14. Moments and equilibrium - Revision history

Ian at 13:09, 27 April 2011

2011-04-27T13:09:59Z

←Older revision		Revision as of 13:09, 27 April 2011
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	{{Selected tab\|[[14. Moments and equilibrium\|Theory]]}}		{{Selected tab\|[[14. Moments and equilibrium\|Theory]]}}
	{{Not selected tab\|[[14. Exercises\|Exercises]]}}		{{Not selected tab\|[[14. Exercises\|Exercises]]}}
		+	{{Not selected tab\|[[14. Video\|Video]]}}
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Ian at 13:51, 8 March 2010

2010-03-08T13:51:26Z

←Older revision		Revision as of 13:51, 8 March 2010
Line 9:		Line 9:
	== '''Key Points''' ==		== '''Key Points''' ==

-	'''In equilibrium''':	+	'''For a body in equilibrium''':

-	• The resultant force on a body must be zero.	+	• The resultant force on the body must be zero.

	'''and'''		'''and'''

-	• The resultant moment on a body must be zero.	+	• The resultant moment of the forces on the body about all points must be zero.
		+
		+	Sometimes it is more convenient to solve a problem just using moments.

Ian at 10:56, 17 February 2010

2010-02-17T10:56:25Z

←Older revision		Revision as of 10:56, 17 February 2010
Line 7:		Line 7:
	\|}		\|}

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

	'''In equilibrium''':		'''In equilibrium''':
Line 14:		Line 13:
	• The resultant force on a body must be zero.		• The resultant force on a body must be zero.

-	'''~~And~~'''	+	'''and'''

	• The resultant moment on a body must be zero.		• The resultant moment on a body must be zero.
Line 26:		Line 25:

	Find the tension in each rope.		Find the tension in each rope.
-

	'''Solution'''		'''Solution'''
Line 40:		Line 38:
	& {{T}_{2}}=\frac{3\times 588}{5}=352\textrm{.}8=353\text{ N (to 3sf)} \\		& {{T}_{2}}=\frac{3\times 588}{5}=352\textrm{.}8=353\text{ N (to 3sf)} \\
	\end{align}</math>		\end{align}</math>
-

	Take moments about the point where T2 acts to give:		Take moments about the point where T2 acts to give:
-		+
-		+
	<math>\begin{align}		<math>\begin{align}
	& 5\times {{T}_{1}}=2\times 588 \\		& 5\times {{T}_{1}}=2\times 588 \\
	& {{T}_{1}}=\frac{2\times 588}{5}=235\textrm{.}2=235\text{ N (to 3sf)} \\		& {{T}_{1}}=\frac{2\times 588}{5}=235\textrm{.}2=235\text{ N (to 3sf)} \\
	\end{align}</math>		\end{align}</math>
-

	Finally for vertical equilibrium we require		Finally for vertical equilibrium we require
	<math>{{T}_{1}}+{{T}_{2}}=588</math>		<math>{{T}_{1}}+{{T}_{2}}=588</math>
-	,which can be used to check the tensions. In is the case we have:	+	, which can be used to check the tensions. In is the case we have:

	<math>352\textrm{.}8+235\textrm{.}2=588</math>		<math>352\textrm{.}8+235\textrm{.}2=588</math>
-
-


Line 67:		Line 60:

	Find the maximum mass that could be placed at either end of the beam if it is to remain in equilibrium.		Find the maximum mass that could be placed at either end of the beam if it is to remain in equilibrium.
-

	'''Solution'''		'''Solution'''
Line 76:		Line 68:

	Taking moments about the point where R1 acts gives:		Taking moments about the point where R1 acts gives:
-

	<math>\begin{align}		<math>\begin{align}
Line 82:		Line 73:
	& {{R}_{2}}=\frac{1\textrm{.}5\times 490}{2}=367\textrm{.}5=368\text{ N (to 3sf)} \\		& {{R}_{2}}=\frac{1\textrm{.}5\times 490}{2}=367\textrm{.}5=368\text{ N (to 3sf)} \\
	\end{align}</math>		\end{align}</math>
-

