PROBLEMS.

125

T x sin /3,sin (a 2 + /3 2 )

7 1 sin ai sin /3 2sin (3 X sin (a 2 + #>) ’

and so on.

Hence all the forces P x P 2 ... are known in terms of T.

We shall now solve a few problems of forces acting on arigid body in the same plane : see Art. 52, 53. When thesystem consists of more than one rigid body, we shall considereach body separately.

Piton. 3. A uniform beam passing freely through a holeII in a wall rests with one end on an inclined plane: find theposition of equilibrium : fig. 62.

AH horizontal h, / A = «, PH-oc , PG=a: z AHP=9,pressure at P=R perpendicular to the plane, pressure at H per-pendicular to beam and = Q : resolving the forces vertically andhorizontally

W — R cos a — Q cos 0 = 0 . (l),

R sin a — Q sin 0 = 0. (2),

taking the centre of moments at P,

Wa cos 0 — Qw = 0 . (3),

these equations involve four unknown quantities R, Q, 0, oo,we must search for a relation between w and 0: this is

w sin ah sin (a + 0)

(4).

Our object is to determine the position of equilibrium: thatis, to find no and 0: we have one equation (4), we must there-fore obtain another between oo and 0 by eliminating R and Qfrom (l) (2) (3).

By (]) (2), elim g . R,

W

~Q'

sin (a + 0) W _ so

sin a * ' J Q a cos 0

sin (a + 0) sc

a cos 0

sin a