4. Forces and vectors - Revision history

Ian at 13:32, 23 March 2010

Ian — Tue, 23 Mar 2010 13:32:53 GMT

←Older revision		Revision as of 13:32, 23 March 2010
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	\end{align}</math>		\end{align}</math>

-	[[Image:TF.~~teori.GIF~~]]	+	[[Image:teori4.gif]]
		+

	<math>F</math> is the magnitude of the force.		<math>F</math> is the magnitude of the force.
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	<math>V</math> is the verical component of the force		<math>V</math> is the verical component of the force
-
-	[[Image:teori4.gif]]

	The figure shows		The figure shows
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	and also how to calculate the angles <math>\alpha</math> and <math>\beta</math>, for example,		and also how to calculate the angles <math>\alpha</math> and <math>\beta</math>, for example,

-	<math>\~~cos~~ \alpha =\frac{H}{F}</math> if <math>H</math> and <math>F</math> are known.	+	<math>\tan \alpha =\frac{V}{H}</math> if <math>H</math> and <math>V</math> are known.
		+
		+	'''Note'''
		+
		+	This equation has infinitely many solutions as if
		+	<math>\alpha </math>
		+	is a solution then
		+	<math>\alpha \pm 180</math>
		+	is also a solution.
		+
		+	In this course it is assumed the solution chosen is between
		+	<math>-180<\alpha \le 180</math>.
		+	Then this equation always has two solutions for
		+	<math>\alpha </math>,
		+	giving two distinct directions. The solution in the quadrant that corresponds to the direction of the vector must be chosen.


		+	[[Image:teori4a.gif]]

Ian at 16:56, 22 March 2010

Ian — Mon, 22 Mar 2010 16:56:52 GMT

←Older revision		Revision as of 16:56, 22 March 2010
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	and also how to calculate the angles <math>\alpha</math> and <math>\beta</math>, for example,		and also how to calculate the angles <math>\alpha</math> and <math>\beta</math>, for example,

-	<math>\cos \alpha =\frac{H}{F}</math>.	+	<math>\cos \alpha =\frac{H}{F}</math> if <math>H</math> and <math>F</math> are known.

Ian at 14:47, 22 March 2010

Ian — Mon, 22 Mar 2010 14:47:06 GMT

←Older revision		Revision as of 14:47, 22 March 2010
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	[[Image:TF.teori.GIF]]		[[Image:TF.teori.GIF]]
		+
		+	<math>F</math> is the magnitude of the force.

	<math>F\cos \alpha </math>		<math>F\cos \alpha </math>
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	<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.
-

	<math>F\sin \alpha </math>		<math>F\sin \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.
		+
		+	If <math>H</math> is the horizontal component of the force and
		+
		+	<math>V</math> is the verical component of the force
		+
		+	[[Image:teori4.gif]]
		+
		+	The figure shows
		+
		+	<math>{{F}^{2}}={{H}^{2}}+{{V}^{2}}</math>
		+
		+	and also how to calculate the angles <math>\alpha</math> and <math>\beta</math>, for example,
		+
		+	<math>\cos \alpha =\frac{H}{F}</math>.
		+
		+

Ian at 09:55, 11 March 2010

Ian — Thu, 11 Mar 2010 09:55:50 GMT

←Older revision		Revision as of 09:55, 11 March 2010
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	<math>\begin{align}		<math>\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} \\	+	& \text{Resultant Force }=\text{ }\left( 20\cos 50{}^\circ -25\cos 20{}^\circ -15\cos 30{}^\circ \right)\mathbf{i}+ \\
-	& =-23~~\textrm{~~.}627\mathbf{i}-3~~\textrm{~~.}730\mathbf{j} \ \text{N}	+	& \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} \text{ N}
	\end{align}</math>		\end{align}</math>

Ian at 13:31, 8 March 2010

Ian — Mon, 08 Mar 2010 13:31:46 GMT

←Older revision		Revision as of 13:31, 8 March 2010
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	== '''Key Points''' ==		== '''Key Points''' ==
		+
		+	The force <math>\mathbf{F}</math> can be resolved into components as follows.

