4. Forces and vectors
From Mechanics
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<math>\begin{align} | <math>\begin{align} | ||
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& =F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j} | & =F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j} | ||
\end{align}</math> | \end{align}</math> | ||
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[[Image:TF.teori.GIF]] | [[Image:TF.teori.GIF]] | ||
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<math>F\cos \alpha </math> | <math>F\cos \alpha </math> | ||
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<math>F\sin \alpha </math> | <math>F\sin \alpha </math> | ||
is called the vertical component of the force. | is called the vertical component of the force. | ||
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'''[[Example 4.1]]''' | '''[[Example 4.1]]''' | ||
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Express each of the forces given below in the form a<math>\mathbf{i}</math> + b<math>\mathbf{j}</math>. | Express each of the forces given below in the form a<math>\mathbf{i}</math> + b<math>\mathbf{j}</math>. | ||
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[[Image:TF4.1a.GIF]] | [[Image:TF4.1a.GIF]] | ||
| - | + | b) | |
[[Image:TF4.1b.GIF]] | [[Image:TF4.1b.GIF]] | ||
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'''Solution''' | '''Solution''' | ||
| - | + | a) <math>20\cos 40{}^\circ \mathbf{i}+20\sin 40{}^\circ \mathbf{j}</math> | |
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| - | <math>20\cos 40{}^\circ \mathbf{i}+20\sin 40 | + | |
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| + | b) <math>-80\cos 30{}^\circ \mathbf{i}+80\sin 30{}^\circ \mathbf{j}</math> | ||
Note the negative sign here in the first term. | Note the negative sign here in the first term. | ||
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'''[[Example 4.2]]''' | '''[[Example 4.2]]''' | ||
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[[Image:TF4.2.GIF]] | [[Image:TF4.2.GIF]] | ||
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'''Solution''' | '''Solution''' | ||
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<math>28\cos 30{}^\circ \mathbf{i}-28\sin 30{}^\circ \mathbf{j}</math> | <math>28\cos 30{}^\circ \mathbf{i}-28\sin 30{}^\circ \mathbf{j}</math> | ||
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Note the negative sign in the second term. | Note the negative sign in the second term. | ||
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'''[[Example 4.3]]''' | '''[[Example 4.3]]''' | ||
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Express the force shown below as a vector in terms of | Express the force shown below as a vector in terms of | ||
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'''Solution''' | '''Solution''' | ||
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<math>-50\cos 44{}^\circ \mathbf{i}-50\sin 44{}^\circ \mathbf{j}</math> | <math>-50\cos 44{}^\circ \mathbf{i}-50\sin 44{}^\circ \mathbf{j}</math> | ||
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Note that here both terms are negative. | Note that here both terms are negative. | ||
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| - | The magnitude, | + | The magnitude, <math>F</math> , of the force is given by, |
<math>F=\sqrt{{{4}^{2}}+{{8}^{2}}}=\sqrt{80}=8.94\text{ N (to 3sf)}</math> | <math>F=\sqrt{{{4}^{2}}+{{8}^{2}}}=\sqrt{80}=8.94\text{ N (to 3sf)}</math> | ||
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Find the magnitude and direction of the resultant of the four forces shown in the diagram. | Find the magnitude and direction of the resultant of the four forces shown in the diagram. | ||
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[[Image:TF4.5.GIF]] | [[Image:TF4.5.GIF]] | ||
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'''Solution''' | '''Solution''' | ||
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| 18 N | | 18 N | ||
|valign="top"| <math>-18\mathbf{j}</math> | |valign="top"| <math>-18\mathbf{j}</math> | ||
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|- | |- | ||
| 25 N | | 25 N | ||
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|valign="top"|<math>-15\cos 30{}^\circ \mathbf{i}+15\sin 30{}^\circ \mathbf{j}</math> | |valign="top"|<math>-15\cos 30{}^\circ \mathbf{i}+15\sin 30{}^\circ \mathbf{j}</math> | ||
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<math>\begin{align} | <math>\begin{align} | ||
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& =-23.627\mathbf{i}-3.730\mathbf{j} | & =-23.627\mathbf{i}-3.730\mathbf{j} | ||
\end{align}</math> | \end{align}</math> | ||
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The magnitude is given by: | The magnitude is given by: | ||
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<math>\sqrt{{{23.627}^{2}}+{{3.730}^{2}}}=23.9\text{ N (to 3sf)}</math> | <math>\sqrt{{{23.627}^{2}}+{{3.730}^{2}}}=23.9\text{ N (to 3sf)}</math> | ||
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The angle <math>\theta </math> can be found using tan. | The angle <math>\theta </math> can be found using tan. | ||
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<math>\begin{align} | <math>\begin{align} | ||
& \tan \theta =\frac{3.730}{23.627} \\ | & \tan \theta =\frac{3.730}{23.627} \\ | ||
& \theta =9.0{}^\circ | & \theta =9.0{}^\circ | ||
\end{align}</math> | \end{align}</math> | ||
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[[Image:TF4.5a.GIF]] | [[Image:TF4.5a.GIF]] | ||
Revision as of 13:05, 29 January 2010
| Theory | Exercises |
Key Points
\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, \displaystyle F , 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}
