Solution 4.6a

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(New page: <math>\begin{align} & \mathbf{F}=F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j} \end{align}</math>)
Current revision (11:39, 27 March 2011) (edit) (undo)
 
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<math>\begin{align}
<math>\begin{align}
& \mathbf{F}=F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j}
& \mathbf{F}=F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j}
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\end{align}</math>
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<math>F1=40\ \text{N}</math>
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and
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<math>\alpha 1=20{}^\circ
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</math>
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this gives
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<math>\begin{align}
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& \mathbf{F}1=40\cos 20{}^\circ
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\mathbf{i}+40\sin 20{}^\circ
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\mathbf{j}=40\times 0\textrm{.}94\mathbf{i}+40\times 0\textrm{.}342\mathbf{j} \\
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& =37\textrm{.}6\mathbf{i}+13\textrm{.}7\mathbf{j}\ \text{N}\\
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\end{align}</math>
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<math>F2=50\ \text{N}</math>
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and
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<math>\alpha 2=-30{}^\circ
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</math>
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this gives
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<math>\begin{align}
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& \mathbf{F}2=50\cos \left( -30
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\right) {}^\circ
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\mathbf{i}+50\sin \left( -30 \right) {}^\circ
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\mathbf{j}=50\times 0\textrm{.}866\mathbf{i}-50\times 0\textrm{.}50\mathbf{j} \\
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& =43\textrm{.}3\mathbf{i}-25\mathbf{j} \ \text{N}\\
\end{align}</math>
\end{align}</math>

Current revision

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


\displaystyle F1=40\ \text{N} and \displaystyle \alpha 1=20{}^\circ this gives


\displaystyle \begin{align} & \mathbf{F}1=40\cos 20{}^\circ \mathbf{i}+40\sin 20{}^\circ \mathbf{j}=40\times 0\textrm{.}94\mathbf{i}+40\times 0\textrm{.}342\mathbf{j} \\ & =37\textrm{.}6\mathbf{i}+13\textrm{.}7\mathbf{j}\ \text{N}\\ \end{align}


\displaystyle F2=50\ \text{N} and \displaystyle \alpha 2=-30{}^\circ this gives


\displaystyle \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}

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