10. Newtons andra lag - Versionshistorik

Tek: flyttade 10. Newton’s second law till 10. Newtons andra lag

2010-03-01T14:05:36Z

flyttade 10. Newton’s second law till 10. Newtons andra lag

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Versionen från 1 mars 2010 kl. 14.05

Ian den 16 februari 2010 kl. 18.53

2010-02-16T18:53:40Z

← Äldre version		Versionen från 16 februari 2010 kl. 18.53
Rad 83:		Rad 83:
	First find the resultant force acting on the car.		First find the resultant force acting on the car.

-	<math>~~\text{Resultant Force }~~=\text{ }3000-R</math>	+	Resultant Force <math>=\text{ }3000-R</math>

	Then apply Newton’s Second Law,		Then apply Newton’s Second Law,
Rad 110:		Rad 110:
	First find the resultant force acting on the car.		First find the resultant force acting on the car.

-	<math>~~\text{Resultant Force }~~=\text{ }5000-2000=3000\text{ N}</math>	+	Resultant Force <math>=\text{ }5000-2000=3000\text{ N}</math>

	Then apply Newton’s Second Law,		Then apply Newton’s Second Law,
Rad 142:		Rad 142:
	The resultant force will be directed down the slope, and is found by resolving the forces parallel to the slope.		The resultant force will be directed down the slope, and is found by resolving the forces parallel to the slope.

-	<math>~~\text{Resultant Force }~~=11760\sin 3{}^\circ -400</math>	+	Resultant Force <math>=11760\sin 3{}^\circ -400</math>

	Then apply Newton’s Second Law,		Then apply Newton’s Second Law,
Rad 186:		Rad 186:
	Now resolving horizontally, the resultant force can be found.		Now resolving horizontally, the resultant force can be found.

		+	Resultant Force
		+
	<math>\begin{align}		<math>\begin{align}
-	& ~~\text{Resultant Force }~~=20\cos 30{}^\circ -F \\	+	& =20\cos 30{}^\circ -F \\
	& =20\cos 30{}^\circ -(5\textrm{.}88-4\sin 30{}^\circ ) \\		& =20\cos 30{}^\circ -(5\textrm{.}88-4\sin 30{}^\circ ) \\
	& =20\cos 30{}^\circ -5\textrm{.}88+4\sin 30{}^\circ		& =20\cos 30{}^\circ -5\textrm{.}88+4\sin 30{}^\circ

Ian den 16 februari 2010 kl. 18.47

2010-02-16T18:47:44Z

← Äldre version		Versionen från 16 februari 2010 kl. 18.47
Rad 56:		Rad 56:

	a) If		a) If
-	<math>a=0.2</math>	+	<math>a=0\textrm{.}2</math>
	, then		, then
-	<math>T=2940+300\times 0.2=3000\text{ N}</math>	+	<math>T=2940+300\times 0\textrm{.}2=3000\text{ N}</math>
	.		.

	b) If		b) If
-	<math>a=-0.05</math>	+	<math>a=-0\textrm{.}05</math>
	(since the lift is accelerating downwards then:		(since the lift is accelerating downwards then:

-	<math>T=2940+300\times (-0.05)=2925\text{ N}</math>	+	<math>T=2940+300\times (-0\textrm{.}05)=2925\text{ N}</math>
	.		.

Rad 90:		Rad 90:

	<math>\begin{align}		<math>\begin{align}
-	& 3000-R=1200\times 1.8 \\	+	& 3000-R=1200\times 1\textrm{.}8 \\
	& R=3000-2160 \\		& R=3000-2160 \\
	& =840\text{ N}		& =840\text{ N}
Rad 118:		Rad 118:
	<math>\begin{align}		<math>\begin{align}
	& 3000=2500a \\		& 3000=2500a \\
-	& a=\frac{3000}{2500}=1.2\text{ m}{{\text{s}}^{\text{-2}}}	+	& a=\frac{3000}{2500}=1\textrm{.}2\text{ m}{{\text{s}}^{\text{-2}}}
	\end{align}</math>		\end{align}</math>

