Page 67 - Mechanical Engineer's Data Handbook

Applied mechanics
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2.1 Basic mechanics nents of these forces in the x and y directions and
constructing a triangle of forces.
2. I. I Force F,=Flcos01+F2cos82+. .
~~
Fy=F,sin8,+F,sin0,+. . .
A force may be represented by an arrow-headed line
called a ‘vector’ which gives ‘magnitude’, proportional The resultant force is Fr=Jm
to its length, its ‘point of application’ and its ‘direc- (3
tion’. at an angle 8,=tan-’ to the x axis.
Referring to the figure, the magnitude is 20N, the
point of application is 0, and the line of action is XX.
A F2
X

2. I .2 Triangle of forces

A force may be resolved into two forces at right angles
to one another. The force F shown is at angle 0 to axis
XX and has components:
-7
F,=FcosO and Fy=Fsin8

FA
YA

Polygon of forces
The force vectors may be added by drawing a polygon
Resultant of several forces of forces. The line completing the polygon is the
resultant (note that its arrow points in the opposite
If several forces F,, F,, F,, etc., act on a body, then the direction), and its angle to a reference direction may be
resultant force may be found by adding the compo- found.