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- 11:55, 26 June 2010 (hist) (diff) Solution 13.3
- 11:50, 26 June 2010 (hist) (diff) Image:13.3.gif (top)
- 11:50, 26 June 2010 (hist) (diff) Solution 13.3 (New page: Image:13.3.gif)
- 11:09, 26 June 2010 (hist) (diff) Solution 13.2
- 11:06, 26 June 2010 (hist) (diff) Solution 13.2
- 11:03, 26 June 2010 (hist) (diff) Solution 13.2 (New page: {| width="100%" cellspacing="10px" align="center" |align="left"| Force | valign="top"|Moment (Nm) |- |8 N at <math>D</math> | valign="top"| <math>-8\times 2=16</math> |- |20 N at <math>...)
- 10:58, 26 June 2010 (hist) (diff) 13. Moments
- 10:50, 26 June 2010 (hist) (diff) 13. Moments
- 12:31, 22 June 2010 (hist) (diff) Solution 11.4c (New page: According to Newton´s Third Law the force that the block exerts on the bucket must be equal to the force the bucket exerts on the block. From the previous part this is <math>50 \text{ N...) (top)
- 12:25, 22 June 2010 (hist) (diff) Solution 11.4b (top)
- 12:20, 22 June 2010 (hist) (diff) Solution 11.4b (New page: Image:11.4b.gif)
- 12:20, 22 June 2010 (hist) (diff) Image:11.4b.gif (top)
- 19:44, 20 June 2010 (hist) (diff) Solution 11.4a (top)
- 19:42, 20 June 2010 (hist) (diff) Solution 11.4a
- 19:31, 20 June 2010 (hist) (diff) Solution 11.4a (New page: Image:11.4a.gif)
- 19:31, 20 June 2010 (hist) (diff) Image:11.4a.gif (top)
- 17:09, 20 June 2010 (hist) (diff) Answer 11.3 (top)
- 17:08, 20 June 2010 (hist) (diff) Answer 11.3 (New page: AS the ball becomes more and more compressed the force that the surface exerts on the ball increases and its speed decreases. The maximum force is when the ball has zero speed and starts t...)
- 17:07, 20 June 2010 (hist) (diff) Image:11.3.gif (top)
- 16:38, 20 June 2010 (hist) (diff) Answer 11.2b (New page: These must be equal and opposite as in the following figure where the forces on the support are in green. Image:11.2b.gif) (top)
- 16:37, 20 June 2010 (hist) (diff) Image:11.2b.gif (top)
- 13:34, 20 June 2010 (hist) (diff) Answer 11.2a (New page: Image:11.2.gif) (top)
- 13:34, 20 June 2010 (hist) (diff) Image:11.2.gif (top)
- 17:29, 14 June 2010 (hist) (diff) Solution 10.7 (top)
- 17:11, 14 June 2010 (hist) (diff) Solution 10.7
- 16:55, 14 June 2010 (hist) (diff) Solution 10.7 (New page: Image:10.7.gif We have drawn all the forces acting on the block in the figure. The acceleration is represented by the vector <math>a</math>. Applying Newton´s Second Law <math>F=ma<...)
- 16:45, 14 June 2010 (hist) (diff) Image:10.7.gif (top)
- 16:07, 14 June 2010 (hist) (diff) Solution 10.6
- 12:23, 14 June 2010 (hist) (diff) 10. Newton’s second law
- 12:20, 14 June 2010 (hist) (diff) Solution 10.6 (New page: Image:10.6.gif The figure shows the forces acting on the particle. Assuming the particle has an acceleration <math>a</math> down the plane Newton´s Second Law applied down the plan...)
- 11:33, 14 June 2010 (hist) (diff) Image:10.6.gif (top)
- 13:54, 23 May 2010 (hist) (diff) m Solution 10.5 (top)
- 13:43, 23 May 2010 (hist) (diff) 10. Exercises
- 13:32, 23 May 2010 (hist) (diff) 10. Newton’s second law
- 13:19, 23 May 2010 (hist) (diff) Solution 10.5 (New page: Image:10.5.gif We have drawn all the forces acting on the block in the figure. The acceleration is represented by the vector <math>a</math>.)
- 13:16, 23 May 2010 (hist) (diff) Image:10.5.gif (top)
- 13:08, 23 May 2010 (hist) (diff) 10. Exercises
- 15:18, 21 May 2010 (hist) (diff) Solution 10.4 (New page: We first calculate the particle´s acceleration using Newton´s Second Law <math>F=ma</math>. <math>\begin{align} & 15=6a \\ & \text{giving} \\ & a=2\textrm{.}5\ \text{m}{{\text{s}}^{-...) (top)
- 15:06, 21 May 2010 (hist) (diff) Solution 10.3 (New page: The force on the particle down the plane is <math>mg\sin {{30}^{\circ }}=m\times 9\textrm{.}81\times \frac{1}{2}=m\times 4\textrm{.}9\ \text{N}</math> Using Newton´s Second Law <math>F...) (top)
- 14:47, 21 May 2010 (hist) (diff) 10. Exercises
- 14:45, 21 May 2010 (hist) (diff) Solution 10.2 (top)
- 14:39, 21 May 2010 (hist) (diff) Solution 10.2 (New page: We first obtain the acceleration <math>a</math>. Using the kinematic equation (see section 6) <math>{{v}^{\ 2}}={{u}^{\ 2}}+2as</math> <math>\begin{align} & {{0}^{2}}={{50}^{2}}+2\tim...)
- 11:18, 20 May 2010 (hist) (diff) Solution 10.1c (top)
- 11:15, 20 May 2010 (hist) (diff) Solution 10.1c (New page: Image:10.1.gif Here <math>T</math> is the tension in the cable and <math>a</math> is the acceleration of the packet. The acceleration is in the opposite direction to the motion.This ...)
- 10:35, 20 May 2010 (hist) (diff) Solution 10.1b (top)
- 10:25, 20 May 2010 (hist) (diff) Solution 10.1b
- 10:22, 20 May 2010 (hist) (diff) Solution 10.1a (top)
- 10:03, 20 May 2010 (hist) (diff) Solution 10.1b (New page: Image:10.1b.gif Here <math>T</math> is the tension in the cable and <math>a</math> is the acceleration of the packet. Applying Newton's second law <math>\begin{align} & \uparrow :\ ...)
- 10:03, 20 May 2010 (hist) (diff) Image:10.1b.gif (top)
- 09:57, 20 May 2010 (hist) (diff) Solution 10.1a
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