Rad 150:		Rad 150:
	<math>\begin{align}		<math>\begin{align}
	& 11760\sin 3{}^\circ -400=1200a \\		& 11760\sin 3{}^\circ -400=1200a \\
-	& a=\frac{11760\sin 3{}^\circ -400}{1200}=0.180\text{ m}{{\text{s}}^{\text{-2}}}\text{ (to 3 sf)} \\	+	& a=\frac{11760\sin 3{}^\circ -400}{1200}=0\textrm{.}180\text{ m}{{\text{s}}^{\text{-2}}}\text{ (to 3 sf)} \\
	\end{align}</math>		\end{align}</math>

Rad 171:		Rad 171:

	<math>\begin{align}		<math>\begin{align}
-	& R+20\sin 30{}^\circ =29.4 \\	+	& R+20\sin 30{}^\circ =29\textrm{.}4 \\
-	& R=29.4-20\sin 30{}^\circ \\	+	& R=29\textrm{.}4-20\sin 30{}^\circ \\
	\end{align}</math>		\end{align}</math>

Rad 180:		Rad 180:

	<math>\begin{align}		<math>\begin{align}
-	& F=0.2\times \left( 29.4-20\sin 30{}^\circ \right) \\	+	& F=0.2\times \left( 29\textrm{.}4-20\sin 30{}^\circ \right) \\
-	& =5.88-4\sin 30{}^\circ	+	& =5\textrm{.}88-4\sin 30{}^\circ
	\end{align}</math>		\end{align}</math>

Rad 188:		Rad 188:
	<math>\begin{align}		<math>\begin{align}
	& \text{Resultant Force }=20\cos 30{}^\circ -F \\		& \text{Resultant Force }=20\cos 30{}^\circ -F \\
-	& =20\cos 30{}^\circ -(5.88-4\sin 30{}^\circ ) \\	+	& =20\cos 30{}^\circ -(5\textrm{.}88-4\sin 30{}^\circ ) \\
-	& =20\cos 30{}^\circ -5.88+4\sin 30{}^\circ	+	& =20\cos 30{}^\circ -5\textrm{.}88+4\sin 30{}^\circ
	\end{align}</math>		\end{align}</math>

Rad 197:		Rad 197:

	<math>\begin{align}		<math>\begin{align}
-	& 20\cos 30{}^\circ -5.88+4\sin 30{}^\circ =3a \\	+	& 20\cos 30{}^\circ -5\textrm{.}88+4\sin 30{}^\circ =3a \\
-	& a=\frac{20\cos 30{}^\circ -5.88+4\sin 30{}^\circ }{3}=4.48\text{ m}{{\text{s}}^{\text{-2}}}\text{ (to 3 sf)} \\	+	& a=\frac{20\cos 30{}^\circ -5.88+4\sin 30{}^\circ }{3}=4\textrm{.}48\text{ m}{{\text{s}}^{\text{-2}}}\text{ (to 3 sf)} \\
	\end{align}</math>		\end{align}</math>

Ian den 16 februari 2010 kl. 17.46

2010-02-16T17:46:51Z

← Äldre version		Versionen från 16 februari 2010 kl. 17.46
Rad 46:		Rad 46:
	Consider the resultant force on the particle, taking the upward direction as positive.		Consider the resultant force on the particle, taking the upward direction as positive.

-	<math>~~\text{Resultant Force }~~=T-2940</math>	+	Resultant Force <math>=T-2940</math>

	Applying Newton’s second law gives:		Applying Newton’s second law gives:
Rad 156:		Rad 156:
	'''[[Example 10.5]]'''		'''[[Example 10.5]]'''

-	A block, of mass 3 kg, is pulled across a rough horizontal plane, by a string inclined at ~~30~~ to the horizontal. The tension in the string is 20 N. The coefficient of friction between the particle and the plane is 0.2. Find the acceleration of the particle.	+	A block, of mass 3 kg, is pulled across a rough horizontal plane, by a string inclined at <math>30
		+	^\circ</math> to the horizontal. The tension in the string is 20 N. The coefficient of friction between the particle and the plane is 0.2. Find the acceleration of the particle.

