Solution 13.2

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{| width="100%" cellspacing="10px" align="center"
{| width="100%" cellspacing="10px" align="center"
|align="left"| Force
|align="left"| Force
 +
| valign="top"|Perp. distance(m)
| valign="top"|Moment (Nm)
| valign="top"|Moment (Nm)
|-
|-
|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>
|-
|-

Revision as of 10:39, 2 August 2010

Force Perp. distance(m) Moment (Nm)
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}
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