4. Forces and vectors
From Mechanics
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| - | |align="left"|Force | + | |align="left"| Force |
| valign="top"|Vector Form | | valign="top"|Vector Form | ||
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| valign="top"| <math>20\cos 50{}^\circ \mathbf{i}+20\sin 50{}^\circ \mathbf{j}</math> | | valign="top"| <math>20\cos 50{}^\circ \mathbf{i}+20\sin 50{}^\circ \mathbf{j}</math> | ||
|- | |- | ||
| - | | | + | | 18 N |
| - | |valign="top"| | + | |valign="top"| <math>-18\mathbf{j}</math> |
| + | |||
| + | |- | ||
| + | | 25 N | ||
| + | |valign="top"|<math>-25\cos 20{}^\circ \mathbf{i}-25\sin 20{}^\circ \mathbf{j}</math> | ||
| + | |- | ||
| + | | 15 N | ||
| + | |valign="top"|<math>-15\cos 30{}^\circ \mathbf{i}+15\sin 30{}^\circ \mathbf{j}</math> | ||
|} | |} | ||
| - | |||
| - | 20 N | ||
| - | <math>20\cos 50{}^\circ \mathbf{i}+20\sin 50{}^\circ \mathbf{j}</math> | ||
| - | 18 N | ||
| - | <math>-18\mathbf{j}</math> | ||
| - | 25 N | ||
| - | <math>-25\cos 20{}^\circ \mathbf{i}-25\sin 20{}^\circ \mathbf{j}</math> | ||
| - | + | ||
| - | + | ||
Revision as of 10:02, 15 May 2009
| Theory | Exercises |
\displaystyle \begin{align}
& \mathbf{F}=F\cos \alpha \mathbf{i}+F\cos (90-\alpha )\mathbf{j} \\
& =F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j}
\end{align}
\displaystyle F\cos \alpha is one component of the force. If \displaystyle \mathbf{i} is horizontal, \displaystyle F\cos \alpha is called the horizontal component of the force.
\displaystyle F\sin \alpha
is another component of the force. If \displaystyle \mathbf{j} is vertical,
\displaystyle F\sin \alpha
is called the vertical component of the force.
Express each of the forces given below in the form a\displaystyle \mathbf{i} + b\displaystyle \mathbf{j}.
(a)
(b)
Solution
(a)
\displaystyle 20\cos 40{}^\circ \mathbf{i}+20\sin 40{}^\circ \mathbf{j}
(b)
\displaystyle -80\cos 30{}^\circ \mathbf{i}+80\sin 30{}^\circ \mathbf{j}
Note the negative sign here in the first term.
Express the force shown below as a vector in terms of \displaystyle \mathbf{i} and \displaystyle \mathbf{j}.
Solution
\displaystyle 28\cos 30{}^\circ \mathbf{i}-28\sin 30{}^\circ \mathbf{j}
Note the negative sign in the second term.
Express the force shown below as a vector in terms of
\displaystyle \mathbf{i}
and
\displaystyle \mathbf{j}
Solution
\displaystyle -50\cos 44{}^\circ \mathbf{i}-50\sin 44{}^\circ \mathbf{j}
Note that here both terms are negative.
Find the magnitude of the force (4\displaystyle \mathbf{i} - 8\displaystyle \mathbf{j}) N. Draw a diagram to show the direction of this force.
Solution
The magnitude, FN , of the force is given by,
\displaystyle F=\sqrt{{{4}^{2}}+{{8}^{2}}}=\sqrt{80}=8.94\text{ N (to 3sf)}
The angle, \displaystyle \theta , is given by,
\displaystyle \theta ={{\tan }^{-1}}\left( \frac{8}{4} \right)=63.4{}^\circ
Find the magnitude and direction of the resultant of the four forces shown in the diagram.
Solution
| Force | Vector Form |
| 20 N | \displaystyle 20\cos 50{}^\circ \mathbf{i}+20\sin 50{}^\circ \mathbf{j} |
| 18 N | \displaystyle -18\mathbf{j} |
| 25 N | \displaystyle -25\cos 20{}^\circ \mathbf{i}-25\sin 20{}^\circ \mathbf{j} |
| 15 N | \displaystyle -15\cos 30{}^\circ \mathbf{i}+15\sin 30{}^\circ \mathbf{j} |
\displaystyle \begin{align}
& \text{Resultant Force }=\text{ }\left( 20\cos 50{}^\circ -25\cos 20{}^\circ -15\cos 30{}^\circ \right)\mathbf{i}+\left( 20\sin 50{}^\circ -18-25\sin 20{}^\circ +15\sin 30{}^\circ \right)\mathbf{j} \\
& =-23.627\mathbf{i}-3.730\mathbf{j}
\end{align}
The magnitude is given by:
\displaystyle \sqrt{{{23.627}^{2}}+{{3.730}^{2}}}=23.9\text{ N (to 3sf)}
The angle \displaystyle \theta can be found using tan.
\displaystyle \begin{align}
& \tan \theta =\frac{3.730}{23.627} \\
& \theta =9.0{}^\circ
\end{align}