	Taking moments about the point where R2 acts gives:		Taking moments about the point where R2 acts gives:
-		+
-		+
	<math>\begin{align}		<math>\begin{align}
	& 2\times {{R}_{1}}=0\textrm{.}5\times 490 \\		& 2\times {{R}_{1}}=0\textrm{.}5\times 490 \\
	& {{R}_{1}}=\frac{0\textrm{.}5\times 490}{2}=122\textrm{.}5=123\text{ N (to 3sf)} \\		& {{R}_{1}}=\frac{0\textrm{.}5\times 490}{2}=122\textrm{.}5=123\text{ N (to 3sf)} \\
	\end{align}</math>		\end{align}</math>
-

	For vertical equilibrium we require		For vertical equilibrium we require
Line 98:		Line 86:

	<math>367\textrm{.}5+122\textrm{.}5=490</math>		<math>367\textrm{.}5+122\textrm{.}5=490</math>
-

	First consider the greatest mass that can be placed at the left hand end of the beam. The diagram below shows the extra force that must now be considered. When the maximum possible mass is used,		First consider the greatest mass that can be placed at the left hand end of the beam. The diagram below shows the extra force that must now be considered. When the maximum possible mass is used,
Line 107:		Line 94:

	Taking moments about the point where R1 acts gives:		Taking moments about the point where R1 acts gives:
-		+
-		+
	<math>\begin{align}		<math>\begin{align}
	& 1\times mg=1\textrm{.}5\times 490 \\		& 1\times mg=1\textrm{.}5\times 490 \\
	& m=\frac{1\textrm{.}5\times 490}{9\textrm{.}8}=75\text{ kg} \\		& m=\frac{1\textrm{.}5\times 490}{9\textrm{.}8}=75\text{ kg} \\
	\end{align}</math>		\end{align}</math>
-

	Similarly for a mass placed at the right hand end of the beam:		Similarly for a mass placed at the right hand end of the beam:
Line 123:		Line 108:
	& m=\frac{0\textrm{.}5\times 490}{2g}=12\textrm{.}5\text{ kg} \\		& m=\frac{0\textrm{.}5\times 490}{2g}=12\textrm{.}5\text{ kg} \\
	\end{align}</math>		\end{align}</math>
-

	Hence the greatest mass that can be placed at either end of the beam		Hence the greatest mass that can be placed at either end of the beam
Line 132:		Line 116:

	A ladder, of length 3 m and mass 20 kg, leans against a smooth, vertical wall so that the angle between the horizontal ground and the ladder is 60<math>{}^\circ </math>.		A ladder, of length 3 m and mass 20 kg, leans against a smooth, vertical wall so that the angle between the horizontal ground and the ladder is 60<math>{}^\circ </math>.
-	(a) Find the magnitude of the friction and normal reaction forces that act on the ladder, if it is in equilibrium.	+
-	(b) Find the minimum value of the coefficient of friction between the ladder and the ground.	+	a) Find the magnitude of the friction and normal reaction forces that act on the ladder, if it is in equilibrium.
		+
		+	b) Find the minimum value of the coefficient of friction between the ladder and the ground.


Line 142:		Line 128:
	[[Image:E14.3fig1.GIF]]		[[Image:E14.3fig1.GIF]]

-	(a)	+	a) Considering the horizontal forces gives:
-	Considering the horizontal forces gives:	+
-		+
-		+
	<math>F=S</math>		<math>F=S</math>
-

	Considering the vertical forces gives:		Considering the vertical forces gives:
-		+
-		+
	<math>R=196</math>		<math>R=196</math>
-

	Taking moment about the base of the ladder gives:		Taking moment about the base of the ladder gives:
-

	<math>\begin{align}		<math>\begin{align}
Line 162:		Line 142:
	& S=\frac{196\times 1\textrm{.}5\cos 60{}^\circ }{3\sin 60{}^\circ }=\frac{196\times \cos 60{}^\circ }{2\sin 60{}^\circ }=\frac{196}{2\tan 60{}^\circ }=56\textrm{.}6\text{ N (to 3 sf)} \\		& S=\frac{196\times 1\textrm{.}5\cos 60{}^\circ }{3\sin 60{}^\circ }=\frac{196\times \cos 60{}^\circ }{2\sin 60{}^\circ }=\frac{196}{2\tan 60{}^\circ }=56\textrm{.}6\text{ N (to 3 sf)} \\
	\end{align}</math>		\end{align}</math>
-

	But since		But since

	<math>F=S</math>		<math>F=S</math>
-	, the friction force has magnitude 56\textrm{.}6 N.	+	, the friction force has magnitude <math>56\textrm{.}6 \text{ N} </math>.

