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From Mechanics
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- 15:42, 2 April 2010 (hist) (diff) Image:5.1aNew.gif (top)
- 15:39, 2 April 2010 (hist) (diff) Image:5.1a.gif (uploaded a new version of "Image:5.1a.gif") (top)
- 15:39, 2 April 2010 (hist) (diff) Image:5.1a.gif (uploaded a new version of "Image:5.1a.gif")
- 15:38, 2 April 2010 (hist) (diff) Image:5.1a.gif (uploaded a new version of "Image:5.1a.gif")
- 15:35, 2 April 2010 (hist) (diff) Image:5.1a.gif (uploaded a new version of "Image:5.1a.gif")
- 15:33, 2 April 2010 (hist) (diff) Solution 5.1d (top)
- 15:32, 2 April 2010 (hist) (diff) Solution 5.1b (top)
- 15:28, 2 April 2010 (hist) (diff) Image:5.1a.gif (uploaded a new version of "Image:5.1a.gif")
- 17:50, 1 April 2010 (hist) (diff) Solution 5.1d (New page: Here <math>F\le \mu N</math> <math>\mu</math> is the coefficient of friction. <math>\begin{align} & 67\textrm{.}0<\mu 184 \\ & \\ & \mu >\frac{67\textrm{.}0}{184} \\ & \\ & \mu >0\...)
- 17:39, 1 April 2010 (hist) (diff) Solution 5.1c (New page: The forces on the particle are in equilibrium as the particle is at rest. The sum of the forces up the plane (resolving up the plane) <math>\begin{align} & F-mg\sin {{20}^{\circ }}=0 \\...) (top)
- 17:23, 1 April 2010 (hist) (diff) Image:5.1a.gif (uploaded a new version of "Image:5.1a.gif")
- 17:16, 1 April 2010 (hist) (diff) Image:5.1a.gif (uploaded a new version of "Image:5.1a.gif")
- 17:15, 1 April 2010 (hist) (diff) Image:5.1a.gif (uploaded a new version of "Image:5.1a.gif")
- 17:12, 1 April 2010 (hist) (diff) Image:5.1a.gif (uploaded a new version of "Image:5.1a.gif")
- 17:12, 1 April 2010 (hist) (diff) Image:5.1a.gif (uploaded a new version of "Image:5.1a.gif")
- 17:10, 1 April 2010 (hist) (diff) Image:5.1a.gif (uploaded a new version of "Image:5.1a.gif")
- 16:20, 1 April 2010 (hist) (diff) Solution 5.1b (New page: The forces on the particle are in equilibrium as the particle is at rest. The sum of the forces in the normal direction (resolving in the normal direction) <math>\begin{align} & N-mg\co...)
- 16:10, 1 April 2010 (hist) (diff) 5. Forces and equilibrium
- 16:00, 1 April 2010 (hist) (diff) Answer 5.1a (New page: Image:5.1a.gif) (top)
- 15:59, 1 April 2010 (hist) (diff) Image:5.1a.gif
- 10:36, 1 April 2010 (hist) (diff) Solution 4.9b (top)
- 10:34, 1 April 2010 (hist) (diff) Solution 4.9b (New page: In part a) the resultant <math>\mathbf{R}</math> is seen to be <math>\begin{align} & \mathbf{R} =11\textrm{.}1\mathbf{i}+3\textrm{.}93\mathbf{j} \ \text{N}\\ \end{align}</math> Also ac...)
- 10:20, 1 April 2010 (hist) (diff) Solution 4.9a (top)
- 09:54, 1 April 2010 (hist) (diff) Solution 4.9a
- 09:42, 1 April 2010 (hist) (diff) Solution 4.9a
- 09:26, 1 April 2010 (hist) (diff) Solution 4.9a
- 09:20, 1 April 2010 (hist) (diff) Solution 4.9a
- 11:51, 31 March 2010 (hist) (diff) Solution 4.9a
- 11:33, 31 March 2010 (hist) (diff) Solution 4.9a (New page: <math>\begin{align} & \mathbf{F}=F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j} \end{align}</math> <math>F1=100\ \text{N}</math> and <math>\alpha 1=30{}^\circ </math> this gives <ma...)
