5. Krafter och jämvikt - Versionshistorik

Tek: flyttade 5. Forces and equilibrium till 5. Krafter och jämvikt

2010-03-01T13:59:16Z

flyttade 5. Forces and equilibrium till 5. Krafter och jämvikt

← Äldre version

Versionen från 1 mars 2010 kl. 13.59

Ian den 11 februari 2010 kl. 11.56

2010-02-11T11:56:00Z

← Äldre version		Versionen från 11 februari 2010 kl. 11.56
Rad 32:		Rad 32:
	Resolving vertically gives;		Resolving vertically gives;

-	<math>{{T}_{1}}\sin 60{}^\circ +{{T}_{2}}\sin 60{}^\circ =2940</math>	+	<math>{{T}_{1}}\sin 60{}^\circ +{{T}_{2}}\sin 60{}^\circ =2940 \ \text{N}</math>

	Now solving the equations by substituting		Now solving the equations by substituting
Rad 67:		Rad 67:

	<math>\begin{align}		<math>\begin{align}
-	& {{T}_{1}}\sin 60{}^\circ =58\textrm{.}8 \\	+	& {{T}_{1}}\sin 60{}^\circ =58\textrm{.}8 \ \text{N}\\
	& {{T}_{1}}=\frac{58\textrm{.}8}{\sin 60{}^\circ }=67\textrm{.}9\text{ N (to 3 sf)} \\		& {{T}_{1}}=\frac{58\textrm{.}8}{\sin 60{}^\circ }=67\textrm{.}9\text{ N (to 3 sf)} \\
	\end{align}</math>		\end{align}</math>
Rad 103:		Rad 103:
	'''[[Example 5.4]]'''		'''[[Example 5.4]]'''

-	A child, of mass ~~30kg~~, slides down a slide at a constant speed. Assume that there is no air resistance acting on the child. The slide makes an angle of <math>{{40}^{\circ }}</math> with the horizontal. Find the magnitude of the friction force on the child and the coefficient of friction.	+	A child, of mass 30 kg, slides down a slide at a constant speed. Assume that there is no air resistance acting on the child. The slide makes an angle of <math>{{40}^{\circ }}</math> with the horizontal. Find the magnitude of the friction force on the child and the coefficient of friction.

	'''Solution'''		'''Solution'''
Rad 120:		Rad 120:
	Resolving perpendicular to the slope gives:		Resolving perpendicular to the slope gives:

-	<math>R=294\cos 40{}^\circ </math>	+	<math>R=294\cos 40{}^\circ \ \text{N}</math>

	As the child is sliding		As the child is sliding
Rad 132:		Rad 132:


-	Note – Angle of Friction	+
		+	'''[[Note – Angle of Friction]]'''

	[[Image:AngleFriction.gif]]		[[Image:AngleFriction.gif]]
Rad 169:		Rad 170:
	Resolving horizontally:		Resolving horizontally:

-	<math>F=T\cos 20{}^\circ </math>	+	<math>F=T\cos 20{}^\circ \ \text{N}</math>

	Resolving vertically:		Resolving vertically:
Rad 177:		Rad 178:
	or		or

-	<math>R=1960-T\sin 20{}^\circ </math>	+	<math>R=1960-T\sin 20{}^\circ \ \text{N}</math>

	As the crate is sliding we can use		As the crate is sliding we can use

Ian den 9 februari 2010 kl. 10.51

2010-02-09T10:51:31Z

← Äldre version		Versionen från 9 februari 2010 kl. 10.51
Rad 67:		Rad 67:

	<math>\begin{align}		<math>\begin{align}
-	& {{T}_{1}}\sin 60{}^\circ =58.8 \\	+	& {{T}_{1}}\sin 60{}^\circ =58\textrm{.}8 \\
-	& {{T}_{1}}=\frac{58.8}{\sin 60{}^\circ }=67.9\text{ N (to 3 sf)} \\	+	& {{T}_{1}}=\frac{58\textrm{.}8}{\sin 60{}^\circ }=67\textrm{.}9\text{ N (to 3 sf)} \\
	\end{align}</math>		\end{align}</math>

Rad 75:		Rad 75:
	<math>\begin{align}		<math>\begin{align}
	& {{T}_{1}}\cos 60{}^\circ ={{T}_{2}} \\		& {{T}_{1}}\cos 60{}^\circ ={{T}_{2}} \\
-	& {{T}_{2}}=\frac{58.8}{\sin 60{}^\circ }\cos 60{}^\circ =33.9\text{ N (to 3sf)} \\	+	& {{T}_{2}}=\frac{58\textrm{.}8}{\sin 60{}^\circ }\cos 60{}^\circ =33\textrm{.}9\text{ N (to 3sf)} \\
	\end{align}</math>		\end{align}</math>

Rad 124:		Rad 124:
	As the child is sliding		As the child is sliding
	<math>F=\mu R</math>		<math>F=\mu R</math>
-	so that we can determine .	+	so that we can determine <math>\mu </math>.