-	(b)	+	b) Using the friction inequality,
-		+
-	Using the friction inequality,	+

	<math>F\le \mu R</math>		<math>F\le \mu R</math>
	gives:		gives:
-		+
-		+
-		+
	<math>\begin{align}		<math>\begin{align}
	& \frac{196}{2\tan 60{}^\circ }\le \mu \times 196 \\		& \frac{196}{2\tan 60{}^\circ }\le \mu \times 196 \\

Ian at 18:10, 26 September 2009

2009-09-26T18:10:09Z

←Older revision		Revision as of 18:10, 26 September 2009
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	\| style="border-bottom:1px solid #797979" width="5px" \|		\| style="border-bottom:1px solid #797979" width="5px" \|
-	{{Selected tab\|[[14. Moments and ~~Equilibrium~~\|Theory]]}}	+	{{Selected tab\|[[14. Moments and equilibrium\|Theory]]}}
	{{Not selected tab\|[[14. Exercises\|Exercises]]}}		{{Not selected tab\|[[14. Exercises\|Exercises]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|

Ian at 19:02, 23 September 2009

2009-09-23T19:02:27Z

←Older revision		Revision as of 19:02, 23 September 2009
Line 38:		Line 38:
	<math>\begin{align}		<math>\begin{align}
	& 5\times {{T}_{2}}=3\times 588 \\		& 5\times {{T}_{2}}=3\times 588 \\
-	& {{T}_{2}}=\frac{3\times 588}{5}=352.8=353\text{ N (to 3sf)} \\	+	& {{T}_{2}}=\frac{3\times 588}{5}=352\textrm{.}8=353\text{ N (to 3sf)} \\
	\end{align}</math>		\end{align}</math>

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	<math>\begin{align}		<math>\begin{align}
	& 5\times {{T}_{1}}=2\times 588 \\		& 5\times {{T}_{1}}=2\times 588 \\
-	& {{T}_{1}}=\frac{2\times 588}{5}=235.2=235\text{ N (to 3sf)} \\	+	& {{T}_{1}}=\frac{2\times 588}{5}=235\textrm{.}2=235\text{ N (to 3sf)} \\
	\end{align}</math>		\end{align}</math>

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	,which can be used to check the tensions. In is the case we have:		,which can be used to check the tensions. In is the case we have:

-	<math>352.8+235.2=588</math>	+	<math>352\textrm{.}8+235\textrm{.}2=588</math>


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	<math>\begin{align}		<math>\begin{align}
-	& 2\times {{R}_{2}}=1.5\times 490 \\	+	& 2\times {{R}_{2}}=1\textrm{.}5\times 490 \\
-	& {{R}_{2}}=\frac{1.5\times 490}{2}=367.5=368\text{ N (to 3sf)} \\	+	& {{R}_{2}}=\frac{1\textrm{.}5\times 490}{2}=367\textrm{.}5=368\text{ N (to 3sf)} \\
	\end{align}</math>		\end{align}</math>

Line 88:		Line 88:

	<math>\begin{align}		<math>\begin{align}
-	& 2\times {{R}_{1}}=0.5\times 490 \\	+	& 2\times {{R}_{1}}=0\textrm{.}5\times 490 \\
-	& {{R}_{1}}=\frac{0.5\times 490}{2}=122.5=123\text{ N (to 3sf)} \\	+	& {{R}_{1}}=\frac{0\textrm{.}5\times 490}{2}=122\textrm{.}5=123\text{ N (to 3sf)} \\
	\end{align}</math>		\end{align}</math>

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	, which can be used to check the tensions. In is the case we have:		, which can be used to check the tensions. In is the case we have:

-	<math>367.5+122.5=490</math>	+	<math>367\textrm{.}5+122\textrm{.}5=490</math>


Line 110:		Line 110:

	<math>\begin{align}		<math>\begin{align}
-	& 1\times mg=1.5\times 490 \\	+	& 1\times mg=1\textrm{.}5\times 490 \\
-	& m=\frac{1.5\times 490}{9.8}=75\text{ kg} \\	+	& m=\frac{1\textrm{.}5\times 490}{9\textrm{.}8}=75\text{ kg} \\
	\end{align}</math>		\end{align}</math>

