User contributions
From Mechanics
(Newest | Oldest) View (Newer 50) (Older 50) (20 | 50 | 100 | 250 | 500)
- 17:59, 14 September 2010 (hist) (diff) Answer 15.4c (New page: <math>0\textrm{.}92 \textrm{ Ns}</math>) (top)
- 17:57, 14 September 2010 (hist) (diff) Answer 15.4b (New page: <math>5\textrm{.}24 \text{ m}{{\text{s}}^{-1}}</math>) (top)
- 17:56, 14 September 2010 (hist) (diff) Answer 15.4a (New page: <math>6\textrm{.}26 \text{ m}{{\text{s}}^{-1}}</math>) (top)
- 17:02, 14 September 2010 (hist) (diff) Solution 15.3 (New page: First find the impulse (change in momentum). <math>\begin{align} & \text{Impulse }=mv-mu \\ & 1200\times 15-1200\times 20 \\ & =-6000\text{ Ns} \end{align}</math> Use the formula ...)
- 16:57, 14 September 2010 (hist) (diff) Answer 15.3 (New page: 68000 Ns) (top)
- 16:44, 14 September 2010 (hist) (diff) Solution 15.2 (top)
- 16:44, 14 September 2010 (hist) (diff) Solution 15.2 (New page: <math>\text{100 grams = 0}\text{.1 kg}\ \text{ }\Rightarrow \ \text{ m = 0}\text{.1 }</math> <math>u = -8</math> <math>\text{m}{{\text{s}}^{-1}}</math> <math>v = 5</math> <math>\text{...)
- 16:24, 14 September 2010 (hist) (diff) Solution 15.1c (New page: <math>36\text{ kmph }=\frac{\text{36}\times \text{1000}}{60\times 60}=10\text{ m}{{\text{s}}^{\text{-1}}}</math> <math>\text{Momentum }=</math> <math>1200 \times 10 = 12000 \text{ Ns}</ma...) (top)
- 16:20, 14 September 2010 (hist) (diff) Solution 15.1b (New page: 3 grams = 0.003 kg Momentum = 0.003 × 4 = 0.012 Ns) (top)
- 16:19, 14 September 2010 (hist) (diff) Solution 15.1a (top)
- 15:46, 14 September 2010 (hist) (diff) Solution 14.7b (New page: Image:14.7.gif Using part a, we see from the figure that <math>F=S=\frac{98}{\tan \theta }</math> and <math>R=196</math> <math>\begin{align} & F\le \mu R \\ & \frac{98}{\tan \the...) (top)
- 15:38, 14 September 2010 (hist) (diff) Solution 14.7a (New page: The diagram shows the forces acting on the ladder. Image:14.7.gif Taking moments about the base of the ladder: <math>\begin{align} & 5\sin \theta \times S=2\textrm{.}5\cos \theta \t...) (top)
- 15:37, 14 September 2010 (hist) (diff) Image:14.7.gif (top)
- 15:33, 14 September 2010 (hist) (diff) Image:14.7temp.gif (top)
- 16:02, 13 September 2010 (hist) (diff) Answer 14.7 (top)
- 15:51, 13 September 2010 (hist) (diff) Solution 14.6b (New page: <math>\begin{align} & F\le \mu R \\ & 44\textrm{.}59\le \mu \times 245 \\ & \mu \ge \frac{44\textrm{.}59}{245} \\ & \mu \ge 0\textrm{.}182 \\ \end{align}</math>) (top)
- 15:48, 13 September 2010 (hist) (diff) Solution 14.6a (top)
- 15:46, 13 September 2010 (hist) (diff) Image:14.6temp.gif (top)
- 15:26, 13 September 2010 (hist) (diff) Solution 14.6a (New page: The diagram shows the forces acting on the ladder. Taking moments about the base of the ladder: <math>\begin{align} & 5\sin 70{}^\circ \times S=2\textrm{.}5\cos 70{}^\circ \times 245 \\...)