	[[Image:TF10.5.GIF]]		[[Image:TF10.5.GIF]]

Ian den 16 februari 2010 kl. 17.42

2010-02-16T17:42:50Z

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Ian den 2 september 2009 kl. 08.11

2009-09-02T08:11:47Z

← Äldre version		Versionen från 2 september 2009 kl. 08.11
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\|[[2. ~~Introduction to force and gravity~~\|Theory]]}}	+	{{Selected tab\|[[10. Newton’s second law\|Theory]]}}
-	{{Not selected tab\|[[2. Exercises\|Exercises]]}}	+	{{Not selected tab\|[[10. Exercises\|Exercises]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
	\|}		\|}

Ian den 2 september 2009 kl. 07.56

2009-09-02T07:56:33Z

← Äldre version		Versionen från 2 september 2009 kl. 07.56
Rad 1:		Rad 1:
		+	__NOTOC__
		+	{\| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
		+	\| 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%"\|
		+	\|}
		+
		+
	10. Newton’s Second Law		10. Newton’s Second Law

Ian den 26 augusti 2009 kl. 11.29

2009-08-26T11:29:57Z

← Äldre version		Versionen från 26 augusti 2009 kl. 11.29
Rad 28:		Rad 28:
	The lift and its passengers have		The lift and its passengers have
	been modelled as a single particle.		been modelled as a single particle.
		+
		+	[[Image:TF10.1.GIF]]

	The diagram shows the		The diagram shows the
Rad 78:		Rad 80:


-		+	[[Image:TF10.2.GIF]]


Rad 112:		Rad 114:


-		+	[[Image:TF10.3.GIF]]

	First find the resultant force acting on the car.		First find the resultant force acting on the car.
Rad 143:		Rad 145:
	acting on the van as it rolls		acting on the van as it rolls
	down the slope.		down the slope.
		+
		+	[[Image:TF10.4.GIF]]

	The forces perpendicular to the slope will be in equilibrium.		The forces perpendicular to the slope will be in equilibrium.
Rad 168:		Rad 172:


-		+	[[Image:TF10.5.GIF]]

Ian den 26 augusti 2009 kl. 11.11

2009-08-26T11:11:23Z

← Äldre version		Versionen från 26 augusti 2009 kl. 11.11
Rad 14:		Rad 14:


-	Example 10.1	+	'''[[Example 10.1]]'''
		+
	A lift and its passengers have a total mass of 300 kg. Find the tension in the lift cable if;		A lift and its passengers have a total mass of 300 kg. Find the tension in the lift cable if;
	(a) it accelerates upwards at 0.2		(a) it accelerates upwards at 0.2
Rad 23:		Rad 24:
	.		.

-	Solution	+	'''Solution'''
		+
	The lift and its passengers have		The lift and its passengers have
	been modelled as a single particle.		been modelled as a single particle.
Rad 64:		Rad 66:


-	Example 10.2	+	'''[[Example 10.2]]'''
		+
	A car accelerates at 1.8		A car accelerates at 1.8
	<math>\text{m}{{\text{s}}^{-2}}</math>		<math>\text{m}{{\text{s}}^{-2}}</math>
	along a straight horizontal road. The mass of the car is 1200 kg. A forward force of magnitude 3000 N acts on the car. Find the magnitude of the resistance force that also acts on the car.		along a straight horizontal road. The mass of the car is 1200 kg. A forward force of magnitude 3000 N acts on the car. Find the magnitude of the resistance force that also acts on the car.

-	Solution	+	'''Solution'''
		+
	The diagram shows the horizontal forces that are acting on the car, which has been modelled as a particle. The acceleration has also been included.		The diagram shows the horizontal forces that are acting on the car, which has been modelled as a particle. The acceleration has also been included.

Rad 98:		Rad 102:
	The resistance force has magnitude 840 N.		The resistance force has magnitude 840 N.