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	<math>\begin{align}		<math>\begin{align}
-	& 2\times mg=0.5\times 490 \\	+	& 2\times mg=0\textrm{.}5\times 490 \\
-	& m=\frac{0.5\times 490}{2g}=12.5\text{ kg} \\	+	& m=\frac{0\textrm{.}5\times 490}{2g}=12\textrm{.}5\text{ kg} \\
	\end{align}</math>		\end{align}</math>

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	'''[[Example 14.3]]'''		'''[[Example 14.3]]'''

-	A ladder, of length 3 m and mass 20 kg, leans against a smooth, vertical wall so that the angle between the horizontal ground and the ladder is ~~60~~.	+	A ladder, of length 3 m and mass 20 kg, leans against a smooth, vertical wall so that the angle between the horizontal ground and the ladder is 60<math>{}^\circ </math>.
	(a) Find the magnitude of the friction and normal reaction forces that act on the ladder, if it is in equilibrium.		(a) Find the magnitude of the friction and normal reaction forces that act on the ladder, if it is in equilibrium.
	(b) Find the minimum value of the coefficient of friction between the ladder and the ground.		(b) Find the minimum value of the coefficient of friction between the ladder and the ground.
Line 159:		Line 159:

	<math>\begin{align}		<math>\begin{align}
-	& 196\times 1.5\cos 60{}^\circ =S\times 3\sin 60{}^\circ \\	+	& 196\times 1\textrm{.}5\cos 60{}^\circ =S\times 3\sin 60{}^\circ \\
-	& S=\frac{196\times 1.5\cos 60{}^\circ }{3\sin 60{}^\circ }=\frac{196\times \cos 60{}^\circ }{2\sin 60{}^\circ }=\frac{196}{2\tan 60{}^\circ }=56.6\text{ N (to 3 sf)} \\	+	& S=\frac{196\times 1\textrm{.}5\cos 60{}^\circ }{3\sin 60{}^\circ }=\frac{196\times \cos 60{}^\circ }{2\sin 60{}^\circ }=\frac{196}{2\tan 60{}^\circ }=56\textrm{.}6\text{ N (to 3 sf)} \\
	\end{align}</math>		\end{align}</math>

Line 167:		Line 167:

	<math>F=S</math>		<math>F=S</math>
-	, the friction force has magnitude 56.6 N.	+	, the friction force has magnitude 56\textrm{.}6 N.

	(b)		(b)
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	& \frac{196}{2\tan 60{}^\circ }\le \mu \times 196 \\		& \frac{196}{2\tan 60{}^\circ }\le \mu \times 196 \\
	& \mu \ge \frac{1}{2\tan 60{}^\circ } \\		& \mu \ge \frac{1}{2\tan 60{}^\circ } \\
-	& \mu \ge 0.289\text{ (to 3sf)} \\	+	& \mu \ge 0\textrm{.}289\text{ (to 3sf)} \\
	\end{align}</math>		\end{align}</math>

Ian at 18:51, 23 September 2009

2009-09-23T18:51:38Z

(Difference between revisions)

Ian at 10:33, 23 September 2009

2009-09-23T10:33:08Z

←Older revision		Revision as of 10:33, 23 September 2009
Line 8:		Line 8:


-	. Moments and Equilibrium	+	14. Moments and Equilibrium

	Key Points		Key Points
Line 18:		Line 18:


-	Example 14.1	+
		+	'''[[Example 14.1]]'''
		+
	A uniform beam has length 8 m and mass 60 kg. It is suspended by two ropes, as shown in the diagram below.		A uniform beam has length 8 m and mass 60 kg. It is suspended by two ropes, as shown in the diagram below.

Line 27:		Line 29:
	Find the tension in each rope.		Find the tension in each rope.

-	Solution	+	'''Solution'''
		+
	The diagram shows the forces acting on the beam.		The diagram shows the forces acting on the beam.

Line 60:		Line 63:
	<math>352\textrm{.}8+235\textrm{.}2=588</math>		<math>352\textrm{.}8+235\textrm{.}2=588</math>

-	Example 14.2	+
		+	'''[[Example 14.2]]'''

	A beam, of mass 50 kg and length 5 m, rests on two supports as shown in the diagram. Find the magnitude of the reaction force exerted by each support.		A beam, of mass 50 kg and length 5 m, rests on two supports as shown in the diagram. Find the magnitude of the reaction force exerted by each support.
Line 70:		Line 74:
	Find the maximum mass that could be placed at either end of the beam if it is to remain in equilibrium.		Find the maximum mass that could be placed at either end of the beam if it is to remain in equilibrium.

