Förberedande Mekanik

Versionen från 18 mars 2009 kl. 15.32 (redigera) Ian (Diskussion | bidrag) (New page: 4. Forces and Vectors Key Points <math>\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{al...) ← Gå till föregående ändring		Versionen från 18 mars 2009 kl. 15.48 (redigera) (ogör) Ian (Diskussion | bidrag) Gå till nästa ändring →
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-	4. Forces and Vectors
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-	Key Points
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	<math>\begin{align}		<math>\begin{align}
	& \mathbf{F}=F\cos \alpha \mathbf{i}+F\cos (90-\alpha )\mathbf{j} \\		& \mathbf{F}=F\cos \alpha \mathbf{i}+F\cos (90-\alpha )\mathbf{j} \\
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	is called the vertical component of the force.		is called the vertical component of the force.
-		+	'''[[Example 4.1]]'''
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-	Example 4.1	+
	Express each of the forces given below in the form ai + bj.		Express each of the forces given below in the form ai + bj.
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-		+	'''Solution'''
-	Solution	+
	(a)		(a)
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	Note the negative sign here in the first term.		Note the negative sign here in the first term.
-	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 i and j.
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-	Solution	+	'''Solution'''
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	Note the negative sign in the second term.		Note the negative sign in the second term.
-	Example 4.3	+	'''[[Example 4.3]]'''
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-	Solution	+	'''Solution'''
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	Note that here both terms are negative.		Note that here both terms are negative.
-	Example 4.4	+	'''[[Example 4.4]]'''
	Find the magnitude of the force (4i - 8j) N. Draw a diagram to show the direction of this force.		Find the magnitude of the force (4i - 8j) N. Draw a diagram to show the direction of this force.
-	Solution	+	'''Solution'''
	The magnitude, FN , of the force is given by,		The magnitude, FN , of the force is given by,
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-	The angle, , is given by,	+	The angle, <math>\theta </math>, is given by,
	<math>\theta ={{\tan }^{-1}}\left( \frac{8}{4} \right)=63.4{}^\circ </math>		<math>\theta ={{\tan }^{-1}}\left( \frac{8}{4} \right)=63.4{}^\circ </math>
-	Example 4.5	+	'''[[Example 4.5]]'''
	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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-	Solution	+	'''Solution'''
	Force Vector Form		Force Vector Form
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-	The angle  can be found using tan.	+	The angle <math>\theta </math> can be found using tan.

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Versionen från 18 mars 2009 kl. 15.48

\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 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 j is vertical, \displaystyle F\sin \alpha is called the vertical component of the force.

Example 4.1

Express each of the forces given below in the form ai + bj.

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.

Example 4.2

Express the force shown below as a vector in terms of i and j.

Solution

\displaystyle 28\cos 30{}^\circ \mathbf{i}-28\sin 30{}^\circ \mathbf{j}

Note the negative sign in the second term.

Example 4.3

Solution

\displaystyle -50\cos 44{}^\circ \mathbf{i}-50\sin 44{}^\circ \mathbf{j}

Note that here both terms are negative.

Example 4.4

Find the magnitude of the force (4i - 8j) 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

Example 4.5

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}