Introduction to forces - Revision history

Ian at 14:15, 20 March 2009

2009-03-20T14:15:08Z

←Older revision		Revision as of 14:15, 20 March 2009
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-	<math>F=\frac{Gm_{1}m_{2}}{d^{2}}</math>	+	{\| width="500px" cellspacing="10px"
		+	\| <math>F=\frac{Gm_{1}m_{2}}{d^{2}}</math>


-
-	<math>G=6.67\times 10^{-11}\text{ kg}^{\text{-1}}\text{m}^{\text{3}}\text{s}^{\text{-2}}</math>

		+	<math>G=6.67\times 10^{-11}\text{ kg}^{\text{-1}}\text{m}^{\text{3}}\text{s}^{\text{-2}}</math>
		+	\| valign="top"\|[[Image:Gravitation.gif]]
		+	\|}

-	[[Image:Gravitation.gif]]

	'''Gravity on Earth'''		'''Gravity on Earth'''

Ian at 10:28, 16 March 2009

2009-03-16T10:28:38Z

←Older revision		Revision as of 10:28, 16 March 2009
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	<math>G=6.67\times 10^{-11}\text{ kg}^{\text{-1}}\text{m}^{\text{3}}\text{s}^{\text{-2}}</math>		<math>G=6.67\times 10^{-11}\text{ kg}^{\text{-1}}\text{m}^{\text{3}}\text{s}^{\text{-2}}</math>

		+
		+	[[Image:Gravitation.gif]]

	'''Gravity on Earth'''		'''Gravity on Earth'''

	The force of gravity is often called the weight.		The force of gravity is often called the weight.
		+


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	& g=9.8\text{ ms}^{\text{-2}} \\		& g=9.8\text{ ms}^{\text{-2}} \\
	\end{align}</math>		\end{align}</math>
		+


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	<math>R_{1}+R_{2}+R_{3}+R_{4}=215600</math>		<math>R_{1}+R_{2}+R_{3}+R_{4}=215600</math>
		+
		+
		+
		+	'''[[Example 2.3]]'''
		+
		+	A box of mass 30 kg is at rest on a table.
		+
		+	(a) Calculate the weight of the box.
		+
		+	(b) State the magnitude of the upward force that the table exerts on the box.
		+
		+	'''Solution'''
		+
		+	(a)
		+
		+	<math>\begin{align}
		+	& W=30\times 9.8 \\
		+	& =294\text{ N}
		+	\end{align}</math>
		+
		+	(b)
		+	An upward force of 294 N must act for the box to remain in equilibrium.
		+
		+
		+
		+	'''[[Example 2.4]]'''
		+
		+	A satellite, of mass 400 kg, is at a height of 12 km above the surface of the earth. Find the magnitude of the gravitational attraction on the satellite.
		+	Data:
		+	<math>G=6.67\times {{10}^{-11}}</math>
		+	kg-1m3s-2
		+	Radius of earth
		+
		+	<math>=6.37\times {{10}^{6}}</math>
		+	m
		+	Mass of earth
		+
		+	<math>=5.98\times {{10}^{24}}</math>
		+	kg
		+
		+	'''Solution'''
		+
		+
		+	<math>\begin{align}
		+	& \frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}=\frac{6.67\times {{10}^{-11}}\times 400\times 5.98\times {{10}^{24}}}{{{\left( 6.37\times {{10}^{6}}+12000 \right)}^{2}}} \\
		+	& =3917\text{ N}
		+	\end{align}</math>
		+
		+
		+	'''
		+	[[Example 2.5]]'''
		+
		+
		+	The sun has mass
		+	<math>1.99\times {{10}^{30}}</math>
		+	kg. The mean distance of the earth from the sun is approximately
		+	<math>1.5\times {{10}^{11}}</math>
		+	m.
		+
		+	(a) Calculate the force that the sun exerts on the earth.
		+	(b) State the force that the earth exerts on the sun.
		+	(c) Explain why this force varies.
		+
		+	'''Solution'''
		+
		+	(a)
		+
		+	<math>\begin{align}
		+	& \frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}=\frac{6.67\times {{10}^{-11}}\times 1.99\times {{10}^{30}}\times 5.98\times {{10}^{24}}}{{{\left( 1.5\times {{10}^{11}} \right)}^{2}}} \\
		+	& =3.53\times {{10}^{22}}\text{ N}
		+	\end{align}</math>
		+
		+	(b)
		+	3.53 × 1022 N
		+
		+	(c) The distance between the earth and the sun varies.