	<math>\begin{align}		<math>\begin{align}

Ian at 11:41, 11 February 2010

Ian — Thu, 11 Feb 2010 11:41:03 GMT

←Older revision		Revision as of 11:41, 11 February 2010
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	'''Solution'''		'''Solution'''

-	a) <math>20\cos 40{}^\circ \mathbf{i}+20\sin 40{}^\circ \mathbf{j}</math>	+	a) <math>20\cos 40{}^\circ \mathbf{i}+20\sin 40{}^\circ \mathbf{j}\ \text{N}</math>

-	b) <math>-80\cos 30{}^\circ \mathbf{i}+80\sin 30{}^\circ \mathbf{j}</math>	+	b) <math>-80\cos 30{}^\circ \mathbf{i}+80\sin 30{}^\circ \mathbf{j}\ \text{N}</math>

	Note the negative sign here in the first term.		Note the negative sign here in the first term.
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	'''Solution'''		'''Solution'''

-	<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}\ \text{N}</math>

	Note the negative sign in the second term.		Note the negative sign in the second term.
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	'''Solution'''		'''Solution'''

-	<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}\ \text{N}</math>

	Note that here both terms are negative.		Note that here both terms are negative.
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	{\| width="100%" cellspacing="10px" align="center"		{\| width="100%" cellspacing="10px" align="center"
	\|align="left"\| Force		\|align="left"\| Force
-	\| valign="top"\|Vector Form	+	\| valign="top"\|Vector Form (N)
	\|-		\|-
	\|20 N		\|20 N
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	<math>\begin{align}		<math>\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} \\		& \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\textrm{.}627\mathbf{i}-3\textrm{.}730\mathbf{j}	+	& =-23\textrm{.}627\mathbf{i}-3\textrm{.}730\mathbf{j} \ \text{N}
	\end{align}</math>		\end{align}</math>

Ian at 14:17, 29 January 2010

Ian — Fri, 29 Jan 2010 14:17:00 GMT

←Older revision		Revision as of 14:17, 29 January 2010
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	The magnitude, <math>F</math> , of the force is given by,		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\textrm{.}94\text{ N (to 3sf)}</math>


	The angle, <math>\theta </math>, 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\textrm{.}4{}^\circ </math>


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	<math>\begin{align}		<math>\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} \\		& \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}	+	& =-23\textrm{.}627\mathbf{i}-3\textrm{.}730\mathbf{j}
	\end{align}</math>		\end{align}</math>

	The magnitude is given by:		The magnitude is given by:

-	<math>\sqrt{{{23.627}^{2}}+{{3.730}^{2}}}=23.9\text{ N (to 3sf)}</math>	+	<math>\sqrt{{{23\textrm{.}627}^{2}}+{{3\textrm{.}730}^{2}}}=23\textrm{.}9\text{ N (to 3sf)}</math>

	The angle <math>\theta </math> can be found using tan.		The angle <math>\theta </math> can be found using tan.

	<math>\begin{align}		<math>\begin{align}
-	& \tan \theta =\frac{3.730}{23.627} \\	+	& \tan \theta =\frac{3\textrm{.}730}{23\textrm{.}627} \\
-	& \theta =9.0{}^\circ	+	& \theta =9\textrm{.}0{}^\circ
	\end{align}</math>		\end{align}</math>

	[[Image:TF4.5a.GIF]]		[[Image:TF4.5a.GIF]]

Ian at 13:05, 29 January 2010

Ian — Fri, 29 Jan 2010 13:05:44 GMT

←Older revision		Revision as of 13:05, 29 January 2010
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	\|}		\|}

		+
		+	== '''Key Points''' ==

	<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>
-
-

	[[Image:TF.teori.GIF]]		[[Image:TF.teori.GIF]]
-
-
-
-

	<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.
		+

	'''[[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>.

