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- 11:22, 23 March 2010 (hist) (diff) Solution 4.5c (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=8\ \text{N}</math> and <math>H=-2\ \text{N}</math>....) (top)
- 11:02, 23 March 2010 (hist) (diff) Solution 4.5b (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=-3\ \text{N}</math> and <math>H=4\ \text{N}</math>....)
- 10:42, 23 March 2010 (hist) (diff) Solution 4.5a (top)
- 10:35, 23 March 2010 (hist) (diff) Solution 4.5a
- 09:58, 23 March 2010 (hist) (diff) Solution 4.5a (New page: Image:teori4.gif Using <math>{{F}^{2}}={{H}^{2}}+{{V}^{2}}</math> Substituting the given values <math>\begin{align} & {{F}^{2}}={{4}^{2}}+{ {3}^{2}} =25 \\ & F=5 \text{ N} \\ \end...)
- 17:37, 22 March 2010 (hist) (diff) Answer 4.4 (top)
- 17:36, 22 March 2010 (hist) (diff) 4. Exercises (top)
- 17:32, 22 March 2010 (hist) (diff) Solution 4.4b (New page: Image:teori4.gif <math>\tan \alpha =\frac{V}{H}</math> Here <math>V=-10\ \text{N}</math> and <math>H=9\ \text{N}</math>. Thus <math>\tan \alpha =\frac{-10}{9}=-1\textrm{.}1</mat...) (top)
- 17:23, 22 March 2010 (hist) (diff) Solution 4.4a (New page: Using <math>{{F}^{2}}={{H}^{2}}+{{V}^{2}}</math> Substituting the given values <math>\begin{align} & {{F}^{2}}={{9}^{2}}+{ {{\left( -10 \right)}^{2}} }=181 \\ & F=13\textrm{.}5\ \text{N...) (top)
- 17:14, 22 March 2010 (hist) (diff) Solution 4.3b (New page: Image:teori4.gif <math>\tan \alpha =\frac{V}{H}</math> Here <math>V=8\ \text{N}</math> and <math>H=6\ \text{N}</math>. Thus <math>\tan \alpha =\frac{8}{6}=1\textrm{.}33</math> g...) (top)
- 17:01, 22 March 2010 (hist) (diff) Solution 4.3a (New page: Using <math>{{F}^{2}}={{H}^{2}}+{{V}^{2}}</math> Substituting the given values <math>\begin{align} & {{F}^{2}}={{6}^{2}}+{{8}^{2}}=100 \\ & F=10\ \text{N} \\ \end{align}</math>) (top)
- 16:56, 22 March 2010 (hist) (diff) 4. Forces and vectors
- 14:47, 22 March 2010 (hist) (diff) 4. Forces and vectors
- 14:41, 22 March 2010 (hist) (diff) Image:Teori4.gif (uploaded a new version of "Image:Teori4.gif") (top)
- 14:28, 22 March 2010 (hist) (diff) Image:Teori4.gif
- 11:33, 22 March 2010 (hist) (diff) Solution 4.1 (New page: Using <math>\begin{align} & \mathbf{F}=F\cos \alpha \mathbf{i}+F\sin \alpha \mathbf{j} \end{align}</math> where <math>\alpha</math> is <math>30{}^\circ</math> and <math>F=80\ \text{N...) (top)
- 11:11, 22 March 2010 (hist) (diff) Answer 4.1 (top)
- 10:34, 22 March 2010 (hist) (diff) 3. Exercises
- 10:33, 22 March 2010 (hist) (diff) Solution 3.13bc (New page: Vertically the forces on the tank are in equilibrium <math>\begin{align} & \uparrow \ R-mg=0 \\ & R=800\times 9\textrm{.}8=7840\ \text{N} \\ \end{align}</math> The maximum friction is ...) (top)
- 10:21, 22 March 2010 (hist) (diff) Image:3.13a.gif (top)
- 10:21, 22 March 2010 (hist) (diff) Answer 3.13a (top)
- 09:18, 22 March 2010 (hist) (diff) Answer 3.13a (New page: Image:3.13a.jpg)
- 16:23, 21 March 2010 (hist) (diff) Solution (New page: The normal force <math>R</math> is equal to the weight <math>mg</math> of the sledge. Thus <math>R=mg=117 \textrm{.}6\ \text{N}</math>. The maximum frictional force is <math>F=\mu R</ma...) (top)
- 16:05, 21 March 2010 (hist) (diff) 3. Exercises
- 12:48, 21 March 2010 (hist) (diff) Solution 3.11c (top)
- 12:27, 21 March 2010 (hist) (diff) Image:3.11c.gif (top)
- 12:27, 21 March 2010 (hist) (diff) Solution 3.11c
- 12:23, 21 March 2010 (hist) (diff) Solution 3.11c (New page: Image:11c.gif)
- 12:05, 21 March 2010 (hist) (diff) Solution 3.11b (top)
- 12:03, 21 March 2010 (hist) (diff) Image:3.11b.gif (top)
- 12:02, 21 March 2010 (hist) (diff) Solution 3.11b (New page: Image:3.11b.gif)
- 11:46, 21 March 2010 (hist) (diff) Solution 3.11a (top)
- 11:42, 21 March 2010 (hist) (diff) Image:3.11a.gif (top)
- 11:42, 21 March 2010 (hist) (diff) Solution 3.11a (New page: Image:3.11a.gif)
- 16:19, 20 March 2010 (hist) (diff) Solution 3.10b (top)
- 16:15, 20 March 2010 (hist) (diff) Solution 3.10d (New page: <math>\begin{align} & F\le \mu R \\ & 78 \textrm{.}4\le \mu 117 \textrm{.}6 \\ & 2\le 3\mu \\ & \mu \ge \frac{2}{3} \\ \end{align}</math>) (top)
- 16:08, 20 March 2010 (hist) (diff) Solution 3.10c (top)
- 16:02, 20 March 2010 (hist) (diff) Image:3.10c.gif (top)
- 16:02, 20 March 2010 (hist) (diff) Solution 3.10c (New page: Image:3.10c.gif)
- 15:49, 20 March 2010 (hist) (diff) Solution 3.10b
- 15:42, 20 March 2010 (hist) (diff) Image:3.10b.gif (top)
- 15:41, 20 March 2010 (hist) (diff) Solution 3.10b (New page: Image:3.10b.gif)
- 15:31, 20 March 2010 (hist) (diff) Solution 3.10a (top)
- 15:29, 20 March 2010 (hist) (diff) Image:3.10a.gif (top)
- 15:28, 20 March 2010 (hist) (diff) Solution 3.10a (New page: Image:3.10a.gif)
- 16:08, 19 March 2010 (hist) (diff) Solution 3.9c (New page: The vertical forces cancel out. Horisontally the maximum frictional force is not enough to keep the box in equilibrium and the horisontal resultant force is <math>\to P-F=1000-784=216\ ...) (top)
- 15:57, 19 March 2010 (hist) (diff) Image:3.9b.gif (top)
- 15:57, 19 March 2010 (hist) (diff) Solution 3.9b (New page: Image:3.9b.gif Friction will "try" and prevent motion that is achieve equilibrium. <math>\begin{align} & \text{For}\ \text{equilibrium}\ P-F=0 \\ & \text{that}\ \text{is}\ F=760\ \...)
- 15:39, 19 March 2010 (hist) (diff) Solution 3.9a
- 15:35, 19 March 2010 (hist) (diff) Image:3.9a.gif (top)
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