-	Example 10.3	+	'''[[Example 10.3]]'''
		+
	A van has mass 2500 kg. A forward force of 5000 N acts on the van and a resistance force of 2000 N also acts. Find the acceleration of the van on a horizontal surface.		A van has mass 2500 kg. A forward force of 5000 N acts on the van and a resistance force of 2000 N also acts. Find the acceleration of the van on a horizontal surface.

-	Solution	+
		+	'''Solution'''
		+
	The diagram shows the horizontal forces that are acting on the van, which has been modelled as a particle.		The diagram shows the horizontal forces that are acting on the van, which has been modelled as a particle.

Rad 123:		Rad 130:
	\end{align}</math>		\end{align}</math>

-	Example 10.4	+	'''[[Example 10.4]]'''
		+
	A van, of mass 1200 kg, rolls down a slope, inclined at		A van, of mass 1200 kg, rolls down a slope, inclined at
	<math>3{}^\circ </math>		<math>3{}^\circ </math>
	to the horizontal and experiences a resistance force of magnitude 400N. Find the acceleration of the car.		to the horizontal and experiences a resistance force of magnitude 400N. Find the acceleration of the car.

-	Solution	+	'''Solution'''
		+
	Model the van as a particle.		Model the van as a particle.

Rad 154:		Rad 163:


-	Example 10.5	+	'''[[Example 10.5]]'''
		+
	A block, of mass 3 kg, is pulled across a rough horizontal plane, by a string inclined at 30 to the horizontal. The tension in the string is 20 N. The coefficient of friction between the particle and the plane is 0.2. Find the acceleration of the particle.		A block, of mass 3 kg, is pulled across a rough horizontal plane, by a string inclined at 30 to the horizontal. The tension in the string is 20 N. The coefficient of friction between the particle and the plane is 0.2. Find the acceleration of the particle.

Rad 162:		Rad 172:


-	Solution	+
		+	'''Solution'''
		+
	Model the crate as a particle.		Model the crate as a particle.

Ian: New page: 10. Newton’s Second Law Key Points Newton's Second Law If the resultant force on a particle is not zero, then it will accelerate. $F=ma$ $\mathbf{F}=m\mathbf{a}...$

2009-08-26T11:05:59Z

New page: 10. Newton’s Second Law Key Points Newton's Second Law If the resultant force on a particle is not zero, then it will accelerate. <math>F=ma</math> <math>\mathbf{F}=m\mathbf{a}...

Ny sida

10. Newton’s Second Law

Key Points

Newton's Second Law

If the resultant force on a particle is not zero, then it will accelerate.

<math>F=ma</math>

<math>\mathbf{F}=m\mathbf{a}</math>

Example 10.1
A lift and its passengers have a total mass of 300 kg. Find the tension in the lift cable if;
(a) it accelerates upwards at 0.2
<math>\text{m}{{\text{s}}^{-2}}</math>
,
(b) it accelerates downwards at 0.05
<math>\text{m}{{\text{s}}^{-2}}</math>
.

Solution
The lift and its passengers have
been modelled as a single particle.

The diagram shows the
forces acting on the lift.

Assume that the lift is accelerating
upwards as indicated in the diagram.

Consider the resultant force on the particle, taking the upward direction as positive.

<math>\text{Resultant Force }=T-2940</math>

Applying Newton’s second law gives:

<math>\begin{align}
& 300a=T-2940 \\
& T=2940+300a \\
\end{align}</math>

(a) If
<math>a=0.2</math>
, then
<math>T=2940+300\times 0.2=3000\text{ N}</math>
.

(b) If
<math>a=-0.05</math>
(since the lift is accelerating downwards then:

<math>T=2940+300\times (-0.05)=2925\text{ N}</math>
.

Example 10.2
A car accelerates at 1.8
<math>\text{m}{{\text{s}}^{-2}}</math>
along a straight horizontal road. The mass of the car is 1200 kg. A forward force of magnitude 3000 N acts on the car. Find the magnitude of the resistance force that also acts on the car.

Solution
The diagram shows the horizontal forces that are acting on the car, which has been modelled as a particle. The acceleration has also been included.

