Solution 4.6b
From Mechanics
(Difference between revisions)
(New page: <math>\begin{align} & \mathbf{F}1=37\textrm{.}6\mathbf{i}+13\textrm{.}7\mathbf{j}\ \text{N}\\ \end{align}</math> <math>\begin{align} & \mathbf{F}2=43\textrm{.}3\mathbf{i}+25\mathbf{j} \...) |
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& \mathbf{F}1=37\textrm{.}6\mathbf{i}+13\textrm{.}7\mathbf{j}\ \text{N}\\ | & \mathbf{F}1=37\textrm{.}6\mathbf{i}+13\textrm{.}7\mathbf{j}\ \text{N}\\ | ||
\end{align}</math> | \end{align}</math> | ||
| - | |||
<math>\begin{align} | <math>\begin{align} | ||
& \mathbf{F}2=43\textrm{.}3\mathbf{i}+25\mathbf{j} \ \text{N}\\ | & \mathbf{F}2=43\textrm{.}3\mathbf{i}+25\mathbf{j} \ \text{N}\\ | ||
\end{align}</math> | \end{align}</math> | ||
| + | |||
| + | Resultant force= | ||
| + | <math>F1+F2=\left( 37\textrm{.}6+43\textrm{.}3 \right)\mathbf{i}+\left( 13\textrm{.}7-25 \right)\mathbf{j}=80\textrm{.}9\mathbf{i}-11\textrm{.}3\mathbf{j}\ \text{N}</math> | ||
| + | |||
| + | |||
| + | This vector has magnitude | ||
| + | <math>\sqrt{{{80\textrm{.}9}^{2}}+{{\left( -11\textrm{.}3 \right)}^{2}}} \ \text{=81}\text{.7}\ \text{N}</math> | ||
| + | |||
| + | using | ||
| + | |||
| + | <math>{{F}^{2}}={{H}^{2}}+{{V}^{2}}</math> | ||
| + | |||
| + | where | ||
| + | |||
| + | <math>H=80\textrm{.}9\ \text{N},\ \ V=-11\textrm{.}3\ \text{N}</math> | ||
Current revision
\displaystyle \begin{align} & \mathbf{F}1=37\textrm{.}6\mathbf{i}+13\textrm{.}7\mathbf{j}\ \text{N}\\ \end{align}
\displaystyle \begin{align} & \mathbf{F}2=43\textrm{.}3\mathbf{i}+25\mathbf{j} \ \text{N}\\ \end{align}
Resultant force= \displaystyle F1+F2=\left( 37\textrm{.}6+43\textrm{.}3 \right)\mathbf{i}+\left( 13\textrm{.}7-25 \right)\mathbf{j}=80\textrm{.}9\mathbf{i}-11\textrm{.}3\mathbf{j}\ \text{N}
This vector has magnitude
\displaystyle \sqrt{{{80\textrm{.}9}^{2}}+{{\left( -11\textrm{.}3 \right)}^{2}}} \ \text{=81}\text{.7}\ \text{N}
using
\displaystyle {{F}^{2}}={{H}^{2}}+{{V}^{2}}
where
\displaystyle H=80\textrm{.}9\ \text{N},\ \ V=-11\textrm{.}3\ \text{N}
