4. Forces and Vectors
From Mechanics
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<math>F\cos \alpha </math> | <math>F\cos \alpha </math> | ||
| - | is one component of the force. If i is horizontal, | + | is one component of the force. If <math>\mathbf{i}</math> is horizontal, |
<math>F\cos \alpha </math> | <math>F\cos \alpha </math> | ||
is called the horizontal component of the force. | is called the horizontal component of the force. | ||
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<math>F\sin \alpha </math> | <math>F\sin \alpha </math> | ||
| - | is another component of the force. If j is vertical, | + | is another component of the force. If <math>\mathbf{j}</math> is vertical, |
<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]]''' | ||
| - | Express each of the forces given below in the form | + | Express each of the forces given below in the form a<math>\mathbf{i}</math> + b<math>\mathbf{j}</math>. |
| + | (a) | ||
| + | [[Image:TF4.1a.GIF]] | ||
| + | (b) | ||
| + | [[Image:TF4.1b.GIF]] | ||
'''Solution''' | '''Solution''' | ||
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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]]''' | ||
| - | Express the force shown below as a vector in terms of i and j. | + | Express the force shown below as a vector in terms of <math>\mathbf{i}</math> and <math>\mathbf{j}</math>. |
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| + | [[Image:TF4.2.GIF]] | ||
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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 | ||
| + | <math>\mathbf{i}</math> | ||
| + | and | ||
| + | <math>\mathbf{j}</math> | ||
| + | [[Image:TF4.3.GIF]] | ||
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Note that here both terms are negative. | Note that here both terms are negative. | ||
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'''[[Example 4.4]]''' | '''[[Example 4.4]]''' | ||
| - | Find the magnitude of the force ( | + | Find the magnitude of the force (4<math>\mathbf{i}</math> - 8<math>\mathbf{j}</math>) N. Draw a diagram to show the direction of this force. |
'''Solution''' | '''Solution''' | ||
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| + | [[Image:TF4.4.GIF]] | ||
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The magnitude, FN , of the force is given by, | The magnitude, FN , of the force is given by, | ||
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| + | [[Image:TF4.5.GIF]] | ||
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& \theta =9.0{}^\circ | & \theta =9.0{}^\circ | ||
\end{align}</math> | \end{align}</math> | ||
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| + | [[Image:TF4.5a.GIF]] | ||
Current revision
\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}