First find the resultant force acting on the car.

<math>\text{Resultant Force }=\text{ }3000-R</math>

Then apply Newton’s Second Law,
<math>F=ma</math>
, to give:

<math>\begin{align}
& 3000-R=1200\times 1.8 \\
& R=3000-2160 \\
& =840\text{ N}
\end{align}</math>

The resistance force has magnitude 840 N.

Example 10.3
A van has mass 2500 kg. A forward force of 5000 N acts on the van and a resistance force of 2000 N also acts. Find the acceleration of the van on a horizontal surface.

Solution
The diagram shows the horizontal forces that are acting on the van, which has been modelled as a particle.

First find the resultant force acting on the car.

<math>\text{Resultant Force }=\text{ }5000-2000=3000\text{ N}</math>

Then apply Newton’s Second Law,
<math>F=ma</math>
, to give:

<math>\begin{align}
& 3000=2500a \\
& a=\frac{3000}{2500}=1.2\text{ m}{{\text{s}}^{\text{-2}}}
\end{align}</math>

Example 10.4
A van, of mass 1200 kg, rolls down a slope, inclined at
<math>3{}^\circ </math>
to the horizontal and experiences a resistance force of magnitude 400N. Find the acceleration of the car.

Solution
Model the van as a particle.

The diagram shows the forces
acting on the van as it rolls
down the slope.

The forces perpendicular to the slope will be in equilibrium.

The resultant force will be directed down the slope, and is found by resolving the forces parallel to the slope.

<math>\text{Resultant Force }=11760\sin 3{}^\circ -400</math>

Then apply Newton’s Second Law,
<math>F=ma</math>
, to give:

<math>\begin{align}
& 11760\sin 3{}^\circ -400=1200a \\
& a=\frac{11760\sin 3{}^\circ -400}{1200}=0.180\text{ m}{{\text{s}}^{\text{-2}}}\text{ (to 3 sf)} \\
\end{align}</math>

Example 10.5
A block, of mass 3 kg, is pulled across a rough horizontal plane, by a string inclined at 30 to the horizontal. The tension in the string is 20 N. The coefficient of friction between the particle and the plane is 0.2. Find the acceleration of the particle.

Solution
Model the crate as a particle.

The diagram shows the forces
acting on the crate.

The vertical components of the forces will be in equilibrium.

<math>\begin{align}
& R+20\sin 30{}^\circ =29.4 \\
& R=29.4-20\sin 30{}^\circ \\
\end{align}</math>

As we know R, we can calculate F using
<math>F=\mu R</math>
. This gives:

<math>\begin{align}
& F=0.2\times \left( 29.4-20\sin 30{}^\circ \right) \\
& =5.88-4\sin 30{}^\circ
\end{align}</math>

Now resolving horizontally, the resultant force can be found.

<math>\begin{align}
& \text{Resultant Force }=20\cos 30{}^\circ -F \\
& =20\cos 30{}^\circ -(5.88-4\sin 30{}^\circ ) \\
& =20\cos 30{}^\circ -5.88+4\sin 30{}^\circ
\end{align}</math>

Then apply Newton’s Second Law,
<math>F=ma</math>
, to give:

<math>\begin{align}
& 20\cos 30{}^\circ -5.88+4\sin 30{}^\circ =3a \\
& a=\frac{20\cos 30{}^\circ -5.88+4\sin 30{}^\circ }{3}=4.48\text{ m}{{\text{s}}^{\text{-2}}}\text{ (to 3 sf)} \\
\end{align}</math>

← Äldre version		Versionen från 2 september 2009 kl. 08.11
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\|[[2. ~~Introduction to force and gravity~~\|Theory]]}}	+	{{Selected tab\|[[10. Newton’s second law\|Theory]]}}
-	{{Not selected tab\|[[2. Exercises\|Exercises]]}}	+	{{Not selected tab\|[[10. Exercises\|Exercises]]}}
	\| style="border-bottom:1px solid #797979" width="100%"\|		\| style="border-bottom:1px solid #797979" width="100%"\|
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