9. Newton’s first law - Revision history

Ian at 09:44, 6 March 2011

Ian — Sun, 06 Mar 2011 09:44:48 GMT

←Older revision		Revision as of 09:44, 6 March 2011
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	Resolving parallel to the slope gives:		Resolving parallel to the slope gives:

-	<math>P=735\sin 6{}^\circ =76.8\text{ N}</math>	+	<math>P=735\sin 6{}^\circ =76\textrm{.}8\text{ N}</math>


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	& F+P=588\sin 30{}^\circ \\		& F+P=588\sin 30{}^\circ \\
	& P=588\sin 30{}^\circ -F \\		& P=588\sin 30{}^\circ -F \\
-	& =588\sin 30{}^\circ -58.8\cos 30{}^\circ \\	+	& =588\sin 30{}^\circ -58\textrm{.}8\cos 30{}^\circ \\
	& =243\text{ N}		& =243\text{ N}
	\end{align}</math>		\end{align}</math>

Ian at 09:48, 18 May 2010

Ian — Tue, 18 May 2010 09:48:39 GMT

←Older revision		Revision as of 09:48, 18 May 2010
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	Note that the reverse is also true, that if a particle moves with constant velocity or remains at rest the resultant force on the particle must be zero.		Note that the reverse is also true, that if a particle moves with constant velocity or remains at rest the resultant force on the particle must be zero.

		+	Friction exists where rough surfaces move relative to one another and satisfies the equation
		+
		+	<math>F = \mu R</math>
		+
		+	Here <math>F</math> is the friction force.
		+
		+	The coefficient of friction is <math>\mu</math>.
		+
		+	The normal reaction force is <math>R</math>.

	'''[[Example 9.1]]'''		'''[[Example 9.1]]'''

Ian at 15:45, 16 May 2010

Ian — Sun, 16 May 2010 15:45:06 GMT

←Older revision		Revision as of 15:45, 16 May 2010
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	Newton's First Law		Newton's First Law

-	If the resultant force on a particle is zero, then it will move with constant velocity or remain at rest. ~~Conversely~~, if a particle moves with constant velocity or remains at rest the resultant force on the particle must be zero.	+	If the resultant force on a particle is zero, then it will move with constant velocity or remain at rest.
		+
		+	Note that the reverse is also true, that if a particle moves with constant velocity or remains at rest the resultant force on the particle must be zero.

Ian at 15:37, 16 May 2010

Ian — Sun, 16 May 2010 15:37:08 GMT

←Older revision		Revision as of 15:37, 16 May 2010
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	Newton's First Law		Newton's First Law

-	If the resultant force on a particle is zero, then it will move with constant velocity or remain at rest.	+	If the resultant force on a particle is zero, then it will move with constant velocity or remain at rest. Conversely, if a particle moves with constant velocity or remains at rest the resultant force on the particle must be zero.

Ian at 16:32, 15 February 2010

Ian — Mon, 15 Feb 2010 16:32:14 GMT

←Older revision		Revision as of 16:32, 15 February 2010
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-	9. Newton’s First Law

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

	Newton's First Law		Newton's First Law

	If the resultant force on a particle is zero, then it will move with constant velocity or remain at rest.		If the resultant force on a particle is zero, then it will move with constant velocity or remain at rest.
-


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	A helicopter of mass 880 kg is rising vertically at a constant rate. Find the magnitude of the lift force acting on the helicopter. How would your answer change if the helicopter was descending at a constant rate?		A helicopter of mass 880 kg is rising vertically at a constant rate. Find the magnitude of the lift force acting on the helicopter. How would your answer change if the helicopter was descending at a constant rate?
-

	'''Solution'''		'''Solution'''

	As the helicopter is rising at a constant rate the forces on it must be in equilibrium.		As the helicopter is rising at a constant rate the forces on it must be in equilibrium.
-
-
-

	[[Image:TF9.1.GIF]]		[[Image:TF9.1.GIF]]
-
-
-

	The lift force is 8624 N.		The lift force is 8624 N.

	If the helicopter is descending at a constant rate the lift force will be 8624 N.		If the helicopter is descending at a constant rate the lift force will be 8624 N.
		+

	'''[[Example 9.2]]'''		'''[[Example 9.2]]'''

	A cyclist, of mass 75 kg, freewheels down a slope inclined at to the horizontal at a constant speed. Find the magnitude of the resistance force acting on the cyclist.		A cyclist, of mass 75 kg, freewheels down a slope inclined at to the horizontal at a constant speed. Find the magnitude of the resistance force acting on the cyclist.
-

	'''Solution'''		'''Solution'''
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	Resolving parallel to the slope gives:		Resolving parallel to the slope gives:

-
	<math>P=735\sin 6{}^\circ =76.8\text{ N}</math>		<math>P=735\sin 6{}^\circ =76.8\text{ N}</math>

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	A skier, of mass 60 kg, skis down a slope inclined at to the horizontal at a constant speed. If the coefficient of friction between her skis and the slope is 0.1, find the magnitude of the air resistance force acting on her.		A skier, of mass 60 kg, skis down a slope inclined at to the horizontal at a constant speed. If the coefficient of friction between her skis and the slope is 0.1, find the magnitude of the air resistance force acting on her.
-

