Solution 4.7a

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(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 <...)
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<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
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<math>\begin{align}
<math>\begin{align}
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& \mathbf{F}2=50\cos \left( -30
+
& \mathbf{F}2=90\cos 110 {}^\circ
-
\right) {}^\circ
+
\mathbf{i}+90\sin 110 {}^\circ
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\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} \\
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& =-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>

Revision as of 14:15, 24 March 2010

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


\displaystyle F1=100\ \text{N} and \displaystyle \alpha 1=30{}^\circ this gives


\displaystyle \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}


\displaystyle F2=90\ \text{N} and \displaystyle \alpha 2=180{}^\circ-70{}^\circ=110{}^\circ this gives


\displaystyle \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}

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