5. Forces and equilibrium - Revision history

Tek: Added video tab

Tek — Thu, 07 Apr 2011 12:43:13 GMT

Added video tab

←Older revision		Revision as of 12:43, 7 April 2011
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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]]}}
		+	{{Not selected tab\|[[5. Video\|Video]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
	\|}		\|}

Ian at 11:53, 27 March 2011

Ian — Sun, 27 Mar 2011 11:53:30 GMT

←Older revision		Revision as of 11:53, 27 March 2011
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	<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\textrm{.}~~840~~\text{ (to 3 sf)} \\	+	& \mu =\frac{294\sin 40{}^\circ }{294\cos 40{}^\circ }=\tan 40{}^\circ =0\textrm{.}839\text{ (to 3 sf)} \\
	\end{align}</math>		\end{align}</math>

Ian at 16:10, 1 April 2010

Ian — Thu, 01 Apr 2010 16:10:38 GMT

←Older revision		Revision as of 16:10, 1 April 2010
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	The forces acting on a particle at rest or moving with constant velocity are in equilibrium.		The forces acting on a particle at rest or moving with constant velocity are in equilibrium.
		+
		+	In practical terms this means, '''for forces in equilibrium the sum of the components of the forces in any direction must be zero'''.
		+
		+

Ian at 11:17, 8 March 2010

Ian — Mon, 08 Mar 2010 11:17:50 GMT

←Older revision		Revision as of 11:17, 8 March 2010
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	== '''Key Points''' ==		== '''Key Points''' ==
		+
		+	The sum of the forces acting on a particle is called the resultant.

	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.
		+
		+	The forces acting on a particle at rest or moving with constant velocity are in equilibrium.

Ian at 11:56, 11 February 2010

Ian — Thu, 11 Feb 2010 11:56:00 GMT

←Older revision		Revision as of 11:56, 11 February 2010
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	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
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	<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>
Line 103:		Line 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'''
Line 120:		Line 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
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-	Note – Angle of Friction	+
		+	'''[[Note – Angle of Friction]]'''

	[[Image:AngleFriction.gif]]		[[Image:AngleFriction.gif]]
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	Resolving horizontally:		Resolving horizontally:

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

	Resolving vertically:		Resolving vertically:
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	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 at 10:51, 9 February 2010

Ian — Tue, 09 Feb 2010 10:51:31 GMT

←Older revision		Revision as of 10:51, 9 February 2010
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	<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>

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	<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>

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	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>

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	<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 at 10:39, 9 February 2010

Ian — Tue, 09 Feb 2010 10:39:19 GMT

←Older revision		Revision as of 10:39, 9 February 2010
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	'''[[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
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	[[Image:ex5.3fig1.gif]]		[[Image:ex5.3fig1.gif]]
-

	Model the lorry as a particle.		Model the lorry as a particle.
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	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>

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

	[[Image:ex5.4fig1.gif]]		[[Image:ex5.4fig1.gif]]
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	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
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	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]]
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	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
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	gives:		gives:

-
	<math>\begin{align}		<math>\begin{align}
	& mg\sin \alpha \le \mu mg\cos \alpha \\		& mg\sin \alpha \le \mu mg\cos \alpha \\
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	& \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
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	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 at 10:32, 9 February 2010

Ian — Tue, 09 Feb 2010 10:32:49 GMT

←Older revision		Revision as of 10:32, 9 February 2010
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	'''[[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 at 10:04, 9 February 2010

Ian — Tue, 09 Feb 2010 10:04:40 GMT

←Older revision		Revision as of 10:04, 9 February 2010
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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\|[[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%"\|
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-	~~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.
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	'''[[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.
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	'''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 \\
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	& {{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]]'''
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	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'''
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	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 at 15:49, 17 September 2009

Ian — Thu, 17 Sep 2009 15:49:43 GMT

←Older revision		Revision as of 15:49, 17 September 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\|[[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%"\|

←Older revision		Revision as of 12:43, 7 April 2011
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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]]}}
		+	{{Not selected tab\|[[5. Video\|Video]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
	\|}		\|}

←Older revision		Revision as of 15:49, 17 September 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\|[[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%"\|