Solution 4.6a

From Mechanics

(Difference between revisions)
Jump to: navigation, search
Current revision (11:39, 27 March 2011) (edit) (undo)
 
(One intermediate revision not shown.)
Line 14: Line 14:
& \mathbf{F}1=40\cos 20{}^\circ
& \mathbf{F}1=40\cos 20{}^\circ
\mathbf{i}+40\sin 20{}^\circ
\mathbf{i}+40\sin 20{}^\circ
-
\mathbf{j}=40\times 0.94\mathbf{i}+40\times 0.342\mathbf{j} \\
+
\mathbf{j}=40\times 0\textrm{.}94\mathbf{i}+40\times 0\textrm{.}342\mathbf{j} \\
-
& =37.6\mathbf{i}+13.7\mathbf{j} \\
+
& =37\textrm{.}6\mathbf{i}+13\textrm{.}7\mathbf{j}\ \text{N}\\
\end{align}</math>
\end{align}</math>
Line 31: Line 31:
\right) {}^\circ
\right) {}^\circ
\mathbf{i}+50\sin \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} \\
+
\mathbf{j}=50\times 0\textrm{.}866\mathbf{i}-50\times 0\textrm{.}50\mathbf{j} \\
-
& =43.3\mathbf{i}+25\mathbf{j} \\
+
& =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}

Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecoursemechanics/index.php/Solution_4.6a"