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- 15:35, 19 March 2010 (hist) (diff) Solution 3.9a (New page: Image:3.9a.gif)
- 15:18, 19 March 2010 (hist) (diff) 3. Other types of force
- 13:26, 19 March 2010 (hist) (diff) Solution 3.8 (New page: See diagram. <math>R1</math> is the reaction force holding up the combined weights of the two lighter boxes. Equilibrium of the two boxes gives <math>\begin{align} & \uparrow \ R1-(5+10...) (top)
- 13:04, 19 March 2010 (hist) (diff) Image:3.8.gif (uploaded a new version of "Image:3.8.gif") (top)
- 12:51, 19 March 2010 (hist) (diff) Image:3.8.gif
- 12:51, 19 March 2010 (hist) (diff) Diagram3.8 (New page: Image:3.8.gif) (top)
- 09:46, 19 March 2010 (hist) (diff) 3. Other types of force
- 17:42, 18 March 2010 (hist) (diff) Answer 3.6a (New page: Image:3.6a.gif) (top)
- 17:42, 18 March 2010 (hist) (diff) Answer 3.6c (New page: Image:3.6c.gif)
- 17:41, 18 March 2010 (hist) (diff) Image:3.6c.gif (top)
- 17:40, 18 March 2010 (hist) (diff) Answer 3.6b (New page: Image:3.6b.gif)
- 17:40, 18 March 2010 (hist) (diff) Image:3.6b.gif (top)
- 17:39, 18 March 2010 (hist) (diff) Image:3.6a.gif (top)
- 18:53, 17 March 2010 (hist) (diff) Image:3.5.gif (uploaded a new version of "Image:3.5.gif") (top)
- 18:49, 17 March 2010 (hist) (diff) Image:3.5.gif (uploaded a new version of "Image:3.5.gif")
- 18:42, 17 March 2010 (hist) (diff) Answer 3.5c (top)
- 18:40, 17 March 2010 (hist) (diff) Solution 3.5b (top)
- 18:39, 17 March 2010 (hist) (diff) Solution 3.5c (top)
- 18:38, 17 March 2010 (hist) (diff) Solution 3.5c (New page: From the diagram <math>\begin{align} & \uparrow \ N2-N1-1.8g=0\ \text{gives} \\ & N2=11.76+1.8\times 9.8=29.4\ \text{N} \\ \end{align}</math>)
- 18:34, 17 March 2010 (hist) (diff) Solution 3.5b (New page: From the diagram <math>\begin{align} & \uparrow N1-1.2g=0\ \text{gives} \\ & N1=1.2\times 9.8=11.76\ \text{N} \\ \end{align}</math>)
- 18:08, 17 March 2010 (hist) (diff) Image:3.5.gif (uploaded a new version of "Image:3.5.gif")
- 18:05, 17 March 2010 (hist) (diff) Answer 3.5a (top)
- 18:00, 17 March 2010 (hist) (diff) Image:3.5.gif
- 12:55, 17 March 2010 (hist) (diff) Answer 3.4 (New page: Image:2.4.gif)
- 12:55, 17 March 2010 (hist) (diff) Image:2.4.gif (top)
- 10:17, 17 March 2010 (hist) (diff) Image:2.1.gif (uploaded a new version of "Image:2.1.gif") (top)
- 10:14, 17 March 2010 (hist) (diff) Image:2.1.gif (uploaded a new version of "Image:2.1.gif")
- 10:09, 17 March 2010 (hist) (diff) Image:2.1.gif (uploaded a new version of "Image:2.1.gif")
- 09:59, 17 March 2010 (hist) (diff) Solution 3.3 (New page: Image:2.1.gif First consider the lower mass. If <math>T=T1</math> is the tension in the lower string then <math>mg=7\times 9\textrm{.}8=68\textrm{....)
- 09:25, 17 March 2010 (hist) (diff) Solution 3.2 (New page: Image:2.1.gif If <math>T</math> is the tension in the spring then <math>\uparrow </math> <math>T-mg=0</math> as the forces on the sphere are in equilibrium. <math>T=mg=2\textrm{.}...) (top)
- 09:23, 17 March 2010 (hist) (diff) Solution 3.1 (top)
- 09:16, 17 March 2010 (hist) (diff) 3. Exercises
- 09:44, 16 March 2010 (hist) (diff) Solution 3.1
- 09:43, 16 March 2010 (hist) (diff) Solution 3.1
- 17:22, 15 March 2010 (hist) (diff) Solution 3.1 (New page: Image:2.1.gif If <math>T</math> is the tension in the cable then <math>T-mg=0</math> as the forces on the sphere are in equilibrium. <math>T=mg=20\times 9.8=196\ \text{N}</math...)
- 17:21, 15 March 2010 (hist) (diff) Image:2.1.gif
- 17:19, 13 March 2010 (hist) (diff) Solution 2.10
- 16:58, 13 March 2010 (hist) (diff) Solution 2.10 (New page: <math>F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}</math> where <math>F</math> is the force on a particle on the surface of Mars, <math>{{m}_{1}}</math> is the mass of Mars, <math>{{m}_{2}...)
- 17:41, 12 March 2010 (hist) (diff) m Solution 2.6 (top)
- 17:38, 12 March 2010 (hist) (diff) Solution 2.9b (New page: In part a) the weight of the man at the top of the mountain is found to be 785 N. At sea level <math>g=9\textrm{.}81\text{ ms}^{\text{-2}}</math>, and his weight is <math>mg</math> g...)
- 17:22, 12 March 2010 (hist) (diff) Solution 2.9a (New page: Use the theory of gravitation equation <math>F=\frac{Gm_{1}m_{2}}{{d}^{\ 2}}</math> where, <math>G=6\textrm{.}67\times 10^{-11}\text{ kg}^{\text{-1}}\text{m}^{\text{3}}\text{s}^{\text{-...) (top)
- 15:02, 12 March 2010 (hist) (diff) Solution 2.8 (New page: Consider a particle on the surface of the planet. <math>F=\frac{G{{m}_{1}}{{m}_{2}}}{{{d}^{2}}}</math> where <math>F</math> is the gravitational force on a particle on the surface of t...)
- 14:28, 12 March 2010 (hist) (diff) Solution 2.7 (top)
- 14:23, 12 March 2010 (hist) (diff) Solution 2.6
- 12:47, 12 March 2010 (hist) (diff) Image:Fig2gif.gif (uploaded a new version of "Image:Fig2gif.gif": Reverted to version as of 12:43, 12 March 2010) (top)
- 12:46, 12 March 2010 (hist) (diff) Image:Fig2gif.gif (uploaded a new version of "Image:Fig2gif.gif")
- 12:43, 12 March 2010 (hist) (diff) Image:Fig2gif.gif (uploaded a new version of "Image:Fig2gif.gif": Reverted to version as of 10:54, 12 March 2010)
- 12:41, 12 March 2010 (hist) (diff) Image:Fig2gif.gif (uploaded a new version of "Image:Fig2gif.gif")
- 11:09, 12 March 2010 (hist) (diff) 2. Introduction to force and gravity
- 10:56, 12 March 2010 (hist) (diff) 2. Introduction to force and gravity
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