| Fig. 17. |
The particular case of three forces is of interest. If they are not all parallel they must be concurrent, and their vector-sum must be zero. Thus three forces acting perpendicular to the sides of a triangle at the middle points will be in equilibrium provided they are proportional to the respective sides, and act all inwards or all outwards. This result is easily extended to the case of a polygon of any number of sides; it has an important application in hydrostatics.
Again, suppose we have a bar AB resting with its ends on two smooth inclined planes which face each other. Let G be the centre of gravity (§ 11), and let AG = a, GB = b. Let α, β be the inclinations of the planes, and θ the angle which the bar makes with the vertical. The position of equilibrium is determined by the consideration that the reactions at A and B, which are by hypothesis normal to the planes, must meet at a point J on the vertical through G. Hence
JG/a = sin (θ − α) / sin α, JG/b = sin (θ + β) / sin β,
whence
| cot θ = | a cot α − b cot β | . |
| a + b |
(6)
If the bar is uniform we have a = b, and
cot θ = 1⁄2 (cot α − cot β).
(7)