Ian at 11:04, 4 April 2010

2010-04-04T11:04:55Z

←Older revision		Revision as of 11:04, 4 April 2010
Line 22:		Line 22:
	=161\mathbf{i}-135\mathbf{j}		=161\mathbf{i}-135\mathbf{j}
	\end{align}</math>		\end{align}</math>
		+
		+	If the unknown force is
		+	<math>\mathbf{F}4</math>
		+
		+	the fact that the forces are in equilibrium gives
		+
		+	<math>\mathbf{F}1+\mathbf{F}2+\mathbf{F}3+\mathbf{F}4=0</math>
		+
		+	or
		+
		+	<math>\mathbf{F}4=-\mathbf{F}1-\mathbf{F}2-\mathbf{F}3=-292\mathbf{i}-62\mathbf{j}\ \text{N}</math>

Ian: New page: From the figure the three known forces can be calculated using, $\begin{align} \mathbf{F} =F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j} \end{align}$ We get $\begin...$

2010-04-04T10:46:20Z

New page: From the figure the three known forces can be calculated using, <math>\begin{align} \mathbf{F} =F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j} \end{align}</math> We get <math>\begin...

New page

From the figure the three known forces can be calculated using,

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

We get

<math>\begin{align}
\mathbf{F}1
=131\mathbf{i}+47\textrm{.}9 \mathbf{j}
\end{align}</math>

<math>\begin{align}
\mathbf{F}2
=150 \mathbf{j}
\end{align}</math>

<math>\begin{align}
\mathbf{F}3
=161\mathbf{i}-135\mathbf{j}
\end{align}</math>

Solution 5.3a - Revision history

Ian at 11:04, 4 April 2010

Ian: New page: From the figure the three known forces can be calculated using, \begin{align} \mathbf{F} =F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j} \end{align} We get \begin...

Ian: New page: From the figure the three known forces can be calculated using, $\begin{align} \mathbf{F} =F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j} \end{align}$ We get $\begin...$