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Solution 4.6a

From Mechanics

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Current revision (11:39, 27 March 2011) (edit) (undo)
 
Line 32: Line 32:
\mathbf{i}+50\sin \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} \\
\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}\\
+
& =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=50cos30i+50sin30j=500.866i500.50j=43.3i25j N