-	(a)	+	a)

	[[Image:TF4.1a.GIF]]		[[Image:TF4.1a.GIF]]

-	(b)	+	b)

	[[Image:TF4.1b.GIF]]		[[Image:TF4.1b.GIF]]
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	'''Solution'''		'''Solution'''

-	(a)	+	a) <math>20\cos 40{}^\circ \mathbf{i}+20\sin 40{}^\circ \mathbf{j}</math>
-		+
-	<math>20\cos 40{}^\circ \mathbf{i}+20\sin 40~~{}^\circ \mathbf{j}</math>~~	+
-		+
-		+
-	~~(b)~~	+
-		+
-	~~<math>-80\cos 30{}^\circ \mathbf{i}+80\sin 30~~{}^\circ \mathbf{j}</math>	+

		+	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.


-
	'''[[Example 4.2]]'''		'''[[Example 4.2]]'''

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	[[Image:TF4.2.GIF]]		[[Image:TF4.2.GIF]]
-

	'''Solution'''		'''Solution'''
-

	<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>
-

	Note the negative sign in the second term.		Note the negative sign in the second term.
-


	'''[[Example 4.3]]'''		'''[[Example 4.3]]'''
-

	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'''
-		+
-		+
	<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>
-

	Note that here both terms are negative.		Note that here both terms are negative.
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-	The magnitude, FN , of the force is given by,	+	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.
-

	[[Image:TF4.5.GIF]]		[[Image:TF4.5.GIF]]
-

	'''Solution'''		'''Solution'''
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	\| 18 N		\| 18 N
	\|valign="top"\| <math>-18\mathbf{j}</math>		\|valign="top"\| <math>-18\mathbf{j}</math>
-
	\|-		\|-
	\| 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>
	\|}		\|}
-
-
-
-
-
-
-
-
-

	<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>
-

	The magnitude is given by:		The magnitude is given by:
-		+
-		+
	<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>
-
-

	The angle <math>\theta </math> can be found using tan.		The angle <math>\theta </math> can be found using tan.
-		+
-		+
	<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>
-

	[[Image:TF4.5a.GIF]]		[[Image:TF4.5a.GIF]]

Ian at 10:34, 19 May 2009

Ian — Tue, 19 May 2009 10:34:28 GMT

←Older revision		Revision as of 10:34, 19 May 2009
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	{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"		{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
	\| style="border-bottom:1px solid #797979" width="5px" \|		\| style="border-bottom:1px solid #797979" width="5px" \|
-	{{Selected tab\|[[2. Forces and vectors\|Theory]]}}	+	{{Selected tab\|[[4. Forces and vectors\|Theory]]}}
-	{{Not selected tab\|[[2. Exercises\|Exercises]]}}	+	{{Not selected tab\|[[4. Exercises\|Exercises]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
	\|}		\|}

Ian at 10:02, 15 May 2009

Ian — Fri, 15 May 2009 10:02:24 GMT

←Older revision		Revision as of 10:02, 15 May 2009
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	{\| width="100%" cellspacing="10px" align="center"		{\| width="100%" cellspacing="10px" align="center"
-	\|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>
	\|-		\|-
-	\| ~~[[Image:fig b.GIF]]~~	+	\| 18 N
-	\|valign="top"\| ~~Modelling the drum as a particle can explain what happens if the string is horizontal, but not what happens if the string is vertical.~~	+	\|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>

-	~~15 N~~	+
-	~~<math>-15\cos 30{}^\circ \mathbf{i}+15\sin 30{}^\circ \mathbf{j}</math>~~	+

←Older revision		Revision as of 10:34, 19 May 2009
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	{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"		{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
	\| style="border-bottom:1px solid #797979" width="5px" \|		\| style="border-bottom:1px solid #797979" width="5px" \|
-	{{Selected tab\|[[2. Forces and vectors\|Theory]]}}	+	{{Selected tab\|[[4. Forces and vectors\|Theory]]}}
-	{{Not selected tab\|[[2. Exercises\|Exercises]]}}	+	{{Not selected tab\|[[4. Exercises\|Exercises]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
	\|}		\|}