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Solution 4.6a

From Mechanics

(Difference between revisions)
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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

F=Fcosi+Fsinj


F1=40 N and 1=20 this gives


F1=40cos20i+40sin20j=400.94i+400.342j=37.6i+13.7j N


F2=50 N and 2=−30 this gives


F2=50cos−30i+50sin−30j=500.866i−500.50j=43.3i−25j N

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