	<math>\begin{align}		<math>\begin{align}
	& 294\sin 40{}^\circ =\mu \times 294\cos 40{}^\circ \\		& 294\sin 40{}^\circ =\mu \times 294\cos 40{}^\circ \\
-	& \mu =\frac{294\sin 40{}^\circ }{294\cos 40{}^\circ }=\tan 40{}^\circ =0.840\text{ (to 3 sf)} \\	+	& \mu =\frac{294\sin 40{}^\circ }{294\cos 40{}^\circ }=\tan 40{}^\circ =0\textrm{.}840\text{ (to 3 sf)} \\
	\end{align}</math>		\end{align}</math>

Rad 182:		Rad 182:
	<math>F=\mu R</math>, which gives:		<math>F=\mu R</math>, which gives:

-	<math>F=0.4R</math>	+	<math>F=0\textrm{.}4R</math>

	Using this equation with the horizontal and vertical equations gives:		Using this equation with the horizontal and vertical equations gives:

	<math>\begin{align}		<math>\begin{align}
-	& T\cos 20{}^\circ =0.4\left( 1960-T\sin 20{}^\circ \right) \\	+	& T\cos 20{}^\circ =0\textrm{.}4\left( 1960-T\sin 20{}^\circ \right) \\
-	& T\cos 20{}^\circ =784-T\times 0.4\sin 20{}^\circ \\	+	& T\cos 20{}^\circ =784-T\times 0\textrm{.}4\sin 20{}^\circ \\
-	& T\cos 20{}^\circ +T\times 0.4\sin 20{}^\circ =784 \\	+	& T\cos 20{}^\circ +T\times 0\textrm{.}4\sin 20{}^\circ =784 \\
-	& T\left( \cos 20{}^\circ +0.4\sin 20{}^\circ \right)=784 \\	+	& T\left( \cos 20{}^\circ +0\textrm{.}4\sin 20{}^\circ \right)=784 \\
-	& T=\frac{784}{\cos 20{}^\circ +0.4\sin 20{}^\circ }=728\text{ N (to 3sf)} \\	+	& T=\frac{784}{\cos 20{}^\circ +0\textrm{.}4\sin 20{}^\circ }=728\text{ N (to 3sf)} \\
	\end{align}</math>		\end{align}</math>

Ian den 9 februari 2010 kl. 10.39

2010-02-09T10:39:19Z

← Äldre version		Versionen från 9 februari 2010 kl. 10.39
Rad 51:		Rad 51:
	'''[[Example 5.2]]'''		'''[[Example 5.2]]'''

		+	[[Image:ex5.2fig1a.gif]]

	A particle of mass 6 kg is suspended by two		A particle of mass 6 kg is suspended by two
Rad 85:		Rad 86:

	[[Image:ex5.3fig1.gif]]		[[Image:ex5.3fig1.gif]]
-

	Model the lorry as a particle.		Model the lorry as a particle.
Rad 94:		Rad 94:
	Resolving perpendicular to the slope gives:		Resolving perpendicular to the slope gives:

-
	<math>R=49000\cos 5{}^\circ =48800\text{ N (to 3 sf)}</math>		<math>R=49000\cos 5{}^\circ =48800\text{ N (to 3 sf)}</math>
-

	Resolving parallel to the slope gives:		Resolving parallel to the slope gives:

-
	<math>P=49000\sin 5{}^\circ =4270\text{ N (to 3sf)}</math>		<math>P=49000\sin 5{}^\circ =4270\text{ N (to 3sf)}</math>

Rad 111:		Rad 108:

	Model the child as a particle.		Model the child as a particle.
-

	[[Image:ex5.4fig1.gif]]		[[Image:ex5.4fig1.gif]]
Rad 120:		Rad 116:
	Resolving parallel to the slope gives.		Resolving parallel to the slope gives.

