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Revision as of 10:46, 4 April 2010 (edit) Ian (Talk | contribs) (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...) ← Previous diff		Current revision (11:04, 4 April 2010) (edit) (undo) Ian (Talk | contribs)
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	=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>

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Current revision

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

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

We get

\displaystyle \begin{align} \mathbf{F}1 =131\mathbf{i}+47\textrm{.}9 \mathbf{j} \end{align}

\displaystyle \begin{align} \mathbf{F}2 =150 \mathbf{j} \end{align}

\displaystyle \begin{align} \mathbf{F}3 =161\mathbf{i}-135\mathbf{j} \end{align}

If the unknown force is \displaystyle \mathbf{F}4

the fact that the forces are in equilibrium gives

\displaystyle \mathbf{F}1+\mathbf{F}2+\mathbf{F}3+\mathbf{F}4=0

or

\displaystyle \mathbf{F}4=-\mathbf{F}1-\mathbf{F}2-\mathbf{F}3=-292\mathbf{i}-62\mathbf{j}\ \text{N}