-	Solution	+	'''Solution'''
		+
	The diagram shows the forces acting on the beam.		The diagram shows the forces acting on the beam.

Line 149:		Line 154:


-	Example 14.3	+
-	A ladder, of length 3 m and mass 20 kg, leans against a smooth, vertical wall so that the angle between the horizontal ground and the ladder is ~~60~~.	+	'''[[Example 14.3]]'''
		+
		+	A ladder, of length 3 m and mass 20 kg, leans against a smooth, vertical wall so that the angle between the horizontal ground and the ladder is 60<math>{}^\circ </math>.
	(a) Find the magnitude of the friction and normal reaction forces that act on the ladder, if it is in equilibrium.		(a) Find the magnitude of the friction and normal reaction forces that act on the ladder, if it is in equilibrium.
	(b) Find the minimum value of the coefficient of friction between the ladder and the ground.		(b) Find the minimum value of the coefficient of friction between the ladder and the ground.

-	Solution	+	'''Solution'''
		+
	The diagram shows the forces acting on the ladder.		The diagram shows the forces acting on the ladder.

Ian at 10:25, 23 September 2009

2009-09-23T10:25:33Z

←Older revision		Revision as of 10:25, 23 September 2009
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	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
	\|}		\|}
		+
		+
		+	. Moments and Equilibrium
		+
		+	Key Points
		+
		+	In equilibrium:
		+	• The resultant force on a body must be zero.
		+	and
		+	• The resultant moment on a body must be zero.
		+
		+
		+	Example 14.1
		+	A uniform beam has length 8 m and mass 60 kg. It is suspended by two ropes, as shown in the diagram below.
		+
		+
		+
		+
		+
		+	Find the tension in each rope.
		+
		+	Solution
		+	The diagram shows the forces acting on the beam.
		+
		+
		+
		+
		+
		+
		+
		+
		+	Take moments about the point where T1 acts to give:
		+
		+
		+	<math>\begin{align}
		+	& 5\times {{T}_{2}}=3\times 588 \\ & {{T}_{2}}=\frac{3\times 588}{5}=352\textrm{.}8=353\text{ N (to 3sf)} \\
		+	\end{align}</math>
		+
		+
		+	Take moments about the point where T2 acts to give:
		+
		+
		+	<math>\begin{align}
		+	& 5\times {{T}_{1}}=2\times 588 \\
		+	& {{T}_{1}}=\frac{2\times 588}{5}=235\textrm{.}2=235\text{ N (to 3sf)} \\
		+	\end{align}</math>
		+
		+
		+	Finally for vertical equilibrium we require
		+	<math>{{T}_{1}}+{{T}_{2}}=588</math>
		+	, which can be used to check the tensions. In is the case we have:
		+
		+	<math>352\textrm{.}8+235\textrm{.}2=588</math>
		+
		+	Example 14.2
		+
		+	A beam, of mass 50 kg and length 5 m, rests on two supports as shown in the diagram. Find the magnitude of the reaction force exerted by each support.
		+
		+
		+
		+
		+
		+	Find the maximum mass that could be placed at either end of the beam if it is to remain in equilibrium.
		+
		+	Solution
		+	The diagram shows the forces acting on the beam.
		+
		+
		+
		+
		+
		+
		+
		+
		+	Taking moments about the point where R1 acts gives:
		+
		+
		+	<math>\begin{align}
		+	& 2\times {{R}_{2}}=1\textrm{.}5\times 490 \\
		+	& {{R}_{2}}=\frac{1\textrm{.}5\times 490}{2}=367\textrm{.}5=368\text{ N (to 3sf)} \\
		+	\end{align}</math>
		+
		+
		+	Taking moments about the point where R2 acts gives:
		+
		+
		+	<math>\begin{align}
		+	& 2\times {{R}_{1}}=0\textrm{.}5\times 490 \\
		+	& {{R}_{1}}=\frac{0\textrm{.}5\times 490}{2}=122\textrm{.}5=123\text{ N (to 3sf)} \\
		+	\end{align}</math>
		+
		+
		+	For vertical equilibrium we require
		+	<math>{{R}_{1}}+{{R}_{2}}=490</math>
