Solution 4.7a

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Current revision (15:04, 24 March 2010) (edit) (undo)
 
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& \mathbf{F}=F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j}
& \mathbf{F}=F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j}
\end{align}</math>
\end{align}</math>
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<math>F1=100\ \text{N}</math>
<math>F1=100\ \text{N}</math>
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</math>
</math>
this gives
this gives
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<math>\begin{align}
<math>\begin{align}
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& =86\textrm{.}6\mathbf{i}+50\mathbf{j}\ \text{N}\\
& =86\textrm{.}6\mathbf{i}+50\mathbf{j}\ \text{N}\\
\end{align}</math>
\end{align}</math>
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- 
<math>F2=90\ \text{N}</math>
<math>F2=90\ \text{N}</math>
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</math>
</math>
this gives
this gives
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<math>\begin{align}
<math>\begin{align}
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\mathbf{j}=90\times \left(-0\textrm{.}342 \right)\mathbf{i}+90\times 0\textrm{.}94\mathbf{j} \\
\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}\\
& =-30\textrm{.}8\mathbf{i}+84\textrm{.}6\mathbf{j} \ \text{N}\\
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\end{align}</math>
 +
 +
<math>F3=80\ \text{N}</math>
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and
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<math>\alpha 3=-90{}^\circ
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</math>
 +
this gives
 +
 +
<math>\begin{align}
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& \mathbf{F}3=80\cos \left(-90{}^\circ \right)
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\mathbf{i}+80\sin \left(-90{}^\circ \right)
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\mathbf{j}=80\times 0\mathbf{i}+80\times \left(-1\right)\mathbf{j} \\
 +
& =-80\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=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}

\displaystyle F3=80\ \text{N} and \displaystyle \alpha 3=-90{}^\circ this gives

\displaystyle \begin{align} & \mathbf{F}3=80\cos \left(-90{}^\circ \right) \mathbf{i}+80\sin \left(-90{}^\circ \right) \mathbf{j}=80\times 0\mathbf{i}+80\times \left(-1\right)\mathbf{j} \\ & =-80\mathbf{j}\ \text{N}\\ \end{align}

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