-
	<math>F=294\sin 40{}^\circ =189\text{ N (to 3sf)}</math>		<math>F=294\sin 40{}^\circ =189\text{ N (to 3sf)}</math>
-

	Resolving perpendicular to the slope gives:		Resolving perpendicular to the slope gives:

-
	<math>R=294\cos 40{}^\circ </math>		<math>R=294\cos 40{}^\circ </math>
-

	As the child is sliding		As the child is sliding
Rad 134:		Rad 126:
	so that we can determine .		so that we can determine .

-
	<math>\begin{align}		<math>\begin{align}
	& 294\sin 40{}^\circ =\mu \times 294\cos 40{}^\circ \\		& 294\sin 40{}^\circ =\mu \times 294\cos 40{}^\circ \\
	& \mu =\frac{294\sin 40{}^\circ }{294\cos 40{}^\circ }=\tan 40{}^\circ =0.840\text{ (to 3 sf)} \\		& \mu =\frac{294\sin 40{}^\circ }{294\cos 40{}^\circ }=\tan 40{}^\circ =0.840\text{ (to 3 sf)} \\
	\end{align}</math>		\end{align}</math>
-
-


	Note – Angle of Friction		Note – Angle of Friction
-

	[[Image:AngleFriction.gif]]		[[Image:AngleFriction.gif]]
Rad 150:		Rad 138:
	If a particle of mass <math>m</math> is at rest on a slope		If a particle of mass <math>m</math> is at rest on a slope
	at an angle <math>\alpha </math> above the horizontal, then :		at an angle <math>\alpha </math> above the horizontal, then :
-

	<math>F=mg\sin \alpha </math>		<math>F=mg\sin \alpha </math>

-
-
	<math>R=mg\cos \alpha </math>		<math>R=mg\cos \alpha </math>
-

	Then using		Then using
Rad 163:		Rad 147:
	gives:		gives:

-
	<math>\begin{align}		<math>\begin{align}
	& mg\sin \alpha \le \mu mg\cos \alpha \\		& mg\sin \alpha \le \mu mg\cos \alpha \\
Rad 169:		Rad 152:
	& \mu \ge \tan \alpha \\		& \mu \ge \tan \alpha \\
	\end{align}</math>		\end{align}</math>
-
-


	'''[[Example 5.5]]'''		'''[[Example 5.5]]'''
-

	[[Image:ex5.5fig1.gif]]		[[Image:ex5.5fig1.gif]]
-

	A crate of mass 200 kg is on a horizontal surface. The coefficient of friction between the crate and the surface is 0.4. The crate is pulled by a rope, as shown in the diagram, so that the crate moves at a constant speed in a straight line. Find the tension in the rope.		A crate of mass 200 kg is on a horizontal surface. The coefficient of friction between the crate and the surface is 0.4. The crate is pulled by a rope, as shown in the diagram, so that the crate moves at a constant speed in a straight line. Find the tension in the rope.
-

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

	[[Image:ex5.5fig2.gif]]		[[Image:ex5.5fig2.gif]]
-

	The diagram shows the forces		The diagram shows the forces
Rad 193:		Rad 168:

	Resolving horizontally:		Resolving horizontally:
-		+
-		+
	<math>F=T\cos 20{}^\circ </math>		<math>F=T\cos 20{}^\circ </math>
-

	Resolving vertically:		Resolving vertically:
-		+
-		+
	<math>R+T\sin 20{}^\circ =1960</math>		<math>R+T\sin 20{}^\circ =1960</math>

	or		or
-		+
	<math>R=1960-T\sin 20{}^\circ </math>		<math>R=1960-T\sin 20{}^\circ </math>
-

	As the crate is sliding we can use		As the crate is sliding we can use
	<math>F=\mu R</math>, which gives:		<math>F=\mu R</math>, which gives:

-
	<math>F=0.4R</math>		<math>F=0.4R</math>
-

	Using this equation with the horizontal and vertical equations gives:		Using this equation with the horizontal and vertical equations gives:

-
	<math>\begin{align}		<math>\begin{align}
	& T\cos 20{}^\circ =0.4\left( 1960-T\sin 20{}^\circ \right) \\		& T\cos 20{}^\circ =0.4\left( 1960-T\sin 20{}^\circ \right) \\