- 14:18, 2 September 2010 (hist) (diff) Solution 14.5b (New page: The diagram shows the forces acting on the beam Image:14.5a.gif In this case <math>{{R}_{A}}=0</math>. Taking moments about the right hand support: <math>\begin{align} & 0\textrm{...) (top)
- 14:03, 2 September 2010 (hist) (diff) Solution 14.3
- 14:03, 2 September 2010 (hist) (diff) Image:14.3a.gif (top)
- 13:50, 2 September 2010 (hist) (diff) Solution 14.5a (New page: The diagram shows the forces acting on the beam Image:14.5a.gif Taking moments about the left hand support: <math>\begin{align} & 1\textrm{.}2\times {{R}_{B}}+0\textrm{.}4\time...) (top)
- 13:49, 2 September 2010 (hist) (diff) Image:14.5a.gif (top)
- 13:46, 2 September 2010 (hist) (diff) Image:14.5.gif (top)
- 12:35, 2 September 2010 (hist) (diff) Solution 14.4b (New page: In this case the reaction at the left hand support will be zero. Taking moments about the right hand support: <math>\begin{align} & 1\times 18\times 9\textrm{.}8=2\times m\times 9\textrm...) (top)
- 12:33, 2 September 2010 (hist) (diff) Solution 14.4a (New page: Taking moments about the left hand support: <math>\begin{align} & 3\times {{R}_{2}}=18\times 9\textrm{.}8\times 2 \\ & {{R}_{2}}=\frac{18\times 9\textrm{.}8\times 2}{3}=117\textrm{.}6\t...) (top)
- 12:27, 2 September 2010 (hist) (diff) Answer 14.4 (top)
- 12:21, 2 September 2010 (hist) (diff) Solution 14.3 (New page: The diagram shows the forces acting. Image:14.3.gif Taking moments about the left hand support: <math>\begin{align} & {{R}_{B}}\times 3=49\times 1+196\times 1\textrm{.}5 \\ & ...)
- 12:11, 2 September 2010 (hist) (diff) Image:14.3.gif (top)
- 12:10, 2 September 2010 (hist) (diff) Image:114.3.gif (top)
- 12:04, 2 September 2010 (hist) (diff) Answer 14.3 (top)
- 12:01, 2 September 2010 (hist) (diff) Answer 14.3
- 19:19, 1 September 2010 (hist) (diff) Solution 14.2 (New page: Taking moments about the edge of the quay gives: <math>\begin{align} & 1\times 60\times 9\textrm{.}8=1\textrm{.}5\times m\times 9\textrm{.}8 \\ & m=\frac{60}{1\textrm{.}5}=40\text{ kg} \...) (top)
- 19:17, 1 September 2010 (hist) (diff) Answer 14.2 (top)
- 18:29, 1 September 2010 (hist) (diff) Solution 14.1b (New page: In this case <math>{{R}_{B}}=0</math>. Taking moments about A: <math>\begin{align} & 0\textrm{.}5\times 20\times 9\textrm{.}8=1\times m\times 9\textrm{.}8 \\ & m=\frac{0\textrm{.}5\ti...) (top)
- 12:53, 1 September 2010 (hist) (diff) Solution 14.1a (New page: Taking moments about the point A: <math>\begin{align} & 1 \textrm{.}5\times {{R}_{B}}=0 \textrm{.}5\times 20\times 9 \textrm{.}8 \\ & {{R}_{B}}=\frac{ 0\textrm{.}5\times 20\times 9 \text...)
- 17:49, 31 August 2010 (hist) (diff) Answer 14.1 (top)
- 10:46, 2 August 2010 (hist) (diff) Solution 13.3
- 10:44, 2 August 2010 (hist) (diff) Solution 13.3
- 10:42, 2 August 2010 (hist) (diff) Solution 13.2 (top)
- 10:39, 2 August 2010 (hist) (diff) Solution 13.2
- 10:22, 2 August 2010 (hist) (diff) Answer 13.1 (top)
- 09:46, 2 August 2010 (hist) (diff) 13. Exercises
- 09:44, 2 August 2010 (hist) (diff) Solution 13.1a (New page: Image:13.1a.gif) (top)
- 09:44, 2 August 2010 (hist) (diff) Image:13.1a.gif (top)
- 09:38, 2 August 2010 (hist) (diff) Answer 13.1
- 09:35, 2 August 2010 (hist) (diff) 13. Exercises
- 09:35, 2 August 2010 (hist) (diff) Solution 13.1 (New page: Image:13.1.gif) (top)
- 09:34, 2 August 2010 (hist) (diff) Image:13.1.gif (top)
(Newest | Oldest) View (Newer 50) (Older 50) (20 | 50 | 100 | 250 | 500)
