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>)
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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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 +
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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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 +
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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 0.94\mathbf{i}+40\times 0.342\mathbf{j} \\
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& =37.6\mathbf{i}+13.7\mathbf{j} \\
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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 0.866\mathbf{i}-50\times 0.50\mathbf{j} \\
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& =43.3\mathbf{i}+25\mathbf{j} \\
\end{align}</math>
\end{align}</math>

Revision as of 15:21, 23 March 2010

\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 0.94\mathbf{i}+40\times 0.342\mathbf{j} \\ & =37.6\mathbf{i}+13.7\mathbf{j} \\ \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 0.866\mathbf{i}-50\times 0.50\mathbf{j} \\ & =43.3\mathbf{i}+25\mathbf{j} \\ \end{align}

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