Solution 4.9a

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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 <ma...)
Current revision (10:20, 1 April 2010) (edit) (undo)
 
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\end{align}</math>
\end{align}</math>
-
<math>F1=100\ \text{N}</math>
+
<math>F1=7\ \text{N}</math>
and
and
-
<math>\alpha 1=30{}^\circ
+
<math>\alpha 1=35{}^\circ
</math>
</math>
this gives
this gives
<math>\begin{align}
<math>\begin{align}
-
& \mathbf{F}1=100\cos 30{}^\circ
+
& \mathbf{F}1=7\cos 35{}^\circ
-
\mathbf{i}+100\sin 30{}^\circ
+
\mathbf{i}+7\sin 35{}^\circ
-
\mathbf{j}=100\times 0\textrm{.}866\mathbf{i}+100\times 0\textrm{.}5\mathbf{j} \\
+
\mathbf{j}=7\times 0\textrm{.}819\mathbf{i}+7\times 0\textrm{.}574\mathbf{j} \\
-
& =86\textrm{.}6\mathbf{i}+50\mathbf{j}\ \text{N}\\
+
& =5\textrm{.}73\mathbf{i}+4\textrm{.}02\mathbf{j}\ \text{N}\\
\end{align}</math>
\end{align}</math>
-
<math>F2=90\ \text{N}</math>
+
<math>F2=90 \text{N}</math>
and
and
-
<math>\alpha 2=180{}^\circ-70{}^\circ=110{}^\circ
+
<math>\alpha 2=90{}^\circ
</math>
</math>
this gives
this gives
<math>\begin{align}
<math>\begin{align}
-
& \mathbf{F}2=90\cos 110 {}^\circ
+
& \mathbf{F}2=9\cos 90 {}^\circ
-
\mathbf{i}+90\sin 110 {}^\circ
+
\mathbf{i}+9\sin 90 {}^\circ
-
\mathbf{j}=90\times \left(-0\textrm{.}342 \right)\mathbf{i}+90\times 0\textrm{.}94\mathbf{j} \\
+
\mathbf{j}=9\times 0\mathbf{i}+9\times 1\textrm{.}0\mathbf{j} \\
-
& =-30\textrm{.}8\mathbf{i}+84\textrm{.}6\mathbf{j} \ \text{N}\\
+
& =9\textrm{.}0\mathbf{j} \ \text{N}\\
\end{align}</math>
\end{align}</math>
-
<math>F3=80\ \text{N}</math>
+
<math>F3=6\ \text{N}</math>
and
and
-
<math>\alpha 3=-90{}^\circ
+
<math>\alpha 3=-180{}^\circ+72{}^\circ=-108{}^\circ
</math>
</math>
this gives
this gives
<math>\begin{align}
<math>\begin{align}
-
& \mathbf{F}3=80\cos \left(-90{}^\circ \right)
+
& \mathbf{F}3=6\cos \left(-108{}^\circ \right)
-
\mathbf{i}+80\sin \left(-90{}^\circ \right)
+
\mathbf{i}+6\sin \left(-108{}^\circ \right)
-
\mathbf{j}=80\times 0\mathbf{i}+80\times \left(-1\right)\mathbf{j} \\
+
\mathbf{j}=6\times \left(-0\textrm{.}309\right)\mathbf{i}+6\times \left(-0\textrm{.}951\right)\mathbf{j} \\
-
& =-80\mathbf{j}\ \text{N}\\
+
& =-1\textrm{.}85\mathbf{i}-5\textrm{.}71\mathbf{j} \ \text{N}\\
 +
\end{align}</math>
 +
 
 +
 
 +
<math>F4=8\ \text{N}</math>
 +
and
 +
<math>\alpha 3=-25{}^\circ
 +
</math>
 +
this gives
 +
 
 +
<math>\begin{align}
 +
& \mathbf{F}4=8\cos \left(-25{}^\circ \right)
 +
\mathbf{i}+8\sin \left(-25{}^\circ \right)
 +
\mathbf{j}=8\times 0\textrm{.}906\mathbf{i}+8\times \left(-0\textrm{.}423\right)\mathbf{j} \\
 +
& =7\textrm{.}25\mathbf{i}-3\textrm{.}38\mathbf{j} \ \text{N}\\
 +
\end{align}</math>
 +
 
 +
The resultant <math>\mathbf{R}</math> is the sum of these four forces.
 +
 
 +
<math>\begin{align}
 +
& \mathbf{R} =11\textrm{.}1\mathbf{i}+3\textrm{.}93\mathbf{j} \ \text{N}\\
 +
\end{align}</math>
 +
 
 +
Then if <math>R</math> is the magnitude of this resultant
 +
 
 +
<math>\begin{align}
 +
& {{R\ }^{2}}={{11\textrm{.}1}^{2}}+{{3\textrm{.}93}^{2}} \\
 +
& R=\sqrt{{{11\textrm{.}1}^{2}}+{{3\textrm{.}93}^{2}}}=11\textrm{.}8\ \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=7\ \text{N} and \displaystyle \alpha 1=35{}^\circ this gives

\displaystyle \begin{align} & \mathbf{F}1=7\cos 35{}^\circ \mathbf{i}+7\sin 35{}^\circ \mathbf{j}=7\times 0\textrm{.}819\mathbf{i}+7\times 0\textrm{.}574\mathbf{j} \\ & =5\textrm{.}73\mathbf{i}+4\textrm{.}02\mathbf{j}\ \text{N}\\ \end{align}

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

\displaystyle \begin{align} & \mathbf{F}2=9\cos 90 {}^\circ \mathbf{i}+9\sin 90 {}^\circ \mathbf{j}=9\times 0\mathbf{i}+9\times 1\textrm{.}0\mathbf{j} \\ & =9\textrm{.}0\mathbf{j} \ \text{N}\\ \end{align}

\displaystyle F3=6\ \text{N} and \displaystyle \alpha 3=-180{}^\circ+72{}^\circ=-108{}^\circ this gives

\displaystyle \begin{align} & \mathbf{F}3=6\cos \left(-108{}^\circ \right) \mathbf{i}+6\sin \left(-108{}^\circ \right) \mathbf{j}=6\times \left(-0\textrm{.}309\right)\mathbf{i}+6\times \left(-0\textrm{.}951\right)\mathbf{j} \\ & =-1\textrm{.}85\mathbf{i}-5\textrm{.}71\mathbf{j} \ \text{N}\\ \end{align}


\displaystyle F4=8\ \text{N} and \displaystyle \alpha 3=-25{}^\circ this gives

\displaystyle \begin{align} & \mathbf{F}4=8\cos \left(-25{}^\circ \right) \mathbf{i}+8\sin \left(-25{}^\circ \right) \mathbf{j}=8\times 0\textrm{.}906\mathbf{i}+8\times \left(-0\textrm{.}423\right)\mathbf{j} \\ & =7\textrm{.}25\mathbf{i}-3\textrm{.}38\mathbf{j} \ \text{N}\\ \end{align}

The resultant \displaystyle \mathbf{R} is the sum of these four forces.

\displaystyle \begin{align} & \mathbf{R} =11\textrm{.}1\mathbf{i}+3\textrm{.}93\mathbf{j} \ \text{N}\\ \end{align}

Then if \displaystyle R is the magnitude of this resultant

\displaystyle \begin{align} & {{R\ }^{2}}={{11\textrm{.}1}^{2}}+{{3\textrm{.}93}^{2}} \\ & R=\sqrt{{{11\textrm{.}1}^{2}}+{{3\textrm{.}93}^{2}}}=11\textrm{.}8\ \text{N} \\ \end{align}

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