Solution 4.9a
From Mechanics
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& \mathbf{F}2=9\cos 90 {}^\circ | & \mathbf{F}2=9\cos 90 {}^\circ | ||
\mathbf{i}+9\sin 90 {}^\circ | \mathbf{i}+9\sin 90 {}^\circ | ||
| - | \mathbf{j}=9\times | + | \mathbf{j}=9\times 0\mathbf{i}+9\times 1\textrm{.}0\mathbf{j} \\ |
| - | & = | + | & =9\textrm{.}0\mathbf{j} \ \text{N}\\ |
\end{align}</math> | \end{align}</math> | ||
Revision as of 09:26, 1 April 2010
\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 0\mathbf{i}+6\times \left(-1\right)\mathbf{j} \\ & =-80\mathbf{j}\ \text{N}\\ \end{align}
\displaystyle F4=8\ \text{N}
and
\displaystyle \alpha 3=-25{}^\circ
this gives
\displaystyle \begin{align} & \mathbf{F}3=8\cos \left(-25{}^\circ \right) \mathbf{i}+8\sin \left(-25{}^\circ \right) \mathbf{j}=8\times 0\mathbf{i}+8\times \left(-1\right)\mathbf{j} \\ & =-30\textrm{.}8\mathbf{i}+84\textrm{.}6\mathbf{j} \ \text{N}\\ \end{align}