	'''Solution'''		'''Solution'''
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	forces acting, where the resistance		forces acting, where the resistance
	force has magnitude .		force has magnitude .
-

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

	<math>R=588\cos 30{}^\circ </math>		<math>R=588\cos 30{}^\circ </math>
-

	Then using		Then using
	<math>F=\mu R</math>		<math>F=\mu R</math>
	, because the skier is sliding gives:		, because the skier is sliding gives:
-		+
-		+
	<math></math>		<math></math>
-

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

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

Ian at 08:09, 2 September 2009

Ian — Wed, 02 Sep 2009 08:09:58 GMT

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	\| style="border-bottom:1px solid #797979" width="5px" \|		\| style="border-bottom:1px solid #797979" width="5px" \|
-	{{Selected tab\|[[2. ~~Introduction to force and gravity~~\|Theory]]}}	+	{{Selected tab\|[[9. Newton’s first law\|Theory]]}}
-	{{Not selected tab\|[[2. Exercises\|Exercises]]}}	+	{{Not selected tab\|[[9. Exercises\|Exercises]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
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Ian at 07:56, 2 September 2009

Ian — Wed, 02 Sep 2009 07:56:05 GMT

←Older revision		Revision as of 07:56, 2 September 2009
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		+	__NOTOC__
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		+	\| style="border-bottom:1px solid #797979" width="5px" \|
		+	{{Selected tab\|[[2. Introduction to force and gravity\|Theory]]}}
		+	{{Not selected tab\|[[2. Exercises\|Exercises]]}}
		+	\| style="border-bottom:1px solid #797979" width="100%"\|
		+	\|}
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	9. Newton’s First Law		9. Newton’s First Law

Ian at 10:28, 26 August 2009

Ian — Wed, 26 Aug 2009 10:28:37 GMT

←Older revision		Revision as of 10:28, 26 August 2009
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	The skier is modelled		The skier is modelled
	as a particle.		as a particle.
		+
		+	[[Image:TF9.3.GIF]]

	The diagram shows the		The diagram shows the

Ian at 10:19, 26 August 2009

Ian — Wed, 26 Aug 2009 10:19:36 GMT

←Older revision		Revision as of 10:19, 26 August 2009
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-		+	[[Image:TF9.1.GIF]]


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	The cyclist and cycle is modelled		The cyclist and cycle is modelled
	as a particle.		as a particle.
		+
		+	[[Image:TF9.2.GIF]]

	The diagram shows the		The diagram shows the

Ian: New page: 9. Newton’s First Law Key Points Newton's First Law If the resultant force on a particle is zero, then it will move with constant velocity or remain at rest. '''Example 9.1''' ...

Ian — Wed, 26 Aug 2009 10:03:47 GMT

New page: 9. Newton’s First Law Key Points Newton's First Law If the resultant force on a particle is zero, then it will move with constant velocity or remain at rest. '''Example 9.1''' ...

New page

9. Newton’s First Law

Key Points

Newton's First Law

If the resultant force on a particle is zero, then it will move with constant velocity or remain at rest.

'''[[Example 9.1]]'''

A helicopter of mass 880 kg is rising vertically at a constant rate. Find the magnitude of the lift force acting on the helicopter. How would your answer change if the helicopter was descending at a constant rate?

'''Solution'''

As the helicopter is rising at a constant rate the forces on it must be in equilibrium.

The lift force is 8624 N.

If the helicopter is descending at a constant rate the lift force will be 8624 N.

'''[[Example 9.2]]'''

A cyclist, of mass 75 kg, freewheels down a slope inclined at to the horizontal at a constant speed. Find the magnitude of the resistance force acting on the cyclist.

'''Solution'''

The cyclist and cycle is modelled
as a particle.

The diagram shows the
forces acting, where the resistance
force has magnitude P.

Resolving parallel to the slope gives:

<math>P=735\sin 6{}^\circ =76.8\text{ N}</math>

'''[[Example 9.3]]'''

A skier, of mass 60 kg, skis down a slope inclined at to the horizontal at a constant speed. If the coefficient of friction between her skis and the slope is 0.1, find the magnitude of the air resistance force acting on her.

'''Solution'''

The skier is modelled
as a particle.

The diagram shows the
forces acting, where the resistance
force has magnitude .

Resolving perpendicular to the slope gives:

<math>R=588\cos 30{}^\circ </math>

Then using
<math>F=\mu R</math>
, because the skier is sliding gives:

<math></math>

Resolving parallel to the slope gives:

<math>\begin{align}
& F+P=588\sin 30{}^\circ \\
& P=588\sin 30{}^\circ -F \\
& =588\sin 30{}^\circ -58.8\cos 30{}^\circ \\
& =243\text{ N}
\end{align}</math>

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-	{{Selected tab\|[[2. ~~Introduction to force and gravity~~\|Theory]]}}	+	{{Selected tab\|[[9. Newton’s first law\|Theory]]}}
-	{{Not selected tab\|[[2. Exercises\|Exercises]]}}	+	{{Not selected tab\|[[9. Exercises\|Exercises]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
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