Ian den 9 februari 2010 kl. 10.32

2010-02-09T10:32:49Z

← Äldre version		Versionen från 9 februari 2010 kl. 10.32
Rad 51:		Rad 51:
	'''[[Example 5.2]]'''		'''[[Example 5.2]]'''

-	[[Image:ex5.2fig1.gif]]

	A particle of mass 6 kg is suspended by two		A particle of mass 6 kg is suspended by two

Ian den 9 februari 2010 kl. 10.04

2010-02-09T10:04:40Z

← Äldre version		Versionen från 9 februari 2010 kl. 10.04
Rad 2:		Rad 2:
	{\| 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\|[[5. Forces and ~~Equilibrium~~\|Theory]]}}	+	{{Selected tab\|[[5. Forces and equilibrium\|Theory]]}}
	{{Not selected tab\|[[5. Exercises\|Exercises]]}}		{{Not selected tab\|[[5. Exercises\|Exercises]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
Rad 8:		Rad 8:


-	~~5. Forces and Equilibrium~~	+	== '''Key Points''' ==
-		+
-	Key Points	+

	If the resultant of the forces acting on a particle is zero we say that these forces are in equilibrium.		If the resultant of the forces acting on a particle is zero we say that these forces are in equilibrium.
Rad 16:		Rad 14:

	'''[[Example 5.1]]'''		'''[[Example 5.1]]'''
-
-
-	[[Image:ex5.5fig1.gif]]
-

	The diagram shows an object, of mass 300 kg, that is at rest and is supported by two cables. Find the tension in each cable.		The diagram shows an object, of mass 300 kg, that is at rest and is supported by two cables. Find the tension in each cable.
Rad 26:		Rad 20:

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

	The diagram shows the forces		The diagram shows the forces
	acting on the object.		acting on the object.
-
-
-
-
-
-

	Resolving horizontally or using the horizontal components of the forces:		Resolving horizontally or using the horizontal components of the forces:

-
	<math>\begin{align}		<math>\begin{align}
	& {{T}_{1}}\cos 60{}^\circ ={{T}_{2}}\cos 60{}^\circ \\ & {{T}_{1}}={{T}_{2}}		& {{T}_{1}}\cos 60{}^\circ ={{T}_{2}}\cos 60{}^\circ \\ & {{T}_{1}}={{T}_{2}}
	\end{align}</math>		\end{align}</math>
-

	Resolving vertically gives;		Resolving vertically gives;

-
	<math>{{T}_{1}}\sin 60{}^\circ +{{T}_{2}}\sin 60{}^\circ =2940</math>		<math>{{T}_{1}}\sin 60{}^\circ +{{T}_{2}}\sin 60{}^\circ =2940</math>
-

	Now solving the equations by substituting		Now solving the equations by substituting
	<math>{{T}_{1}}={{T}_{2}}</math>		<math>{{T}_{1}}={{T}_{2}}</math>
		+
	gives:		gives:

-
	<math>\begin{align}		<math>\begin{align}
	& {{T}_{1}}\sin 60{}^\circ +{{T}_{2}}\sin 60{}^\circ =2940 \\		& {{T}_{1}}\sin 60{}^\circ +{{T}_{2}}\sin 60{}^\circ =2940 \\
Rad 63:		Rad 44:
	& {{T}_{1}}=\frac{2940}{2\sin 60{}^\circ }=1700\text{ N (to 3sf)} \\		& {{T}_{1}}=\frac{2940}{2\sin 60{}^\circ }=1700\text{ N (to 3sf)} \\
	\end{align}</math>		\end{align}</math>
-

	And also		And also
-	<math>{{T}_{2}}=1700\text{ N (to 3 sf)}</math>	+	<math>{{T}_{2}}=1700\text{ N (to 3 sf)}</math>.
-	.	+

	'''[[Example 5.2]]'''		'''[[Example 5.2]]'''
Rad 76:		Rad 56:
	strings as shown in the diagram. Note that one		strings as shown in the diagram. Note that one
	string is horizontal. Find the tension in each string.		string is horizontal. Find the tension in each string.
-

	'''Solution'''		'''Solution'''
Rad 87:		Rad 66:
	Resolving vertically:		Resolving vertically:

-
	<math>\begin{align}		<math>\begin{align}
	& {{T}_{1}}\sin 60{}^\circ =58.8 \\		& {{T}_{1}}\sin 60{}^\circ =58.8 \\
	& {{T}_{1}}=\frac{58.8}{\sin 60{}^\circ }=67.9\text{ N (to 3 sf)} \\		& {{T}_{1}}=\frac{58.8}{\sin 60{}^\circ }=67.9\text{ N (to 3 sf)} \\
	\end{align}</math>		\end{align}</math>
-

	Resolving horizontally:		Resolving horizontally:

-
	<math>\begin{align}		<math>\begin{align}
	& {{T}_{1}}\cos 60{}^\circ ={{T}_{2}} \\		& {{T}_{1}}\cos 60{}^\circ ={{T}_{2}} \\
	& {{T}_{2}}=\frac{58.8}{\sin 60{}^\circ }\cos 60{}^\circ =33.9\text{ N (to 3sf)} \\		& {{T}_{2}}=\frac{58.8}{\sin 60{}^\circ }\cos 60{}^\circ =33.9\text{ N (to 3sf)} \\
	\end{align}</math>		\end{align}</math>
-

Ian den 17 september 2009 kl. 15.49

2009-09-17T15:49:43Z

← Äldre version		Versionen från 17 september 2009 kl. 15.49
Rad 2:		Rad 2:
	{\| 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\|[[5. ~~Moments~~\|Theory]]}}	+	{{Selected tab\|[[5. Forces and Equilibrium\|Theory]]}}
	{{Not selected tab\|[[5. Exercises\|Exercises]]}}		{{Not selected tab\|[[5. Exercises\|Exercises]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|

Ian den 17 september 2009 kl. 15.48

2009-09-17T15:48:39Z

← Äldre version		Versionen från 17 september 2009 kl. 15.48
Rad 2:		Rad 2:
	{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"		{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
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-	{{Selected tab\|[[13. Moments\|Theory]]}}	+	{{Selected tab\|[[5. Moments\|Theory]]}}
-	{{Not selected tab\|[[13. Exercises\|Exercises]]}}	+	{{Not selected tab\|[[5. Exercises\|Exercises]]}}
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	\|}		\|}

Ian den 17 september 2009 kl. 15.47

2009-09-17T15:47:31Z

← Äldre version		Versionen från 17 september 2009 kl. 15.47
Rad 2:		Rad 2:
	{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"		{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
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-	{{Selected tab\|[[5. ~~Forces and equilibrium~~\|Theory]]}}	+	{{Selected tab\|[[13. Moments\|Theory]]}}
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	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
	\|}		\|}

Ian den 10 september 2009 kl. 12.52

2009-09-10T12:52:20Z

← Äldre version		Versionen från 10 september 2009 kl. 12.52
Rad 2:		Rad 2:
	{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"		{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
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-	{{Selected tab\|[[5.Forces and equilibrium\|Theory]]}}	+	{{Selected tab\|[[5. Forces and equilibrium\|Theory]]}}
-	{{Not selected tab\|[[5.Exercises\|Exercises]]}}	+	{{Not selected tab\|[[5. Exercises\|Exercises]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
	\|}		\|}

← Äldre version		Versionen från 17 september 2009 kl. 15.49
Rad 2:		Rad 2:
	{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"		{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
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-	{{Selected tab\|[[5. ~~Moments~~\|Theory]]}}	+	{{Selected tab\|[[5. Forces and Equilibrium\|Theory]]}}
	{{Not selected tab\|[[5. Exercises\|Exercises]]}}		{{Not selected tab\|[[5. Exercises\|Exercises]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|

← Äldre version		Versionen från 17 september 2009 kl. 15.48
Rad 2:		Rad 2:
	{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"		{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
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-	{{Selected tab\|[[13. Moments\|Theory]]}}	+	{{Selected tab\|[[5. Moments\|Theory]]}}
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	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
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← Äldre version		Versionen från 17 september 2009 kl. 15.47
Rad 2:		Rad 2:
	{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"		{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
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-	{{Selected tab\|[[5. ~~Forces and equilibrium~~\|Theory]]}}	+	{{Selected tab\|[[13. Moments\|Theory]]}}
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	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
	\|}		\|}

← Äldre version		Versionen från 10 september 2009 kl. 12.52
Rad 2:		Rad 2:
	{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"		{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
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-	{{Selected tab\|[[5.Forces and equilibrium\|Theory]]}}	+	{{Selected tab\|[[5. Forces and equilibrium\|Theory]]}}
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