Solution 13.2

From Mechanics

(Difference between revisions)
Jump to: navigation, search
Current revision (10:42, 2 August 2010) (edit) (undo)
 
(One intermediate revision not shown.)
Line 1: Line 1:
{| width="100%" cellspacing="10px" align="center"
{| width="100%" cellspacing="10px" align="center"
|align="left"| Force
|align="left"| Force
 +
| valign="top"|Perpendicular
| valign="top"|Moment (Nm)
| valign="top"|Moment (Nm)
 +
|-
 +
|
 +
| valign="top"| distance (m)
 +
| valign="top"|
|-
|-
|8 N at <math>D</math>
|8 N at <math>D</math>
 +
| valign="top"|2
| valign="top"| <math>-8\times 2=-16</math>
| valign="top"| <math>-8\times 2=-16</math>
|-
|-
|20 N at <math>D</math>
|20 N at <math>D</math>
 +
| valign="top"|0
|valign="top"| <math>20\times 0=0</math>
|valign="top"| <math>20\times 0=0</math>
|-
|-
|18 N at <math>C</math>
|18 N at <math>C</math>
 +
| valign="top"|5
| valign="top"| <math>-18\times 5=-90</math>
| valign="top"| <math>-18\times 5=-90</math>
|-
|-
|16 N at <math>C</math>
|16 N at <math>C</math>
 +
| valign="top"|2
| valign="top"| <math>16\times 2=32</math>
| valign="top"| <math>16\times 2=32</math>
|-
|-
-
|24 N at <math>B</math>
+
|24 N at <math>B</math>
 +
| valign="top"|0
| valign="top"| <math>24\times 0=0</math>
| valign="top"| <math>24\times 0=0</math>
|-
|-

Current revision

Force Perpendicular Moment (Nm)
distance (m)
8 N at \displaystyle D 2 \displaystyle -8\times 2=-16
20 N at \displaystyle D 0 \displaystyle 20\times 0=0
18 N at \displaystyle C 5 \displaystyle -18\times 5=-90
16 N at \displaystyle C 2 \displaystyle 16\times 2=32
24 N at \displaystyle B 0 \displaystyle 24\times 0=0
Total Moment \displaystyle -16+0-90+32+0=-74\text{ Nm}
Retrieved from "http://wiki.sommarmatte.se/wikis/2009/bridgecoursemechanics/index.php/Solution_13.2"