} \\
		+	& =-30\textrm{.}8\mathbf{i}+84\textrm{.}6\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
2010-03-31T11:33:17Z

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 <ma...

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=90\ \text{N}</math>
and
<math>\alpha 2=180{}^\circ-70{}^\circ=110{}^\circ
</math>
this gives

<math>\begin{align}
& \mathbf{F}2=90\cos 110 {}^\circ
\mathbf{i}+90\sin 110 {}^\circ
\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}\\
\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>