- 16:01, 24 March 2010 (hist) (diff) Solution 4.8c (New page: <math>{\mathbf{F}_{1}}</math> = (<math>9\mathbf{i}+12\mathbf{j}</math>) N <math>{\mathbf{F}_{2}}</math> = (<math>-4\mathbf{i}+\mathbf{j}</math>) N Their sum is the resultant force <mat...) (top)
- 15:54, 24 March 2010 (hist) (diff) Solution 4.8b (New page: Image:4.8b.gif <math>\tan \alpha =\frac{12}{9}=1\textrm{.}33</math> giving <math>\alpha =53\textrm{.}1{}^\circ</math>) (top)
- 15:50, 24 March 2010 (hist) (diff) Image:4.8b.gif (top)
- 15:43, 24 March 2010 (hist) (diff) Solution 4.8a (New page: <math>{\mathbf{F}_{1}}</math> = (<math>9\mathbf{i}+12\mathbf{j}</math>) N <math>\sqrt{{{9}^{2}}+{{12}^{2}}}=15\ \text{N}</math>) (top)
- 15:37, 24 March 2010 (hist) (diff) Solution 4.7b
- 15:26, 24 March 2010 (hist) (diff) Solution 4.7b
- 15:09, 24 March 2010 (hist) (diff) Solution 4.7b (New page: <math>\begin{align} & \mathbf{F}1= & =86\textrm{.}6\mathbf{i}+50\mathbf{j}\ \text{N}\\ \end{align}</math> <math>\begin{align} & \mathbf{F}2= & =-30\textrm{.}8\mathbf{i}+84\textrm{.}6\mat...)
- 15:04, 24 March 2010 (hist) (diff) Solution 4.7a (top)
- 14:15, 24 March 2010 (hist) (diff) Solution 4.7a
- 13:37, 24 March 2010 (hist) (diff) Solution 4.7a (New page: <math>\begin{align} & \mathbf{F}=F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j} \end{align}</math> <math>F1=100\ \text{N}</math> and <math>\alpha 1=30{}^\circ </math> this gives <...)
- 16:02, 23 March 2010 (hist) (diff) Solution 4.6c
- 16:01, 23 March 2010 (hist) (diff) Image:4.6c.gif (top)
- 15:58, 23 March 2010 (hist) (diff) Solution 4.6c (New page: Resultant force <math>=80\textrm{.}9\mathbf{i}-11\textrm{.}3\mathbf{j}\ \text{N}</math> Thus <math>\tan \alpha =\frac{-11\textrm{.}3}{80.9}=-0\textrm{.}14</math> giving <math>\alpha =-...)
- 15:48, 23 March 2010 (hist) (diff) Solution 4.6b (top)
- 15:27, 23 March 2010 (hist) (diff) Solution 4.6b (New page: <math>\begin{align} & \mathbf{F}1=37\textrm{.}6\mathbf{i}+13\textrm{.}7\mathbf{j}\ \text{N}\\ \end{align}</math> <math>\begin{align} & \mathbf{F}2=43\textrm{.}3\mathbf{i}+25\mathbf{j} \...)
- 15:22, 23 March 2010 (hist) (diff) Solution 4.6a
- 15:21, 23 March 2010 (hist) (diff) Solution 4.6a
- 15:16, 23 March 2010 (hist) (diff) Solution 4.6a (New page: <math>\begin{align} & \mathbf{F}=F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j} \end{align}</math>)
- 13:32, 23 March 2010 (hist) (diff) 4. Forces and vectors (top)
- 13:20, 23 March 2010 (hist) (diff) Image:Teori4a.gif (top)
- 12:33, 23 March 2010 (hist) (diff) Solution 4.5d (New page: Image:teori4.gif Using <math>{{F}^{2}}={{H}^{2}}+{{V}^{2}}</math> and <math>\tan \alpha =\frac{V}{H} </math> Here <math>V=-7\ \text{N}</math> and <math>H=-5\ \text{N}</math>...)
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