13. Moments
From Mechanics
(Difference between revisions)
| Line 50: | Line 50: | ||
Solution | Solution | ||
| - | Force Moment | + | {| width="100%" cellspacing="10px" align="center" |
| - | 5N at O | + | |align="left"| Force |
| - | <math>5\times 0=0</math> | + | | valign="top"|Moment |
| + | |- | ||
| + | |5N at O | ||
| + | | valign="top"| <math>5\times 0=0</math> | ||
| + | |- | ||
| + | |8 N | ||
| + | |valign="top"| <math>-8\times 1.2=-9.6</math> | ||
| + | |- | ||
| - | 8 N | ||
| - | <math>-8\times 1.2=-9.6</math> | ||
| - | 7 N | ||
| - | <math>7\times 0=0</math> | ||
| - | 6 N | + | |7 N |
| - | <math>-6\times 0.5=-3</math> | + | | valign="top"| <math>7\times 0=0</math> |
| + | |- | ||
| + | |6 N | ||
| + | | valign="top"| <math>-6\times 0.5=-3</math> | ||
| + | |- | ||
| + | |5 N | ||
| + | | valign="top"| <math>5\times 1.2=6</math> | ||
| + | |- | ||
| + | |4 N | ||
| + | | valign="top"| <math>4\times 0.5=2</math> | ||
| + | |- | ||
| + | |Total Moment | ||
| + | | valign="top"| <math>0-9.6+0-3+6+2=-4.6\text{ Nm}</math> | ||
| - | + | |} | |
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Revision as of 11:09, 18 September 2009
| Theory | Exercises |
Key Points
The moment of the force about the point O is the product of the force and the perpendicular distance to the line of action of the force from O.
\displaystyle \text{Moment }=Fd
\displaystyle \text{Moment }=Fd\sin \theta
Clockwise moments are negative.
Anti-clockwise moments are positive.
Example 13.1
Example 13.2
Example 13.3
For the rectangular lamina shown below, find the total moment of the forces acting, about the corner marked O.
Solution
| Force | Moment |
| 5N at O | \displaystyle 5\times 0=0 |
| 8 N | \displaystyle -8\times 1.2=-9.6 |
| 7 N | \displaystyle 7\times 0=0 |
| 6 N | \displaystyle -6\times 0.5=-3 |
| 5 N | \displaystyle 5\times 1.2=6 |
| 4 N | \displaystyle 4\times 0.5=2 |
| Total Moment | \displaystyle 0-9.6+0-3+6+2=-4.6\text{ Nm} |
