13. Moments
From Mechanics
| Line 15: | Line 15: | ||
| - | + | [[Image:T13.GIF]] | |
| - | + | ||
| - | + | <math>\text{Moment }=Fd</math> <math>\ \ \ \ \ \ \ \ \ \text{Moment }=Fd\sin \theta </math> | |
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | + | ||
| - | <math>\text{Moment }=Fd</math> | + | |
| - | + | ||
| - | <math>\text{Moment }=Fd\sin \theta </math> | + | |
Revision as of 16:13, 17 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 11.1
Find the moment of each force shown below about the point O.
(a) (b)
Solution
(a) (b)
\displaystyle 20\times 0.8=16
\displaystyle 12\times 1.2=14.4
Moment is -16 Nm Moment is 14.4 Nm
Example 13.2 Find the moment of each force shown below about the point O. (a) (b)
Solution
(a) (b)
\displaystyle 40\times 3\sin 60{}^\circ =104
\displaystyle 100\times 2\sin 20{}^\circ =68.4
Moment is 104 Nm (to 3 sf) Moment is 68.4 Nm (to 3 sf)
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}