Ian at 15:03, 13 March 2009

2009-03-13T15:03:20Z

←Older revision		Revision as of 15:03, 13 March 2009
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	The diagram shows the lorry and its weight.		The diagram shows the lorry and its weight.

		+	[[Image:truck.gif\|\|center]]

Ian at 14:59, 13 March 2009

2009-03-13T14:59:04Z

←Older revision		Revision as of 14:59, 13 March 2009
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	<math>\text{5}.\text{98}\times \text{1}0^{\text{24}}\text{ }</math>		<math>\text{5}.\text{98}\times \text{1}0^{\text{24}}\text{ }</math>
	kg		kg
		+
		+
		+	'''[[Example 2.1]]'''
		+
		+	Describe whether or not the forces acting on the following objects are in equilibrium:
		+
		+	(a) A passenger in a train that travels at a constant speed.
		+
		+	(b) A hot air balloon rising at a constant rate.
		+
		+	(c) A stone dropped into a very deep well full of water.
		+
		+	'''Solution'''
		+
		+	(a) Yes, if it is travelling in a straight line.
		+
		+	(b) Yes, if it is travelling in a straight line.
		+
		+	(c) Yes, if it reaches a terminal velocity, so that it is travelling in a straight line at a constant speed.
		+
		+
		+	'''[[Example 2.2]]'''
		+
		+	Find the magnitude of the force of gravity (weight) acting on a lorry of mass 22 tonnes.
		+
		+	'''Solution'''
		+
		+	This is calculated using the fact that the weight is given by mg.
		+
		+
		+	<math>\begin{align}
		+	& mg=22000\times 9.8 \\
		+	& =215600\text{ N}
		+	\end{align}</math>
		+
		+
		+	The diagram shows the lorry and its weight.
		+
		+
		+
		+	Note that reaction forces also act upwards on each wheel.
		+
		+
		+	<math>R_{1}+R_{2}+R_{3}+R_{4}=215600</math>

Ian: New page: ==Newton's First Law== A particle will move with a constant velocity or remain at rest if the resultant force on the particle is zero. '''Equilibrium''' If the resultant force on a par...

2009-03-13T14:32:46Z

New page: ==Newton's First Law== A particle will move with a constant velocity or remain at rest if the resultant force on the particle is zero. '''Equilibrium''' If the resultant force on a par...

New page

==Newton's First Law==

A particle will move with a constant velocity or remain at rest if the resultant force on the particle is zero.

'''Equilibrium'''

If the resultant force on a particle is zero, then the forces acting on the particle are said to be in equilibrium.

'''The Universal Law of Gravitation'''

<math>F=\frac{Gm_{1}m_{2}}{d^{2}}</math>

<math>G=6.67\times 10^{-11}\text{ kg}^{\text{-1}}\text{m}^{\text{3}}\text{s}^{\text{-2}}</math>

'''Gravity on Earth'''

The force of gravity is often called the weight.

<math>\begin{align}
& F=mg \\
& g=9.8\text{ ms}^{\text{-2}} \\
\end{align}</math>

'''Data'''

Radius of Earth is
<math>\text{6}.\text{37}\times \text{1}0^{\text{6}}\text{ }</math>
metres

Mass of Earth is
<math>\text{5}.\text{98}\times \text{1}0^{\text{24}}\text{ }</math>
kg