		+	,
		+
		+	which can be used to check the tensions. In is the case we have:
		+
		+	<math>367\textrm{.}5+122\textrm{.}5=490</math>
		+
		+
		+	First consider the greatest mass that can be placed at the left hand end of the beam. The diagram below shows the extra force that must now be considered. When the maximum possible mass is used,
		+	<math>{{R}_{2}}=0</math>
		+	.
		+
		+
		+
		+
		+
		+
		+
		+
		+	Taking moments about the point where R1 acts gives:
		+
		+
		+	<math>\begin{align}
		+	& 1\times mg=1\textrm{.}5\times 490 \\
		+	& m=\frac{1\textrm{.}5\times 490}{9\textrm{.}8}=75\text{ kg} \\
		+	\end{align}</math>
		+
		+
		+	Similarly for a mass placed at the right hand end of the beam:
		+
		+
		+
		+
		+
		+
		+
		+
		+
		+
		+
		+	<math>\begin{align}
		+	& 2\times mg=0\textrm{.}5\times 490 \\
		+	& m=\frac{0\textrm{.}5\times 490}{2g}=12\textrm{.}5\text{ kg} \\
		+	\end{align}</math>
		+
		+
		+	Hence the greatest mass that can be placed at either end of the beam
		+	is 12\textrm{.}5 kg.
		+
		+
		+	Example 14.3
		+	A ladder, of length 3 m and mass 20 kg, leans against a smooth, vertical wall so that the angle between the horizontal ground and the ladder is 60.
		+	(a) Find the magnitude of the friction and normal reaction forces that act on the ladder, if it is in equilibrium.
		+	(b) Find the minimum value of the coefficient of friction between the ladder and the ground.
		+
		+	Solution
		+	The diagram shows the forces acting on the ladder.
		+
		+
		+
		+
		+
		+
		+
		+
		+
		+	(a)
		+	Considering the horizontal forces gives:
		+
		+
		+	<math>F=S</math>
		+
		+
		+	Considering the vertical forces gives:
		+
		+
		+	<math>R=196</math>
		+
		+
		+	Taking moment about the base of the ladder gives:
		+
		+
		+	<math>\begin{align}
		+	& 196\times 1\textrm{.}5\cos 60{}^\circ =S\times 3\sin 60{}^\circ \\
		+	& S=\frac{196\times 1\textrm{.}5\cos 60{}^\circ }{3\sin 60{}^\circ }=\frac{196\times \cos 60{}^\circ }{2\sin 60{}^\circ }=\frac{196}{2\tan 60{}^\circ }=56\textrm{.}6\text{ N (to 3 sf)} \\
		+	\end{align}</math>
		+
		+
		+	But since
		+	<math>F=S</math>
		+	, the friction force has magnitude 56\textrm{.}6 N.
		+
		+	(b)
		+
		+	Using the friction inequality,
		+	<math>F\le \mu R</math>
		+
		+	gives:
		+
		+
		+	<math>\begin{align}
		+	& \frac{196}{2\tan 60{}^\circ }\le \mu \times 196 \\
		+	& \mu \ge \frac{1}{2\tan 60{}^\circ } \\
		+	& \mu \ge 0\textrm{.}289\text{ (to 3sf)} \\
		+	\end{align}</math>

Ian at 15:51, 20 September 2009

2009-09-20T15:51:06Z

←Older revision		Revision as of 15:51, 20 September 2009
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-	{{Selected tab\|[[14. Equilibrium\|Theory]]}}	+	{{Selected tab\|[[14. Moments and Equilibrium\|Theory]]}}
	{{Not selected tab\|[[14. Exercises\|Exercises]]}}		{{Not selected tab\|[[14. Exercises\|Exercises]]}}
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Ian: New page: NOTOC {| border="0" cellspacing="0" cellpadding="0" height="30" width="100%" | style="border-bottom:1px solid #797979" width="5px" | {{Selected tab|Theory}} ...

2009-09-20T15:49:20Z

New page: __NOTOC__ {| border="0" cellspacing="0" cellpadding="0" height="30" width="100%" | style="border-bottom:1px solid #797979" width="5px" |   {{Selected tab|Theory}} ...

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{{Selected tab|[[14. Equilibrium|Theory]]}}
{{Not selected tab|[[14. Exercises|Exercises]]}}
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|}