9. Newtons första lag - Versionshistorik

Tek: flyttade 9. Newton’s first law till 9. Newtons första lag

2010-03-01T14:05:01Z

flyttade 9. Newton’s first law till 9. Newtons första lag

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Versionen från 1 mars 2010 kl. 14.05

Ian den 15 februari 2010 kl. 16.32

2010-02-15T16:32:14Z

← Äldre version		Versionen från 15 februari 2010 kl. 16.32
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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?
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	'''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.
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	[[Image:TF9.1.GIF]]		[[Image:TF9.1.GIF]]
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	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.
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	'''Solution'''		'''Solution'''
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	Resolving parallel to the slope gives:		Resolving parallel to the slope gives:

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	<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.
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	'''Solution'''		'''Solution'''
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	forces acting, where the resistance		forces acting, where the resistance
	force has magnitude .		force has magnitude .
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	Resolving perpendicular to the slope gives:		Resolving perpendicular to the slope gives:
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	<math>R=588\cos 30{}^\circ </math>		<math>R=588\cos 30{}^\circ </math>
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	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:
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-		+
	<math></math>		<math></math>
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	Resolving parallel to the slope gives:		Resolving parallel to the slope gives:
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	<math>\begin{align}		<math>\begin{align}

Ian den 2 september 2009 kl. 08.09

2009-09-02T08:09:58Z

← Äldre version		Versionen från 2 september 2009 kl. 08.09
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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]]}}
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Ian den 2 september 2009 kl. 07.56

2009-09-02T07:56:05Z

← Äldre version		Versionen från 2 september 2009 kl. 07.56
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		+	__NOTOC__
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		+	{{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 den 26 augusti 2009 kl. 10.28

2009-08-26T10:28:37Z

← Äldre version		Versionen från 26 augusti 2009 kl. 10.28
Rad 61:		Rad 61:
	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 den 26 augusti 2009 kl. 10.19

2009-08-26T10:19:36Z

← Äldre version		Versionen från 26 augusti 2009 kl. 10.19
Rad 21:		Rad 21:


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


Rad 39:		Rad 39:
	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''' ...

2009-08-26T10:03:47Z

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''' ...

Ny sida

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>

← Äldre version		Versionen från 2 september 2009 kl. 08.09
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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]]}}
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